Week 0: Introduction

What we're trying to achieve in this course.

Imagine that it is ten years after you graduate. You may be an engineer, a scientist, or in some quite different job. But let's say you have a problem to solve. This course aims to equip you with some very powerful mental tools to help you solve that problem. Some of these tools are specific bits of physics knowledge. But mostly, it's a different way of looking at the world.

This approach goes by many names - "Going to back to first principles", "Problem solving strategies", "Modelling", and even "Thinking like a physicist".

Why does this matter? Because it will affect the whole future of the human race. Will the future look like this:

Or like this? Ninja Physics/Paul Francis 2

Will science and technology continue to advance, leading us into an unimaginable interstellar future, or will progress slow down and stop, leaving an overpopulated and polluted earth to fight over finite resources?

Is scientific progress slowing down?

Many very respectable researchers believe that scientific progress is slowing down.

Others disagree.

Why might you think scientific progress is slowing down? We think that science is advancing rapidly now - with such wonders as the internet, smartphones and DNA sequencing.

But compare them to the discoveries of 100-200 years ago, like radio, electricity, cars, jet engines, artificial fertiliser and antibiotics. It's hard to argue that the recent changes have had anything like such a big effect on people’s lives.

Why might things be slowing down? Possibly because science is becoming harder.

Galileo could revolutionise science by rolling balls across a table - nowadays it takes the multi- billion-dollar large hadron collider, run by teams of thousands of researchers. You could argue that the "low hanging fruit" has been picked. There were a certain number of scientific discoveries which could rapidly be turned into very useful inventions - things like electricity

(light bulbs, telegraph, TV) and thermodynamics (steam engines, cars, jet engines). But they were all discovered a long time ago - more recent discoveries like the standard model of particle physics, or General Relativity, have had little or no effect on people's day-to-day lives. The Ninja Physics/Paul Francis 3 number of scientists in the world keeps growing rapidly, yet the growth in economic productivity due to science seems to be constantly dropping. The cumulative volume of human knowledge doubles every ten years, yet humans are not getting ten times smarter or able to learn stuff ten times faster every ten years - can our brains really cope?

On the other hand, there are people who believe that scientific progress is accelerating.

They point out that exponential growth (like we are seeing in computer speeds) has as increasing effect with time. Doubling something that is already very large has a much bigger effect than doubling something small. And even though the easier scientific problems have been solved, we have much more powerful tools to help us solve future problems. An artificial-intelligence- supported cyborg using quantum metrology can solve problems that would have baffled Einstein.

And we know that there is a lot more to learn because nature can produce marvels (like a mosquito) that our technologies cannot even come close to matching.

So which is it going to be - science grinding to a halt, or accelerating into the future?

That's up to you. If you don't do it, as the scientists and engineers at one of the world's greatest universities, who will?

References:

https://en.wikipedia.org/wiki/The_Great_Stagnation - a book arguing the progress is slowing down.

https://en.wikipedia.org/wiki/Race_Against_the_Machine - a book arguing the opposite.

Ninja Physics

In this course, we will be focusing on what I like to call "Ninja Physics", but is also called "Back-of-the-envelope" or "Fermi problems". Ninja Physics/Paul Francis 4

The basic idea is to very quickly come up with an approximate answer to a problem, using lots of approximations and simplifications to get to the essential physics quickly. I call it

Ninja physics because it is very fast and very powerful, but often doesn't stand up well in the light of day. It's an alternative to standard "Samurai" physics, which is also very powerful, and is all about playing by the laws, but is much slower and more cumbersome.

Why do Ninja Physics?

It seems rather strange to deliberately make approximations. Why don't we just do things properly and get the "right answer"?

Let's go back and think why we might need to do physics in the first place. We are probably trying to solve a problem, or come up with some new technology to improve life.

So let's imagine that ten years from now, you have been given a problem to solve. It will probably be a very difficult problem (top students like you won't be given the easy problems).

You will probably have no idea where to start.

That's where Ninja physics comes in. It allows you to play with ideas quickly, and get your mind around what are the key factors. You can try out ideas and quickly see if they are at all sensible. For even the most brilliant researchers, only 1 in 10 of their ideas turns out to work

(if all your ideas work, you are clearly having unimaginative, unambitious ideas). Ninja physics can rapidly eliminate most of the ideas that won't work, allowing you to focus on the few that will.

Once you've identified the idea or ideas that have a real chance of working, Ninja physics helps you work out what factors are crucial to its success, and which are just minor details. You can then focus your efforts on better understanding the most important factors. Ninja Physics/Paul Francis 5

And only then, once you've found an idea that will work, and you've identified what are the most important factors, is it time to put in the details and try to get a very accurate "right answer".

Basic Principles

There are three basic principles of Ninja Physics:

● Go to first principles

● Build a simplified model

● Focus on what matters most

What are "First Principles"? This means trying to understand the deep fundamental reasons behind something. Often it means trying to understand a situation in terms of the most fundamental laws of physics. It can mean asking "why" repeatedly, as you try to understand a situation in terms of very basic principles.

What is a "simplified model"? Scientists and engineers can never calculate the full complexity of any real world situation. We always have to simplify it somehow to make it tractable. The trick is to choose the best simplifications - simplifications that make a problem solvable but don't mean that your model is no longer like reality.

How do you focus on what matters most? To begin with, only worry about things that would change your answer by more than an order of magnitude (a factor of roughly 10 or more).

At this stage, don't bother yourself about relatively unimportant things like factors of two, pi etc.

Once you are sure you have the big stuff nailed down, then you can start worrying about factors of three etc.

I'll show you lots of examples of how to do all this as the course goes on! Ninja Physics/Paul Francis 6

Week 1: The Physics of Normal Cars

Flying Cars

Venture capitalist Peter Thiel famously said "We wanted flying cars, instead we got 140 characters". He was bemoaning what he sees as a slow-down in technological progress. While there has been rapid progress in computers and smartphones, many other technologies like transport and energy are advancing very slowly.

Ninja Physics/Paul Francis 7

(graphs from http://foundersfund.com/the-future/ )

Note the immense advance in computer chip power, and the complete lack of recent major advances in either productivity or travel speed across the North Atlantic (though this immense advance in computing power was only achieved by a vast increase in research resources - http://marginalrevolution.com/marginalrevolution/2016/12/depressing-paper-great- stagnation.html, and many think that this rapid advance in computing power is coming to an end

- see - https://www.theguardian.com/technology/2017/jan/26/vanishing-point-rise-invisible- computer)

Flying cars would be amazing. They would solve congestion problems in our cities

(though they wouldn't solve parking problems unless they could fly themselves home after dropping us off), and transform the way we live and the shapes of our cities. Countless science- fiction shows have featured flying cars, ranging from Star Wars to Back to the Future. And yet we don't have them. Every year or two someone claims to be on the brink of launching a commercial flying car, yet it never quite seems to "take off".

Why? Is it just that researchers are being slack and unambitious? Or is there some fundamental physical reason why flying cars are hard? And if so, what is this fundamental reason and can we get around it some-how? Ninja Physics/Paul Francis 8

This is a good job for Ninja physics - trying to understand what the fundamental physics of a flying car might be, and see what, if anything, is limiting it. This is what "Going back to first principles" is all about.

How can we go about analysing something like the feasibility of flying cars?

One approach is to try and figure out particular possible designs, and how feasible they are. This is valuable, but it doesn't really give you a general answer to the question "are flying cars feasible"?

In his situation, the best approach is often to "follow the energy". This can often give you a very broad and general picture of what is and is not really important.

Coming up with a model

So how can we "go back to first principles" and figure out why cars are as efficient as they are?

We need to come up with a simplified model of a car. A model that allows us to focus on what really matters. Modern cars are incredibly complicated - if we tried to analyse every component and how it works, we'd rapidly be lost in the detail. We need a simplified model.

We are going to follow the energy - so a good starting point might be to ask "Where does the energy go"? We know cars need energy to move. Why?

Our simple model of a car might be a box on wheels that moves. To make things concrete, let's imagine our car driving from Canberra to Sydney. Why would moving a box need energy?

A few possibilities come to mind. The box needs to start and stop. When it starts from rest and accelerates, it's gaining kinetic energy. Ninja Physics/Paul Francis 9

Another possibility - friction. You will have to exert energy to overcome the friction of a car.

But if you think about this, there will actually be two types of friction here. One is

"rolling resistance" - the friction between the wheels and ground. The other is wind resistance - the force needed to push the moving box through the air. If you had a magnetic levitating car you might be able to get rid of rolling resistance. And if you were driving on the Moon, there would be no atmosphere and hence no air resistance. But on Earth, on wheels, both will apply.

There will also be energy wasted in the car - friction in the gears, heat dissipated in the engine etc., but for the moment that's not part of our model - we're just going to consider what is needed to get a box on wheels to Sydney.

Finally, the road to Sydney is not flat. Our car will have to overcome gravitational potential energy every time it goes up a hill.

Let's consider these in turn and see which is most important.

#

Starting and stopping

How much energy does it take to accelerate a car?

Let's consider accelerating a car from 0 to 110 km/hr (110 km/hr = 110x1000/(60x60) ~

30 m/s). It must acquire a kinetic energy of . The mass of a typical car is 1-2 tonnes (1000-

2000 kg), so let's call it 1500 kg, so the energy needed is

.

So if you have to stop and start a few times, we'd be looking at roughly a MJ or two. Ninja Physics/Paul Francis 10

Now this is a pretty approximate calculation. Different cars will have different masses and will probably drive at different speeds. But this gives us an "order of magnitude" estimate which is quite good enough for now.

Rolling Resistance

How about rolling resistance? This is caused by the tyres deforming as they touch the road then springing back as the wheel rotates. Hard wheels (like the steel wheels on a train, or the narrow high-pressure wheels on a racing bike) have low rolling resistance, while large fat under-inflated tyres would have a larger value.

How can we estimate this?

One way is to think about how much force it takes to push a car along a flat road. As the speed is low, wind resistance isn't important, so the force required is mostly just overcoming rolling resistance.

Think back to when you've had to push a broken-down car, or watched other people do it.

It is clearly possible for one or two people to push a car along, but it's certainly hard work - the people pushing are grunting and cursing. So the force needed to overcome rolling resistance and push a car along is clearly close to the maximum pushing force a human can exert.

But what is this maximum force a human can exert? 1N? 10N? 100N? 1000N? 10000N?

Clearly it will vary from person to person and depending on the angle and direction of the push, but we should at least be able to estimate a rough number.

For myself, I know I can lift up my kids, but not really an adult. When I go to the gym, I can move ~30 kg weights but not ~100kg weights (you are probably fitter than me!) So an estimate of the maximum force my muscles can exert is lifting ~ 30 kg, so a force of mg ~ 300 Ninja Physics/Paul Francis 11

N. I could estimate this other ways - e.g. from how fast I can cycle up hill, or from how heavy a box I can push along the floor, but this is probably as good as any.

So this suggests that the rolling resistance of a car is around 300N.

Notice I've approximated g as 10 rather than 9.8. This introduces as 2% error, but almost certainly that's the least of our worries here - almost certainly my estimate of a 30kg weight is wrong by far more than 2%. And lifting a weight uses different muscles from pushing a car - another possible source of error that is also probably much bigger than 2%.

How good is this estimate? One way to test it is to ask - could the answer be an order of magnitude larger or smaller? Could we push a car with an order of magnitude less force - i.e.

30N rather than 300N? That's equivalent to lifting a 3kg weight - i.e. a large 3L milk bottle. I can do that with one finger. I can't push a car along with one finger, so that's definitely too small.

How about 3000N? That is like lifting three people (100 kg) people. Only really strong circus performers, Olympic athletes and superheros can do that. But you don't need someone like that to push a car along a flat road. So that's too high. So it seems that to within an order of magnitude,

300N is probably about right.

If you look on the internet, you find that rolling resistance has been extensively studied

(of course), and is normally calculated as a fraction of the weight of the car. This fraction is around 0.02 for typical car tyres, so the force is 0.02 times mg, or 0.02x1500x10 ~ 300N. So both approaches match, giving me some confidence that this isn't too far off.

That's the force, but how do we convert this into energy? Well, energy applied to an object is called "Work" and is equal to force times distance (as long as the force and distance are in the same direction). So in this case we are applying a force of around 300N over a distance of Ninja Physics/Paul Francis 12 300 km, so the work is 300x300,000 which comes out to around 108 J. So 100 MJ - two orders of magnitude more than the energy needed to accelerate the car up to the speed limit.

So for this sort of drive, we can basically ignore the accelerating energy - it's irrelevant compared to the energy needed to overcome rolling resistance. Though this might not be true for urban driving where you aren't going a large distance, and where you have to stop and start a lot.

Drag

How about wind resistance (also known as drag)?

Experience suggests that it is important. Cars are designed to be streamlined. When you stick your hand out the window of a car moving at highway speeds, the wind applies a strong force to it.

So how can we estimate the drag force on a moving object, such as a car?

As an object moves through a fluid, it traces out a volume and all the fluid in this volume ha to get out of its way. At low velocities, the fluid can flow calmly and smoothly around the moving object (so-called laminar flow). But at higher velocities, the flow becomes turbulent and the air swirls around in a messy way as an object pushes through.

Which situation applies here? The normal way to tell is to compute something called the

Reynolds number:

where is the density (~1.2 kg m-3 for air), v the velocity, L the length scale and is the dynamic viscosity - a measure of how viscous a fluid is (around 2x10-5 Pa S for air). If the

Reynolds number is more than around 1000, the flow is generally turbulent.

In the case of a speeding car, L~2m (rough thickness of a car), v~ 30 m/s, so the

Reynolds number is around 2x105, well into the turbulent regime. Ninja Physics/Paul Francis 13

In this case, we can crudely approximate that the fluid pushed out of the way of the car ends up travelling forwards at roughly the speed of the car (actually swirling around in some complicated pattern, but with roughly the same speed. Not all the air will do this, and not all will be at the car speed, and we can allow for this by putting in a fudge factor, as we will see).

So we need to accelerate the swept up air to the car speed v. How much air is in the volume swept out by the car? Crudely speaking, in a time t the car moves a distance vt, so the swept out volume is equal to vt times the cross-sectional area of the car A. So the mass of air swept up in time t is .

This air must be accelerated up to speed v in time t, so the acceleration is v/t. Using

Newton's second law (F=ma) this means that the car must apply a force

So the drag force is equal to the cross-sectional area of the car, times the density of air, times the driving speed squared. Which all makes sense - you'd expect a bigger object to have more trouble moving, and a dense medium would also take more effort to push it out the way.

And the great importance of speed makes sense too. Ninja Physics/Paul Francis 14 In reality not all the air swept up is accelerated so much - some flows smoothly around the car. But even air not directly in front of the car will be disturbed. To allow for all this, a fudge-factor C, known as the drag coefficient, is normally used (and for various obscure reasons a factor of 1/2 is also put in), so the equation for the drag force on an object moving at turbulent speeds is:

Is this any use given it has an unknown fudge-factor C in it? Well luckily C is reasonably constant for most objects at different speeds. For streamlined objects like most modern cars,

C~0.3. Incredibly streamlined objects (like solar racers and aircraft) might have C~0.1 or even less, while trucks have C~0.6 and deliberately anti-streamlined objects like parachutes might have C>1. But if you assume C~0.5 for most objects you won't be far wrong.

So what is this force on a car driving to Sydney? A Toyota Corolla is about 1.5m wide and 1.5m tall. You can look up drag coefficients for cars (which come from wind-tunnel testing), and they are around 0.3 for a car like a Corolla. So the drag force comes out at

1/2x0.3x1.52x1.2x302~360 N.

Which is about the same force we got from rolling resistance. So the energy needed to overcome drag while driving to Sydney is also around 108 J.

So it looks at this stage as if rolling resistance and drag are roughly equal, while starting and stopping is negligible in comparison.

Skin Drag

A small complication - this equation for drag depended only on the cross-sectional area of an object, not on how long it is. In reality there is a bit of so-called "skin friction" caused by the air sliding past the sides of an object, that depends on the length of an object. It is pretty Ninja Physics/Paul Francis 15 irrelevant for this calculation, but can be important for long and thin objects like trains and planes...

Hills

But how about going up hills? The road to Sydney is far from flat.

A typical hill might involve climbing around 100m. The energy needed to climb is thus mgh ~1500x10x100 ~ 106 J. Which is a hundred times less than the energies needed to overcome either the rolling resistance or the drag. So unless the route involves a lot of hills, this can safely be ignored. And, of course, if there are uphill climbs you also get much of that potential energy back when you roll down the other side of the hill!

Petrol Consumption

So, it seems that if you drive to somewhere like Sydney, the bulk of your energy is spent overcoming rolling resistance and atmospheric drag, and your total energy consumption will be around 2x108 J.

Let's double-check this. We know roughly how much petrol a car uses to get to Sydney, and the chemists can tell us the amount of energy you get by burning a litre of petrol. So we can work out how much energy a car really uses to get to Sydney and compare it to our estimate of how much it should need!

Modern cars use around 7 litres per hundred kilometres in freeway driving (you can find this figure on numerous web sites). Petrol contains around 3x107 J/L, so going to Sydney (around

300 km) uses 3x7x3x107 ~ 6x108 J. Which is roughly three times what we calculated.

That's actually very good agreement for a rough estimate like this! We got the order of magnitude quite correct. And you would expect a car to be less than 100% efficient in turning petrol into motion. Ninja Physics/Paul Francis 16 In particular, a car engine is a heat engine - a device that burns petrol to create heat, and then uses this heat to create motion. And ever since the 19th century, it has been known that heat engines are fundamentally limited in their efficiency.

Any heat engine takes a temperature difference and converts it into useful work that can drive motion. In the case of a car engine, the temperature difference is between the heat inside a cylinder during a fuel explosion, and the outside air. It is possible to show using the laws of thermodynamics that the most efficient possible engine uses the "Carnot cycle" and has an efficiency of

where TC is the cold temperature (the outside temperature in this case) and TH is the hot temperature (the temperature inside the cylinder during the explosion in this case). The efficiency tells us the fraction of the heat energy that is converted into a usable form and can be used to drive the car.

So to get an efficient car engine, you need to have the inside of the engine as hot as possible. But if you make it too hot, you get two problems. Firstly, the materials in the engine probably can't survive very great heat. But secondly, if the engine gets too hot, the fuel will explode prematurely when it is injected - this is called "knocking" and tends to be the fundamental limit in practice (diesel engines are more efficient because the fuel self-ignites here so they can run hotter, without knocking).

If we assume our car engine runs at around 220 C (~ 500 K) and that the air outside is at around 20C (~300 K), then the efficiency will be less than or equal to 1-300/500 = 40%. So generating 6x108 J of heat will only result in 40%x6x108 ~ 2.4x108 J, which is very close to our estimate of how much energy should be needed. The rest of the heat energy obtained from Ninja Physics/Paul Francis 17 burning the petrol ends up as waste heat - the car's radiator will transfer this heat to the air around the car.

Heating

It would be an interesting puzzle to calculate how much all this heat will warm up the air around a road. So let's do it quickly (just for fun). One way would be to compare the warming due to cars with the natural warming - say the sunlight. Sunlight deposits around 1000 W per square metre (when the Sun is shining directly on the surface without too much cloud - this is known as the solar constant). We'll need a surface area to work out the total solar heating. The

Hume Highway to Sydney is around 300km long. How wide is it? It has two lanes in each direction, each of which is a bit wider than a car - let's say 2.5 m wide per lane, so the total highway is around 10m wide, and hence has a surface area of around 300x1000x10 ~ 3x106 m2.

So the total power deposited on the Hume is 1000 times this per second - i.e. around 3x109 W.

Which is a lot!

Now we're trying to compare an energy per second (from the Sun) with an energy per car per 3 hours. We could divide the energy from each car by the number of seconds in three hours

(3x60x60 ~ 104 s). Each car is burning 6x108 J over 3 hours (some of which goes straight to waste heat, the rest goes to overcoming drag and rolling resistance, which ends up as heat), so a car is heating up the Hume Highway and environs by around 6x104 W.

But that's just one car. How many are there on the highway to Sydney at a given instant?

One way would be to mentally picture the road an estimate how many cars there are in (say) a 1 km part of the highway. When I see a "turn off in 1km" sign, there might typically be around 10 cars between me and the turn-off. So if there are ~ 10 cars per kilometre, there would be around Ninja Physics/Paul Francis 18

3000 on the whole Hume Highway, so the total power generated would be around 3000x6x104 ~

2x108 W.

Another way to estimate the number of cars would be to use the cars per second passing a given point. Typically there might be a car passing you as you stand by the side of the road every few seconds - let's say one car per 5 sec. At 110 km/hr (~ 30 m/s) that means there are around

5x30 = 150 metres between cars, which gives around 7 per km - close to what I estimated above.

So it looks plausible. Clearly if there is a traffic jam there could be a lot more cars per kilometre, but then they probably wouldn't be going so fast and hence burning energy at a smaller rate!

