Classical Physics “Inertial Reference Frame” (Section 5.2): a reference frame in which an object obeys Newton’s Laws, i.e. F = ma and if F = 0 (object does not interact with other objects), its velocity = constant (a = 0). Any reference frame moving with constant velocity with respect to an inertial reference frame is also an inertial reference frame. (e.g. “a frame moving at constant velocity with respect to distant stars”) Consider the coordinates (x,y,z,t) and (x’,y’,z’,t’) of some object in two inertial reference, S and S’, where S’ has constant velocity v in the x-direction with respect to S. Suppose their origins coincide at t = t’= 0. The object may have a force acting on it. Then Newton/Galileo would state: x’ = x – vt, y’ = y, z’ = z, and t’ = t. Since t = t’: dx’/dt’ = dx/dt – v, dy’/dt’ = dy/dt, dz’/dt’ = dz/dt ax’ = d2x’/dt’2 = d2x/dt2 = ax ay’ = d2y’/dt’2 = d2y/dt2 = ay az’ = d2y’/dt’2 = d2z/dt2 = az Therefore F’ = ma’ = ma = F So both observers will not only agree on Newton’s Laws, but will agree on the value of F.

Galileo’s Relativity: The laws of mechanics (i.e. all of physics in Galileo’s time) are the same in all inertial reference frames. Einstein’s (Special) Relativity: 1) All the laws of physics are the same in all inertial reference frames. One of those laws is that the speed of electromagnetic waves (i.e. light) in vacuum = c = 3.00 x 108 m/s in all inertial frames, regardless of the speed of the source of the light or the speed of the observer. (c invariance postulate) #1 means that there is no absolute reference frame, so no experiment you can do that tells you that you are moving (with constant velocity) – all motion is relative. “Special”: As stated by Einstein, the 1905 theory only applied to inertial reference frames. In 1915, Einstein extended his theory to get rid of this limitation and considered any frame, even accelerating frames.: “General Relativity”. For now, we’ll ignore whether frames are inertial and restate #1 as “All laws of physics are the same in reference frames which are moving at constant velocities with respect to each other”; i.e. you cannot tell if you are moving at constant velocity.

Consider a light in a train traveling at v with respect to observer B on the ground. If observer A on the train measures the speed of the light to be = c. Galileo: Observer B will measure the speed of the light to be c+v. Einstein: Observer B will measure the speed of the light to be c.

Assume origins coincide at t = t’=0: x(0) = x’(0) = y(0) = y’(0) = z(0) = z’(0) = 0 Einstein/Lorentz Galileo/Newton x’ =  (x – vt) y’ = y z’ = z t’ =  (t – vx/c2) where   1/(1 - v2/c2)1/2 x’ = x – vt y’ = y z’ = z t’ = t

Lorentz Transformation Galilean Transformation x’ =  (x – vt) y’ = y z’ = z t’ =  (t – vx/c2) where   1/(1 - v2/c2)1/2 x’ = x – vt y’ = y z’ = z t’ = t Lorentz (1892) showed that all of Maxwell’s Eqtns. (not only the wave equation) , which are inconsistent with the Galilean Transformation, are consistent with the “Lorentz Transformation”, but there was no convenient explanation of this. Einstein provided the explanation in 1905 in his Theory of Special Relativity simply by his two assumptions: a) physics is same in all “inertial” frames and b) speed of light is invariant. Exercise (homework): invert the above Lorentz Eqtns. to show: x =  (x’ + vt’), y = y’, z = z’, t =  (t’ + vx’/c2) (as expected) Instead of going through Einstein’s proof of the Lorentz Transformation, we will consider a few special cases: simultaneity, time dilation, length contraction.

If v << c,   1. Then if also have dx/dt << c, so x << ct, then Einstein/Lorentz  Galileo/Newton. That is, relativistic effects only become noticeable if either the reference frame (S’) or the object (x) is moving at very high speeds (i.e. approaching c). If |v| << c and |dx/dt| << c, relativistic effects become very negligible.

x’ = (x - vt) y’ = y z’ = z t’ =  (t - vx/c2) where   1/(1 - v2/c2)1/2 Lorentz transformation The easiest to understand is that distances perpendicular to the direction of motion are the same for all observers: y’ = y and z’ = z: Suppose moving perpendicular distances were smaller (or larger) than stationary perpendicular distances. Observer S’ can drag a meter stick, with pencils on each end, drawing lines as she travels with respect to S. At any time afterwards, with no need to synchronize clocks, S can take his meter stick and measure the distance between the lines. They can compare their measurements: if yS  yS’, they can use this to determine “who was moving” and “who was stationary”, in contradiction to the relativity postulate. Clearly, the same argument cannot be used for distances parallel to the motion!

