Euclidean Motion

Aug. 15, 2017

In the last lecture, we got oriented with orientations of space while you simultaneously got oriented with your first classes at a university. I imagine it was pretty disorienting! As a reminder, tonight there will be office hours and hopefully if you are feeling confused about the first lecture, you will come and ask questions. Tomorrow, your first problem set is due, and it will cover material from today and yesterday so hopefully you have started.

1 One dimension

Today we will discuss motion in Euclidean space. We start with one dimensional motion. After all the hubub about vectors and multiple dimensions yesterday, you may feel too sophisticated to go back to the simple world where only scalars exist. However, that would be a mistake, because there is another dimension lurking around that we have not discussed–time! When we speak of one dimensional motion, we mean one spatial dimension. This can mean considering just one axis of a more complicated multiple dimensional motion–a concept we will find useful in later lectures, or motion along a one-dimensional curve. In fact these are not distinct. At the end of your first quarter at UCSB, you will encounter the theory of special relativity. This is the modification of physics when we include time as a dimension along which we may move1 There it will be very useful to think of motion as a one dimensional curve in the time-space (t − x) plane, called a worldline. For that reason, we will learn one dimensional motion from that perspective, and maybe throw in some special relativity. Consider a curve in the t − x plane 1. This curve defines a trajectory in the one-dimensional world. Such a curve describes not only where you ended up, but also the path you took and the speed at which you were travelling at all times. For instance, figure 1 could describe the events that transpired as I left the house this morning (assuming my house is one dimensional). The origin is when I looked at my watch in my kitchen. I start by walking towards my door, when I realize I left my phone on the charger and turn around. After picking up my phone, I walk back to the door only to realize that I left my wallet on the bed and turn around again. When I return to the door, I realize I don’t have my keys and have to turn around again. I am now sprinting to get my keys and get out the door so I am not late to class. As you follow my world line, what do you notice? At first, the line is very steep–I am not moving very quickly, so a large amount of time passes as I meander to the door. As I pause at the door, my worldline is vertical–time passes but I do not move in space. Next, my worldline becomes less steep. My speed has picked up as I realize I may be late. By the end of the trip, my worldline is the least steep because I am moving over more space in a given time. How do we phrase this in the language of math? To start, my worldline is a smooth curve that has only one value for each moment of time. Such a curve mathematically is described by a function. We start by writing the worldline as the function x(t). Because of conventions, t is always the vertical axis, but to align with your experiences with functions, you have to turn your head and put it on the horizontal axis. Points on this curve are called events. A simple example of such a worldline is the height of my diaphragm as I breathe 2. If I define the origin as the point where I have fully exhaled, and call the maximum height of my diaphragm h, then the

1Unfortunately, only in one direction.

1 worldline of my diaphragm is h 2πt x (t) = 1 − cos . L 2 T After every period of time T , my diaphragm returns to the same location, and so this defines the period of my breath. A complicated example of a worldline is the height of my diaphragm when I snore. I won’t write an equation for this. The relationship between slope and speed in figure 2 should be awakening the part of your brain where you hid all of your calculus knowledge. To begin, the slope of your worldline has the right units to be a speed– ∆x meters speed ∼ ∝ . (1) ∆t seconds However, there are two ways in which we can define speed. The first is an average speed. Pick two points on the worldline, (t1, x1) and (t2, x2) and calculate x − x v¯ = 2 1 . (2) t2 − t1 This is occasionally a useful quantity in physics, but it doesn’t really capture how fast an observer is going at an instant. In fact, because an observer can go backwards and forwards (in the positive and negative xˆ direction), the average speed between two events can be zero even if the instantaneous speed never vanishes. Clearly this doesn’t capture the nuances of motion that one may hope to describe. In fact, the average speed of my diaphragm when I do and do not snore is very similar. To capture instantaneous speed, one needs to consider points that are very close together

x2 − x1 dx v = lim ≡ . (3) (t2−t1)→0 t − t dt 2 1 t=t1 It’s calculus time. This v is typically called the velocity, but in one dimension things are simple and it can also be called the speed. In higher dimensions, these will be distinct as velocity will be a vector but speed is a scalar. One can invert the derivative (a.k.a integrate) to find the displacement, or difference in positions of two different events