So this is interesting - it's only ten times less than the solar energy. And that's assuming direct sunlight. On a cloudy day, when there is heavy but fast-moving traffic, the heat from the cars might rival the heat from the Sun. If it's a windy day this heat will be rapidly spread over a huge volume, but on a calm still day or night, it may stay close to the road for a while, so you might expect the road to be measurably warmer than the surrounding countryside!

Conclusions

Anyway - back to our main discussion - what is the energy budget of a normal car?

Remarkably enough, it seems that normal cars are pretty close to their theoretical maximum efficiency. 60% of the energy is wasted as heat - but you can't do much better unless you develop some sort of engine that runs very hot. The rest is split fairly evenly between overcoming drag and overcoming rolling resistance. There isn't a huge amount wasted anywhere else. Our calculations were pretty approximate, so maybe there is a factor of two missing somewhere, but not a factor of ten.

So basically there is no way you could have a car that gets ten or a hundred times the fuel efficiency of current ones, unless you could: Ninja Physics/Paul Francis 19

● Reduce air resistance by a factor of ten or more. An ultra-streamlined teardrop might improve

things by a factor of 2-3, but to do better you'd have to reduce the cross-sectional area -

say by having a narrow car (i.e. a tandem motorbike...). Or put the road in a tunnel and

pump out the air. Or have computer-controlled cars driving immediately behind each

other to take advantage of slip-streaming. But even if you reduce the drag to zero, that

will only double the fuel economy unless you reduce the rolling resistance.

● Reduce the rolling resistance by a factor of ten or more. Rolling resistance comes from the

deformation of the wheels where they touch the road. Having narrower tyres with higher

tyre pressure will reduce that. The lowest rolling resistance is if you have metal wheels

on a metal track (i.e. a train...) and is about ten times lower than typically inflated car

tyres on a typical road. More realistically you could have narrow tyres with much higher

pressure (like road bicycle tyres). But that comes at the cost of lower grip and worse

cornering, and more wear and tear on the road due to the higher tyre pressure.

● Increase the temperature of the inside of the engine to several thousand Kelvin. Which would

melt steel - but maybe some sort of silicon carbide fusion reactor (petrol wouldn't survive

at that temperature)?

So basically there are good fundamental physics reasons why we won't get cars that are ten or a hundred times more fuel efficient that current ones very easily! We can expect computers to get ten times better every few years but we really can't expect the same of cars without either violating the laws of physics or doing something radical like changing them into trains...

Lessons Learned Ninja Physics/Paul Francis 20

The most important things I was demonstrating in these notes is not the physics of cars - it is the Ninja physics approach I took. Here are some specific things I did which are important:

● Don't worry about small factors and things that only make a small difference. So for example

the exact size and number of the hills en-route to Sydney - I just guessed and in the end it

made very little difference.

● Often you end up comparing two numbers - say the energy used to overcome drag and the

actual combustion energy. This is very powerful.

● Numbers matter - 108 and 107 J are very different things. Pay attention to numbers, not just

equations.

● Simplify. Air flowing past a car doesn't behave as we said, but it will probably be close

enough that our estimate isn't ridiculous.

● Go to first principles if you really want to understand something. Ninja Physics/Paul Francis 21

Week 2: Flying cars

What's different with flying?

So now let us do a similar analysis with flying cars. Where must the energy go? And to make it concrete, let's use the example of flying from Canberra to Sydney once more.

What is our model here? We need to work out how much energy is needed to fly. But that's not simple - because flying involves several different processes. One is rising up to a particular altitude, where the energy needed can be calculated using potential energy. This energy will be the same whether you fly up, go up in an elevator or are fired up by a catapult.

Then to fly somewhere through the atmosphere, you will need to overcome drag (just like a car on the ground must overcome drag).

And thirdly, you need to stay in the air while you get to wherever you are travelling.

We'll address these different processes in turn.

Flying cars have one major advantage over normal cars - no rolling resistance.

How about drag? If a flying car is roughly the same size and shape as a normal car, and goes at the same speed, then the drag will be about the same.

There are a couple of possible differences here, however. One is that flying cars might be able to go much faster than normal cars. However, because drag is proportional to v2, this will Ninja Physics/Paul Francis 22

only make the energy consumption worse. So for the moment let's assume that a flying car goes at the same speed as a ground car (~100 km/hr), but if we find the energy losses aren't too bad, we can consider increasing the speed.

Another possible difference is that a flying car (if pressurised) could travel at a high altitude, where the air is thinner. As drag is proportional to the density of air, this would reduce the energy needed to overcome air resistance.

How thick is the Earth's Atmosphere?

So how thick is the atmosphere - how high would a flying car need to go to significantly reduce the drag? We know that it's very hard to breath on mountain tops (4-5 km up) and nearly impossible on top of Mt Everest (~ 9 km up), so clearly we are looking at heights of 5-10 km to get above quite a lot of the atmosphere.

One way to estimate the height of the atmosphere is to use the air pressure at the Earth's surface, which is a whopping ~105 Pa at sea level (we'll talk much more about air pressure and where it comes from later in the course). The SI units of pressure are Pascals and are equal to a newton per square metre. So that means that a cubic metre of air at the Earth's surface is exerting a force of around 105 N per square metre on everything around it, including a column of air going all the way up from that cube to the top of the atmosphere. Ninja Physics/Paul Francis 23

As the column of air is not (on average) accelerating up or down, the forces on it must balance. So the downward weight force, mg, will be equal to the upwards pressure force P. If we make the (clearly wrong but useful) assumption that all the air has the same density as sea-level air, around 1 kg/m3, and that gravity doesn't change much over the thickness of the atmosphere, then the weight of a column with cross-sectional area 1 m2 is just the density times the height times g - i.e. 1xhx10 and is equal to the pressure 105 Pa. So the thickness of the atmosphere must be h~104 m = 10 km. Which matches our expectation from above.

In reality, of course, the atmosphere doesn't abruptly end at a particular height; it just gets thinner and thinner until it is imperceptible. We will derive an equation for this later, but roughly speaking, the density of the atmosphere halves every 5 km you go up. To be precise, the density goes as

where is the density at sea level, h is your height, and H is the scale height of the atmosphere, which is around 7km. Ninja Physics/Paul Francis 24

So if your flying car could go up 10km, the density would only be 25% that at sea level, so the drag would be four times less.

Climbing energy

So is it worth climbing up to this sort of altitude? Or should your flying car stay near the ground?

We already saw that the energy needed to climb a few hundred metres is negligible compared to the drag. But if we want to climb to (say) 10km altitude, we'd use up mgh ~

103x10x104 ~ 108 J, which is comparable to the drag energy loss. So what we'd save in drag, we'd use up in climbing.

So for the trip to Sydney at 110 km/hr, it doesn't make much difference. For shorter/slower trips, it would make sense to stay low, as the climbing energy would greatly outweigh the drag losses. But if you wanted to go further, or travel faster, then it would make sense to get high.

Catapult?

So now we come to the crucial point - how can a flying car stay in the air? And how much energy must it use to stay there? A car has weight (mg) which applies a downward force.

To stay in the air, this must be balanced by some upwards force?

Or must it? One possible way to fly would be projectile motion - you could fire a car out of some sort of gun or catapult and just let it fly through the air like an artillery shell. Is this feasible? Well - we know that it is possible to fire artillery shells out to distances of up to 30km if you have a big enough gun (like the Amiens rail gun used by the Germans in World War 1). Ninja Physics/Paul Francis 25

(public domain image from the Australian War Memorial)

How feasible is this? Not very - guns like this fire their shells with supersonic muzzle velocities (for the Amiens gun, around 700 m/s), so the noise of cars breaking the sound barrier would be distracting, to say the least. Accelerating a car up to these speeds would need to be done gradually to avoid crushing the car and its occupants with the acceleration. If we accelerate at no more than g (~10 m s-2), it would take over a minute to get a car up to this launch speed

(700/10 = 70 sec) and a distance of

So a launch track 24km long! And the car would need to be very strong, heat resistant and highly streamlined to withstand the supersonic atmospheric drag. And I've no idea how you would safely land it at the end of its trajectory...

I suppose something much smaller scale might work - like firing cars over intersections, but projectile motion for long distance flight doesn't seem very feasible.

Staying in the air Ninja Physics/Paul Francis 26

So it looks like we will need to supply some force to oppose gravity and keep our car in the air. First of all - can we think of some argument to work out in a general way the energy involved in staying in the air?

Energy used (work done) is force times the distance moved in the direction of the force.

So to hover or fly horizontally, theoretically no work needs to be done, because the direction of the force (vertical) and the motion (horizontal) are not the same. And indeed this makes sense - it is possible to hover in the air using no energy if (for example) you are suspended from a wire or on a monorail.

So that's certainly one way to have flying cars - build elevated roads and drive along them. In this case the normal force from the road keeps the car in the air and no energy is required. But elevated freeways aren't quite what we had in mind when we talk about flying cars!

Another possibility would be to use buoyancy - suspend the car from a balloon or fill it with hydrogen or some such.

Buoyancy

How does Buoyancy work? To understand this, we have to ask a very basic question - why doesn't the air in your room fall down? Take a particular cubic metre of air in your room.

Any cubic metre will do. It weighs around a kilogram (similar to a water bottle). If you hold a water bottle in the middle of your room and let go, it will fall. Why doesn't the cubic metre of air? There must be some upward force balancing its weight.

What is this upward force? The air below our imaginary cube is applying an upwards pressure to the cube. And the air above it is applying a downward pressure. The upward pressure must be slightly greater than the downward pressure to balance the weight (about mg~10 N Ninja Physics/Paul Francis 27 greater, in fact!). Which is a very tiny percentage change, given that pressure at the Earth's surface is a whopping 100,000N.

So the pressure of the atmosphere must decrease as you go up, to balance gravity and keep the air from falling down. Later in the course we'll use this fact to derive an equation for the density of the atmosphere at different altitudes. But for the moment, consider what this means for the force on a solid object (like a car or a person) immersed in a fluid (like water or air).

There will be a pressure force on all sides of the object, but the force on the bottom will be slightly greater than the force on the top. So there will be a small net upwards force. This is called buoyancy.

How can we calculate it? Well, there is a sneaky trick:

Take an object immersed is a fluid (which has a pressure gradient). What is the net force on it?

Very hard to calculate - I suppose you could write down some equation for the shape of the object and integrate the vector pressure force over the surface - but sooner you than me!

But here's the trick - replace the object with an identical shaped bit of the fluid... Ninja Physics/Paul Francis 28

The laws of physics apply to any part of the world, so let's pick as our part of interest (our system)this rather odd shaped bit of fluid. This bit of the fluid isn't going to move, rise or sink, so the net pressure force on it must be exactly enough to balance its gravity.

So the net pressure force (i.e. the buoyancy) acting on this surface:

Must be equal and opposite to the weight of the fluid in this shape: Ninja Physics/Paul Francis 29

That's how you work out the buoyancy force - work out the weight of the fluid displaced

(i.e. the fluid that isn't there! The fluid that would have been there if the object wasn't there. The buoyant force is equal and opposite to the weight of this displaced fluid.

So let's work out what it would take to get a car to fly by buoyancy. Let's have a balloon filled with hydrogen or helium or something similar. To get a buoyant force capable of sustaining a 1000 kg car (mg ~ 104 N), we would need to displace the same mass of air, i.e. ~ 103 m3 of air. Of course whatever you are using to displace the air (helium? Hot air?) would also have mass, but the density of helium at atmospheric temperatures and pressures is about ten times less than that of air, so we can probably ignore it for a quick calculation.

We won't fit 103 cubic metres of air in a car - at most a car's interior would be 2m wide,

2m tall and perhaps 4m long, so giving 16 cubic meters and an upward force of 16 times g times the density of air - i.e. around 16x10x1=160N, which only contributes around 1.6% of the weight of the car.

So really we need a balloon of some sort. How big? Well, it needs to displace 103 m3 of air, so a cube 10m on a side would do. Probably a bit bigger because it would have to support the Ninja Physics/Paul Francis 30 weight of the balloon, and because that gas inside is not zero density. So our flying car would look something like this:

Hard to park - but perhaps not totally impossible?

But such a big balloon will have major problems with drag. If you remember, drag is proportional to cross-sectional area times velocity squared. The cross sectional area has increased from roughly 2x2 m2 (for the car) to roughly 10x10 m2 for the balloon - i.e. it's 25 times greater. So if you go at the same speed the drag will be 25 times greater - so the energy consumption will be ruinous. Alternatively, you could go five times slower (i.e. around 30 km/hr) to reduce the drag by 52 and compensate for the larger surface area.

This all matches what we know about airships, dirigibles, blimps and zeppelins - they are very large and very slow. So it's plausible.

Would a buoyant flying car be a sensible idea? The slow speed is a bit of a killer, except maybe in very congested cities, where floating through the air at 30 km/hr may well be preferable to sitting in a traffic jam below. Parking would still be a problem, but perhaps you could develop some way to pump the air out of the balloon and stow it away when it's not needed? Operating the flying car on windy days would also be tricky. Ninja Physics/Paul Francis 31

Pushing air down

So if buoyancy isn't looking too good, how else would we generate a force to keep our car in the air? The most common approach, used by planes and helicopters, is to apply a downward force to the surrounding air. By Newton's third law, the air must therefore apply an upward force to you, keeping you in the air. As you are applying a downward force to the air, it will be accelerated downwards.

In a helicopter, the blades push air down, and the force needed to accelerate the air downwards has an equal and opposite force upwards on the blades.

For an aeroplane, the wings deflect the airflow downwards, producing the required downward flow of air, and hence the equal and opposite upwards force to keep the plane in the air.

Note that an incorrect explanation of how planes fly is often taught in schools. This wrong explanation says that wings have curved upper surfaces and flat lower ones, so the air needs to go faster over the top. Faster moving fluid has a lower pressure (Bernoulli's effect) and so sucks the plane up. But that explanation is wrong. Firstly, it assumes that the air going over Ninja Physics/Paul Francis 32

the top and bottom of the wing has to arrive at the end at the same time, which it doesn't. Not all planes have curved upper wing surfaces, and stunt planes can fly upside-down (with the curved surface pointing towards the ground) which is clearly inconsistent with this explanation.

So perhaps we could have something like this for our flying car - perhaps some fans attached which blow air downwards and hence help keep the car in the air?

How much air?

So how much air would you need to push downwards to keep a flying car in the air?

Let's imagine we push a mass m down in a time t at a velocity v. How much force must we apply to the air to achieve this?

Newton's second law tells us that F=ma, and we know that the acceleration is the rate of change of speed - i.e. a=v/t. So the force we are applying to the air (which is the same as the force the air is applying to us, by Newton's third law) is

where m/t is the mass pushed down per unit time. So basically we have a choice - we can push a small mass down at a high speed or a large mass down at a slow speed.

Let's imagine we have four fans, one at each corner of the car, similar to many drones and science fiction flying cars. We'll give each fan an area of one square metre. How much mass of air can they push down? Ninja Physics/Paul Francis 33

The blue cylinders are the volume of air blown through each fan in time t - a cylinder of cross-sectional area A and length velocity times time (vt).

The mass blown down per unit time is thus the density times the volume for each fan. So

for each fan.

To support a car (mass ~ 103 kg) with four such fans each of area 1 square metre, we find that

so

The mass being pushed down per second is 4x1x50=200 kg per second.

So we can estimate the energy needed - the kinetic energy of the gas is

every second, so 250000 kW. Ninja Physics/Paul Francis 34

So - to get to Sydney, we'd need to stay in the air for 3 hours (3x60x60 ~ 104 s), so the energy needed would be around 2.5x109 J!

If we remember, driving to Sydney required around 2x108 J, so a flying car of this design would need around ten times more energy.

Feasibility?

So is this feasible? The very high energy consumption is certainly a problem. But we could have said the same thing in the 19th century when people first proposed replacing horses with cars. "They will use much more energy!" And they do. But much more energy was available, thanks to fossil fuels, and cars took over.

So if we come up with some plentiful, easily stored, clear energy source (pocket fusion reactors anyone?) then flying cars would certainly be possible. We'll come later on in the course to the question of whether there are fundamental limits on energy supplies.

But if we're stuck with current energy supplies, this huge energy consumption is certainly a problem.

But is there anything we can do about it?

Well - one thing to note is that the power needed to stay in the air depends on the time you spent in the air. If you make your flying car faster, you don't need to spend so much time in the air. If we go at around 110 km/hr, we use around 109 J to stay in the air and around 108 J to overcome drag, for a total of around 1.1x109 J. But if we double our speed, we only need to stay in the air for half the time, so our "staying in the air" power usage drops to only 5x108 J. On the other hand, drag is proportional to v2, so doubling the speed increases the drag energy to 4x108 J, for a total of 9x108 J. A little better (and a faster car is definitely good), but still pretty prohibitive. Going faster still, and the drag becomes very big and makes things much worse. Ninja Physics/Paul Francis 35 Perhaps we don't need to stay in the air the whole way to Sydney? You could drive on the ground most of the way and only take to the air when there's a traffic jam, or to fly over traffic lights. That might be feasible if you can make a car that can convert between driving and flying easily.

How about the size of the fans? If you remember, the upwards force is equal to the mass per unit second pushed down times the velocity it is pushed down. So we could have larger fans and a slower speed, or smaller fans and a higher speed, and still keep our flying car in the air.

But the energy needed is proportional to mass times velocity squared. So if you have fans that are ten times smaller but blowing the air ten times as fast, the energy changes by (1/10)x102

- i.e. becomes ten times greater, the change in speed outweighing the change in mass, because the speed is squared.

So what this is telling us is that to minimise energy usage, it is best to push the maximum possible amount of mass downwards at a lower speed. An ideal flying car would pull hundreds of elephants downwards. Because they are lots of them and they are heavy, you wouldn't need to pull them down fast, so your energy usage would be very low. An African elephant has a mass of around 6 tonnes (6000 kg) so if you could push a hundred elephants down per second, m/t would be 6x105 kg/s, so to support a car you'd only need to push them down at around 1.7 cm/sec, and your total energy usage would be

So if there were large number of African elephants floating gently in the air between

Canberra and Sydney, you could have a flying car with large robot arms that gently pushed a hundred of them down at a gentle pace every second, and keep yourself in the air with a power consumption of less than 100 W (less than 106 J for the whole trip to Sydney). Ninja Physics/Paul Francis 36 But alas there aren't that many flying elephants around (probably just as well - imagine all the dung, and how much farmers would complain when a flock of elephants landed on their crops...). But the point remains true - if we could push more mass down, we'd only need a lower speed and our energy usage would be less.

Helicopter?

One way to push more air down would be to have bigger fans. Consider a really large single fan - i.e. basically a helicopter of mass M, with rotors of radius r (~ 5 m for typical light civilian helicopters). Let us assume that all the air that goes through the disk traced out by the rotor tips is pushed down at some speed v. This gives us a cylinder of air of radius r and length vt pushed down in time t, very similarly to what we did for the flying car. So the mass pushed down per unit time

and the upward force is

so

The power (energy per unit time) is

Ninja Physics/Paul Francis 37 (i.e. it is proportional to r square and v cubed. But in order to generate enough thrust to keep us in the air, we saw that v is inversely proportional to r. Put that in and we find that

So basically the bigger the rotor, the smaller your power consumption.

So if we want to reduce the power needed to keep our flying car in the air to around 108 J

(I.e. comparable to the drag and ten times less than our calculated value), instead of four fans with areas of one square metre (so radii of around 50cm) we'd need four fans with radii ten times as large - i.e. around 5m in radius. Or a single fan with a radius big enough to give the same area as these four fans (10m). So it would look like a helicopter. But a helicopter with a rotor radius around twice as large as typical civilian helicopters. And that's assuming 100% efficiency for the rotor blades at pushing the air down that comes within the disk traced out by their tips.

Does this make sense? It suggests that typical helicopters would be at best around 2 times less efficient that cars, and probably worse (they have to carry all that fuel and so would be more heavy, and they need more safety features, a tail etc). Which sounds about right - helicopters are clearly more expensive to run than cars, or everyone would use them, but not prohibitively more expensive, as they are widely used by rich people, the military, and in emergencies.

Doing this calculation, it seems obvious that you should build helicopters with longer rotors. If you can extend the rotor length of 10m they would be as efficient as cars, and if you could go to 20 or 30m they would be an excellent method of transport (though too big to really be described as a car and fit in normal parking spots).

Why don't we do this? I'm not sure, but a few possibilities spring to mind. Firstly, there will be drag as the blades rotate, which we haven't considered. Long rotor blades have more cross-sectional area and probably have to move faster at the tips in order to push enough air Ninja Physics/Paul Francis 38

down, so their drag will be high. If you make the rotors too long, the tips might need to go supersonic, which would probably be a bad idea. Helicopters with longer blades will need larger places to land, and the longer blades would presumably have worse aerodynamics and would need to be stronger and stiffer to avoid snapping under the stress of operation. This would probably be a good Ninja physics project in itself - what are the fundamental reasons why helicopter blades are not longer?