Clearly, the same argument cannot be used for distances parallel to the motion! v If I want to compare the length of a moving meter stick with that of a stationary one, I have to be sure I’m measuring the front and back of the moving meter stick at the same instant, i.e. simultaneously, but this is tricky!

Notice that if two events occur simultaneously in the S frame (i. e Notice that if two events occur simultaneously in the S frame (i.e. t1 = t2), they are not necessarily simultaneous in S’ frame if they occur at different places: t’1 = (t1-vx1/c2)  (t2-vx2/c2) = t’2 if x1  x2. Einstein explained this with the following “Gedanken” (i.e. thought experiment):

L Consider train traveling at speed v. At time t = t’, the center of the train car (O’) coincides with point O on the ground. A little later, at times tA = tB = L/2c, where L is the length of the car, an observer O sees the light from lightning bolts that hit the ground at A and B, so observer O concludes that the lightning must have hit A and B simultaneously at tA = tB = 0.

L Suppose observer O’ has her head out the window so she can also see the lightning flashes. Observer O points out that since O’ is moving toward B, observer O’ will see the lightning flash from B before she sees the one from A. Since observer O’ also knows that light from both places travels at speed c, she concludes that the light took times t’A = t’B = L’/2c to reach her [where L’ is the distance between A and B that she measures which we’ll see is different from L]. Therefore, she concludes that the lightning must have hit B before it hit A: t’B < 0 < t’A. t’ =  (t – vx/c2), where   1/(1 - v2/c2)1/2 So t’B = (tB – vL/2c2) = -vL/2c2 and t’A = (tA + vL/2c2) = +vL/2c2

Note that the same arguments don’t hold for sound (i.e. the thunder): Observer O will hear the thunder at times tA(thunder) = tB(thunder) = L/2vs, where vs is the speed of sound in air, but will point out that observer O’, whose head is sticking out the window, will hear the thunder from B before she hears it from A, because she is moving toward B and away from A. Observer O’ will say “of course, I will hear B’s thunder before A’s because the sound from B is traveling toward me at speed (vs+v) while the sound from A is traveling toward me at speed (vs-v). ” That is: observer O’ does not need to conclude that the lightning hit B before A on the basis of when the thunder reached her. The difference between sound and light is that light travels at the same speed (c) for all observers while sound is carried by the air, so observers who are moving at different speeds with respect to the air will measure different speeds of sound.

Here’s another Gedanken experiment that explains why “simultaneity is relative”. Observer O wants to exactly synchronize clocks on two different walls in his house, without moving the clocks. This is tricky, because they are in different places. How can he do this synchronization? Here is one technique. He initially turns the clocks off but sets them at exactly the same time. Then he goes half way between them and simultaneously fires lasers at the two clocks. When the light rays hit the clocks, they start. If he is sure that he is exactly in the center and that he fires the lasers simultaneously, he (correctly) believes that the clocks will be in sync.

Richard Feynman calls this “failure of simultaneity at a distance.” It is why events that are at different places may be simultaneous in one frame but not in another.

Lorentz Transformation x’ =  (x – vt) y’ = y z’ = z t’ =  (t – vx/c2) where   1/(1 - v2/c2)1/2 Problem: A rocket is traveling in the positive x-direction away from earth at speed 0.3c; it leaves earth (x = x’= 0) at t = t’= 0. An observer on earth measures the rocket to be at x1 at time t1= x1/(0.3c). What are the corresponding positions and times in the rocket’s frame?

Lorentz Transformation x’ =  (x – vt) y’ = y z’ = z t’ =  (t – vx/c2) where   1/(1 - v2/c2)1/2 Problem: A rocket is traveling in the positive x-direction away from earth at speed 0.3c; it leaves earth (x = x’= 0) at t = t’= 0. An observer on earth measures the rocket to be at x1 at time t1=x1/(0.3c). What are the corresponding positions and times in the rocket’s frame?  = 1/(1- 0.32)1/2 = 1/(1-0.09)1/2 = 1/0.911/2 = 1.048 x1’ = 1.048 [x1- (0.3c)(x1/0.3c)] = 0 (obviously!) t1’ = 1.048 [t1 – (0.3c)x1/c2] = 1.048 t1 [1- 0.32] = 0.95t1 t1’ < t1 : Less time passed in the “moving frame” (i.e. the frame which moved with the event so that “x’ didn’t change in”) than in the “stationary frame” (i.e. the frame which observed x to change).