A Derivation of =

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Additional study material

Meitner Day

Contents Page 2 Meitner Day additional material 4 Introduction to Lise Meitner 5 The road to Meitner’s discovery 7 The published note in Nature 9 Special Relativity 10 Preliminary considerations 11 The two postulates of Special Relativity 12 Inertial reference frames 14 The invariant space-time interval 17 The Galilean transformation 19 The Lorentz transformation 24 The Lorentz velocity transformation 28 Momentum 37 Bibliography and further reading

Meitner day additional material

This additional material is to be studied on and after the day itself. The purpose is to provide interested students with an understandable and accessible derivation of the famous E=mc2 formula in terms of mathematics that is appropriate to those students studying A level Physics and/or A level Maths.

As an A level student myself, I remember my own frustrating attempts to teach myself the maths and physics of relativity theory. I looked at a variety of books and found that they were either of the popular science kind, with no mathematical content at all, or else they were academic texts that took too much of the maths for granted and as a consequence I stumbled at the point of trying to derive the Lorentz transformation. This material is an attempt to remedy this by presenting a single continuous mathematical argument leading from first principles to the famous formula E=mc2 itself. Mathematical content and requirements.

It is my view that the average well prepared sixth form student who studies for an A-level in mathematics will find that the Special Theory of Relativity is not significantly more difficult to learn than the methods of the standard practice exercises in the textbooks that are taught in secondary schools every day of the week. The only difference is that the mathematical tools learned during GCSE and A level must be applied together with some additional concepts to form one continuous mathematical argument. My hope is that this material will bridge the gap between the thinking of A- level mathematics and the level of thinking required in university. It presents one continuous mathematical argument which uses many of the tools learnt in school mathematics but no more.

This derivation is a composite of various sources. It relies heavily on that given in the excellent book “Beyond the mechanical universe” by Olenic, 2007. I have tried to present the material in such a way that I balance two conflicting criteria; first I assume the minimum of prerequisite mathematical knowledge and only a level that uses the maths and physics taught in secondary school; and second I try to avoid watering down the subject matter. Students are not required to have already studied in detail

2 electro-magnetism, the theory of electromagnetic waves or any physics beyond A-level maths and physics. I have included only those physical theories and concepts that are directly related to the mathematical narrative.

The mathematical derivation presented here is fairly close to the way in which it was first understood by physicists at the turn of the 20th century. This is an algebraic approach based on the Lorentz transformation to convert from one observer’s coordinates to those of another who moves at a fixed speed with respect to the first. The derivation will lead you along the following route. First the need for special relativity will be outlined and the relevant parts of Newtonian physics will be explained. The mathematical argument will then pass through the derivation of the Lorentz transformation and the gamma factor followed by the velocity transformation, the considerations of the conservation of momentum and the conservation of mass and finally by use of the binomial theorem to the E=mc2 formula itself.

You are not expected to understand it all at once. It may well take you all the way to the end of year 13 and into university to fully get it all. As you learn something about each of the topics mentioned above you should be able to get a little further. Maths takes practice and you need to develop a way of embracing the difficulties that you encounter along the way. To all students, good luck and if the going gets tuff my advice is to write it out by hand and check out additional material on the internet or from other sources or even ask your teacher. Hopefully this little booklet will spur you on to ask more questions in class at the very least.

I am indebted to many people for their help, direct and indirect, in the preparation of this material. I would like to especially thank colleagues at QE Academy Trust School in Crediton, Devon for their help and guidance during the preparation of the text.

Lise Meitner An Introduction

Lise Meitner, was the first person to account for and to explain the atomic fission of uranium. During the early part of the 20th century, in collaboration with others she participated in the development of a series of experiments designed to bombard uranium with neutrons. The expectation was that some of the uranium nuclei would absorb one or more of the incoming slowly moving neutrons and form heavier elements new to science. They expected to find elements with a greater atomic mass than uranium, known as transuranes. Instead they found barium, an element roughly half as massive as uranium. Lise Meitner provided the staggering explanation that the uranium atom had split in two roughly equal parts, releasing energy as it did so.

It is important to remember that up until this moment (the winter of 1938) no one had thought that it would be possible to do anything more than chip a small piece, perhaps an alpha particle or two off an atomic nucleus. Meitner calculated that as each uranium nucleus is split into two, a large amount of energy was released. This was a momentous discovery, a scientific breakthrough of the first magnitude. Meitner had discovered a method of accessing a source of energy that was new to science. It was a moment that would change the world. I and many others believe she never received the credit she deserved and as a consequence she is largely unknown to this day.

We are still living with the consequences of what the world has done with the discovery that Meitner made. On the one hand we have nuclear weapons of mass destruction capable of releasing nuclear energy in such gigantic amounts and with such catastrophic after effect that our very existence on the planet may be threatened if ever they are used again in an act of war. On the other hand the discoveries of Lise Meitner may yet turn out to be part of the solution to the equally dangerous threat of global warming that looms large over us all. This is because Lise Meitner’s discovery of nuclear fission constitutes the basis of the technology that is

4 behind the generation of electricity by the nuclear power industry. It may yet come to be the case that Lise Meitner’s discovery plays its part in saving us all from the consequences of continued use of fossil fuels by providing the whole world with base level electricity supply to enable us the time to develop and deploy the full range of carbon neutral reusable resources that are required to power a decarbonised world economy.