Yet another approach would be to push down something heavier than air. One interesting approach is "flyboards" (https://en.wikipedia.org/wiki/Flyboard), where you "fly" over water, connected via a 20m hose to a water pump in a water scooter on the surface. This water scooter pumps high pressure water up to you, which is then squirted out of jets attached to your feet, allowing you to hover. Because the flyboard is squirting water (rather than air) downwards, and the power source is on the water scooter and not the flying object, this is perfectly feasible and energetically quite affordable.

You could presumably do this with a car, but it could only travel over water and would be connected to a boat on the surface, so I'm not sure how useful it would be.

Aeroplane?

The most common way to push air down (and hence fly) is to have big wings which deflect air-flow downwards. This is how planes work, and judging by how common they are, it looks like they must be the most efficient for of air travel. So can we work out the key physics here to see if this approach could work for a flying car?

The basic idea is that the wings deflect air downwards at some angle . Ninja Physics/Paul Francis 39

The aerodynamics of how this happens will be complicated, but basically air that passes close to the wind will get deflected, while air passing further away won't. In reality things will gradually shift from one to the other, but to make our estimate simple, lets just assume that the air is either deflected (within height range h) or not.

So how much air will be deflected?

All the air in a slab of width W (the wingspan) and height h (the range over which the wing affects the air) will be deflected down. In time t, the length of this slab will be vt, as usual. Ninja Physics/Paul Francis 40

Now let's guess that h is 10% of W - that sounds roughly right to me. And that the deflection angle is about 10 degrees. Both these guesses are highly uncertain, so let's test them using a known aircraft.

The calculation is basically the same as we've seen before, except that the lift force is given by the vertical acceleration of the air, which instead of being mv/t, is . The mass in the slab is

So the upward force is

We can use this to calculate how fast a plane must travel to be able to support its own weight and hence take off:

Let's test this, using a known figure. According to the internet, the Airbus A380 has a maximum take-off weight of 575 tonnes, a wingspan of 80m and a take-off speed of 150 Knots

(~100 m/s). But by our formula and making our assumptions, we get a take-off speed of around

190 m/s.

So the agreement isn't bad, but it looks like real planes do better than we estimated - if we increase h to be 20% of W, and increase the deflection angle to 20 degrees, we get good agreement.

So we can now use this (with our new values of h and the deflection angle) to estimate what a winged flying car might look like. If we want to take off on a typical suburban street, we Ninja Physics/Paul Francis 41 will need a wingspan of no more than around 6m. Plug that in for a 1000 kg car and we find a take-off speed of around 60 metres per second (216 km/hr).

Which seems a bit extreme to be practical. You could take off on a long freeway OK, but you'd have to drive there first.

How about the efficiency? If you have a long wing, you would be directing a lot of air down, so your efficiency is probably fairly good, but a short wing will hurt you here - you'd need to go faster to direct enough air downwards, which would reduce your efficiency.

Conclusions

So what have we learned about the physics of flying cars? Is there a good physical reason why we don't have flying cars?

Basically, yes. The only sensible way to stay in the air seems to be by pushing air downwards. And for this to be reasonably efficient, you need to push a large amount of air down at a slow speed (rather than a small amount at a high speed). For this to work, your flying car needs to be big. There are basically three options:

● Jet pack model - pushes a small amount of air down very fast. So incredibly inefficient, which

is why jetpacks have never really caught on.

● Helicopter model - use rotors to push a medium amount of air down at medium speed. Still a

bit big to really replace a car, and still pretty expensive to run, but they do exist for rich

people and special occasions.

● Aeroplane model - use wings to push a large amount of air down at a low speed. Efficient but

so large they need special facilities (called airports). Very streamlined to reduce drag, and

they take advantage of the thin atmosphere at altitude by flying high to reduce drag. Ninja Physics/Paul Francis 42

As an interesting aside, this idea that to be efficient you want to push lots of air back at a low speed is fundamental in the design of jet engines. The most efficient ones are so-called "high bypass" engines which use a small turbine in the middle to drive a very large fan, hence pushing a large volume of air back at a slow speed.

So it is really no mystery why we don't have flying cars - there are good physical reasons why it’s very hard to have a car-sized flying machine that is reasonably efficient.

How might this change?

One possibility would be if we had some new cheap compact energy source, as we mentioned earlier. Portable nuclear reactors would make many things possible.

Without that, what could we do? Our best hope comes from the fact that fans only burn energy while they are on. We could perhaps have some type of car that only used fans briefly.

Perhaps it takes off using them but then deploys some long (efficient) wings to fly to your destination. Most of the flying cars currently under development seem to use some variant of this. Or you might drive most of the way on the ground, then only take off to skip traffic jams.

Are there some more radical solutions? Perhaps we could use something other than

"pushing air downwards" to keep our cars in the sky? Perhaps a car that swung from cables, similar to Spiderman? Perhaps you could have water jets on the ground that squirt water up to keep cars in the air? Generate really strong electric or magnetic fields and use electromagnetic forces to keep a car up (it works for maglev trains but is very hard if you are more than a few centimetres from the magnetic field source).

Here are some current attempts to build flying cars -

http://www.vox.com/new-money/2016/12/30/14105960/flying-car-future-explained

Here's a NASA engineer's look at the possible science of feasible flying cars: Ninja Physics/Paul Francis 43

https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20110011311.pdf

Another quite different approach is to get 3D motion not by flying, but by going underground...

https://www.wired.com/2017/01/inside-tunnel-elon-musk-already-digging-los-angeles/y

Lessons Learned

What new Ninja physics techniques did I use in this worked example?

● Use general knowledge. When working out the thickness of the atmosphere, I used the fact

that you get short of breath on high mountains, combined with my knowledge of

mountain heights.

● There is more than one way to solve almost every problem. And it's usually good to use

several methods as a check. So for working out the thickness of the atmosphere I used

two methods and compared them to a formula.

● Compare to real-world situations. When working out the feasibility of a catapult, I compared

to the Amiens rail gun, while with flying I compared to planes and helicopters. This is

both a short-cut and a check.

● Look at what equations depend on (e.g. force due to pushing air down depends on mass and

velocity) and use this to work out behaviour for different cases.

● One word ("flying") may contain a myriad of different physical concepts (e.g. rising up,

hovering, buoyancy, blowing air down). Unless you tease them all apart with some hard

thinking, you'll be in trouble Ninja Physics/Paul Francis 44

Appendix: Ninja Ticks

Orders of Magnitude

It is often hard to stop worrying about details and focus on the big picture. Using "Orders of magnitude" is one mental trick that can help.

The order of magnitude of a number is that number rounded to the nearest power of ten.

So 1.7x103 would have an order of magnitude of 103. 9.6x10-4 would have an order of magnitude of 10-3. When do you round up as opposed to rounding down? By whichever ratio is closest, which means values of greater than the square root of 10 (~ 3.1) are rounded up, while values below that are rounded down. So 3.5x109 rounds to 1010, while 2.7x109 rounds to 109.

Another way to think of this is rounding to the nearest integer logarithm (to base 10). So

3 3 log10(2.7x10 ) is 3.4, round it to 3, so the order of magnitude is 10 .

We say two numbers have the same order of magnitude of a number if the big one divided by the little one is less than 10. For example, 23 and 82 have the same order of magnitude, but 23 and 820 do not.

-- John C. Baez

Ninja Physics/Paul Francis 45

Why do this? It forces you to focus on the biggest effects and ignore the details. And it makes calculations in your head really easy. Once you've got a rough answer this way, if it is interesting you can go back and add back in factors of 3 etc to get a more accurate answer.

And please don't take this as a set of rules to be followed. It's a helpful hint - but don't take it too seriously.

#

Guesstimation and Bracketing

Traditionally an important part of Ninja physics was guesstimating numbers. This is less important now with the internet - if you want to know the drag coefficient of a cricket ball or the magnetic field of Jupiter, just look it up.

But often you need to estimate the order of magnitude of a number which you can't look up, or you want to check that a number you got off the internet is roughly right (often they are totally wrong).

Here are some tricks to help you estimate a number.

● Compare it to something you do know. For example, if I asked you how tall a 10 story

building is (in metres) you probably don't have a figure off the top of your head. But you

know roughly how tall a person is (1.5-2m), so each floor must be a bit bigger than that

(2.5-3m?), so a 10 storey building will be something like 25-30m tall.

● Geometric bracketing. Work out the highest and lowest plausible values and then take the

geometric mean. For example, if I asked you to estimate the width of Sydney Harbour,

step through orders of magnitude of distance and see what is plausible. Is a width of 1m

plausible? Clearly not - you could touch the other side. 10m? Still too small - that's

something like 10-15 paces, and it takes a lot more than that to walk over the bridge. Ninja Physics/Paul Francis 46

100? Still sounds a bit small, but getting closer. Now let's try on the upper side. 1000km

is much too big - that's the whole way to Brisbane. 100km? That's bigger than all of

Sydney - if it was that big you couldn't see the other side. 10km - also sounds a bit big.

So we're saying it has to be somewhere between 100m (102 m) and 10 km (104 m). So

take the geometric mean, which is 103 m - one kilometre. And you're probably not too far

off.

● Work backwards from its effects. Instead of trying to work out a number directly, work

backwards from what it can do. For example, how much energy is there in a punch? No

idea - but I do know that a really big punch to the chin can lift someone off the ground by

a centimetre or two. A person weighs ~ 102 kg, g~10, 1cm is 10-2m, so mgh ~10J.

● Subdivide the problem. And keep subdividing until you have something you can estimate.

This is often done with a tree diagram - a diagram with branches, where you break down

each number into sub-numbers, which are themselves broken down. For example, how

many grains of sand are there on Earth? Here's one possible tree diagram for this: Ninja Physics/Paul Francis 47

#

Convert to familiar units

Many numbers in physics and astrophysics are much too large, too small or too abstract to easily interpret. This means that typically you will have no idea if they are right or not. If I told you that a star had a mass of 1024 kg - would that look sensible to you? It's a big number, stars are big - but the difference between 1020 big and 1024 big is like the difference between a metre and ten kilometres - quite important! How about if I told you that a molecule was 10-7 m big. That's a small number, molecules are small, but is it really sensible? Or if I told you that a bar magnet produced a magnetic field of 5 T. Plausible? Or a person running at 10 metres per second? Ninja Physics/Paul Francis 48

One good way to check is to compare the number to something more familiar. Perhaps change the units to something you know better.

Consider the runner. Is 10m/s a sensible speed? Most of us don't think of speeds in metres per second. But if you convert this into km per hour, it comes out as 36 km/hr. Which seems remarkably fast!

How about a molecule being 10-7 m across? You can compare this to the typical size of an atom (which is something we derive later in this course, and which you can look up), which is around 0.1 nm (10-10 m). So you can see that this molecule is around 1000 times bigger than an atom. That's a big molecule! Might be OK if you were talking about some complicated protein or long-chain polymer, but if you were telling me a water molecule was that big, you'd need to go back and find out what went wrong in your calculation.

And how about a star having a mass of 1024 kg? Compare it to the mass of our Sun, which is 2x1030 kg. Suddenly 1024 kg doesn't seem so big any more - it's two million times less massive than the Sun. In fact it's ~ 6 times less massive than the Earth. A star that's less massive than a small rocky planet? I think something must have gone wrong in that calculation!

Ratios

This is a very useful trick that is surprisingly rarely used by students.

Basic idea - let's say you are trying to work out some property, like the brightness of a star or the amount of energy generated by a wind turbine. The equation to work it out is long and complicated, and depends on several parameters you don't really know. So what do you do?

Often, you will know this property for some other case. So you might want to work out the brightness of the star Vega, and you know the brightness of the star the Sun. Or you are Ninja Physics/Paul Francis 49

trying to work out how power a 10m tall wind turbine near Canberra will generate, but you know how much power a 5m high wind turbine near Adelaide generates.

In situations like this, you can often use ratios to get your answer very quickly and easily.

But only if you know how your property scales.

For a star, it is known that the brightness is proportional to the 4th power of the mass of the star (with some caveats we won't go into here...). Vega is 2.1 times more massive than the

Sun. So Vega should have a luminosity 2.14 ~ 20 times greater than the Sun.

For a wind turbine, the power roughly depends on the square of the height, and on the cube of the wind velocity. If the average wind speed near Adelaide is 30% greater than near

Canberra, then the lower wind speed should decrease the power generated by a factor of 0.73 ~

0.34, but the greater height should increase it by a factor of 22~4. So the total Canberra wind turbine power will be 0.34x4 ~ 1.36 times that of the Adelaide turbine.

Swept Volume

This is a surprisingly useful trick that crops up all over the place in a wide variety of situations - definitely one to have in your mental tool box.

You can use it if something is moving and you want to know what it will hit. Or if a fluid is moving past an object and you want to know how much is going to interact with that object.

Or you are squirting out some fluid and you want to know how much you have squirted out.

The basic idea is to work out an imaginary volume (the "swept volume") which contains everything you are going to hit or interact with. The amount of material you are going to interact with in a particular time is the amount of material in this volume. Ninja Physics/Paul Francis 50

We already used this method to work out the drag equation (by seeing the volume of gas that will hit the front of a car) and the lift force due to fans (by seeing the volume of air blown out by the fans).

The swept volume in a time t is equal to the cross-section area A times vt, where v is the relative velocity of object and fluid. Ninja Physics/Paul Francis 51

Chapter 7

Cameras

Why aren't Smartphone Cameras Better?

Smartphone cameras have become vastly more powerful over the last few years - they are a major selling point for many phone buyers, and so the rival smartphone companies are frantically researching ways to make them better.

They currently have two main drawbacks. One is lack of much zooming capability. If you

"pinch to zoom", your image rapidly becomes blurry and pixelated. The other is low-light capabilities - often when you take pictures in low light conditions, they are very grainy. Ninja Physics/Paul Francis 52

If we "pinch to zoom" on this image...

we start hitting the resolution limit - note that the writing is too blurred to be readable.

Why aren't cameras better? What's limiting their performance? Is there something we could do to improve things? Let's see if we can figure this out. Ninja Physics/Paul Francis 53

Note that this is of deep interest to astronomers. It would be great to be able to take pictures of planets around other stars, photograph the aliens as they relax in their gardens. But we can't with our current telescopes. Why can't be just build telescopes with more magnification?

This is also of immense interest to many other people, particularly the military. Huge amounts of military equipment depends on various sensors, and it's crucial to understand their limitations and what (if anything) can be done about it.

This is another example of "going to first principles". It will also introduce some key ideas in optics, waves and quantum mechanics.

Digital Camera Basics

At its simplest, a camera consists of a lens and a detector (typically a Charge Coupled

Device or CCD).

The lens is made of some substance with a refractive index greater than air. When the light hits the boundary between the lens and the air, it is deflected (according to Snell's law).

Optical designers will optimise the shape of the lens, using ray-tracing software, so that an ideal lens should take the light from any given point on the object being photographed, and focus it on a particular point on the detector.

Real lenses are more complicated - they typically consist of a whole series of bits of glass, designed to give sharp images at a range of wavelengths, across the whole field of view, but this simple model gives you the basic idea. Telescopes and the human eye work similarly. Ninja Physics/Paul Francis 54

(exploded view of iPhone camera, showing the large number of nested lenses.)

If you need to, you can look up a large number of equations relating the distance to the object, the focal length and the shape of the lens. But if we assume that the lens designers have done their job correctly, the main thing we need to know is that a ray drawn through the centre of the lens is not deflected - it continues in the same direction after passing through the lens.

Pixel size limit

The detector consists of a number of pixels. Whenever a photon of light lands on a pixel, it has some chance of exciting an electron, creating an electron-hole pair. The electron is then attracted to an electrode. When you read out the detector, the number of electrons is counted, and these electron counts are used to make up the image.

What limits how much detail you have in your image? One limitation is the number of pixels in your detector. Let's investigate this limit.Let's consider some light rays that pass through the centre of the lens - they are not deflected:

Ninja Physics/Paul Francis 55

All the light that falls on one pixel is recoded only as being in that pixel. It doesn't matter whether the light falls near the top of the pixel or near the bottom - that information is lost. So this places a fundamental limit on how much detail you can see - detail that shifts where the light falls on the detector by less than the size of a pixel is lost.

For objects that are a reasonable distance from the camera (i.e. r >> D, where D is the diameter of the lens), the best measure of how much details you can see is the angle which shifts the position on the detector by the width of one pixel.

So what detail size d corresponds to what angle and what pixel size p?

Small Angle Approximation

It is easy to get bogged down in trigonometry working this out, but luckily there is a widely-used approximation you can use to greatly simplify this.

If you have a long thin triangle and you know one of the long lengths, what is the end length? I show above three possible interpretations of this problem.

In the first, the triangle has a right-angle at the top right. L is thus the hypoteneuse and p is the opposite. The ratio of opposite to hypoteneuse is known as the sine, so Ninja Physics/Paul Francis 56

and rearranging:

In the second case, however, the right-angle is at the bottom, so L is the adjacent and p the opposite. The ratio of opposite to adjacent is known as the tangent of an angle, so

And so

In the third case, p is part of the circumference of a circle of radius r. The whole circumference would have a length , but the arc is only a fraction

of the whole circle, if the angle is measured in degrees, or

if the angle is measured in radians (there are two pi radians in a circle).

So if we measure the angle in radians,

So we have three possible equations for p (and you could come up with more - for example dividing the triangle into two with right-angles in the centre.

Luckily, it doesn't make any difference which one we use, as long as the angle is small.

That's because of the small angle approximation - for , (try it out on your calculator if you don't believe me). Ninja Physics/Paul Francis 57

So let's use the simplest -

Angles

We can use this to work out the angle subtended by a pixel.

To be specific, let's use the camera in my smartphone, a Samsung Galaxy S6. This uses a

Sony IMX240 sensor, which has a focal length L=4.3mm, and has 1.12 micron pixels and a f- number (Focal length divided by the diameter of the lens) of f1.9.

So what range of angles of light will fall into a single pixel? P = 1.12 microns and

L=4.3mm, so

Radians are not a familiar unit to most of us, so let's convert this into degrees, remembering that a radian is 180/pi degrees. It comes out as 0.0149 degrees - still too small to easily comprehend. There are special units for very small angles:

An arcminute, which is 1/60 of a degree

An arcsecond, which is 1/60 of an arcminute.

This angle is just a little bit less than an arcminute.

Now what could we see with this. Let's say I'm looking at an object r=5 metres away.

What detail size d corresponds to this angle? Using the small angle approximation once more,

i.e. around a millimetre.

Is this right? Here's a picture of the far side of my office.

Ninja Physics/Paul Francis 58

Now if we zoom in on that old exam paper: Ninja Physics/Paul Francis 59

You see that we can clearly separate out different words, but not really separate out different letters. So a minimum detail size of around a millimetre is about right!

Limits to Angular Resolution

This minimum angle that allows you to resolve details is called the angular resolution: the smaller the better!

So can we make it smaller (better) and see more details?

One way to do this would be to increase the focal length: remember that

Ninja Physics/Paul Francis 60

Indeed most dedicated zoom lenses are very long - and telescopes are much longer still.

But that won't easily work for a smartphone - you don't want a long lens sticking out the back.

So how about making the pixels smaller? Detectors are basically specialised silicon chips, and the technology exists to make silicon components down to around 10nm size - 100 times smaller than the pixels in my smartphone. If you used a detector with pixels 100 times smaller, wouldn't that be like having a 100 times zoom? You could take a photo of a book on the opposite side of a football pitch and read the words!

What could stop this working?

One possibility is imperfections in the lens. If the lens is not polished to precisely the right shape, some of the light may not end up where it is supposed to. But this is seldom the major problem with modern high quality optics.

Another possibility is chromatic aberration. The refractive index of most glasses depends on the wavelength, so a lens that focuses (say) blue light perfectly will probably not focus red light very well, or vice versa. Camera designer try to overcome this using multiple lenses made of different glasses, and mostly succeed, though it can remain a problem. And it introduces a new problem - every time light passes from material with one refractive index to another, some is reflected. The bigger the difference in refractive index, the more the reflection: for light passing from a substance with refractive index n1 to one with n2, the fraction reflected is given by:

which for light passing from air (n1=1) to glass (n2~1.5) gives 4% reflection. Have too many air-glass reflections and you can be losing serious light, as well as getting ghost images Ninja Physics/Paul Francis 61 from all the reflected light bouncing around. More expensive lenses have thin coatings on their surface with intermediate refractive indexes, which minimise this loss.

Yet another possibility is that a lens has to be designed to take light from anywhere in the field of view and focus it perfectly. It is possible to design a lens to focus light from one direction perfectly, but being able to handle all possible directions is impossible. Designers use ray tracing software - which sends imaginary rays of light from sources at different distances and different directions though different parts of the lens - they try to optimise the performance so that objects of all colours, at all focus distances, in all parts of the field of view, all give sharp images.

Once upon a time, this was regarded as impossible. Back then, all lenses had spherical surfaces, because of the way they were polished. By combining several spherical lenses you could do a pretty good job of balancing all these requirements, but not a perfect one.

A few decades ago, however, Japanese optical designers decided that it was now possible with computer milling machines to polish a lens to any mathematical shape they like, not just a sphere. By combining several of these "aspheric" lenses, you can make extremely good lenses - at a price!

So lens design isn't the fundamental limit, at least so far. The fundamental limit is diffraction.