Z t2 Z t2 dx δx = vdt = dt = x(t2) − x(t1). (4) t1 t1 dt The velocity defines how quickly an observer’s position changes with time. For my diaphragm, one finds that the velocity is dx (t) πh 2πt v (t) = L = sin . L dt T T Clearly the velocity is not constant and so one can also ask how quickly the velocity changes with time. This defines the acceleration and is found by applying an additional derivative, dv d2x a(t) = = . (5) dt dt2 According to a previous instructor’s lecture notes (available online), the third, fourth, and fifth derivatives of position are called the “snap, crackle, and pop,” respectively. Typically, these are not discussed, as they are not directly related to forces and energy, which are the main governing concepts of Newtonian mechanics. Assuming the Rice Krispies vanish, then the highest non-zero derivative of the position is the acceleration. Because its derivative vanishes, it is a constant. Thus we can integrate it once to find, Z t v(t) − v0 = adt = at. (6) t=0

2 Here we defined v(0) ≡ v0. We can integrate this again and find Z t 0 0 1 2 x(t) − x0 = v(t )dt = at + v0t. (7) t=0 2 Now, for problem solving, sometimes we are giving quantities appearing in separate equations. Let’s say we wanted to find the velocity at which someone would have to jump to beat the world record high jump. This was set in 1993 by Javier Sotomayor from Cuba and was 2.45 meters. As soon as Sotomayor left the ground, there was only one force acting on him, gravity, which gives a constance acceleration of 9.8 m/s2 toward the ground. To find the velocity, we have a distance and an acceleration but no time. In the two equations we have derived, equations 6 and 7, time appears in both. Hence we need to eliminate it. This is easiest in the first equation. Solving for time and inserting into the second equation gives

1 v(t) − v 2 v(t) − v x(t) − x = a 0 + v 0 (8) 0 2 a 0 a 2 2 ⇒ 2a (x(t) − x0) = v(t) − v0 (9)

At the top of the jump, Sotomayor was not moving in the vertical direction, so that v(ttop) = 0. Then we ﬁnd that Sotomayor left the ground at roughly 6.93 m/s or 15.5 mph.

2 More dimensions

The equations just derived are suﬃcient to describe trajectories in one dimension. However, we live in more dimensions, and so how do we describe motion in more than one direction at once. We use vectors! As stated earlier, one way to think about one dimensional motion is to think of motion restricted to one coordinate axis. We can write the one dimensional worldline in terms of a three dimensional worldline as

x(t) = ~x(t) · xˆ (10)

where we deﬁned

~x(t) = (x(t), y(t), z(t)). (11)

For motion only along the x-axis, y(t) and z(t) are both zero. However, we know that we can change reference frames by translating and rotating. This means that what we thought of as motion along one direction can be thought of as motion along any of the other directions. For instance, by rotating my coordinate axes about the z-axis by an angle φ, the worldline of my lungs may be written

~xL(t) = xL(t)(cos φ, sin φ, 0). (12)

Thinking about higher dimensions in terms of rotating coordinate frames illuminates the higher dimensional versions of velocity and acceleration. In the velocity, the time derivative derivative acts on each component of the position vector separately, d~x dx dy dz ~v(t) = = ( , , ) (13) dt dt dt dt and similarly for the acceleration, d~v dv dv dv ~a(t) = = ( x , y , z ). (14) dt dt dt dt Here we deﬁned the notation

vx = ~v · x.ˆ (15)

3 As stated earlier, in higher dimensions, there is a distinction between speed and velocity. The velocity is the derivative of the position vector. The speed is the norm of this vector and is a scalar. Given a constant acceleration vector a, we can simply modify 6 and 7 to higher dimensions,

~v(t) − ~v0 = ~at (16) 1 ~x(t) − ~x = ~at2 + ~v t. (17) 0 2 0 An important example of these equations is the equation of projectile motion, in which our axes are aligned such that the zˆ direction is anti-aligned with the gravitational acceleration. There is no acceleration in the xˆ or yˆ directions. Then we have   x0 + v0,xt ~x(t) =  y0 + v0,yt  (18) 1 2 z0 + v0,zt + 2 gt where g = −9.8m/s2.