Scientists are optimists by nature and Lise Meitner should be regarded as foremost amongst them. It will fall to the next generation of physicists to solve these problems. That is you! I hope that you read on and are inspired to follow in Lise Meitner’s footsteps and go on to study maths and physics at A level and university.

The road to Meitner’s discovery

Lisa Meitner is unusual for making her discovery at the fairly advanced age of 60. This was after a long career in physics working at the Kaiser Wilhelm Institute for Chemistry in Berlin. Like most scientists she worked as part of a team, most notably with Otto Hahn, with whom she collaborated for many years. Fritz Strassmann and Otto Frisch also played key roles in her scientific life. Hahn was a specialist in chemistry, Meitner was a specialist in physics and together they joined forces in their study of radioactive elements. In 1938 they wanted to discover and understand the effects of bombarding uranium with neutrons. Hahn would use his chemistry expertise to separate and isolate the resulting products and Lise would design and prepare the experimental apparatus and provide the physical explanation and interpretation. Their work drew on the discoveries of other great experimentalists including J.J. Thomson, Marie-Curie, Rutherford and Chadwick. It also draws on the theoretical and mathematical advances attributable to Newton, Leibnitz, Coulomb, Maxwell and Einstein. All of these are represented in the set of cards given to you as part of the Meitner day. The published work of contemporaries such as Fermi and Chadwick provided some of the experimental methods and fine- tuned their approach.

The basis of the experimental technique was to irradiate uranium with neutrons from a neutron source provided by a sealed mixture of radium

5 and beryllium surrounded by a layer of solid paraffin. The radium, an alpha emitter effectively knocked neutrons out of the beryllium nuclei and the paraffin moderator acted to slow the emitted neutrons down. The slow neutrons were then capable of being captured by the uranium nuclei placed adjacent to the source. The resulting products were then separated and their chemical properties determined by a complex series of experiments involving known carrier elements. The formation of products which were chemically inseperable from the barium carrier was the final experimental finding. The uranium had been split into two large fragments, barium and krypton. It took Meitner’s genius to piece it all together and to draw the conclusion that they had created barium from uranium, they called the process nuclear fission.

Meitner made the discovery after having been sent details of the experimental findings in Berlin while working with Otto Frish in Sweden. She was Jewish and, having narrowly escaped to Sweden from Nazi Germany, was unable to collaborate in producing the scientific paper written by Hahn and Strassmann in Berlin that lead to the award of the Nobel Prize. She was only able to send a short note to the scientific publication Nature. On the next pages we provide a copy of the note she published in Nature telling of her interpretation of the results of the experiments that she had designed and worked on for so long. This is the discovery of nuclear fission.

Disintegration of Uranium by Neutrons: A New Type of Nuclear Reaction

Lise Meitner and Otto R. Frisch

Nature, 143, 239-240, (Feb. 11, 1939)

On bombarding uranium with neutrons, Fermi and collaborators1 found that at least four radioactive substances were produced, to two of which atomic numbers larger than 92 were ascribed. Further investigations2 demonstrated the existence of at least nine radioactive periods, six of which were assigned to elements beyond uranium, and nuclear isomerism had to be assumed in order to account for their chemical behavior together with their genetic relations.

In making chemical assignments, it was always assumed that these radioactive bodies had atomic numbers near that of the element bombarded, since only particles with one or two charges were known to be emitted from nuclei. A body, for example, with similar properties to those of osmium was assumed to be eka-osmium (Z = 94) rather than osmium (z = 76) or ruthenium (z = 44).

Following up an observation of Curie and Savitch3, Hahn and Strassmann4 found that a group of at least three radioactive bodies, formed from uranium under neutron bombardment, were chemically similar to barium and, therefore, presumably isotopic with radium. Further investigation5, however showed that it was impossible to separate those bodies from barium (although mesothorium, an isotope of radium, was readily separated in the same experiment), so that Hahn and Strassmann were forced to conclude that isotopes of barium (Z = 56) are formed as a consequence of the bombardment of uranium (Z = 92) with neutrons.

At first sight, this result seems very hard to understand. The formation of elements much below uranium has been considered before, but was always rejected for physical reasons, so long as the chemical evidence was not entirely clear cut. The emission, within a short time, of a large number of charged particles may be regarded as excluded by the small penetrability of the 'Coulomb barrier', indicated by Gamov's theory of alpha decay.

On the basis, however, of present ideas about the behaviour of heavy nuclei6, an entirely different and essentially classical picture of these new disintegration processes suggests itself. On account of their close packing and strong energy exchange, the particles in a heavy nucleus would be expected to move in a collective way which has some resemblance to the movement of a liquid drop. If the movement is made sufficiently violent by adding energy, such a drop may divide itself into two smaller drops.

In the discussion of the energies involved in the deformation of nuclei, the concept of surface tension has been used7 and its value has been estimated from simple considerations regarding nuclear forces. It must be remembered, however, that the surface tension of a charged droplet is diminished by its charge, and a rough estimate shows that the surface tension of nuclei, decreasing with increasing nuclear charge, may become zero for atomic numbers of the order of 100.