Huygen's Principle

To understand the diffraction limit, we'll have to stop thinking of light as little particles

(photons) that go in straight lines (rays), and instead think of it as a wave. Visible light is an electromagnetic wave, with wavelength ranging from around 0.4 microns (blue light) to around

0.7 microns (red light). Ninja Physics/Paul Francis 62

Now the funny thing about waves - they don't really move. Think of a water wave. The wave moves, but the water only goes in little circles, ending up back where it started. If you pile up water in some place and let go, it spreads out in all directions. And the same applies for electromagnetic waves. If you have an appropriate electric and magnetic field in one location, it will spread out in all directions.

Why then does light travel in a straight line? The answer to this comes from Huygen's

Principle, which says that waves spread out in every direction, but only move in the direction in which all the peaks and troughs line up in phase.

Diffraction Limit

So - if you have an infinitely long straight wave, it will continue to move in a straight line.

And if you have a point-like wave, it will spread out like ripples from a stone in all direction. Ninja Physics/Paul Francis 63

But what happens if you have a long straight wave but it had to go through a hole (like a lens) so it is still straight, but is now short. What happens then?

Consider the spherical waves from the top and bottom of the hole. Going straight ahead, they add up in phase - peaks like up with peaks, so the wave definitely goes that way. But if we shift the angle a little bit, do they still add up in phase? Ninja Physics/Paul Francis 64

If you look at the waves going off at an angle , the wave from the top has to go an extra distance .

If this extra distance is comparable to the wavelength , then the waves from top and bottom (and all the places in-between) will not all line up - there will be destructive interference, but if , the waves will mostly add up in phase, so you will get a strong signal.

This means that light going through an aperture of size D will spread out over an angle of roughly

(measured in radians).

And this really happens: Ninja Physics/Paul Francis 65

The narrower the slit, or the longer the wavelength, the more the waves will spread out.

This is called diffraction. Ninja Physics/Paul Francis 66

#

Camera Diffraction Limit

So what is this diffraction limit for a smartphone camera?

For my Samsung, the lens diameter is 2.26mm. For visible light, the wavelength is around 5x10-7 m. So the diffraction limit:

Now if you remember, that was about the angle subtended by the pixels.

So this tells us that diffraction spreads out the light coming in a given direction by about the width of a pixel - so the limit due to the pixel size and the limit due to diffraction are about the same.

This is not a coincidence - there would be no point in having smaller pixels, because diffraction would just spread the light over more of them - you wouldn't see any more detail.

This is why most smartphone cameras have ~ 1 micron pixel sizes: given their short focal Ninja Physics/Paul Francis 67 lengths, there is no point in having smaller ones, as diffraction would blur your image anyway by this amount.

How could we avoid this limit? The obvious way is to have a wider lens (larger D). But that would make the camera larger and harder to fit in a smartphone. It is really hard to design a set of lenses whose focal length is less than its diameter, and still get a good image quality everywhere.

Most professional photographers who need to zoom in a lot (like wildlife photographers) will have very big and very wide zoom lenses.

Astronomers, who need to zoom in even more, use really big telescopes!

The other alternative is to use shorter wavelengths. If you took a picture at ultra-violet wavelengths (~ 0.3 microns) it could be twice as sharp. But nobody really wants to see what things look like in the ultra-violet, so that's not very useful for commercial cameras (and light with wavelengths less than around 0.32nm can’t penetrate the ozone layer…).

Magnification

People often ask what "magnification" a camera or telescope has.

Magnification can mean different things. If you take a small photo and photocopy it up to the size of a poster, the magnification is enormous. But you won't see any more detail in the big print than in the small - probably less, in practice.

So what really matters is not how big an image is, but how much detail you can pick out.

This is called the angular resolution.

We've seen that good smartphone cameras have angular resolutions of around 2x10-4 radians. How does this compare with the human eye? Ninja Physics/Paul Francis 68

I just drew two dots on a piece of paper, roughly one mm apart. I propped the piece of paper up on my desk, and then walked backwards, still looking at the two dots. I wanted to see how far back I could go and still pick out the fact that there were two dots, not just one. The answer was about 5m. Similarly, I can just about read printed material in a letter out to a few metre distance.

So the angular resolution of my eye is roughly 1mm/5m ~ 2x10-4 m ~ 1 arcminute. Worse if you need glasses.

So this means that the resolution of a good modern smartphone camera is about the same as the human eye. So if you like, you can say that these cameras have a magnification of one - you see the same level of detail as you would by eye.

Photons

So we've seen that the wave nature of light places a fairly severe limit on the angular resolution of smartphone cameras. You can't zoom in just by having smaller pixels - they would just give you a larger blurrier image, with no more detail visible.

Now how about the other challenge to smartphone photography - low light performance?

If you take pictures in dim conditions, they usually come out grainy, and sometimes completely black. Why is this? Why do cameras need lots of light to take pictures?

To understand this, we need to think about light as particles - photons, rather than as waves. Light is in reality both a wave and a particle, and mind-bending quantum mechanical weirdness that we will cover later in this course. To understand diffraction, you have to think of light as a wave, but to understand the limitations of low-light photography, you have to think of light as a particle. Ninja Physics/Paul Francis 69 Quantum mechanics tells us that the energy of an individual quantum of light (a photon) is , where h is Planck's constant 6.63x10-34 m2 kg s-1, and (nu) is the frequency.

What is the frequency of visible light?

For any wave, its speed is equal to its frequency times its wavelength. Why? Imagine watching a wave go past. You count waves going past each second, each of wavelength . So the total length of light-wave that has gone past must be , and is equal to the speed of the wave.

So for light, which always travels at the speed of light (duh) in a vacuum,

Visible light has a wavelength of around 5.5x10-7 m, and c=3x108 m/s, so we calculate a frequency of around 5.5x1014 Hz. Extremely high!

If we multiply this by Planck's constant, we find that each visible light photon has an energy of around 3.6x10-19 J. Not big.

Lightbulb Power

OK - we've seen that photons are pretty weak. But how many of them would you get in a typical low-light photography situation?

What is a typical photographically challenging situation? I imagine trying to take a picture of someone in my garden, lit only by the light-bulb on our porch, which is around 10m away. I know from experience that this is about the limit of my smartphone camera.

So how many photons are involved in a situation like this?

First of all, let's work out how many photons are put out by the lightbulb. We then need to work out how many hit the person, bounce off and reach the camera, and then land on a given pixel on the detector. Ninja Physics/Paul Francis 70

Let's start with the power put out by the light-bulb. A typical incandescent (tungsten filament) light bulb has a power-rating of around 60W, while an equally bright LED light has a power-rating of around 6-9W. But we know that light-bulbs get hot, so much of the energy is being wasted as heat, not turned to light. If you search the web on this, you find that LEDs convert around 10% of the power into light, while incandescent bulbs convert around 2%. So either way, we are getting around 1W of light energy out.

Another way to work this out is to look up the brightness of a typical bulb, which is measured in lumens. One lumen at the wavelength the human eye is most sensitive to (0.55 microns, yellow) is roughly 1/700 W, and a typical 6-9 W light-bulb puts out around 450 lumens.

So that gives us around 0.5 W, comparable to the figure above.

So let's run with 1W as the power of a light-bulb.

Inverse Square Law

But how much of this 1W of power reaches the person we are photographing?

If we assume there is no fog, the photons will not be absorbed until they hit something.

But as they fly out from the light-bulb in all directions, they are spread over a bigger and bigger area, so the number of photons hitting a particular unit area goes down.

Imagine you have a light-source that emits a total power L uniformly in all directions.

This total power is called the luminosity in astronomy, and might represent the radiation coming from a star, or the light power emitted by a light-bulb. Now imagine a sphere of radius r centred on the light-source. All this power L must pass through this sphere (if there is no fog to absorb it inside the sphere). As the light is assumed to spread out equally in all directions, that means that the power is spread uniformly across the sphere, so the power per unit area at the sphere (known as the flux f) is just the luminosity divided by the surface area of the sphere: i.e. Ninja Physics/Paul Francis 71

This is an example of an inverse-square law - something that decreases as one over the distance squared. Other examples include gravity and electric field.

So the flux of light hitting someone 10m from the light-bulb will be

Some of this will be absorbed when it hits the person, and some will be reflected off. It it hits a white tee-shirt, most would be reflected, while if it hits dark hair, rather little will be reflected. For the moment, to keep things simple, let's assume it's all reflected (say shining off someone's gleaming white teeth...).

How much hits each pixel?

So the light bounces off the person - let's assume it bounces off evenly in all directions

(probably not true, but close enough for this sort of Ninja calculation). Some fraction will hit the camera lens, and land on a particular pixel.

Ninja Physics/Paul Francis 72 There are two effects going on here. Firstly, only the light from a small square region of the person will fall in a given pixel on the camera. The further the person is from the camera, the larger will be the area of the person per pixel. And because the area is larger, more light from the light-bulb will be hitting it and bouncing off.

The light from this square region will then spread out in all directions, and as it spreads over a larger and larger sphere, there will be less light per unit area.

So let's say the amount of light reflecting off the person per unit area is I.

The square on the person that will be focused on a given pixel has length l, where by the small angle approximation, . Now if you remember,

where p is the pixel size and L is the focal length of the camera. So factoring this in, the total amount of light power reflected from this square is

This light will spread out over a sphere, as per the usual inverse-square-law. Well, actually it will bounce off more in some directions than others, but this should be good to within a factor of two or so. So the power per unit area (flux) reaching the camera lens will be

How much of this will go through the camera lens? The power going into the camera will be the power per unit area times the area of the camera lens. If the lens has a diameter D, then it's area is

So the power entering the camera lens that will fall on his particular pixel is given by Ninja Physics/Paul Francis 73

A few of things to note about this. Firstly, the amount of power does not depend on how far the camera is from the target! Which sounds strange - surely you'd get more light if you were closer in? The reason is that we are dealing with the light on each pixel, and the further away you go, the more of your target fits into each pixel.

Secondly, it depends on D/L. The diameter of the lens divided by your focal length. This parameter has a name - or to be precise, its reciprocal has a name, the f-number. The f-number is

L/D, so we see that the amount of light energy hitting each pixel depends on 1/f2. So you really want a small f-number to get lots of light on each pixel. Indeed lenses with small f-numbers are called "fast" lenses. Current smartphone camera adverts boast of the small f-numbers of their lenses, as this increases their low-light sensitivity.

Thirdly, other things being equal, small pixels means less light per pixel. This might be a problem - but how serious is it?

Count the Photons

We now have all the clues we need to work out how many photons are landing on each pixel. We know that I ~ 10-3 W m-2. For my Samsung Galaxy S6, the pixel size p = 1.12x10-6 m, and the camera has a f-number of 1.9. So the energy landing on each pixel per second, when I point at the person lit by a 10m distant light-bulb, is

Not very much power.

Now we calculated earlier that each visible light photon has an energy of around 3.6x10-

19 J. Which means we are getting around 60 photons per second per pixel. Ninja Physics/Paul Francis 74

But if our exposure time is 1 second, camera shake will blur our photo (unless we are using a tripod). To get a sharp image, we need an exposure time of something like 1/60 of a second (though we may be able to get away with longer if we use image stabilisation). So we are looking at around 1 photon per pixel! And that's assuming every photon is detected.

This is a very rough calculation, but you can see that we are photon-starved in this situation. You could count the number of photons per pixel in each exposure on the fingers of one hand.

How does this affect the image quality?

Root n statistics

The small number of photons hitting each pixel gives us two problems.

To get a nice image of the person, we'd like to know the relative brightness of their face, their hair, see the shadows under their chin etc. So the expected amount of light from different parts of their body (falling on different pixels in the camera) should be different - otherwise they would look like a cartoon.

So let's say we should be getting 20% more light in one pixel than another. In bright daylight, you might get 10000 photons in one pixel and 12000 in the other, which is fine. But in low light conditions, you might expect to get 1 photon and 1.2 photons. And therein lies the first problem. You can't get fractional photons. They are quanta of energy, and the whole idea of a quantum is that you can't divide it up. You either get one or you don't.

So most likely you will get one photon in both pixels, and the difference in brightness is lost. Or you might (roughly 1/5 of the time) get two photons in the second pixel, in which case the difference in brightness will be grossly exaggerated. Either way, your image won't look good. Ninja Physics/Paul Francis 75

But it's even worse than this. The arrival of photons is a random process. If you expect

(say) 3 photons per exposure in a given pixel, sometimes you will indeed get 3. But sometimes you will only get 2 while other times you may get 4.

Think of it light counting raindrops landing in a particular paving stone in a rain shower.

On average each paving stone may get 1 drop per second. But the arrival of each raindrop is random - some stones may get 3 or 4 drops in one second, and none the next.

We cover this situation in the labs, when we talk about counting experiments.

Mathematically the number of photons you expect is given by the Poisson distribution. And roughly speaking, if you expect n photons, you will actually get photons - i.e. an uncertainty of root n.

So let's say you were getting 100 photons per pixel on average. The uncertainty would be the square root of 100, i.e. 10. Which means that if you took a picture of a perfectly blank sheet of paper, it would not look blank - pixels would randomly look ~10% brighter of fainter than the average. Which would look pretty speckled. To bring the fluctuations down to 1%, you'd need

10000 photons per pixel.

And this is photon noise - a fundamental limit to low-light photography.

Thermal noise

In reality, things are even worse than this. A photon hitting a detector creates an electron- hole pair. The electron is attracted to an electrode, and the charge counted.

But all the atoms in your detector are jiggling around due to their thermal energy. And sometimes this jiggling will hit an electron hard enough to create an electron-hole pair, even in the total absence of light. This is called "dark current". On average you can subtract it off, because it should be the same for each pixel, but once again the root-n statistics will come in to Ninja Physics/Paul Francis 76 play and introduce an uncertainty n how much dark current is in each pixel. The only cure for this is to cool down your detector. This is what astronomers do, but it's not practical for smartphone cameras.

Then your electrode has to measure the charge. But the charge on one electron is pretty small - it must be amplified before it can be counted. But the electrons in the circuit doing the amplification are jiggling around due to their thermal energy, and this to introduces noise.

In practice, with a short exposure time and carefully designed electronics, these effects aren't too bad. But the photon noise is a fundamental problem and no amount of clever engineering can do away with it.

So how can you get better photographs in low light conditions? One answer is a smaller f-number, as we discussed before. But it's hard to get good optics in a very low f-number lens across the whole field of view.

Another possibility is to make the pixel size bigger. But that then reduces the fine detail in your image, unless you make your focal length longer (and hence your camera bigger). This is why my digital SLR camera takes better low-light pictures than my smart-phone: it has a f-1.4 lens, and 4 micron pixels. The f-1.4 lens gives a (1.9/1.4)2 advantage, and the 4 micron pixels gives a (4/1.12)2 advantage.

Another approach is to cheat. Most things you photograph are going to have flat expanses of uniform brightness. Such a region will be spread over many pixels. If you average the signal detected by all these pixels, the averaging will reduce the noise, as we discuss in the labs. Most modern cameras and smartphones use some sort of smoothing to reduce noise in low-light conditions, though this comes at the expense of reducing detail.

# Ninja Physics/Paul Francis 77

Conclusions

So - can we make smartphone cameras better? Not easily? We could make their pixels smaller and have more mega-pixels, but that wouldn't give us much more detail due to the diffraction limit. And in low light conditions, there are too few pixels to give good images through small pixels.

One solution is to build a really big camera with a small f-number and large pixels. And that it what astronomers do. Astronomical telescopes can have angular resolutions of better than

0.1 arcseconds (600 times better than my smartphone), and detect distant galaxies 1011 times too faint to see with the human eye. But these telescopes are limited by the same factors that limit a smartphone camera - the diffraction limit and a shortage of photons.

Another solution is being clever with the software - this can't bring back details lost in noise, but it can do a lot -

https://www.fastcompany.com/3067252/the-future-of-photography-is-about-computation Ninja Physics/Paul Francis 78

Week 4: Quantum Mechanics and the Properties of Matter

Quantum Mechanics and the Properties of Matter

Is it possible to develop materials that are twice as strong as our current best? Ten times as strong? A million times as strong? Could we develop fuels or batteries that store ten times as much energy as our current best? A hundred times more? Or are we stuck with only incremental improvements?

To answer these questions, we will "go to first principles" - i.e. try to see what are the fundamental physical principles behind the properties of matter. And that means delving into quantum mechanics, as without it the fundamental properties of matter are incomprehensible.

Spectrographs

The first evidence for quantum mechanics came from spectroscopy. If you heat up some gas, it will glow. If you take the glowing light and put it through a prism or a diffraction grating, you can split it up into its component wavelengths. Ninja Physics/Paul Francis 79

Schematic diagram explaining how a simple spectrograph works.

The WiFeS spectrograph on the ANU's 2.3m telescope at Siding Spring Observatory.

We normally plot the output from a spectrograph as a spectrum - a plot of the intensity of light at different wavelengths. Here, for example, is a spectrum I took of a quasar (a giant black Ninja Physics/Paul Francis 80

hole in the middle of a galaxy, eating the surrounding gas - this particular one is about 12 billion light years away):

What you can see is that at some wavelengths this quasar emits light very strongly (most strongly of all at around a wavelength of 514 nm) while at other wavelengths it emits very little light (e.g. 380 nm).

How does a spectrum correspond to what you'd see with the human eye? Basically the human eye contains three different types of receptor cells (called cone cells). Each type of cone cell is sensitive to light with a variety of different wavelengths: Ninja Physics/Paul Francis 81

So the "S" cones (S for short wavelength) are most sensitive to light at around 440nm, but have some sensitivity to light over the range 400-520nm.

When you look at most objects, you will be picking up light at a variety of wavelengths.

This light will stimulate all three types of cones to some extent, and the combined signals are processed in the brain to give you the perception of colour. The background colours in the above plot show you what a beam of mono-chromatic light (i.e. light with only one wavelength) would look like. But most objects will give light with a variety of wavelengths.

What would you expect the spectrum of hot gas to look like? Light (and all electromagnetic radiation) is emitted whenever you accelerate charge, as we will see later in this course. If you heat up a gas, the atoms will be bumping into each other regularly, which might disturb the electrons and make them accelerate. But this would be a very chaotic process, causing many different amounts of acceleration, and hence light with all sorts of different wavelengths. Ninja Physics/Paul Francis 82 Instead, what you get is quite different:

This is a spectrum of some heated copper and argon vapour (a combination often used to calibrate telescope spectrographs). What you see is that this gas emits very strong radiation at some very specific wavelengths, and none at all at other wavelengths.

Everything in the above graph is perceived as blue by the human eye - so this level of detail is invisible to us without the aid of a spectrograph.

This is incredibly useful in practice - by looking at what exact wavelengths some gas emits, you can deduce the composition of the gas, and sometimes its motion, temperature and pressure.

But what's the reason for these ridiculously specific wavelengths? Why would the gas above emit so strongly at (say) 454.6 nm wavelength, but not at 454.7 nm?

One clue - exactly the same thing occurs with the sound spectra from musical instruments. If you pluck a guitar string and record the sound, you can plot air pressure versus time - it looks like this:

Ninja Physics/Paul Francis 83

You can break this pattern up into its component frequencies. Basically any pattern can be reproduced out of a combination of pure sine-waves. Take this pattern, for example:

Looks pretty complicated. But it is just the sum of the following pure sine-waves:

This can be done mathematically using a technique known as a Fourier Transform. You will learn this in 2nd year, though many software packages will do this for you ("frequency analysis").

If we decompose the sound of a plucked guitar string into its component frequencies, we get the following sound spectrum: Ninja Physics/Paul Francis 84

(obtained using the sox program: http://sox.sourceforge.net/)

What you can see is that just like the light spectrum of the gas, there are sharp peaks with not much in-between - the peaks are around 50db above the troughs, which is a factor of 105 difference in energy.

Why does the sound emitted by a plucked guitar string have peaks at certain frequencies, and very little power at others? This happens because sound is a wave, and because the guitar string is fixed at both ends. This means it can only vibrate with wavelengths that allow both ends to remain fixed: Ninja Physics/Paul Francis 85

The lowest frequency (about 90Hz for the guitar string - note ~F2, the F two octaves below middle C) is the pattern shown on the left, with maximum vibration in the centre and none at the ends. In this case, the wave on the string has a fixed point (node) at each end and a point of maximum oscillation (antinode) in the centre. So the wavelength of this string waves is twice the length of the guitar string.

Then you get the next pattern, with a node in the middle as well as the ends, and a wavelength half that of the previous one (which means a frequency twice as great - we'll examine the relationship between wavelengths and frequencies later in the course). This corresponds to the spectral peak at around 180 Hz (F one octave below middle C). And so on - all the wavelengths must allow both ends of the string to not move, which means they must have integer ratios of frequency to the lowest peak (known as the fundamental). These integer frequency ratios were discovered by Pythagorus originally and are the basis of chords and harmony in western music. These string waves trapped on each end are called standing waves. We'll cover this much more in the waves and optics part of the course. Ninja Physics/Paul Francis 86

Most musical instruments have some equivalent of this. For wind instruments, you have sound waves inside the tube, which must have zero amplitude at closed ends, and maximum amplitude at any open ends. This once again gives a series of different frequencies with integer ratios of frequency. Drums and bells also radiate sound at particular frequencies, but due to their complex shapes these different frequencies may not be at integer ratios, and so don't sound harmonic.

