PHYS 113: Relativity IIA Simple Experiment with LightAs I’ve mentioned before, you are in the primed frame (the one inside the train). This is avery special train, one set up with mirrors and lasers, and sophisticated chronometers andthe like.Light (in the form of lasers) is particularly important because of an experiment done in1887 by Michelson and Morely, which showed that light travels at a constant speed for allobservers. Using that fact, and the fact that at low speeds our picture of the universe shouldbe the same as that suggested by Galileo, will be our basis of special relativity.So, consider the following apparatus within your tra in.h123In this “experiment” there are 3 events:1) A pulse is fired (at the speed of light) fro m the bottom of the train.2) The pulse is reflected from the top of the train.3) The pulse is detected at the bottom of the train again.View from the primed frame:Despite the small offset in the picture, events “1 ” and “3” take place at the same position(according to you), and thus:∆x′= 0However, since the height of the train is h, the light must have traveled a distance of 2hbetween the beginning and end of the experiment, and thus:∆t′=2hcAll we’ve used here is the fact that light travels at the speed of light.View from the unprimed frame:Now, from the unprimed frame, the situation looks somewhat different. The detector hasmoved from when the laser was fired, and thus, from my perspective, the experiment lookslike:Relativity II: Lorentz Transforms– 1PHYS 113: Relativity II1hv32The solid tr ain represent s the position o f the train at the beginning of the experiment, andthe dotted line represents the position of the train at the end of the experiment. The detectormust have moved a distance:∆x = v∆tfrom event “1” to “3”. This is the same result we would have gotten from Galilean relativity.Well, except f or one thing. We are no longer assuming that ∆t = ∆t′.In this case, we note that the total trip length is:dtot= 2sv∆t22+ h2In case it’s not obvious where I got that, each light beam can be written as an x-component(v∆t/2, where ∆t is the time o bserved by me), and a y-component, h. I’m just using thePythagorean theorem to add them.We also note that the light travel distance can be related to the full time interval between“1” and “3” via the relation:∆t =dtotc= 2sv∆t2c2+h2c2=vuutv∆tc2+ 2hc!2=sv∆tc2+ (∆t′)2where I’ve used ∆t′= 2h/c explicitly.Relativity II: Lorentz Transforms– 2PHYS 113: Relativity IIInteresting... I now have ∆t in terms o f ∆t′. Squaring the expression, I get:∆t2=v∆tc2+ (∆t′)2 1 −v2c2!∆t2= (∆t′)2∆t2=(∆t′)21 − v2/c2∆t =∆t′q1 − v2/c2Thus, we’ve found an interesting relation:∆t = γ∆t′+ f(v)∆x′(1)whereγ ≡1q1 − v2/c2(2)Note that I put a function f(v) in there. Why? Because we don’t know what happens if∆x′6= 0. We’ll figure it out eventually.Also, by the same a rgument, since we found:∆x = v∆tthen∆x = vγ∆t′+ g(v)∆x′(3)These equations need to be symmetric. In other words, there shouldn’t be anything which candistinguish between the “moving” (primed) frame and the “stationary” (unprimed frame).To you, I should just appear to be moving to the left. And thus, we should have the relations:∆x′= −vγ∆t + g(−v)∆x (4)and∆t′= γ∆t + f(−v)∆x (5)To figure out what f(v) and g(v) are, we first note, t hat by definition, ∆x = ∆x. Thus,plugging equations 4 and 5 into 3 we get:∆x = vγ(γ∆t + f(−v)∆x) + g(v)(−vγ∆t + g(−v)∆x)= (vγf(−v) + g(v)g(−v)) ∆x +vγ2− vγg(v)∆tSince the left has to equal the right, the terms in front of ∆x have to add to 1, and the termsin front of ∆t have to add to 0. Thus:vγ2= vγg(v)vγf(−v) = 1 − g(v)g(−v)Relativity II: Lorentz Transforms– 3PHYS 113: Relativity IIThe first of these yield:g(v) = g(−v) = γwhile plugging in gamma (after a bit o f algebra) yields:f(v) = γv/c2We ( finally!) have our Lorentz tr ansforms:∆x′= γ∆x − vγ∆t (6)∆t′= γ∆t − v/c2∆x (7)∆x = γ∆x′+ v/cγ∆t′(8)∆t = γ∆t′+ v/c2γ∆x′(9)(10)Note that if γ ≃ 1, then these are simply ∆x′= ∆x − v∆t, and ∆t′= ∆t, just as with theGalilean transforms.Moreover, as with our Galilean transforms, we can combine these to figure out how velocitytransforms:u′=u − v1 −uvc2(11)u =u′+ v1 +u′vc2(12)(13)Some ExamplesA Lightbeam:Consider what happens if you shine a flashlight (u′= c) on your train which is moving, say,at half the speed of light (v = c/2). The speed I observe the lightbeam to be traveling is:u =c + c/21 +c/2 cc2=1.5c1.5= cA Meter Stick:Now, what happens if you have a meter stick in your ship (which is lying on the floor in theˆi direction)? You observe it as having a length, of course, of 1 m.I, however, notice something different. Since I’m measuring the length at a particular mo-mentum, I measure ∆t = 0 (the front and back are measured at the same time). Thus,according to equation 7, ∆x′= γ∆x. Or, if I am measuring the length of the rod in myframe:∆x =∆x′γ=LγRelativity II: Lorentz Transforms– 4PHYS 113: Relativity IISince γ is always greater than or equal to 1, I always measure moving meter sticks to beshorter than stationary o nes. Not just meter sticks – everything. You, your ship, yourcontrol panels, a nd so on.This is known as length contraction. Now here’s the wacky thing – because of the sym-metry of the equations, you will find that a meter stick in my frame (again, oriented alongthe x-axis) looks short to you.A Clock:What about a clock? Well, imagine that a clock ticks on your ship ∆t′= 1s. For you, thetwo ticks of the clo ck happen in the same place ∆x′= 0 . Thus, according to equation 10,we have ∆t = γ∆t′. In other words your clock (and your heartbeat, and the oscillations ofyour molecules and everything else, indeed time itself) seems to run slow on your ship.But again – to you, it appears that my clock (and heart, etc.) are running slow!This is known as time dilation.A Spacetime ExampleHere’s a snapshot of a stationary lab in which a light beam is bouncing back and forth asseen from within the lab, and also as seen from an o bserver moving at 1/2 the sp eed of light:Relativity II: Lorentz Transforms–