It seems therefore possible that the uranium nucleus has only small stability of form, and may, after neutron capture, divide itself into two nuclei of roughly equal size (the precise ratio of sizes depending on finer structural features and perhaps partly on chance). These two nuclei will repel each other and should gain a total kinetic energy of c. 200 Mev., as calculated from nuclear radius and charge. This amount of energy may actually be expected to be available from the difference in packing fraction between uranium and the elements in the middle of the periodic system. The whole 'fission' process can thus be described in an essentially classical way, without having to consider quantum-mechanical 'tunnel effects', which would actually be extremely small, on account of the large masses involved.

After division, the high neutron/proton ratio of uranium will tend to readjust itself by beta decay to the lower value suitable for lighter elements. Probably each part will thus give rise to a chain of disintegrations. If one of the parts is an isotope of barium8, the other will be krypton (Z = 92 - 56), which might decay through rubidium, strontium and yttrium to zirconium. Perhaps one or two of the supposed barium-lanthanum-cerium chains are then actually strontium-yttrium-zirconium chains.

It is possible8, and seems to us rather probable, that the periods which have been ascribed to elements beyond uranium are also due to light elements. From the chemical evidence, the two short periods (10 sec. and 40 sec.) so far ascribed to 239U might be masurium isotopes (Z = 43) decaying through ruthenium, rhodium, palladium and silver into cadmium.

In all these cases it might not be necessary to assume nuclear isomersim; but the different radioactive periods belonging to the same chemical element may then be attributed to different isotopes of this element, since varying proportions of neutrons may be given to the two parts of the uranium nucleus.

By bombarding thorium with neutrons, activities are which have been ascribed to radium and actinium isotopes8. Some of these periods are approximately equal to periods of barium and lanthanum isotopes resulting from the bombardment of uranium. We should therefore like to suggest that these periods are due to a 'fission' of thorium which is like that of uranium and results partly in the same products. Of course, it would be especially interesting if one could obtain one of those products from a light element, for example, by means of neutron capture.

It might be mentioned that the body with the half-life 24 min2 which was chemically identified with uranium is probably really 239U and goes over into eka-rhenium which appears inactive but may decay slowly, probably with emission of alpha particles. (From inspection of the natural radioactive elements, 239U cannot be expected to give more than one or two beta decays; the long chain of observed decays has always puzzled us.) The formation of this body is a typical resonance process9; the compound state must have a life-time of a million times longer than the time it would take the nucleus to divide itself. Perhaps this state corresponds to some highly symmetrical type of motion of nuclear matter which does not favor 'fission' of the nucleus.

1. Fermi, E., Amaldi, F., d'Agostino, O., Rasetti, F., and Segré, E. Proc. Roy. Soc., A, 146, 483 (1934). 2. See Meitner, L., Hahn, O., and Strassmann, F., Z. Phys., 106, 249 (1937). 3. Curie, I., and Savitch, P., C.R., 208, 906, 1643 (1938). 4. Hahn, O., and Strassmann, F., Naturwiss., 26, 756 (1938). 5. Hahn, O., and Strassmann, F., Naturwiss., 27, 11 (1939). 6. Bohr, N., NATURE, 137, 344, 351 (1936). 7. Bohr, N., and Kalckar, F., Kgl. Danske Vis. Selskab, Math. Phys. Medd. 14, Nr. 10 (1937). 8. See Meithner, L., Strassmann, F., and Hahn, O., Z. Phys. 109, 538 (1938). 9. Bethe, A. H., and Placzec, G., Phys. Rev., 51, 405 (1937). Glossary

Radiactive periods: The length of half life of a radioactive source.

Nuclear isomerism: The ability of nuclei to occur in different metastable states.

Special Relativity

The Special Theory of Relativity was first presented to the world by Albert Einstein in 1905. It is a theory of space, time and motion. Special Relativity neglects the effects of mass and energy on space and time and is a special case of the somewhat more complicated General Theory of Relativity. The Newtonian world view interprets space and time as separately invariant and in particular it assumes space to be geometrically Euclidean and time to be universal. Special Relativity on the other hand dispenses with the idea that space and time are separately invariant and combines them into space-time that is geometrically hyperbolic. You have already experienced the effects of relativity, but you may not know it. Imagine you are looking out of the window of a train that has stopped at the station. Right outside the window is another train, which has stopped to let people off at the other platform. Suddenly, you notice your train is moving, it slowly moves off. You begin to pass the carriages of the train on the other platform as it gets left behind. All of a sudden the end of the other train passes by and you are looking out across the line to an empty platform. To your surprise you realise that you are not moving. You are in fact still at rest at the station and it was the other train that was moving. This is relativity! Lise Meitner’s discoveries were one of the first practical applications of special relativity and Meitner’s calculations were some of the first proof of the validity of the famous E=mc2 formula. As part of the Lise Meitner day you will have had the opportunity to perform a version of this calculation.

Preliminary considerations

In order for us to tackle the derivation of Einstein’s formula we must define a few terms and make sure we understand some of the principles of physics according to Newton. These include inertial reference frames, speed, velocity, momentum and kinetic energy and certain assumptions about the nature of space and time. One of the first things to get your head around is why we can’t just use Newton’s physics anyway. After all it seems to describe the world around us pretty well. Why do we need special relativity? Newtonian physics and the need for Special Relativity

According to the principles of the Newtonian physics the velocity addition law tells us that if two objects approach each other at speed, then the relative speed of the two objects can be found by adding the speeds together. This seems to be common sense. If you move towards me at 5 mph and I move towards you at 5 mph, then we must move towards each other at a closing speed of 10mph. It turns out that this common sense notion is only an approximation and it lies at the heart of the theory of special relativity. At the normal speeds at which we move around the correction is too small to notice, as we deal with faster speeds near the speed of light the correction needed becomes much greater.