So, we see that we can get narrow spectral lines if we confine waves in some way, like on a string or in a tube.

But how can that apply to spectral lines of light from atoms? Atoms consist of a nucleus with electrons moving around them, but these are particles, not waves. You can't have standing waves, nodes and hence spectral lines for particles, can you?

This is the central weirdness of quantum mechanics - you need particles to behave like waves some times. But surely waves and particles are quite different things?

Two Slit Experiment

There is a famous experiment that brings this paradox into sharp focus: the two slit experiment.

Imagine that you have a source of electrons in a vacuum chamber. It emits electrons in random directions every so often, which then fly off until they hit the walls. The walls are covered with phosphor, so when an electron hits the wall it produces a tiny flash of light which can be seen and recorded. This is very much like what you have in an old-fashioned cathode ray tube TV. Ninja Physics/Paul Francis 87

If we recorded all the flashes of light for a while, we'd expect to get a pattern that might look something like this:

Now - let's put a barrier between the electron source and the phosphor coated wall - a metal sheet that blocks the electrons. But we'll cut two slits in the the barrier to allow some of the electrons through:

Ninja Physics/Paul Francis 88

What would you expect the pattern or recorded flashes on the phosphor-coated wall to look like now? You'd expect most electrons to be blocked, so you'd only see flashes from those that went through the slits, so you'd think we'd see something like this:

But if you do the experiment, what you actually get is a pattern like this:

The most common place to get flashes is actually half way between the two slits! You get a series of bands of flashes, with bands with no flashes in-between.

Here is some data from an actual experiment that does this: Ninja Physics/Paul Francis 89

(from https://en.wikipedia.org/wiki/Double-slit_experiment#/media/File:Double- slit_experiment_results_Tanamura_2.jpg )

This is weird! How can an electron end up at the point on the phosphor wall directly behind the central blocking? Why do you get bands of electrons arriving?

This pattern seems completely inexplicable if electrons are particles, but would make perfect sense if they were waves. A wave would diffract through each slits, spreading out. And in some directions the waves would add up in phase and reinforce, giving a strong signal, while in Ninja Physics/Paul Francis 90 others the peaks coming from one trough would line up with the troughs coming from the other, you'd get destructive interference and no signal.

But how can an electron be a wave? It produces a single flash when it hits the phosphor screen. That means it has a definite time and place - which sounds like a particle, not a wave.

And there's worse. You can reduce the intensity of the electron source so low that there will typically be only one electron flying around at a time. And you still get these bands! But you only get interference patterns like this by combining the waves going through both slits. How can a single electron go through both slits?

So - what can we do? We need electrons to behave both like a particle (have a definite time and place when they hit the screen) but also behave like a wave (go through both slits at once, interfere with itself, diffract through the slit).

Probability Waves

The solution to this apparent paradox is a very strange one - the idea of probability waves.

The basic idea is - electrons are waves, but these are waves of probability. The amplitude of the wave tells you the probability that the electron is at a particular place, when you make a measurement.

As long as you don't make a measurement that might pinpoint the location of the electron, it remains as a wave, and can interfere with itself, diffract, etc. But as soon as you make a measurement that ties its position down (like using a phosphor screen), the waveform

"collapses", and you will measure a definite position. Where this position might be is random, but the probability is given by the amplitude of the wave function. Ninja Physics/Paul Francis 91

Which is seriously weird. When you don't look, electrons behave like waves. They diffract around corners, go through two slits at once, and generally are spread-out with no definite position. But as soon as you look, everything changes. Suddenly they have a definite position. But what position? That's random. But not totally random - the odds of finding it somewhere depends on the whole spread-out wave business.

There are many serious objections to this whole picture. It introduces randomness and probability into the fundamentals of the universe. Before quantum mechanics came along, the laws of physics were laws - they told you what would happen. "The particle will be here - no ifs, no buts." But now the laws are more like guidelines: "There is a 37% chance of the particle being over there, and a 16% chance of it being here...". And this uncertainty is not just because our measurements are poor, or because there is something going on we haven't accounted for. Even with perfect measurements and perfect theory, the world is still fundamentally uncertain. This really annoyed Einstein, who famously said "God does not play dice". But it seems that God did not agree...

And there's worse! The probability wave collapses to give you a particle with a definite position when you make a measurement. But what counts as a measurement? And why should it collapse the waveform?

This paradox was most famously highlighted with Schrodinger's cat. Imagine that you have a cat in a box, together with a radioactive source. This radioactive source has a 50% chance of decaying, and if it decays, there is some mechanism that will release poison and kill the cat.

Common sense would tell us that the cat will be either alive or dead, even though we might not know which until we open the box. But common sense tells us that the electron must go through one or the other slit, but it doesn't. And according to quantum mechanics, what you Ninja Physics/Paul Francis 92 really get inside the box is a probability wave consisting of a partially alive and a partially dead cat, which can interfere with each other. It's only when you open the box that you are making an observation and the waveform collapses into either an alive or a dead cat.

Many people find this deeply unsatisfying. Quantum mechanics works extraordinarily well experimentally, but has these paradoxes at its heart. They have been debated for decades, and many ideas have been floated for how to understand them. But for now, let's just accept that particles can behave like waves, and see where it gets us.

Explaining Spectral Lines

So - if electrons are waves as well as particles, can that explain spectra lines? Yes - it does, because in an atom, these waves are confined by their electrostatic attraction to the nucleus.

Which makes them very much like guitar string waves or sound waves in a flute - they are confined, which means that only certain frequencies are allowed. For atoms things are a bit more complicated because the waves are confined in fuzzy-sided 3D boxes, and because for most atoms there are multiple electrons confined there. But the basic principle is the same - waves confined can only have certain discrete frequencies. The details keep physical chemists in work!

It is possible to solve the quantum mechanics wave equation (Schrodinger's equation) and derive the electron wave patterns using just the laws of quantum mechanics, though in practice this can only be done for fairly simple atoms and molecules.

But there is a crucial difference to sound waves from a musical instrument. If a guitar string is plucked so that it vibrates with a particular pattern, it will shake the guitar with that same frequency, which in turn will shake the air and radiate away energy at this pitch. Which means the vibration will steadily weaken until it's stopped vibrating, unless you pluck the string again. Ninja Physics/Paul Francis 93

But for electron waves confined in an atom, they do not normally radiate any energy away (that would be equivalent to an electron vanishing into thin air, which would violate any number of conservation laws). Instead, they can change from one wavelength state (say having two wavelengths across the atom) to another (say having only half a wavelength across the atom). Each wavelength state has a different energy - the shorter the wavelength, the greater the energy. So when an electron goes from a short wavelength state to a longer one, it must emit the difference in energy - and that's where the photons in spectral emission lines come from.

If left to themselves, most electrons will settle down into the lowest energy stare, known as the “Ground State”, which has half a wavelength across the diameter of the atom, with a node at each side and an anti-node in the middle. This is very similar to the fundamental note of a guitar string.

If the gas is heated up, atoms will collide with each other, and this may knock some of the electrons up to “excited states”, perhaps with one wavelength across the atom - notes at either side and in the middle, with anti-nodes half way out on both sides. Given time the electron will decay back to the ground state, releasing a photon with the difference in energy. The spectral lines coming from a given gas thus tell us the differences between its various electron energy levels. The strength of the different lines can also tell us how many electrons have been excited to the different energy levels, which will depend on the number and violence of collisions between the gas atoms, and hence on the temperature and pressure of the gas.

de Broglie Hypothesis

In 1924, Louis de Broglie hypothesised that all objects made of matter (including electrons) have a wave nature, with wavelength given by the equation , where p is the momentum of the object in question, and h is Planck's constant 6.626x10-34 Js. This was Ninja Physics/Paul Francis 94 experimentally confirmed within three years, and incorporated as one of the fundamental principles of quantum mechanics.

What sorts of wavelengths are we talking about? Consider a person walking down the street. They would have a mass of around 70kg and a speed of around 1 metre per second, so their momentum p=mv ~ 70 kg m/s. So the wavelength is Planck's constant divided by 70, which is around 10-35 m. Far too small to be measurable or of any practical consequence - this is why people don't diffract around corners (or to be precise, why the diffraction is so small that it can't be measured).

But if you consider much smaller bits of matter, like electrons, the momentum is much smaller. Consider an electron travelling at 1m/s - it's momentum will be around

, so its wavelength will be around 6x10-4m - only a bit less than a millimetre! In practice most electrons travel much faster than this, but if you can cool one down enough to sub metre-per-second speeds, its wavelength can get long enough to have some very interesting effects. This is one of the major research areas here in Physics at the ANU.

Size of an atom

We now know enough to estimate the most fundamental property of materials - the size of an atom! Let's work with hydrogen for simplicity sake. It consists of an electron and a proton.

The electron is attracted to the proton by an electrostatic force: Coulomb's law. If it weren't for quantum mechanics, the electron would just fall in to the proton and we wouldn't have any atoms. But something stops the electron from falling in - instead it forms a fuzzy cloud around the proton.

What stops the electron from falling in? Once it was thought that the electrons could orbit around the proton, with centripetal force balancing gravity (much like the planets orbit around Ninja Physics/Paul Francis 95 the Sun). But electromagnetic theory showed that whenever you accelerate a charge, it radiates electromagnetic radiation. Motion in a circle is acceleration, and the time needed for an electron to radiate away its orbital energy would be very short, so that can't work.

But quantum mechanics gives us the answer - the electron is actually a probability wave.

The shorter the wavelength, the large its momentum, and hence its energy. That means that if you try to squash an electron into a smaller and smaller space, its energy gets larger and larger.

Let's try to put a number on this. Let's assume that an electron in an atom is spread over a region of size ~a. We won't worry too much about whether this is a radius or diameter - it's just an order-of-magnitude length scale. If you ever tried to measure the electron's position (say by bouncing some x-rays off it) you would collapse the waveform and find a definite momentary position, but as soon as you stopped looking it would go back to fuzzyness, and next time you measured where it was, it would some somewhere quite different.

This means that the wavelength of the electron wave cannot be much bigger than ~a, so it's momentum will be roughly p~h/a. Does this mean the electron is going somewhere? On average no - but if you chose to measure the momentum of the electron at some moment, it might be going one way or another, but this is an estimate of how much momentum it would have at that time. If you measured its momentum later, you'd get a different answer.

What is its energy? Well, kinetic energy is . What is the velocity? Well, we know that the momentum p=h/a, and momentum is classically defined as mass times velocity p=mv.

So v=p/m, and kinetic energy is

So the smaller we make the electron cloud, the more the energy. Ninja Physics/Paul Francis 96

But we also need to factor in the electrostatic attraction between proton and neutron. The electrostatic potential energy between two charges q1 and q2 separated by distance r is:

which you can get by integrating Coulomb's law. It's negative because an electron dropped towards a proton gains kinetic energy as it accelerates in, which must be coming at the expense of losing potential energy. In this case, both charges are the same (e=1.6x10-19 C), and we'll assume that the distance between the electron and the proton is roughly a. The electron is, of course, a fuzzy spread-out wave, so defining its position is tricky, but most of the fuzz will be roughly this far out.

The total energy is thus:

What does this look like? The first term is positive, and goes as one over a squared, so it will dominate for small values of a. The second term only goes as one over a, and is negative, and will dominate at larger distances, so combined, the curve will look something like this:

Ninja Physics/Paul Francis 97

Natural systems like to settle down to the lowest energy state - so we expect atoms to have a typical size at the minimum of this curve. Where is this minimum? Differentiate the energy equation and set this to zero to find the flat part of the curve - the minimum:

so

So according to this calculation, atoms should have a size a of around 10-10 m. Which is about right - this is the typical size of atoms.

So - we now have fundamental way of understanding why atoms are about 10-10 m big.

The electrostatic attraction of the proton and electrons tries to shrink the atom, but because the electrons are waves, if you want to squeeze them into a smaller space their wavelength gets small, so their momentum and energy become very large.

Density

So now we know how big atoms are, we can use this to estimate the typical density of matter.

Let's take Copper. It has a density of 8960 kg m-3. From a periodic table you can see that copper has an atomic number of 29 and an atomic weight of around 64. This means it has 29 protons and (64-29) = 35 neutrons. Protons and neutrons both weigh roughly one atomic mass unit u=1.66x10-27 kg (neutrons actually weigh slightly more - but that difference is negligible at the level of approximation we're working at). So a copper atom has a mass of roughly

65x1.66x10-27 ~ 10-25 kg. Ninja Physics/Paul Francis 98

And how many atoms are there in a cubic metre? We know that the size is around 10-10 m, but we didn't specify whether this was a radius or a diameter (our calculation was much too approximate to worry about details like that). Let's treat this size as a radius, and assume that copper atoms sit in a rectangular grid (they actually have a face-centred cubic structure, but that's not so different for current purposes). So each atom is alone in a block of size 2x10-10m on a side. Thus the total number of atoms in a one cubic metre block of copper will be around:

So the density of copper should be roughly

which is actually pretty close to the measured value - well within a factor of two.

So we see that there are fundamental limits to the densities of normal matter. Atoms can't get much smaller due to the wave nature of electrons. Not all atoms are exactly the same size, but you're not going to get atoms ten or a hundred times smaller than this. So if you want a really dense material, your best bet is to make it out of a substance with a high atomic mass, like osmium or uranium. You can't make something the thickness of tissue paper that weighs tonnes.

But how about the material in neutron stars and white dwarfs? A teaspoon full of neutron star material would weigh a billion tonnes (the density of neutron-star material is around 1017 kg m-3). Doesn't that go against what we just calculated?

Well - we were assuming that atoms are surrounded by clouds of electrons, and that the typical separation of atoms is given by the radius of this electron cloud. But in a neutron star the gravity is so intense that the electrons have been forced into the nuclei, where they've combined with the protons to form neutrons. White dwarf stars still have electrons, but the energy equation Ninja Physics/Paul Francis 99 we used to work out the size of each atom needs a new term - a gravitational term, which turns out to be much stronger than the electrostatic term and causes electrons to be squeezed into much smaller spaces.

There are a number of practical applications that could benefit from higher density materials. The best known in armour-piercing shells to destroy tanks - or alternatively tank armour to resist these shells. Depleted uranium is widely used in both tank shells and tank armour.

Energy Storage

We've calculated the size of an atom - but how about the energy it can store? We've seen that a hydrogen atom has a size of around 10-10 m. The electrostatic potential energy you get by moving an electron from infinity down to a=10-10 m from a proton is:

This is a very small number as expressed in Joules - physicists typically quote atomic energies in electron volts. An electron volt is the energy an electron acquires when accelerated over a potential difference of one volt. They are written as eV, and have a numerical value of

1.602x10-19 J. The energy we calculated here is -14 eV.

This is actually very close to the correct value of the binding energy of an electron in a hydrogen atom, which is 13.6 eV (also known as the Rydberg energy). This is the energy you need to hit a neutral hydrogen atom with to ionise it - to knock its electron clean out. And it's the energy you get if you combine an electron with a proton.

How would you expect this energy to differ for other atoms? I'd expect it to be similar to the energy needed to add or remove the last electron from a heavier atom, for example, to remove the 12th electron from a carbon atom. The exact figure would differ, depending on the Ninja Physics/Paul Francis 100 precise size of each atom, but it should be roughly the same. If you had a bare heavy nucleus

(e.g. an iron nucleus with no electrons) and added a first electron to it, I imagine you'd get quite a bit more energy, because of the larger charge of the nucleus. But that's not a common situation - most ions are only missing one or two electrons.

How about chemical bond energies? Roughly speaking, an ionic bond involves one atom losing an electron and another one gaining it, so the energy should be broadly comparable to the figure we calculate. The exact details are very complicated, and are the subject of most of chemistry, but I'd imagine strong bonds would have a bit less energy than what we calculated, and weak bonds would have much less energy.

Is this correct? Let's check. Chemists have measured and tabulated typical bond energies.

For the bond between hydrogen and carbon, this energy is 411 kJ/mol. A mole is, by definition,

Avogadro's number worth of atoms or molecules - i.e. 6.02x1023 atoms.

So the energy per bond is:

which is pretty close to our calculation.

In practice, chemical reactions do not just involve breaking a bond - they involve replacing it with another, typically lower energy bond, so the energy you get out should be the difference between the starting and finishing bond energies. Which will be less than the bond energies themselves.

How much less? Let's look at a practical example, the combustion of petrol. This involves the reaction 2C8H18+25O2->16CO2+18H2O, which experimentally liberates around

4x107 J/kg. Ninja Physics/Paul Francis 101

A molecule of C8H18 has 8 carbon atoms (each of mass 12 atomic mass units) and 18 hydrogen atoms (mass 1 atomic mass unit). So the total molecular mass is ~ (8x12+18)x1.66x10-

27 ~ 2x10-25 kg. So there are roughly 5x1024 molecules in a kilogram of petrol, each of which has

8+18=26 bonds.

Thus the energy change per bond during the chemical reaction (when the C-H bonds are replaced with C-O and H-O bonds)is:

which is a little less than the ~ 10-18 J/bond we estimated earlier, but not too far off.

So what is this telling us? That there is a fundamental reason for the amount of energy you can get from fuel. You can't get much more than 10-18 J from a given atom. This gives you around 40-50 MJ/kg from petrol or similar hydrocarbons - it can go as high as around 55 MJ/kg for methane and down to around 10-20 MJ/kg for burning wood or digesting food.

The best possible chemical energy storage would involve the maximum number of bonds but the lightest atoms - which you get by burning hydrogen with oxygen. But even this only gets you up to around 140 MJ/kg. Some rocket engines use this, due to its excellent energy to mass ratio, though the practical problems of keeping liquified hydrogen in safe storage aboard the rocket mean that hydrocarbon fuels are often more practical. Hydrogen fuel cells are also used to supply electricity and are one possible future power source for cars.

By comparison, the lithium-ion batteries used in most portable electrical devices are pretty pathetic - even the best lithium-ion cells only have energy densities of around 0.9 MJ/kg.

Why so low? The actual energy you get out of a battery per kilogram of lithium used is 41.7

MJ/kg - i.e. very comparable to petrol. The problem is that most of the mass in your battery is not the lithium ions. A lithium ion battery has the active ingredient (Lithium) dissolved in an Ninja Physics/Paul Francis 102 organic solvent, sandwiched between electrodes (typically of carbon and metal oxide). Then the whole thing needs to be enclosed to stop it reacting with air or exploding/leaking. So in the end, most of the mass is in the solvent, electrodes and case - only ~2% is actually the lithium ions.

The same is really true for internal combustion engines. While the petrol itself has a very high energy density, it needs an engine to turn it into useful work, and typical car engines have masses of around 150kg. If you fill up a car with 50 litres of petrol (one litre of petrol has a mass of around 1kg), this means you need 3kg of engine for every kg of fuel, reducing the total energy density by a factor of 4. And because a petrol engine is a heat engine, it has a maximum efficiency of no more than around 50%. So realistically a hydrocarbon engine only generates around 1/2x50/4~6 MJ/kg. Which is still a lot better than lithium-ion cells. Makes you wonder if it might be feasible to build really small petrol-driven engines to power laptops...

So - there is a fundamental limit to how much energy you can get from atoms. But most current power sources are quite a way off this limit, so there is room for improvement.

And we haven't discussed nuclear power - that's a whole different matter, and can provide

MUCH higher energy densities, because it relies on the binding energy of atomic nuclei, not atoms.

Tensile strength

How strong can matter be? Tensile strength is how much you can pull something before it snaps - it's measured in newtons per square metre - i.e. the force that a cable of cross-sectional area one square metre can withstand. Note that these are the same units as pressure (Pascals - equal to newtons per square metre).

How much force does it take to break a chemical bond? We've seen that the energy required is around 10-18 J, and that atoms are around 10-10 m in radius. We can work out the force Ninja Physics/Paul Francis 103 using the fact that work done (change in energy) is force times distance. So the force is roughly the change in energy divided by the distance over which the force is applied. Our length scale here is around 10-10 m, so the force must be roughly:

Another way to think of this is that force is the gradient of the potential energy diagram.

You can see that the gradient of this curve is zero (it's flat) at the bottom. That means that very little force is needed to move an atom when it's near its equilibrium position. But then as you pull them further and further apart, the slope steepens, so the force gets stronger. This is normal behaviour for a spring. The blue line shows the maximum slope - the slope here is roughly 10-18 divided by 10-10, as above. Note that the above curve and values are for one electron in a single atom, not a chemical bond, but we're assuming that strong bonds behave similarly, and as we saw in the last section, they do!

And then as we pull the atoms further apart still, the curve flattens out and the force goes away - this happens when a material has broken. Ninja Physics/Paul Francis 104 So - we need a force of around 10-8 N to pull atoms apart. This can actually be measured directly - you can use atomic force microscopes or optical tweezers to stretch and pull apart individual atomic bonds!