Aside: A little relativity The equations defined above are true for an observer in one reference frame. One of the most bizarre and awe-inspiring discoveries we have made in the last century is that when we change reference frames, the above equations must be modified when we move close to the speed of light. This is because the speed of light is a cosmic speed limit. A simple though experiment (called a gedankenexperiment shows that such a speed limit modifies physics: Consider a clock made by photons bouncing off two mirrors at rest. The time it takes for the photon to go from one mirror to another is one second. Now consider moving the two mirrors at the same speed to the left. This is a choice of reference frame. If an observer’s reference frame moves along with the mirrors, they say that the mirrors are “at rest.” When the reference frame is such that the mirrors are moving (or is the reference frame moving), the physics should not change, because this is a symmetry of flat space. Consider the perspective that the mirrors are moving and so the observer is at rest. Here, the path of light rays obeys the equations we derived earlier. In particular, since the velocity is constant, there is no acceleration and the average speed and instantaneous speed are the same. In a fixed amount of time, one can calculate the distance light must travel. Because we only changed the reference frame in one direction, let’s say xˆ, the worldline of the lightray is

~x(t) = (vmirt, vvert, 0) (19)

and

~v(t) = (vmir, vver, 0). (20)

2 2 2 Because light travels at the same speed in all reference frames, c = vmir + vver. Now, in the frame at which the mirrors are at rest, the worldline of the light ray is

~x0(t) = (0, ct0, 0). (21)

The two frames have to agree in the yˆ direction and hence q 0 2 2 t = t 1 − vmir/c . (22)

Clocks dont agree! The moving observer has time moving slower (one second in t frame is less in t0 frame). From now on, deﬁne 1 γ = (23) p 2 2 1 − vmir/c

4 so that

t0 = t/γ. (24)

What if we instead orient the mirror to be aligned with the motion? A fixed observer watching the train move sees the photons bounce between the mirrors. When the train is at rest, the distance between the mirrors is l. In the time it takes for photons to bounce between the mirrors, t1, the second mirror has moved a distance ∆x = vmirT . Then, in the fixed observer reference frame, the distance between the mirrors defining the clock is

x1 = l + vmirt1. (25)

Since light travels at the speed of light,

ct1 = l + vmirt1 (26)

so that l t1 = (27) c − vmir

On the return trip, the distance is shorter by the same amount, and it takes a time t2 to make the trip

x2 = l − vmirt2 (28)

and l t2 = . (29) c + vmir The total trip is

2 vmir ∆x = 2l + vmir(t1 − t2) = 2l(1 + 2 2 ) (30) c − vmir and we know that (or we can solve this equation)

2lc2 ∆x = c(t1 + t2) = 2 2 X. (31) c − vmir The total time for the trip is 2l T = t + t = γ2 = γT 0 (32) 1 2 c where in the last equality, we used our result for time dilation. Now, T 0 is the period on the train which is

l0 T 0 = 2 (33) c and so we ﬁnd that l0 l = . (34) γ While times dilate, lengths contract! When we include time as a coordinate in four-dimensional space, changing reference frames between the train and observer at rest is analogous to a rotation in the plane spanned by the time and one spatial coordinate (the coordinate along which the train moves). This “rotation” is called a boost

5 because we speed up on set of axes. Under a boost along the xˆ direction by a velocity v, the new coordinate frame is

t0 = t/γ, x0 = γx, y0 = y, z0 = z. (35)

Notice that

s2 = −c2t2 + x2 = −c2t02 + x02. (36)

This is called the invariant length of the event because it does not change under a boost. In fact, we can define a new set of coordinates, s t = sinh(η), x = s cosh(η). (37) c Along this curve (fixed s, but varying η) in the x − t plane, the invariant length doesn’t change. This should remind you of a circle on a plane. Along any point of the circle, the distance from the origin does not change. Just like we can label a point on the circle by an angle θ, we can label a point on the hyperboloid defined in eq. 37 by its rapidity, η. The relationship between the circle and the functions in equation 37 is deep. It has to do with complex numbers and it leads to many surprising relations in physics.

6 Figure 1: My worldline this morning.

7 4

t 2

0 0.0 0.2 0.4 0.6 0.8 1.0 h

Figure 2: The worldline of my diaphragm as I breathe.

8 9 10 1.0

0.5

t 0.0

-0.5

-1.0

1.0 1.1 1.2 1.3 1.4 1.5 x

Figure 3: All points on this curve have s = 1. The red dots are intervals of η = .2.