The two postulates of Special Relativity

The Special Theory of Relativity is a theory of space, time and motion. It is based on two fundamental postulates. These are the two facts upon which the whole theory is built. They have been checked by many experiments and no experiment has ever found evidence to contradict them. The student needs to accept them as fact, as Einstein did when he originally devised the theory.

Postulate 1: The constancy of the speed of light

The speed of light is always measured to be the same by any observer, no matter what the speed of the source of the light or that of the observer. We give the speed of light the symbol c, it has the value of 3.0 x108 m/s.

Postulate 2: The principle of relativity

The laws of physics must be written in the same form in all reference frames. All inertial reference frames are equivalent.

ν c

Figure 1

The figure represents a driver in a car, which moves to the right at the speed v, and a man sitting in a chair at rest. Both will measure light approaching them to be moving at the speed c. Let us consider the situation in figure 1, above. The car moves to the right at a speed v, and a light beam moves to the left at a speed c. According to Newtonian physics we would expect the observer in the car to measure the speed of the oncoming light as the speed v + c. For the person sitting still in the chair however, we would expect them to measure the oncoming light to have a speed c. In other words common sense tells us that we would expect the two observers to measure different speeds for the speed of light. However, this is not the case. Both will measure the speed of the oncoming light to be c. This is the essential thought experiment that Einstein considered and which ultimately led him to suggest that Newton’s description required modification. That modification is the Special Theory of Relativity.

Inertial reference frames and standard configuration

An inertial reference frame is a set of coordinate axes that is at rest or that moves with a constant velocity without rotation in which Newton’s laws apply. There are many assumptions in Newtonian physics and most of Newtonian physics is best understood from the point of view of inertial frames. An important assumption is that in the Newtonian world view time and distance are the same in all reference frames. To begin with let us accept this idea, you will later see that this is in fact an approximation and needs to be corrected for. The correction is very small at slow speeds but becomes larger at faster speeds. The genius of Newton was to take the work of Kepler, who studied how the planets move around the sun, and the work of Galileo who studied how things move on and near the surface of the Earth and combine them to form laws that were universal in their applicability. Laws that worked equally well for the motion of planets and for the motion of falling apples near the surface of the Earth. I state Newton’s laws below

Newton’s first law: A particle will remain at rest or continue in a state of uniform motion in a straight line unless acted upon by unbalanced forces.

Newton’s second law: When an object is acted upon by a force there will be an acceleration of the body in the direction of the force.

Newton’s third law: To every action there is an equal and opposite reaction.

/ / y y S / S B B v

U W W

O x O/ x/ A A/

Figure 2. A linear transformation between two inertial reference frames. The two reference frames S and S/ are in standard configuration. In reference frame S a particle moves from A to B with a velocity U. The reference frame S/ moves to the right at velocity v, the same velocity as the horizontal component of the velocity U. An observer in reference frame S/ finds the particle moves vertically from A/ to B/ with velocity W. From the inertial frameS/ the frame S moves to the left with velocity –v.

Consider figure 2, above. Imagine that at time t=0 both of the origins coincide and that the x and y axes of inertial frame S are parallel to the x/ and y/ axes of the inertial frame S/. At t = 0 a particle begins to move at a constant velocity from A to B. On the left the observer in reference frame S considers his or herself to be at rest. According to an observer in this frame a particle moves from A to B with a velocity U. This has a horizontal component of velocity V, and a vertical component of velocity, W. At t=0 the observer in frame S/ begins to moves to the right along the x axis at a constant velocity V and thus measures the particle to move in a straight line vertically with a velocity W. Thus straight lines map between reference frames linearly, straight lines go to straight lines.

The invariant space-time interval

In Newtonian physics both distance and time are separately invariant. This means that space and time are independent and separate things that can have no influence on each other. This means that the distance between two points is always measured to be the same by any observer, no matter what their position or speed of motion or orientation. Time is also considered to be an invariant. Time in Newtonian physics always passes at the same rate in any place as measured by any observer no matter what their location, orientation or speed of movement.

In special relativity this is not the case. This is because the speed of light is a constant for all observers. As you will see, the Lorentz transformation shows us that space and time get mixed up when observers compare distances and times from different frames of reference related by a boost (when one frame is moving at a constant speed with respect to another) and it is the space-time interval (dS)2 that is invariant. Take a look at equations of the Lorentz transformation in the boxes below. You can see that in frame S both position and time are a combination of the position and time coordinates in S/. We will derive these below.

푥 = 훾(푥/ + 푣푡/)

푣푥/ 푡 = 훾(푡/ + ) 푐2

In special relativity time passes at different rates in frames that are moving with respect to each other, and distances between the same two points are observed to be different in different frames. The space time interval in frame S is given by (dS)2 = (ct)2 – x2 – y2 – z2. Take a look at figures 3 and 4 below.