Now let's consider our cable, with cross-sectional area one square metre. Imagine slicing through the cable. How many atoms will you see on the cut slice? If each atom has a radius of ~

10-10 m, then each atom occupies a box 2x10-10 m on a side. So over the whole cross section, you will have:

So if we have this many atoms, and each can exert a force of 10-8 N, the total force that the cable can supply must be around 10-8x2.5x1019 ~2x1011 Pa.

How does this compare to actual materials? High quality steel has a tensile strength of around 1000 MPa (109 Pa) which is considerably worse than this limit. This is because steel usually fails because of imperfections and/or cracks propagating with it. Some synthetic polymers, like Aramid (Kevlar) are five times stronger, which is still not close to the theoretical limit (these fibres are used in bullet-proof vests). The strongest materials yet produced are carbon nanotubes, with tensile strengths of up to 63 GPa (6x1010 Pa) which is actually getting pretty close to the theoretical value. This is presumably because they have no imperfections and have strong atomic bonds lined up perfectly. Unfortunately, they have only been manufactured in tiny lengths so far...

So - in theory we could have materials with tensile strengths of up to around 1011 Pa.

These would be incredible for all sorts of technological applications, from space elevators to tennis rackets. Carbon nanotubes actually get close to this limit, but all other materials are one to two orders of magnitude worse, due to imperfections and propagating cracks. Ninja Physics/Paul Francis 105

Conclusions

So, what have we learnt? We are pretty much stuck with using atoms to make things, and the properties of atoms are set by the laws of electric fields and by quantum mechanics. The size of atoms is pretty much fixed at around 10-10 m. This in itself places a limit on how small you can make things. Microelectronics is already pushing close to this - components in microprocessors are now only ~ 10 nm (10-8 m) in size, meaning they are only around 100 atoms across.

The energy needed to pull an electron out of an atom is around 10-18 J, and this is also a rough upper limit on the energy you can get per chemical bond. This is pretty close to what you actually get from burning hydrocarbons and eating a ham sandwich, but is much higher than what you get from a rechargeable battery. So there can't be any super-fuels with drastically more energy, but there is a lot of potential to improve batteries.

To force needed to pull an electron out of an atom (which should be roughly the force needed to break a chemical bond) is around 10-8 N, which corresponds to a tensile strength of around 1011 Pa (100 GPa). Most practical materials are 10-100 times worse than this, because of defects, but if you can make perfect chemical bonds (as with carbon nano-tubes) you can get close to this theoretical limit.

You should also pay attention to many of the methods used in these calculations. In addition to introducing the key concepts of quantum mechanics, we looked for an energy minimum to find the size of at atom - a very common and powerful technique you will see many times again in physics. And we used the slope of the energy diagram to work out the force - also another very common and powerful technique you will see many times again. Ninja Physics/Paul Francis 106

Week 5: The Nano-World and Statistical Thermodynamics

Plenty of Room at the Bottom

We've seen that there are good physical reason why we don't have immensely better cars and flying cars - it's not just that scientists haven't been smart enough.

But could this be true more widely? Could science be coming up against hard limits, where future major breakthroughs are not possible?

This may possibly be the case for big things - like faster planes. But we know it's not true for small things, because nature has evolved technological marvels (like a mosquito) that are far more sophisticated than anything humans can produce. Just think of it - a machine with a mass of

10-6 kg that can fly for hours without refuelling, contains a sophisticated sensor package capable of tracking down food and mates, can make copies of itself (reproduce) without the aid of a factory, repair damage to itself. Pretty amazing.

Richard Feynman gave a talk in 1959

(https://en.wikipedia.org/wiki/There's_Plenty_of_Room_at_the_Bottom ) in which he said

"There's plenty of room at the bottom". There may or may not be room for progress with big things (spaceships, aircraft etc) but there is certainly plenty of room for technological advances in small things. And he's been proven right since then - there has only been limited progress with Ninja Physics/Paul Francis 107 large things since then, but there has been fantastic exponential progress in small things like microprocessors.

This has led to the whole field of nano-technology. Many people are very sceptical about this term - it often seems to be purely a way of re-badging bits of physics, chemistry and biology to make them seem trendy to governments. We know that nature was able to evolve incredible nano-machines, but it did take nature 4 billion years to get there - is it really feasible for us to manipulate the nano-world, what are the factors that might limit us, and what could we hope to achieve?

In this section we'll investigate the strange properties of the nano-realm. In particular, I'll focus on thermodynamics and some macro-scale properties that depend on it, as this turns out to be crucial to understanding what happens at nano-scales.

We'll start off by thinking about how physics varies on different scales.

Electricity vs Gravity

There are many books and movies in which heroes are shrunk to tiny sizes or enlarged to enormous size (e.g. Ant-man, Gulliver's Travels, "Honey I Shrunk the Kids", Fantastic Voyage).

In most of these cases, the shrunk or enlarged people walk around just like normal. But that's actually not very realistic - when you change size by a lot, different laws of physics become important.

It is possible to sketch out some different size regimes, where different laws of physics are most important. All the laws of physics apply at all scales - it's just that some are more important than others in particular size ranges.

One issue is the relative importance of gravity compared to electromagnetic forces (such as chemical bonds). Normally, the electromagnetic force is MUCH stronger than gravity. To see Ninja Physics/Paul Francis 108 this, consider two people, each with a mass of around 100 kg, standing 1m apart. The gravitational force between them is given by Newton's law of gravity:

(it's not going to push you over...). Note I've approximated G as 10-10 (its actual value is

6.67x10-11 N m2 kg-2).

But now imagine that both people had 1% too many electrons. What would the electromagnetic force between them be? Now a proton weighs around 10-27 kg, so there are roughly 102/10-27 = 1029 electrons in a 100 kg person (neglecting the presence of neutrons - which is probably fine for this sort of very rough estimate). Charge on an electron is around 10-19

C, so your total charge would be ~108 C each (1% of 1010).

The electrostatic repulsion between the two people would be given by Coulomb's law -

- big enough to rip the Earth in two!

So if electromagnetic forces are so very much stronger than gravity, why is gravity so important on a day-to-day basis? The reason is that electric charges can be both positive and negative, and across the universe as a whole, the positive and negative charges are exquisitely balanced. But masses are always positive, so even though gravity is very weak, if you add up enough mass, it will eventually beat everything else. But on small scales, where charges are not perfectly balanced, electric forces will totally overwhelm gravity.

# Ninja Physics/Paul Francis 109

Square-Cube Law

Another reason why physics is different on different scales is the square-cube law.

Imagine some object (like a human), of length r. If you make it bigger, while keeping the proportions the same, the surface area (and any other area, like the cross-section of the legs) increases proportional to r2. So physics that depends on the areas, like drag forces, heat exchange with the surroundings, strength of limbs and diffusion of oxygen in the lungs, increase as the square of the length.

The volume, however, increases proportional to r3. So things that depend on the volume, like the mass and hence the gravity and momentum, and the total heat capacity, will increase as the cube of the length.

So both increase as you get bigger, but the volume increases faster. As an example of how important this can be, consider the ratio of weight to drag forces. Imagine our object is a cube of size r and with the density of water (1000 kg m-3), moving at v=1m/s through air. The drag force is roughly

(neglecting small factors like 1/2 and the drag coefficient). The weight force is F = mg ~

10x1000r3 = 104 r3. So the ratio weight/drag is 104 r.

So for a one metre length object (like a fridge), weight is ten thousand times stronger than drag. So don't push a fridge off the roof - it will break.

By the time you're looking at a grain of pollen (r ~ 10-5 m), drag is ten times greater than weight - so it will mostly just float around on the wind and not fall.

Different Size Regimes Ninja Physics/Paul Francis 110

So factoring this in, along with some other physics, we can define some rough size regimes, along with the physics that dominates at them.

1: Quantum world: r<10-9 m or so. On this scale, the wavelengths of the quantum mechanical probability waves are important, so you are in the world of quantum weirdness, with interference, standing waves, diffraction of matter etc.

2: Nano-world: 10-9 - 10-7 m or so. This is the world we will cover in this section - where random thermal motions typically dominate over everything else.

3: Micro-world: 10-7 - 10-4 m or so. Dominated by friction and drag. Fluid flows are laminar, gravity mostly irrelevant. Everything in thermal equilibrium.

4: Normal-world: 10-4 - 104 m or so. What you've spent your life in.

5: Astronomy-world: 104 - 1022 m or so. Gravity dominates, friction and drag are irrelevant. Objects move in orbits for ever.

6: Cosmology-world: > 1023 m or so. On really big scales (tens of mega-parsecs and up, even gravity becomes irrelevant, due to the inverse square law. Instead, the expansion of space and dark energy are the most important factors.

Don't take this division too seriously! The boundaries are not hard and fast, and depend on many things other than just size. But this is a good guideline to which physics you should thing about first!

In this section, we're going to focus on the special physics of the nano-world, and in particular thermal energy. And we'll show how it also has big implications on larger scales. Ninja Physics/Paul Francis 111

Internal Thermal Energy

Nano-scales are dominated by thermal motions. So let's spend some time analysing them.

Heat is often described as "atoms jiggling around". The faster they jiggle, the higher the temperature. But it's a bit more subtle than that. If atoms are moving like this:

then even though the atoms are all moving, we'd describe this as an object moving, not as heat. If, on the other hand, the atoms are moving like this:

then we would call this internal thermal energy, not bulk motion.

There's actually an interesting intermediate pattern that looks something like this: Ninja Physics/Paul Francis 112

this is disordered motion, but not totally disordered - this is called turbulence, and is very hard to study. If you blow a jet of air out of your mouth, it starts off as bulk motion, but then turbulence kicks in and the flow breaks up into smaller and smaller eddies (a process known as a

Kolmogorov cascade), until eventually every atom is doing its own thing, and the energy is now purely heat.

In reality, there is no fundamental difference between internal thermal energy and other motion. The distinction is purely a matter of convenience. If all the atoms in an object are moving the same way, we can lump them together and treat them as one combined object. But if they are all jiggling in different direction, while in principle we could calculate the motion of each atom, in practice this is impossible (even a small object can have 1020 atoms). So we treat their motions as random and statistical: as internal thermal energy.

Internal thermal energy can technically defined as all the energy of the atoms and molecules that is in small scale motions, vibrations and stresses. Motions that are too small to analyse one-by-one, and random enough that we can treat them statistically.

Ninja Physics/Paul Francis 113

In a monatomic gas (one where each atom is not bound to any other atom), this internal thermal energy is entirely in the form of kinetic energy of the atoms, as they race around in random directions. Diatomic gasses (like air) can also have some internal thermal energy in the form of vibrations in the bonds joining the atoms in each molecule together. This energy will alternate between potential energy in the bond and kinetic energy in the relative motion of the two atoms. You will also have internal thermal energy in the form of rotation of these molecules.

In a solid, the energy will mostly be in the form of vibrations in the bonds joining neighbouring atoms (once again oscillating between potential energy in the bonds and kinetic energy in the random relative motions).

Equipartition and Boltzmann's Constant

The fundamental idea behind statistical thermodynamics is "Equipartition". This principle states that on average, the internal thermal energy in a system will be shared equally between all possible forms.

For example, consider a monatomic gas. The thermal energy here will be in the form of motion. We can break down the motion into three components - motion along the x, y and z axes, however we define them. The principle of equipartition says that on average, there will be the same amount of thermal energy in each of these. If you started off with all the atoms racing back and forth along (say) the x axis, but no motion along the y and z directions, that wouldn't last long. Atoms would bounce off each other, and before long the average motions in all three axes would be comparable.

For a diatomic gas, something similar would happen. If you somehow started off with the molecules moving but not vibrating or rotating, collisions would soon spread the energy equally between rotation, vibration and motion along the three axes. Ninja Physics/Paul Francis 114

The principle of equipartition basically says that so much random stuff and so many collisions are going on in most materials at the microscopic level that you are pretty sure to have the energy spread evenly between all possible forms. For any individual atom or molecule, most of the energy may be in only some of the forms at a particular moment.

The energy associated with each form of thermal energy is equal to:

where T is the temperature (in Kelvin) and k is Boltzmann's constant,

This equation links the macroscopic quantity (temperature) to the microscopic properties

(mean thermal energy of atoms and molecules). And it's all we need to deduce an awful lot!

Specific Heat Capacity

The specific heat capacity of a material is the amount of energy needed to heat up one kilogram of the material by one degree.

Can we work this out from first principles?

Let's start with a simple material, like a block of pure metal. Copper has an experimentally measured specific heat capacity of 385 J/kg, while Iron has a slightly higher specific heat capacity (450 J/kg). Can we explain why it takes this much energy to heat up these metals?

An atom in a solid can vibrate back and forth in the x, y and z directions, but that's all it can do. Along each axis the energy will alternate back and forth between kinetic and potential

(energy in stretched bonds), but the both the kinetic and the potential energy should be 1/2 kT on average. Thus the total energy per molecule should be 6x1/2 kT = 3kT. Ninja Physics/Paul Francis 115

How many molecules are there in a kilogram of metal? That's just the weight (1kg) divided by the weight of each molecule, which is to a good approximation its atomic mass a times the atomic mass unit u. So the total thermal energy in a block of metal should be:

That's the total thermal energy, and it's pretty big. Consider a 1kg lump of iron at room temperature (~300 K). The atomic mass of iron is 56. So the energy is

Compare that to the kinetic energy of this lump of iron if you threw it at someone. Let's say you can throw it pretty hard - at say 5 m/s. The kinetic energy would be

So the energy in the random thermal motions of a block of iron (or of anything else at room temperature) contain vastly more energy than any plausible motion of this blow.

Now to work out the specific heat capacity of an object, we need to calculate how much extra energy dE is needed to increase the temperature by an amount dT. We get this by differentiating the above equation:

Plug in the numbers, and you get a predicted specific heat capacity for copper of 393 J/kg and for iron, 445 J/kg. Which values are scarily close to the measured ones! This actually seems to work! Ninja Physics/Paul Francis 116

Now let's try it for water. Water has a very much greater specific heat capacity - around

4184 J/kg - i.e. about ten times that of the metals. Which is very lucky for the Earth - this allows the oceans to damp out seasonal changes in temperature, giving cooling sea breezes in summer and keeping Europe warm in winter due to the heat carried by the gulf stream.

But why is its specific heat capacity so incredibly high? If we can figure this out, perhaps we could have more of whatever makes water so special and have some super substance that can absorb incredible amounts of heat?

The amount of thermal energy in a material depends on the number of molecules, and not on how heavy each molecule is. This gives water an advantage, as its molecular mass is much less than the atomic mass of iron or copper (because hydrogen and oxygen are relatively light atoms), but not enough to explain the difference.

More of the difference comes from the fact that a water molecule can absorb energy in many more ways. It can move along three axes, and stress the bonds binding it to other water molecules along three axes, just like a metal atom. But in addition, it can rotate around three axes, and the bonds between the hydrogen and oxygen atoms can vibrate. All of which gives you a lot of different degrees of freedom, each of which will have its 1/2 kT. Though even factoring all this in, you don't predict as value as large as is observed. Most likely the difference is due to transient hydrogen bonds between broken. These bonds are much weaker than the bonds between the hydrogen and oxygen atoms, but as you heat a liquid up more of them break, mopping up energy.

So you'd expect the highest specific heat capacities for liquids of big but lightweight molecules, which can store energy in lots of different ways but not weigh too much, and which have hydrogen bonds holding them together as liquids. And this turns out to be true - the only Ninja Physics/Paul Francis 117 common substance with a higher specific heat capacity than water is liquid ammonia, which has all these same factors.

Boiling Point

Water boils at 100 C (by definition). Can we explain this number from first principles?

We see in the quantum mechanics section that a typical bond energy is around 10-18 J, but that applies to strong bonds, like the covalent bond between the hydrogen and oxygen atoms in water - not to the much weaker hydrogen bonds binding one water molecule to another.

What is the energy needed to break one of these weaker bonds? We can work this out from the observed latent heat of vaporisation. It takes a whopping 2.26x106 J to boil a litre of water (the latent heat of vaporisation of water). If we assume one bond per water molecule

(which is probably too small, but only by a factor of 2 or so) then the number of bonds in a kilogram of water is equal to the number of molecules, which is the mass of water (1 kg) divided by the mass of one molecule (18 times the atomic mass unit - two from the hydrogen and 16 from the oxygen). Doing this calculation, we find a bond energy of around 10-20 J, which is indeed somewhat weaker than that for a strong bond.

What breaks this bond when something boils? That must be the thermal energy of the molecules jiggling around, which is ~3/2 kT. This allows us to estimate the boiling point as:

Which is not bad - the true figure is 373 K high! It is, however, a little high. Why might that be?

The crucial point is that kT is the average energy of a molecule. Some will have more, and some will have less. It is possible to calculate the probability distribution of energies - i.e. the probability that a given molecule will have a particular energy at a particular moment (bear in Ninja Physics/Paul Francis 118 mind that molecules are constantly colliding with each other and exchanging energy). This is called the Maxwell-Boltzmann distribution, and you'll see how it's derived in 2nd year.

But the crucial point is - a small fraction of the molecules will at any particular time have considerably more than the average amount of energy. So the boiling point of water is not the temperature at which the average molecule has enough thermal energy to break its bonds and escape - it's the temperature at which a small fraction of particularly energetic molecules can escape. They fly away, taking their energy with them, leaving the slower-moving molecules behind (i.e. cooling the remaining water). But if you keep applying heat to the water, more will get hot enough to escape until the water all boils.

This is how sweat keeps you cool. Even though it's at well below boiling point, a small fraction of the fastest moving molecules can break their bonds and escape. Losing the fastest molecules drops the average speed of the remaining ones, cooling down the sweat and hence you. Without this mechanism, people would be unable to live anywhere on Earth where the temperature ever went much above body temperature (~37C), as there would be no way to cool our bodies down. Which means Australia would have been uninhabitable until the invention of air conditioning...

I actually wonder what would happen to people if the temperature ever went above 37C in a place with 100% relative humidity. With this humidity your sweat could not evaporate, so there would be no way to cool yourself down - your body temperature would presumably rise to be at least as great as the outside temperature, with nasty consequences - similar to a bad fever, or even death if the temperature went high enough. Luckily Earth at present seems to have places with 100% humidity, and places with temperatures well over 37C, but they are not the same places, at least not at the same time. Ninja Physics/Paul Francis 119

How fast are air molecules moving?

You are sitting (I hope) in a room filled with air. How fast are the molecules in the air moving?

They can move in three directions (x, y and z axes) and their motion along each axis will, on average, have an energy of . So the total average motion energy of each molecule will be around .

This then will be equal to the average kinetic energy - i.e.

So rearranging, the mean velocity is

What's the mass of an air molecule? Air is mostly nitrogen, and nitrogen is diatomic at atmospheric temperatures and pressures, so a typical molecule has two nitrogen atoms.

Nitrogen has an atomic mass of 14 (i.e. is has a total of 14 nucleons, protons plus neutrons). The mass of a nucleon is roughly the atomic mass unit:

So the mass of a typical air molecule will be ~ 2x14xu = 4.6x10-26 kg. If we plug this into the above equation, along with an assumed temperature of around 300 K (27 C), we find that the typical velocity of an air molecule is around 500 m/s. Which is very fast - that is 1800 km/hr!

That's the speed of a rifle bullet. Ninja Physics/Paul Francis 120

This is why the nano-scale world is dominated by thermal motions - these motions are fast and contain a lot of energy. If you are a nano-scale machine, you are constantly being bombarded by these very fast moving molecules.

So at room temperature, molecules move fast. As each molecule at a given temperature has the same kinetic energy, this means that lighter molecules (like hydrogen) will be going faster than the heavier ones.

Mean Free Path

We've seen that air molecules are racing around at around 500 m/s. But how far do they get before they hit another molecule? Let's try and estimate this distance (known as the mean free path).

How close does a molecule need to get to another one for it to collide? This is not a straight-forward question to answer - molecules are not like billiard balls - they are fuzzy and don't have sharp edges.

For a simple model - let's assume all molecules are spherical. Let only one molecule move - assume all others are stationary, and let's see how far it could go before hitting one.

Assuming the others are stationary shouldn't make much difference - their movement may cause them to move into or out of the path of the moving molecule, which should roughly balance out. Ninja Physics/Paul Francis 121

We can draw an imaginary cylinder around the path of this moving molecule. Its radius r is double the radius of the individual molecule - because if it passes within this distance of the centre of another molecule, there will be a collision. What is the average number of collisions in time t? If there are n molecules per unit volume, then in time t, the volume of the cylinder swept out is its cross-sectional area (pi r2) times its length (velocity times time). A collision will happen on average when there is one atom in this volume - i.e. when

So the average distance between collisions l is

We can work out N from the density of air (~1 kg m-3)divided by the mass of one molecule (4.6x10-26 for N2), which comes out as around 2x1025 m-3. But what is the effective r of an air molecule? Ninja Physics/Paul Francis 122

We show in the quantum mechanics section that the radius of an atom is around 10-10 m.