Z S Expanding spherical P(x,y,z) wave-front Z

ct y

x Origin x

Figure 3 A flash of light occurs at the origin of the S frame. It spreads out at the speed c as a spherical wave-front. Using the formula that distance is speed multiplied by time, we can see that the distance the light has travelled is, distance = ct. The distance travelled by the light is given by Pythagoras theorem and this is (ct)2 = x2 + y2 + z2. z/ z v / Expanding S S spherical y y/ wave-front. ct ct/

x/ x Origin Origin of S of S/ Expanding spherical wave-front in S/

Figure 4 Frame S has coordinate axes x, y, z and 푆/ the axes x/, y/, z/. The frames are in in standard configuration with axes parallel. Frame 푆/ moves to the right along the coincident x and 푥/ axes at speed v. A flash of light occurs at the origin of both frames when they coincide. Both S and 푆/ observe the spherical wave-front to move away from their origins at the speed c. In frame S the distance to the spherical wave-front from O is given by Pythagoras theorem as (ct)2 = x2 + y2 + z2. In frame 푆/ by (ct/)2 = (x/)2 + (y/)2 + (z/)2. Both observers measure light to travel at speed c. The space- time interval travelled by the light front in S is the same as the space-time interval travelled by the light as observed in S/. The space-time interval in frame (dS)2 = (dS/)2 it is the same in both frames. (dS)2 = (ct)2 – x2 – y2 – z2 and this is equal to the space-time interval observed in S/ as (dS/)2 = (ct/)2– (x15/)2 – (y/)2 – (z/)2. We get the key equation (ct)2 – x2 – y2 – z2 = (ct/)2– (x/)2 – (y/)2 – (z/)2

So we arrive the following equation; (ct)2 – x2 – y2 – z2 = (ct/)2– (x/)2 – (y/)2 – (z/)2.

Inertial reference frames and the Principle of relativity

This is one of the most important ideas to grasp about special relativity. An inertial reference frame is what you are in if you are standing still or are moving at a constant speed in a straight line. All inertial frames are equivalent. There is no experiment that can be carried out inside of an inertial frame will allow the determination of the absolute velocity of that frame. You can only tell how fast you are going relative to something else.

You might say that’s rubbish, for a start off I know that I’m not moving because I am standing still. Now that sounds fine, but consider this, you are standing on the surface of a planet that is rotating on its axis at about 1000 miles an hour. The planet revolves in an orbit around the Sun at a speed of about 60 000 miles an hour. The Sun moves around the galaxy at several hundred thousand miles an hour. The galaxy is orbiting the centre of mass of a cluster of galaxies called the Local Group and this cluster is accelerating away from all other galaxy groups. You think you are at rest, but in reality you are moving in an extremely complex way at a phenomenal speed while rotating and revolving through a whole series of complex orbital motions. We can only measure motion relative to something else. It is the constant value of the speed of light that is an invariant. The speed of light is always measured to be the same in any reference frame and that is 299, 792, 458 mph

We will also need the following concepts: Newtonian Momentum

Momentum this is the product of the mass of a body (m) multiplied by its velocity (v). P = mv Kinetic Energy

This can be thought of as the energy of motion. It is half of the mass multiplied by the velocity squared.

E = ½ mv2

The Galilean transformation.

According to Newton and Galileo we have the following situation. Let us imagine we have two inertial frames in standard configuration. At some time after t = 0 we have the situation as shown in figure 5. Remember that as this is Newtonian Physics the time is the same in both frames.

/ y y

S S/

/ x P (x, y) vt U

P/(x/, y/) U/

x x/ O O/ x/

Figure 5. Two inertial reference frames in standard configuration. The observer in frame S regards his or herself at rest. Frame S/ moves to the

right parallel to the x axis at a speed v. A particle moves parallel to the x axis at a speed U as observed in the S frame. The particle has the

coordinates P (x, y) in the S frame and P/(x/,y/) in the S/ frame.

We now need to derive a set of equations relating the co-ordinates of the point P in the two frames. That is to say if an observer in frame S gives the point P the coordinates (x, y ), what are the coordinates given to the same point by an observer in frame S/?

From figure 5 We can form the following equations relating the two sets of coordinates.

푥/ = 푥 − 푣푡

푦/ = 푦

푡/ = 푡

The velocity transformation

Differentiating with respect to t/ (remember in Newtonian physics that t/ is the same as t so differentiating with respect to t/ is the same as differentiating with respect to t ).

푑푥/ 푑푥 푑(푣푡) / = − 푑푡 푑푡 푑푡 / 푑푥 / / 푑푥 Now is the speed of the particle in the S frame, and that is U and is 푑푡/ 푑푡 the speed of the particle in the S frame, and that is U. So rewriting and remembering that v is a constant we have

푈/ = 푈 − 푣

In other words the speed of the particle in the S/ frame is the speed of the particle as measured in the S frame minus the speed of the S/ frame itself. This is what our common sense would expect us to find.

However if the speed of light is a constant in both the S and S/ frames then this must only be an approximation and a new transformation is needed, this is the Lorentz-Fitzgerald transformation, named after its two discoverers. We will simplify matters by concentrating on only one space coordinate, x and the time coordinate, t.

The Lorentz Transformation We form two linear equations where x and t are coordinates in the S frame and x/ and t/ are coordinates in the S/ frame, with the unknown constants A, B, C and D. Our task is to find these constants 푥 = 퐴 푥/ + 퐵 푡/ ①

푡 = 퐶 푥/ + 퐷푡/ ②

Consider the origin of the S/ frame where 푥/ = 0

Substitute into equation ① the value 푥/ = 0 and rearrange to make 푡/ the subject.

푥 = 퐵 푡/ 푥 푡/= 퐵 / / Substitute for 푡 in equation ② remembering that 푥 = 0 and rearrange. 푥 푡 = 퐷 퐵 퐵푡 = 퐷푥

Differentiate this equation with respect to t. 푑(퐵푡) 푑(퐷푥) = 푑푡 푑푡 푑푥 퐵 = 퐷 푑푡

푑푥 Now is v, the velocity of frame S/ as observed by S 푑푡

So we have 퐵 = 퐷푣 ③

Now consider the origin of the S frame where x =0

Substitute for x = 0 into Equation ①

0 = 퐴푥/ + 퐵푡/

Differentiate this with respect to t/

푑(퐴푥/ ) 푑(퐵푡/) 0 = + 푑푡/ 푑푡/ 푑푥/ 푑푡/ 0 = 퐴 + 퐵 푑푡/ 푑푡/

푑푥/ = -ν, this is the speed of frame S as observed by S/. 푑푡/ 0 = 퐴 (−휈) + 퐵

0 = −퐴푣 + 퐵

Rearrange to make B the subject

퐵 = 퐴휈

So 퐵 = 퐴휈 ③

And 퐵 = 퐷휈 ④

Comparing the two equations

So 퐴 = 퐷

Rewrite ① and ② substituting for 퐴 = 퐷 푎푛푑 퐵 = 퐴휈 this gives

푥 = 퐴푥/ + 퐴휈푡/

푡 = 퐶푥/ + 퐴푡/

Let us factorise, taking out a factor of A

푥 = 퐴(푥/ + 휈푡/)

퐶 푡 = 퐴( 푥/ + 푡/) 퐴 퐶 Let us call 퐸 = 퐴 푥 = 퐴(푥/ + 휈푡/) ⑤

푡 = 퐴(퐸푥/ + 푡/) ⑥

Substitute the expressions for 푥 푎푛푑 푡 from ⑤ and ⑥ into the equation describing the spherical wave-front (see figures 3 and 4 above) rewritten below, gives

푐2푡2 − 푥2 = 푐2푡/2 − 푥/2 ⑦

푐2퐴2( 퐸푥/ + 푡/)2−퐴2(푥/ + 휈푡/)2 = 푐2푡/2 − 푥/2 ⑧

Now consider the point with the coordinates 푥/ = 0, 푡/ = 1 in the 푆/ frame and substitute these values into equation ⑧.

This gives us

푐2퐴2 − 퐴2휈2 = 푐2

Factorising by taking the factor A out of the bracket gives

퐴2(푐2 − 휈2) = 푐2

Rearranging to make 퐴2 the subject gives

푐2 퐴2 = (푐2−휈2)

Divide top and bottom of the fraction by 푐2

푐2 2 2 푐 퐴 = 푐2 휈2 ( − ) 푐2 푐2

2 1 퐴 = 휈2 (1− ) 푐2 Take the square root of both sides

1 퐴 = 휈2 √(1− ) 푐2

This is a very important finding. A is a function of v, the velocity of the observer’s frame and it is known as the gamma factor and it is given the symbol 훾.

1 훾 = 휈2 √(1− ) 푐2

Now consider the point in the S/ frame with the coordinates 푥/ = 1, 푡/ = 0 and substitute these into ⑧ written below.

푐2퐴2( 퐸푥/ + 푡/)2−퐴2(푥/ + 휈푡/)2 = 푐2푡/2 − 푥/2 ⑧

This gives us

푐2퐴2퐸2 − 퐴2 = −1

1 2 2 Rewriting A using 퐴 = 휈2 gives us (1− ) 푐2 푐2퐸2 1 − = −1 휈2 휈2 1 − 1 − 푐2 푐2

휈2 Multiplying both sides by 1 − gives us 푐2

휈2 푐2퐸2 − 1 = −1(1 − ) 푐2 휈2 푐2퐸2 − 1 = − 1 푐2 Cancelling the -1 on both sides gives us

휈2 푐2퐸2 = 푐2 22

And dividing both sides by c2 gives

휈2 퐸2 = 푐4 Taking the square root of both sides we find that 휈 퐸 = 푐2

Rewriting ⑤ and ⑥ and substituting for A and E where

1 푣 퐴 = 훾 = and E = 휈2 2 √(1− ) 푐 푐2

푥 = 퐴(푥/ + 휈푡/) ⑤

푡 = 퐴(퐸푥/ + 푡/) ⑥

1 푥 = (푥/ + 휈푡/) 휈2 √1 − 푐2 1 휈푥/ 푡 = ( + 푡/) 휈2 푐2 √(1 − ) 푐2

These can be written as below and finally we have the Lorentz transformation equations.

푥 = 훾(푥/ + 휈푡/) ⑨

/ 휈푥/ 푡 = 훾(푡 + 2 ) ⑩ 푐

The reverse transformation is found by substituting –v for v as below

/ 푥 = 훾(푥 − 휈푡) ⑪

휈푥 푡/ = 훾(푡 − ) ⑫ 푐2

The velocity transformation

We have the following situation. We have two inertial frames in standard configuration. At some time after t=0 we have the situation as shown in figure 6 below. Remember that as this is relatavistic physics and so the position and time coordinates are different in each frame.