So a diatomic molecule like N2 or O2 is about 4x10-10m long and 2x10-10 m wide. Let's approximate it as a sphere of radius around 3x10-10m - which means r~6x10-10m (i.e. if the centres of two molecules get within this distance of each other, there will be a collision). Plug this in to the above equation and you get a mean free path l ~ 44nm (44x10-9 m).

Which is actually about right! The real mean free path of an air molecule is actually around 68nm. Which is a very small number, but still around 200 times the size of each molecule.

How does this compare to the typical spacing of molecules? If we have 2x1025 molecules per cubic metre, then if w put each in a box of side d, 2x1025 x d3 ~ 1, so d~36nm - i.e. around

100 times the size of each molecule. If you imagine each molecule as being the size of a golf ball, they would be separated from their neighbours by around two metres - the height of a tall person. So air is mostly empty space!

How long does a molecule take between collisions? If vt~68nm and v~500 m/s, then the average time between collisions t~10-10 s - 0.1 nS. A very short time period. So a typical air molecule endures around ten billion collisions per second! Ninja Physics/Paul Francis 123

Air Pressure

This constant bombardment by molecules is a big problem (or opportunity) for nano- machines. But even on bigger scales, the cumulative effect of all these impacts is a big one - it's known as "Air Pressure". It's an enormous force - around 100,000 newtons per square metre (Pa) at sea level. That's about ten tonnes of force applied to your skin. Luckily it doesn't crush you because the pressure on all sides of you, and inside you, are all exquisitely balanced. Even a tiny pressure imbalance would be enough to knock you over.

Actually - there a few situations in which this immense pressure is not balanced and can be seen. One would be trying to open the outside door of a spaceship, or a submarine, where the pressure on one side is much greater than on the other. Another occurs when scuba diving. The pressure doubles when you go 10m underwater - so if your tank fed you air at surface pressure you would be unable to breath - your lung muscles wouldn't be strong enough to pull the air in.

So in practice you are fed air through a "regulator", which gives you the air at the ambient pressure. But that can be a problem if you panic while deep underwater, hold your breath and bolt for the surface. The air inside your lungs will be at much higher than atmospheric pressure, and if you swim rapidly up into lower pressure regions, the imbalance between the inside and outside pressure can rip you apart (and you have no nerves on the inside of your lungs to warn you of the damage you are doing, until too late). So keep breathing as you go up, to equalise the pressure inside and outside...

Let's calculate how big this pressure force is from first principles. Let's imagine that we have n molecules per unit volume. Their kinetic energy along any one axis will be 1/2 kT - let's use the axis pointing towards a wall, and call it the x axis. So this will be equal to their kinetic energy. So Ninja Physics/Paul Francis 124

so

This is an average - some will be going towards the wall, and some away from it, but on average this is their typical speed.

How many molecules will hit the wall in a given time t?

Think of a box as drawn above. And consider the components of the velocities in the x direction (towards and away from the wall). Roughly speaking, half the atoms in this volume will be moving towards the wall and half away. To begin with, let's assume that the atoms do not collide with each other. In this case, half the atoms within distance vt of the wall will hit it during time t. So the number hitting the wall in time t will be:

where N is the number of air molecules per unit volume. Ninja Physics/Paul Francis 125

How realistic is this? In practice air molecules will frequently hit other air molecules before reaching the wall. But if t is small enough, most molecules won't hit each other during that time, so it will work. Also, collisions won't really affect the impact rate - they will mean that molecules far from the wall can't reach it, but those close to the wall will hit it multiple times as the bounce off more distant molecules - so this should more-or-less cancel out.

Let's assume that molecules bounce off the wall elastically. Realistic? Well, if collisions are inelastic that would remove energy from the gas and give it to the wall. Which would be realistic if the gas was hotter than the wall, but wouldn't make sense if they are at the same temperature. And a small inelasticity will make very little difference - so let's assume things are elastic.

In this case, a particle moving towards the wall bounces back with an equal and opposite speed. So the change in velocity is double the initial velocity. The average acceleration over time t of each molecule is thus 2v/t. Use f=ma, and factor in the number of molecules hitting, and we find that:

Now F/A (force per unit area) is called pressure. And N is the number of atoms n in a volume V, so we can re-write this in terms of pressure P and number n (rather than number per unit volume) to get:

Ninja Physics/Paul Francis 126

which is the ideal gas law equation. (it's often quoted as V=mRT, where m is the number of moles of gas present, rather than the number of molecules. M is thus n divided by Avogadro's constant, so to compensate, R is equal to k times Avogadro's constant).

Does this equation work? Consider one cubic metre of air - it weighs 1.2kg, which means it has around 1.2/4.6x10-26 = 2.6x1025 molecules in it (we worked out the mass of a typical air molecule in the previous section).

So at 300K, the pressure is 2.6x1025x1.23x10-23x300=107,000 Pa, in good agreement with actual atmospheric pressure!

Cellular Machinery

So the nano-world looks like a pretty hostile place! Imagine that you had been shrunken down to be only 100 atoms long (10 nm). The molecules in the air around you would appear the size of golf balls (~ 1% of your height). Every part of your body is hit by golf-balls travelling at the speed of rifle bullets roughly ten billion times a second. Sound painful? This is clearly going to jerk you around - as different parts of your body are hit in different directions this will distort you in all sorts of ways.

It sounds like this constant bombardment would make building nano-machinery very hard. But in fact, biology has evolved molecular machines that actually use this constant bombardment to make them work! Without this ceaseless molecular storm, our cells would cease to work and we'd die instantaneously. There's quite a good book about this if you are interested called "Life's Ratchet, How Molecular Machines Extract Order from Chaos, by Peter M

Hoffmann" - http://lifesratchet.com/ Read it if you'd like the details.

But basically, our cells contain several molecular machines that walk around our cells, transporting chemicals from place to place, assembling and disassembling structure - all the jobs Ninja Physics/Paul Francis 127 that human builders do on a normal construction site, only incredibly shrunken. You can see an animation of one of these molecules (kinesin) walking through a cell at https://www.youtube.com/watch?v=y-uuk4Pr2i8 .

It basically has two legs. But these legs do not contain muscles, or any sort of motor.

Instead, the constant molecular bombardment shifts them around randomly until they are where they want to be, at which point a chemical reaction locks them in place and releases the other leg, which will randomly wiggle around until it's in the place where you want it. You can try walking like this - raise one leg and wiggle it around at random. When it's randomly got to a forward position, drop it to the floor, then raise your other leg and repeat.

Indeed it is molecular machines like this that are responsible for human muscles. The fundamental components in your muscles that make them contract are called sarcomeres. They consist of long fibrous proteins that slide past each other, causing them to contract. And what makes these fibres slide past each other? Some of these molecular machines, attached to one fibre and walking along the other one.

You can see an animation of this process at https://www.youtube.com/watch?v=zQocsLRm7_A

Or for more details see

https://www.youtube.com/watch?v=YAJ-9nPSqwA

So next time you move any part of your body (hopefully right now, as your heat beats and your lungs breath), this is caused by molecular chaos helping molecule machines pull protein fibres along past each other. Ninja Physics/Paul Francis 128

Week 6: Human Performance

Human Performance

What limits the performance of human athletes? This is a very active area of research, particularly here in Canberra with the Australian Institute of Sport. But we can think of this more generally too.

Is it possible to have animals that are much stronger or faster than currently exist?

We as humans operate on time-scales of seconds. It takes seconds to think about things and to do things. Growing and repairing damage is slower - it takes days (~ 104 seconds).

There are life-forms on Earth that are much slower than animals. We call them plants.

They take days or months to respond to changes in their environment.

Imagine that aliens landed on earth, and they operated 104 times faster than we do (i.e. to them, we would look as slow as trees). Would we even notice they had landed? They could enter your room, carefully inspect every part, put it all back and leave in a blink of an eye. Perhaps they did while you were reading this!

Alternatively, this super-fast motion could be possessed by a super-hero ("The Flash")?

My question is - could something like this be possible? Or more practically, could we build robot warriors with this level of performance Ninja Physics/Paul Francis 129

One clue - we know that the cells in a human body do not operate at anything like their full potential. Each cell in a mouse burns energy at a rate roughly 10 times greater than a cell in a human. To supply enough blood to power these mighty cells, a mouse heat beats 300-800 per minute (compared to around 80 in a resting human). Just imagine if you could get the muscle cells in a human to burn energy like this - you would definitely be in superhero territory (see the book "Why Size Matters" by J.T. Bonner http://press.princeton.edu/titles/8241.html).

So let's look at what the capabilities of the human body are, and what physics might be limiting them.

We often start off a puzzle like this by brainstorming - trying to come up with all possible ideas. We can then look at them in turn and try to work out which are sensible.

One possible limit - forces. Is there some limit to how much force w can exert? Another possible limit - energy. How much do we need, where does it come from, how does it get where it is needed? Let's look at these in turn.

How much force can humans exert?

Clearly the bigger your muscles, the more force you can exert. So perhaps a sensible measure would be the "stress" your muscles can produce - the force per unit area.

Let's think about my legs. I can lift my body weight on one leg, just about. I weigh 70kg, so that's a force of mg ~ 70x10 = 700 N.

The muscles in my thigh have a cross-section of roughly 10cm by 10cm. So the stress is force over area - i.e. 700/(0.12) = 70,000 ~ 105 N/m2.

How about a more elite athlete? Here is Usain Bolt starting off a run.

Ninja Physics/Paul Francis 130

From http://www.dailymail.co.uk/news/article-2186096/London-2012-Olympics-Usain-

Bolt-grabs-camera-snaps-historic-frame-winning-200m-final.html

How much force is he exerting? If you measure his speed during a run, you get a plot like this:

(from http://iopscience.iop.org/article/10.1088/0143-

0807/34/5/1227;jsessionid=3D9E9592D97BC7614B4A5D107E911567.ip-10-40-1-105) Ninja Physics/Paul Francis 131

This graph is steepest at the start, so that's where his acceleration is greatest. Then it flattens out after the first ~ 3 seconds, at a speed of around 12 m/s.

We can use F=ma to estimate the force he is applying at the start. After 2.5 seconds he's moving at around 10m/s, so the average acceleration over this time is around 10/2.5 = 4m/s. This will be an underestimate of his true maximum acceleration - you can see from the graph that the curve is actually steeper right at the start. What is this maximum speed? We can estimate it from the graph:

We draw a tangent line - tangent o the slope at t=0. The slope of this green dashed line will give his starting acceleration. The line crosses v=10m/s at around 1s, so the initial acceleration must be around 10m s-2 (i.e. ~ g!). He obviously can't sustain this for more than a tiny fraction of a second, which is why it actually takes him around 2.5 sec to get up to this speed.

So - using F=Ma, the force required is around his mass (95 kg) times 10 m s-2, to roughly

1000 N.

But this isn't the total force he is exerting - it's just the horizontal component of his total force. Ninja Physics/Paul Francis 132

Force is a vector - it has both magnitude (how strong it is) and direction.

In this case we know roughly what direction the total force must be pointing - along his body. If there was only a vertical force, he would spin in an anti-clockwise way, and if there was only a horizontal force, he would spin in a clockwise sense. As he isn't spinning very much, the total force must point along his body.

If there is a vertical force, why isn't he accelerating upwards? I suppose he did a bit right at the beginning, but by the point when this photo was taken he's already standing up. So the vertical force isn't there to make him accelerate upwards, it's there to oppose gravity and stop him sinking into the ground. So the vertical force will be around Mg ~ 950 N. Which is about the same as the horizontal force. Which is about right - if the two parts of the force were the same, he would be at a 45 degree angle, which isn't too far wrong. In fact, it looks to me as if he's at an angle which is a bit less than 45 degrees, which suggests that the vertical force is a bit less than the horizontal.

But anyway, if we assume they are both around 1000 N, then the total force will be given by Pythagoras's theorem - i.e.

Ninja Physics/Paul Francis 133

which is not that different from the force I exert getting up off the ground on one leg.

"Not that different" in the Ninja sense, of course - the little bit of difference is why he is an

Olympic medallist and I'm not... His leg muscles are much thicker than mine, so the stress (force per unit cross-sectional area) is probably around the same.

In general, most muscles in most animals have about the same performance - a stress of around 2x105 N m-2, in pretty good agreement to what we estimated here.

Torque

So we've estimated the stress exerted by muscles. But in general this will not be the force you actually exert when you push something. Your muscles typically pull on your bones, which then push your hand/leg/whatever to apply the actual force. And this can considerably change the actual force exerted.

Consider the exercise known as a "bicep curl". You hold a weight in your arm, and lift it up by bending at the elbow. How much weight should you be able to lift?

You raise the weight by shrinking the bicep muscle, which is attached up near the shoulder and down past the elbow. It causes the forearm to rotate and hence lift the weight.

So how much should we be able to lift in this way? To work this out, we need a new bit of physics, we need to know about rotational forces, which are called Torques.

A torque tells you how much you are trying to make something rotate. It's equal to the force you apply multiplied by how far you apply it from the hinge. Ninja Physics/Paul Francis 134 In this case the hinge is the elbow. The bicep muscle is applying an upward force a distance r from the hinge, and the weight is applying a downward force a distance L from the hinge.

Now let's imagine the weight is just sitting there - you are not raising or lowering it. In this case, the torques must balance. If they didn't, the weight would be accelerating up or down.

Let us (arbitrarily) decide that clockwise torques are positive. So the torque caused by the weight is positive and the torque caused by the muscle is negative. The total torque is thus

(so each torque is just the force times its distance from the hinge).

Rearranging, we find that

So how much should a typical person be able to lift in this way? My biceps are roughly

5cm by 5cm, so the force they exert will be 0.052x2x105 ~ 500N. Plugging this into the above equation, we get a weight M~ 8kg. Which is about right, judging by what I see at the gym.

Force Limit

So we've seen that muscles generally produce a stress of around 2x105 N m-2. As this applies to all muscles in all animals, perhaps there is a fundamental reason why it isn't greater?

We can compare this to the force between atoms. We worked out earlier that if every atom pulled every adjacent atom with the full intermolecular force, the total stress would be around 1011 n m-2. Which is a million times greater than muscles can exert. So the strength of atomic bonds isn't the reason why muscles aren't stronger.

In fact, it is quite easy to build machines that can exert much more force (per unit cross sectional area) than muscles. Many excavating machines use hydraulics in place of muscles. Ninja Physics/Paul Francis 135

These are pistons pushed by high pressure oil. And the oil is these systems routinely works at pressures of up to 500 bar (a bar is 105 Pa, which is roughly one atmosphere). So this is around

250 times more stress than muscles can exert.

So it doesn't seem that muscles are up against any fundamental limit. They are in fact very puny compared to some quite simple machinery! And muscles are not just slightly worse - they are drastically worse.

This seems rather odd to me. Why hasn't four billion years of evolution come up with something better, like hydraulic powered legs?

One possibility is that evolution is stuck in a dead end with the way muscles currently work. Or perhaps the way muscles work confers some other benefit?

Or it could be that force is not the limiting factor to animal performance. If something else limits you, then there is no evolutionary benefit in having better muscles.

Energy Consumption

So - it doesn't seem that the force exerted by animal muscles is up against any fundamental physical limit. So could it be that energy, rather than force, is what limits human performance?

To investigate this, let’s think about how much energy a human uses, and where it goes.

One way to investigate this is to look at how much energy you consume. You could keep track of how much you eat, for example. The normal dietary recommendation is that someone eat around 8000 kJ per day if they are not to put on or lose weight (higher for men, lower for women). This corresponds to around 8x106/(24x60x60) = 93 W.

So roughly speaking, humans consume energy at the rate of ~ 100 W, as measured by how much you eat (and factoring in the combustion energy of your food). Which is pretty Ninja Physics/Paul Francis 136 economical - remember that a car was using around 2x108 J driving to Sydney, which is about

20,000W. You could power 200 people for the energy cost of one car.

Another way to measure this human energy expenditure is to look at how much oxygen you breath in, and factor in the combustion energy per unit weight of oxygen.

(figure from MH Laughlin (1999): Cardiovascular response to exercise, American

Journal of Physiology, 277, S244.

This plot is based on people pedalling on a bicycle ergometer. People who don't get much exercise (sedentary in the graph above) can produce up to around 200W when they try hard, while elite endurance athletes can manage over 400W.

The slope of the above line tells us the efficiency of human exercise. You work this out by comparing how much work is being done by your muscles on the bicycle ergometer to the combustion energy of the oxygen you are consuming.

What you find is that only about 25% of the combustion energy is going into mechanical work. The rest is presumably ending up as heat.

Work Done

How much mechanical work does a human body need to do when exercising? Ninja Physics/Paul Francis 137 Let's consider an easy case - my morning exercise - a walk up Mt Ainslie. This is about a

300m elevation gain, so the energy needed is the increase in potential energy mgh - which for my mass (70 kg) is 300x70x10 = 210 kJ.

That's not very much! It's about the energy in one slice of dry bread. My exercise tracker logs a brisk 30 min up-hill walk at more like 1000 kJ. Which is probably about right if you factor in the 25% efficiency of converting food into muscular work.

How about walking or running on the flat? Why do you spend energy (say) walking around Lake Burley Griffin? Or running the 100m sprint at the olympics? There is no height gain. There is no friction - your feet are not sliding along the ground. There is wind drag - but it's pretty small. Walking around the inner basin of Lake Burley Griffin takes me around an hour to cover the 5km, so my speed is 5000/3600 = 1.4m/s. If we assume I have a cross-sectional area of

1m2 and a drag coefficient of 1, then the drag force is

So the energy is force times distance - i.e. 5 kJ. Tiny.

Running the 100m at Usain Bolt's speed (12 m/s), the drag force is larger due to the greater speed - around 72N. But that's still only a small fraction of the 1000 N force Usain exerts.

And over 100m, the work done is only 7kJ - minuscule. It is a lot more significant at bicycle speeds - indeed drag is often the limiting factor for cycling races. But for running, it will only make tiny fraction of a second differences in time - enough maybe to establish world records, but not the dominant energy drain.

So where does the energy go when running or walking? You have to put in kinetic energy to accelerate up to your running or walking speed - but that's also pretty small, for the Usain Bolt sprint the kinetic energy is: Ninja Physics/Paul Francis 138

So where does the energy go when running?

I've seen a lot of different answers written down in various places, but most of them don't make sense. One common answer is that when a runner puts the front foot down, their body has to rise up. But if you look at movies of runners, it is clear that the centre of mass of their body doesn't move up or down very much.

Let's say Usain Bolt's centre of mass moves up 20cm every stride (which is almost certainly too big). His mass is around 100kg, and he runs the 100m sprint in around 50 paces. So he has to raise his body by 0.2m a total of 50 times, so the energy is 50 mgh = 10 kJ. Still pretty small.

I think that the energy is mostly used in accelerating the back leg forward each stride.

Let's imagine that the bottom part of his leg has a mass of around 5kg (probably an underestimate), and he needs to accelerate it from rest (at the back of his stride) up to 24 m/s

(double his speed, to bring it forward in time for the next stride.

So the kinetic energy per stride is

But he has to do this 50 times, hence requiring around 70 kJ. An order of magnitude more than the other forms of energy loss.

Many books analyse the swing of legs differently - they model it as a pendulum. A pendulum of length L swings with a period of

Ninja Physics/Paul Francis 139

which for a leg, with its centre of mass ~ 0.5m below the hips, gives a period of around

1.3 sec. That feels about right - if I relax my leg muscles and swing one leg while standing on tip-toe with the other, it takes ~ a second to swing forward.

So if you move at around 1 stride per second, this natural pendulum motion will do the work for you. This would therefore be a very economical mode of transport. But it's not fast - if each stride is around 1m, this gives you a speed of roughly 1m/s which is 3.6km/hr. Which is indeed a typical walking pace. So normal walking uses the natural pendulum swing to make your stride very efficient. But it won't win any Olympic races - for that, you need to take strides far more rapidly (5 per second for Usain Bolt), and that burns energy.

So for the 100m sprint, Usain Bolt is exerting around 100 kJ in 10 sec, which is 10 kW power. Factor in the 25% efficiency, and the power consumed in a whopping 40 kW. This is comparable to the power of a car driving to Sydney at 110 km/hr! It's vastly more than the power consumed in any longer lasting activity.

But bear in mind that one chocolate bar contains around 1000 kJ. A world record in the

100m consumes about 40% of a chocolate bar of energy...

Limiting Factors

So we've worked out the sort of energies required to power human activity. Are these a limiting factor?

Brainstorming, there are a few possible ways in which this energy could be limiting.

1. It's hard to get enough oxygen to the cells (out of breath)

2. You overheat (can't get rid of the waste heat fast enough) Ninja Physics/Paul Francis 140

There are some other possibilities, like inability to store enough food, or get rid of waste prodyucts fast enough, or insufficient circulation to get the food and oxygen into the cells fast enough. But let's focus on these two.

Diffusion

To get oxygen into the muscle cells, there is a three-step process. Firstly, it diffuses from the air in the lungs into the bloodstream nearby. Secondly, the blood is pumped into a vast number of tiny capillaries. And thirdly, the oxygen diffuses from the blood into the cells. A similar process applies to food and the waste products.

So the first and last steps are diffusion. What is this, and could it be a limiting factor?

Let's say we open a bottle of perfume in a room with no air currents. How could the smell spread? To begin with, you'd have a high concentration of perfume molecules just near the mouth of the bottle, and none anywhere else. But these perfume molecules will move off at high speed in both directions. They will rapidly crash into air molecules and head off in random directions. Some will head back towards the perfume bottle, but some will keep on moving outwards.

This sort of motion is known mathematically as a "random walk". On average the molecules don't go anywhere - as many go from left to right as go from right to left. But they Ninja Physics/Paul Francis 141 spread out. You can use statistics to estimate how much the distribution spreads out - the

(sigma) width in the graph above. If each step is a distance l, and there has been time for n steps, then

Why the square root of n? You might think that with n steps of length l, you would go a distance nl. And that's true if all those steps were in the same direction. But the odds of all the steps being in the same direction become very small if you have a lot of steps. In one second, for example, you'd have 1010 steps - if there is a 50% chance of each step being in either direction (to stick in 1D for simplicity sake), the odds of all being in the same direction are

which is a very small number!

On the other hand, the odds of a molecule having exactly the same number of positive and negative steps and hence ending up where it began is also very low. In 1D you can show mathematically that you end up with a Gaussian distribution (bell curve) of standard deviation sigma, which is equal to the step length times the square root of the number of steps. This is actually the same maths behind when averaging multiple experimental measurements, the uncertainty goes down as the square root of the number of measurements (i.e. if you had 100 opinion polls each with an uncertainty of 3% and you average them, the resultant average will have an uncertainty of 3% divided by the square root of 100 - i.e. 0.3%. That is, if the errors are random and not systematic). We will talk about this in the labs.

So how far will our perfume go? If each step is around 10-7 m, and you have 1010 steps in a second, then after one second, Ninja Physics/Paul Francis 142

So the smell will spread out by about 1cm in a second. How about an hour? That is 3600 seconds, so 3.6x1013 steps, so the distance is around 0.6m. Not very far. And because the distance depends on the square root of the time, waiting ten hours only increases the distance by the square root of ten - around a factor of three.

In reality smells spread around rooms much faster than this - which tells us that diffusion is not the dominant effect - usually convection or other air currents are much faster.

In water, because the density is roughly 1000 times greater, the mean free path is 1000 times smaller, so in a second, you will only go a distance of about 10 micro-metres.

Lung Size

This small diffusion constant is a big problem for life-forms. It means that oxygen will only diffuse over distances of a micro-metres on reasonably short time-scales.

For single-celled life-forms, this isn't a problem. Diffusion can supply them with all the oxygen they need, because they are only a few micro-metres in size. But for anything bigger than a microbe, it is a real problem. Larger organisms need specialist organs such a gills or lungs to get enough oxygen into their bloodstream.

The oxygen supply in a human involves three steps. Firstly, the oxygen must diffuse through the surface of the lungs into the bloodstream. The blood must then carry it in a timely fashion to within ~ 10 microns of every cell. And then diffusion will do the last step, carrying oxygen into the cells and around them to where it is needed.

The biggest problem here is the first step - getting enough oxygen to diffuse through the surface of the lung into the bloodstream. The human body requires more than 1020 oxygen molecules per second, and to get that much to diffuse through a membrane, it needs a surface Ninja Physics/Paul Francis 143 area of tens of metres! See the book Zoological Physics by B.K. Ahlborn (Springer 2004) for the full derivation.

But how can we possibly fit a surface area of tens of square metres into a human body that's only one or two metres tall?

And in fact, the internal surface area of the human lungs is around 80 square metres!

How can you get such an enormous surface area into the volume of the lungs (less than 0.1 cubic metres?) It's done by having the interior surface of the lungs covered with vast numbers of alveoli - little circular air sacs only around 100 microns in radius. That means each has a surface area of which is around 10-7 m2. So we need more than 108 of them. But as each only occupies a volume of around

you can easily fit this into the lung volume.

In fact, the square-cube law is at work here - if you made the alveoli smaller, you could presumably fit an even bigger surface area in, and hence get more oxygen.

So - getting enough oxygen into the body is a formidable challenge, but there is a solution - lungs with incredibly convoluted interior surfaces - and there is no really obvious reason why you couldn't make these surfaces more convoluted still, and hence get even more oxygen in. But it will be challenging.

At the other end of the process, how do you get oxygen close enough to every muscle cell

(i.e. within a few tens of microns) so it can diffuse in? The body's answer is a vast number (~

1010) of tiny veins called capillaries, each only around 1 microns thick and 1mm long. There are so many of these that they pass close to every cell, so diffusion can transport oxygen in and waste products out.

# Ninja Physics/Paul Francis 144 Compost Piles

This isn't directly relevant to human performance, but it's cool - this diffusion calculation also explains how to prevent a compost pile from smelling. In a compost pile, bacteria and fungi break down organic material, consuming oxygen and generating heat in the process.

A compost pile will have a network of tiny air gaps throughout it.

Heat generated by the decomposition inside heats the air, which causes it to rise and flow out through the top. This in turn sucks cool air in at the bottom, bringing fresh oxygen into the compost pile. But if you get closed-in passages, they will fill with stagnant air. For these passages, only diffusion will allow oxygen to get in. And we've just seen that oxygen can only diffuse over distances of centimetres. So if you have dead-end crevices longer than this, the decomposition can use up all the oxygen, leading to anaerobic areas, which are the source of bad smells.

So the art of producing effective compost piles is to make sure there are plenty of continuous air passages through it. This is normally done by turning it over regularly, and making sure that it is not too wet - as water tends to block the air passages.

Getting rid of heat

So now we get to the last potential limiting factor - getting rid of heat. Remember - no machine is 100% efficient. Most (including muscles) are only 20-30% efficient, so 70-80% of Ninja Physics/Paul Francis 145 the energy ends up being wasted as heat. If this heat builds up in a body, eventually it would kill you.

We know that athletes get very hot and sweaty, so this is clearly a potential issue. It's also an issue in many other situations. For example, many modern computers are limited by cooling, not by microprocessor speed.

When you supply heat to an object, first of all it will heat up. You can calculate how much using the specific heat capacity. But as the object gets hotter, it will start to lose more and more heat to its surroundings. Eventually it will get so hot that the heat lost balances the heat supplied, and you will be in equilibrium.

When doing a heating calculation, the first thing to do is work out which limiting case you are in. Are you just starting to warm something up, in which case you need the physics of specific heat capacities. Or has it come into equilibrium, in which case you should balance heat in and heat out. We'll see later in the course how to deal with the intermediate situation where both bits of physics apply - it's complicated!

Which regime are we in for the human body?

A human has a mass of around 70 kg, and is mostly made of water, which has a specific heat capacity of around 4200 J/kg/K. So the energy needed to raise your temperature by 1 degree is 70x4000 ~ 300 kJ.

Now if you remember, walking up Mt Ainslie required around 1000 kJ. If 75% of that ends up as heat, that would raise a body's temperature by 2-3 degrees. Which is enough to make you very sick indeed - raising the body's temperature from 37C to 40C is very serious. A jog could well generate twice that amount of energy, which would raise your temperature to around Ninja Physics/Paul Francis 146

43 degrees, which would kill you. So for any prolonged exercise like this, the human body must clearly get rid of most of this heat. So you are probably in an equilibrium situation.

For a very short burst of exercise, like a 100m spring, the energy used up is a few hundred kJ (~ 100 kJ of work done, multiply by 3-4 to get the heat generated). Which might be bearable - but a one degree rise in core temperature is not a good thing. So even for short activity, you really want to lose some of the heat.

So how can a human body lose heat? There are basically three ways:

● Radiation

● Heat is conducted through the skin and clothes and into the air around the body, and then

carried away by wind or convection.

● Evaporation - water/seat on your skin evaporates and carries away heat.

Let's look at these in turn.

Radiation

Everything (that's at a temperature above absolute zero, which means everything...) emits electromagnetic radiation. Why?

Remember - the fundamental idea of statistical thermodynamics is equipartition - on average, energy will be shared between every possible form. Some energy will be in motion, some in vibration, some in rotation etc, and on average, as the atoms jiggle around and collide, the energy in each will even out. It turns out that this can be used to calculate thermal emission.

The basic idea: empty space is be full of electric and magnetic fields. And electric and magnetic fields can contain energy. If you put some empty space near an object of temperature

T, then energy can pass back and forth between the fields in this empty space and the jiggling Ninja Physics/Paul Francis 147 atoms in the object. For example, an electric field can accelerate electrons and protons in the object, and a jiggling charge in the object will produce electric and magnetic fields.

Now let us imagine that these interactions are really strong. In this case, energy is easily shared between the thermal energy of the atoms and the electric and magnetic field energy of the empty space. So each form of electric and magnetic energy will have the same 1/2kT that each type of motion and vibration of the atoms has.

This situation is called a "Black Body". Black, because a black object is one that interacts strongly with light (for example, by absorbing all the light that falls on it). And in this situation, you can calculate the radiation field in the empty space. You'll see this calculation in 2nd year thermodynamics.

For our current purposes, the key result from this calculation is the Stefan-Boltzmann equation, telling you the total power of the electromagnetic radiation emitted by a black body:

where P is the power emitted, A the surface area, (the Stefan-

Boltzmann constant), and T is the temperature in Kelvin.

What wavelengths does this radiation come out at? That's given by the Wien's displacement law:

where b is Wien's displacement constant, 2.9x10-3 m K. For a human body at 36C

(273+36 K), this corresponds to a wavelength of around 9 micro-metres, in the mid-IR.

Now this is the emission from a black body - an object that interacts strongly with the surrounding electric and magnetic fields - so strongly that it is in equilibrium with them. How realistic is that? Ninja Physics/Paul Francis 148

Well - very few objects are true black bodies, but many come fairly close, at least at some wavelengths. You can crudely approximate the non-black-body nature of real objects by putting in an efficiency factor, so

where the efficiency is called the emissivity, and would be 1 for a dark black object, and as low as 0.01 for a highly polished mirror. So for Ninja purposes you can probably approximate it as 0.5 or 1, except for very light shiny objects. For human skin, at the infra-red wavelengths where we mostly radiate, the emissivity is about 0.93 for both light and dark skinned people (Watmough & Oliver 1968, http://www.nature.com/nature/journal/v219/n5154/abs/219622a0.html) so we are pretty close to being black bodies (at visible light wavelengths dark skinned people have emissivities of around

0.8 while pale skinned people have values of around 0.4, but this doesn't really affect radiation as humans are not hot enough to emit anything at those wavelengths).

Human radiation

So - how much radiation would a naked human emit? An average adult man has a surface area of 1.9 m2, while for a woman the typical area is around 1.6 m2. Let's call it two square metres for ninja purposes. Using the Stefan-Boltzmann equation and assuming an emissivity of 1 and a skin temperature of 35C, the power radiated comes out as 1020W.

But that's too simple. Because we will also be absorbing radiation from our skin. If we are surrounded by objects at some temperature T, they will all be emitting radiation which will hit us. How can we calculate the energy we gain via radiation? Ninja Physics/Paul Francis 149

Well, there's a trick we can use! If we were surrounded by objects at the same temperature as us, then then energy we receive would have to balance the energy we radiate - we'd all be in thermal equilibrium with our surroundings and with the energy field.

So - if we're surrounded by an environment at temperature Ten, the heat we gain from it is just what we would lose if we were at the same temperature - i.e.

So how much heat do we gain from the environment? If the environment is at 20C, this comes out as 836W, so on average we're losing 1020-836 = 184W.

If it's colder outside, say 0C, then we lose 391W via radiation, while if the outside temperature approaches 35C, we won't lose anything.

So on a 20C day, we lose more energy than our resting metabolic rate from radiation.

Indeed radiation could keep us cool with moderate exercise. But even if it is freezing outside, it isn't enough to cool down someone exercising really hard.

It is a problem if you are trying to keep warm in a cold climate - radiation losses can be substantial. The solution is clothing. Your clothing will also radiate energy, but if the outside layers of your clothes are at a lower temperature than your skin, they will radiate less. Some coats and emergency blankets have reflective coating on their insides, to reflect the thermal radiation from the body back, which will also help.

Conductivity

Conduction is one of the ways heat can move through an object (the others are convection and radiation), and is physically very similar to diffusion. Like diffusion, heat is conducted from regions with a lot of it to regions with a little, and the rate of flow is proportional to the temperature gradient, so: Ninja Physics/Paul Francis 150

(in 1D), where H is the heat flow, k the thermal conductivity, A the area and T the temperature.

In a gas, conductivity is virtually identical to diffusion - fast moving atoms from the hot region diffuse through the slower moving colder atoms. In liquids and solids things are more complicated - coordinated vibrations (photons) can carry heat around, and in electrical conductors, free electrons knocked by fast moving atoms in one part of the metal carry their energy large distances before bouncing off other colder atoms and passing on their energy.

It is possible to estimate the thermal conductivity of air in much the same way as we calculated the diffusion constant, and you get k~0.025 w m-1 K-1. Feel free to do this calculation - it's not hard. Though in practice this only applies to tiny bits of air - if you have a large quantity of air, any temperature difference will cause convection to take place, spreading the heat around much more rapidly. However, this value of k is close to what you get with many insulators, such as the batts used in roof insulation, which are indeed mostly made up of air divided up into small spaces to prevent convection. It's also what you get with double-glazed windows, where the cavity between the two sheets of glass must be small enough to stop convection. Or with much clothing, which traps air between the fibres.

How good would a 10cm thick layer of roofing insulation be in keeping a house warm in winter? Let's say we have a 100 square metre house, and a temperature inside of 20C, and outside of 0C. dT/dx is thus 20/0.1 = 200 K/m.

The heat flow through A=100 m2 of this is thus:

Ninja Physics/Paul Francis 151 Is this a plausible amount? Most electric heaters seem to have power-ratings of 1000-

2000 W, which is comparable, and we know heaters can keep a house warm, so this looks like a sensible figure.

Over a night, how much heat do we lose? Let's say 10 hours, 3600 seconds in an hour, so the heat loss through the insulated roof will be around 500x10x3600 = 18 MJ. Gas in Canberra costs around 2.5 cents per mega-Joule, so it costs around 45c to offset the heat leak through the roof. Of course there will be other heat leaks - through windows, walls, floor etc. But this suggests that insulation of this thickness can greatly reduce heating bills.

Metal has a much greater thermal conductivity due to the movement of free electrons carrying heat from hot areas to cool - for steel is is around k~ 50 W m-1 K-1 while for copper it is k ~ 385 W m-1 K-1.

Imagine that you put a teaspoon in your tea/coffee (at around 100C). How hot will the handle get? We know from experience that it is possible to stir your tea without your fingers getting burned, but we also know that if you leave a spoon in your tea for a while, the end will be too hot hot to touch comfortably. So we are expecting it to take some tens of seconds to heat up.

Is this what the physics predicts?

So what's our model here? One end of the spoon is at 100C, while the other starts at room temperature. Heat will be conducted down the handle - which we'll assume is made of steel and is around 1mm thick and 10mm wide - i.e. A~ 0.001x0.003 = 3x10-6 m2. The handle of the spoon is around 10cm long, so dT/dx ~ (100-20)/0.1 = 800 K/m. That's the initial figure - as the handle warms up the temperature gradient will decrease.

So how fast is heat going to flow up the spoon handle?

Ninja Physics/Paul Francis 152 So that's not a lot - but is it enough to heat up the handle of the spoon? Let's say the top of the handle is 1mm thick, 10mm wide and 3cm long. It's volume is thus a whopping ~10-7 m3.

The density of steel is around 8000 kg/m3, so the mass is ~8x10-4 kg. As we saw earlier, the specific heat capacity of iron is around 500 J kg-1 K-1. So it takes around 500x8x10-4 = 0.4 J to warm up the handle by one degree. So if we're supplying 0.12W, that means it will take around 3 seconds to warm up by one degree - so perhaps a minute or two to heat it up to uncomfortable temperatures. Which seems about right, in my experience with teaspoons!

How effective is conduction at cooling down athletes? Conduction through the skin is quite fast, as blood flow brings the heat up to very close to the surface. The major barrier to heat loss is conduction into the surrounding air.

If you were naked on a calm day, the heat will be conducted into the air, until the air is hot enough that it begins to convect - i.e. the hot air near the skin rises and flows away, being replaced by cooler air. If there is some wind, or a fan blowing, or you are running, the heat might only need to be conducted through a thin layer of stagnant air right against your body.

If we assume the air needs to be conducted 1cm before being blown or convected away, then the energy loss rate for a person with a two square metre surface area is

(assuming a 35 degree skin temperature and 20 degree air temperature). A windy day would increase this by reducing the distance the heat needs to move, as would a colder day.

That's pretty small, compared to even the energy consumed by a resting person, let alone someone exercising.

Clothes are effectively layers of stagnant air trapped against your body, and so their effect can be calculated similarly. Ninja Physics/Paul Francis 153

So the conduction heat loss depends a lot on environmental conditions. Meteorologists have models of a typical human in typical clothes that they use to calculate wind chill factors - how much more heat you lose when the wind reduces the thickness of the stagnant air layer.

But basically, this heat loss rate is not going to be sufficient for anyone exercising, unless it is very cold and windy outside. We will need some other mechanism to keep us cool.

Evaporation

So radiation and conduction are probably not going to carry away enough heat to keep an athlete alive through any sustains exercise, especially on a hot day. So what else can we do?

For human, the answer is sweat - cover our skins with a layer of water which can evaporate off to cool us.

The latent heat of vaporisation of water is 2.3x106 J/kg, which is how much heat every litre of evaporating sweat carries away.

How fast does water evaporate? The physics of this is complicated - it depends on the relative humidity of the air, the temperature and the wind speed. You can find tables of evaporation rates for different conditions on the web and in engineering handbooks.

But for a rough figure, we can look at experimental results. Meteorologists measure evaporation rates all the time (they are an important figure for agriculture). They do this by putting a pan of water outside and measuring how rapidly the level goes down. In Canberra the pan evaporation is 110mm per month (200 mm per month in summer). In the outback it can be up to 420 mm/month in summer, while in hot but humid places like Singapore the evaporation rate is only around 120 mm/month. 1mm per month on a one square metre area would be 10-

3x1x1x1000 = 1 kg m-3 per month, which is 4x10-7 kg m-2 s-1. Ninja Physics/Paul Francis 154

So let's say you are going for a run in Canberra in summer. The evaporation rate here is around 8x10-5 kg m-2 s-2. If you have a 2 square metre surface area, and your skin is entirely covered in sweat, then you could lose 2x8x10-5 ~ 1.6x10-4 kg/s, or 0.6 litres per hour of water.

Which sounds about right for an average day or night in summer - you'd lose more in mid-day on a hot sunny day, and less on a cool humid night. How much energy does this carry away?

Multiply by the latent heat and you get 368 W.

So this is the biggest heat loss - enough to cope with reasonably strenuous exercise, just about. But if you were in Singapore on a hot humid day, your heat loss would be much less.

Conclusions

It seems to me that inability to get rid of surplus heat is the major factor limiting animal performance. It seems clear we could have stronger muscles. Oxygen flow is a limit, but there seems no obvious reason why we couldn't redesign our lungs and circulation system to get more oxygen in, if we needed to.

But any strenuous exercise really pushes the limits of our ability to get rid of heat.

Radiation can be effective if you are naked and doing your exercise on a cold day, but won't work if the outside temperature approaches body temperature. Conduction similarly requires special conditions to be very effective (particularly a cold wind), and won't work if its hot and calm.

Sweating is the most effective way to get rid of heat, but is limited by the rate at which water will evaporate from out surface area. In hot humid places, even sweating can't get rid of the heat that would be worked up by strenuous exercise.

So what would happen if we had a super-hero like the flash operating at ten times the normal metabolic rate (much as mice do). Mice can get away with this because they are small - Ninja Physics/Paul Francis 155

the square-cube law means that the ratio of surface area to body mass is much higher for mice, so they can get rid of energy efficiently. But for the flash, he is limited to the same 2 square metres that we have.

A normal human doing heavy exercise might be burning 400W. So let's say a superhero

(the Flash?) burns ten times as much - 4000 W. He can radiate away around 200 on a 20C day, conduct away around 10W, and sweat away around 400, leaving around 3600 W stuck in his body. It takes 300 kJ to raise the body's temperature by a degree, so his temperature would increase by 1 degree roughly every minute. He'd die in about 7 minutes, and spontaneously combust in an hour or two of activity. SO super-human feats are only possible in very short bursts, followed by cool-down time. Which is very much what Usain Bolt does.

So if we want super-fast aliens or robots, we need to find some more effective way to get rid of heat. Elephants use radiators (their ears) which they pump heat into via blood flow. But despite this, elephants need a low metabolic rate to stop combusting.

This same problem applies to modern computers. They are usually run slower than they are capable of, because the limiting factor is how fast the heat can be dissipated. High end gaming computers will use water cooling and radiators, while standard computers use fans. But it’s not easy - especially for smartphone or tablet processors where you really don't want to use a fan.