/ y y

اS S

/ x P (x, y) vt U / / / P (x , y ) U/ x

x x/ O O/ x/

Figure 6. Two inertial reference frames in standard configuration. The observer in frame S regards his or herself at rest. Frame S/ moves to the right

parallel to the x axis at a speed v. A particle moves parallel to the x axis at a speed U as observed in the S frame and has the coordinates P (x, y).In the S/

frame the particle has coordinates P/(x/, y/) and moves with a speed U/.

We now need to derive a set of equations relating the velocity of the particle in the two frames. That is to say if an observer in frame S observes the particle to have the velocity U, how does this compare to the velocity U/ as measured by the observer in S/? To do this we must differentiate the Lorentz transformation equations for position and time.

Let us begin with the Lorentz transformation equations that relate the position and time coordinates in S/ reference frame. 푥/ = 훾(푥 − 휈푡)

휈푥 푡/ = 훾(푡 − ) 푐2

The velocity of the particle in the S/ frame is given by the change in position (x/) divided by the change in time (t/) so to find the velocity we must differentiate the first equation with respect to t/.

푑푥/ 푑푥 푑푡 / = 훾( / − 푣 /) 푑푡 푑푡 푑푡

/ 푑푥 / / Now / is the velocity of the particle as observed by S and that is U 푑푡 푑푥 푑푡 So we have 푈/ = 훾( − 푣 ) 푑푡/ 푑푡/ Now let’s use the chain rule and re-write the first term inside the bracket.

푑푥 푑푥 푑푡 = . 푑푡/ 푑푡 푑푡/

푑푥 푑푡 푑푡 푈/ = 훾( . − 푣 ) 푑푡 푑푡/ 푑푡/

푑푡 Now let’s factorise by taking out the factor 푑푡/

푑푥 푑푡 푈/ = 훾( − 푣) 푑푡 푑푡/

25 The first term in the bracket is 푈, so re-writing we get

푑푡 푈/ = 훾(푈 − 푣) ⑬ 푑푡/

푑푡 1 And recognising the reciprocal = 푑푡/ 푑푡/ 푑푡 This is the reciprocal of the derivative of the second Lorentz transformation equation. / 휈푥 푡 = 훾(푡 − ) 푐2

Differentiating the second of the Lorentz transformation equations we get

푑푡/ 푑푡 푣 푑푥 = 훾( − ) 푑푡 푑푡 푐2 푑푡

푑푡 푑푥 Now we know that = 1 and that = 푈 so we can rewrite the equation above as 푑푡 푑푡

푑푡/ 푈푣 = 훾(1 − ) 푑푡 푐2 Substituting this into equation ⑬ below

푑푡 푈/ = 훾(푈 − 푣) 푑푡/

훾(푈 − 푣) 푈/ = 푈푣 훾( 1 − ) 푐2

Cancelling the γ’s and removing the brackets we have.

푈 − 푣 푈/ = 푈푣 1 − 푐2

This is the velocity transformation equation. And replacing 푣 for −푣 gives the reverse transformation

푈/ + 푣 푈 = 푈/푣 1 + 푐2

Momentum

The fact that velocities between reference frames do not add has some far reaching consequences one of which is that we have to redefine the law of momentum to allow for this. Consider the situation in figure 7, below. This is the frame S, the frame you are sitting in. Before After S At rest 푢 −푢

FOOTCANDLE PRODUCTIO NS

푚 푚 2푚

Figure 7.Two particles each of mass 푚 approach each other and are involved in an inelastic collision. After the collision they stick together and remain at rest.

Looking at this situation from the frame S . We will consider the conservation of 2 momentum. We have A derivation of =

푚푢 + 푚(−푢) = 2푚. 0 A practical guide

The total momentum before the collision is equal to zero, as is the momentum after the collision and so momentum is conserved.John Teasdale 8/22/2016

Now let us consider the collision from a different frame of reference. We will look at it from the reference frame of the mass moving to the left with the / velocity – 푢 . This is the frame S

The need for special relativity is outlined and concepts of Newtonian physics are outlined.

At rest 푢/ 푢

푚(푢) 푚0 M(u)

Figure 8. This is the frameS/ of the left moving mass. According to Newton we would have the velocity of the right moving mass equal to 2u, the sum of the velocities of the two masses. With the Lorentz transformation we must use u/.

There is a problem now though because the Lorentz velocity transformation tells us that 푢/ ≠ 2푢

This is telling us that in order for momentum to be conserved, the mass must be a function of the velocity.

This is telling us that because velocities do not add, instead due to the fact that speed of light is a constant, that is invariant in all frames, the velocity is given by the Lorentz velocity translation rule which we found in order to account for this fact. This means that we have arrived at the conclusion that in order to conserve momentum mass must be a function of velocity. So we have followed a line of reasoning that went.

푐 → 푣 → 푚

In words; because of the constancy of the speed of light velocities cannot add we must use the Lorentz velocity transformation so that in

order to conserve momentum we must multiply the mass by a factor which is a function of the velocity.

With reference to figures 7 and 8, and from the conservation of mass we have

/ 푚(푢 ) + 푚0 = 푀(푢) ①

From the conservation of momentum

/ / 푚(푢 )푢 + 푚0 . 0 = 푀(푢)푢 ②

Substituting for 푀(푢) from ① into ②

/ / / 푚(푢 )푢 = (푚(푢 ) + 푚0) u

Multiplying out the bracket gives

/ / / 푚(푢 )푢 = 푚(푢 )푢 + 푚0 u

Which we can rearrange to give

/ / / 푚(푢 )푢 − 푚(푢 )푢 = 푚0 u

And taking out a factor of 푚(푢/)

푚(푢/)(푢/ − 푢) = 푚 u 0

And a final rearrangement gives

푚0푢 푚(푢/) = (푢/ − 푢)

And dividing the right hand side by 푢 we get

푚0 푚(푢/) = ③ 푢/ ( − 1) 푢

Now let us examine the expression in the denominator

We get the value of 푢/ from the velocity transformation equation

/ 푢 − 푣 푢 = 푢푣 1 − 푐2

So referring to figure 7 and 8 where the velocity of the S/ frame is -u

푢 − (−푢) 푢/ = 푢(−푢) 1 − 푐2

2푢 푢/ = 푢2 1 + 푐2

And multiplying top and bottom of the fraction by 푐2

2푢푐2 푢/ = 푐2 + 푢2

푢/ Now going back to our equation for ( − 1) and substituting for 푢/ 푢

2푢푐2 푐2 + 푢2 − 1 푢

Simplifying gives

2푐2 − 1 푐2 + 푢2

Which is equivalent to

2푐2 푐2 + 푢2 − 푐2 + 푢2 푐2 + 푢2

Which we can rewrite over one denominator as

2푐2 − 푐2 − 푢2 2 2 푐 + 푢

Therefore we have

푢/ 푐2−푢2 ( − 1)= 2 2 푢 푐 +푢

Now let us consider the expression inside the square root of the denominator of the factor γ from the Lorentz transformation equations. 1 훾 = 푣2 √1 − 푐2

This time let us consider replacing 푣, which is the

speed of one reference푣2 frame with respect to another, / 1 − with 푢 , which푐 is2 the speed of the frame of the reference frame of the left moving mass.

2푢푐2 Where 푢/ = (푐2+ 푢2)

Let us substitute(2푢푐2) for2 푣 (푐2 + 푢2) 2 1 − 푐2

4푢2푐4 1 1 − × (푐2+ 푢2)2 푐2

Which by cancelling the 푐2 and rearranging by re expressing 1 we get

( 푐2 + 푢2)2 4푢2푐2 − ( 푐2 + 푢2)2 (푐2 + 푢2)2

2 2 2 2 2 4 (푐 + 푢 )( 푐 + 푢 ) 4푢 푐 − ( 푐2 + 푢2)2 (푐2 + 푢2)2

푐4 + 2푐2푢2 + 푢4 − 4푢2푐2

( 푐2 + 푢2)2

This gives

푐4 − 2푐2푢2 + 푢4

( 푐2 + 푢2)2

Which can be factorised to give

(푐2 − 푢2)2

( 푐2 + 푢2)2

When we compare this to equation ③ where

푢/ 푐2−푢2 ( − 1)= 푢 푐2+푢2

We see that

/ 푢 (푐2−푢2)2 ( − 1)=√ 푢 ( 푐2+푢2)2

And therefore from equation ③

/ 2 푢 푢/ ( − 1)= √1 − 푢 푐2

And so we can rewrite equation ③ below

푚0 푚(푢/) = 푢/ ( − 1) 푢 34

To give

푚0 푚(푢/) =

푢/2 √1 − 푐2 Which we can rewrite as

/ 푚(푢 ) = 훾푚0

Changing notation we can say that for any reference frame

푚(푢) = 훾푚0

1 Now 훾 = 푣2 √1− 푐2

Which we can write in index form as

2 1 푣 − 훾 = (1 − ) 2 푐2

Rewriting equation in this way we have

2 1 푣 − 푚(푢) = 푚 (1 − ) 2 0 푐2 Expanding using the binomial theorem we have using

푛(푛 − 1)푥2 푛(푛 − 1)(푛 − 2)푥3 (1 + 푥)푛 = 1 + 푛푥 + + + ⋯ 2! 3!

2 1 3 푣2 2 (− ) (− ) (− ) 1 푣 2 2 푐2 푚(푢) = 푚 (1 + (− ) (− ) + + ⋯ ) 0 2 푐2 2!

푣2 3푣4 푚(푢) = 푚 (1 + + … ) 0 2푐2 8푐4

2 Multiplying out the bracket we get both sides by 푐 to clear fractions we get

1 푣2 3 푣4 푚(푢) = 푚0 + 푚0 + 푚0 + ⋯ 2 푐2 8 푐4

Multiplying both sides by 푐2 to clear fractions we get

1 3 푣4 푚(푢)푐2 = 푚 푐2 + 푚 푣2 + 푚 + ⋯ 0 2 0 8 0 푐2

We can ignore all terms after the first two if 푣 is small compared to 푐, which it is at these everyday speeds.

This gives us

1 푚(푢)푐2 = 푚 푐2 + 푚 푣2 0 2 0

Dimensional analysis shows that the units of mc2 are kgm2s-2 and these are

the units of energy. This tells us that the total energy content of a body is equal to the sum of its rest energy and its kinetic energy.

In all reference frames E = mc2

Bibliography and further reading

Katz, R. An introduction to special relativity. Van Nostrand Company

Olenick, R.P. et al. Beyond the mechanical universe. Cambridge