1 Relativity notes Shankar Let us go over how the Lorentz transformation was derived and what it represents. An event is something that happens at a definite time and place, like a firecracker going off. Let us say I assign to it coordinates (x, t) and you, moving to the right at velocity u,assigncoordinates(x,t). It is assumed that when you, sitting at x =0passedme(sittingat x =0),wesetourclockstozero: t = t =0.Thusouroriginsin space-time coincide. Let us see how the coordinates would have been related in pre- Einstein days. First, once we synchronize clocks at t = t =0,theyremain synchronized for all future times t = t.Thisisthenotionofabsolute time we all believe in our daily life. You will assign to the event the spatial coordinates x = x ut. (1) − This relationship is easy to understand: your origin is ut meters to the right of mine (Figure (2)) because you have been moving to the right at velocity u for time t so the x-coordinate you assign to an event will be less by ut compared to what I assign. You can invert S S' ut x x' Figure 1: The same event (solid dot) is assigned coordinates (x, t) by me (S) and (x,t)by you (S’). In the pre-Einstein days t = t not only initially (when our origins crossed) but always. 2 this relation to say x = x + ut. (2) How are these two equations modified post-Einstein? If the ve- locity of light is to be same for both you and me, it is clear we do not agree on lengths or times or both. Thus if I predict you will say the event is at x = x ut,youwillsaythatmylengthsneedtobe − modified by a factor γ so that the correct answer is x = γ(x ut). (3) − Likewise when you predict I will say x = x + ut Iwillsay,”No, your lengths are off, so the correct result is x = γ(x + ut).” (4) Note two things. First, I leave open the option that the time elapsed between when we synchronized clocks and when the fire- cracker went off is t for you and t for me, with two times being pos- sibly different. Next, the ”fudge factor” for converting your lengths to mine and mine to yours are the same γ.Thiscomesfromthe postulate that both observers are equivalent. So let us look at the equations we have: x = γ(x + ut) (5) x = γ(x ut). (6) − We proceed to nail down γ as follows. The event in question was afirecrackergoingoff.Supposewhenouroriginscoincidedwesent off a light pulse that this pulse set off the firecracker. Since the light pulse took t seconds to travel x meters according to me and took t seconds to go x meters according to you and we both agree on the value of c,itmustbetrueforthis particular event that x = ct and x = ct. (7) 3 Let us multiply the LHS of Eqn 5 by the LHS of 6 and equate the result to the product of the RHS’s to get 2 2 xx = γ (xx + xut xut u tt), and upon setting x = ct, x = ct we get(8) 2 2 2 − − 2 c tt = γ (c tt + uctt uctt u tt)andnowuponcancellingtt (9) 1 − − γ2 = (10) 1 u2 − c2 1 γ = . (11) u2 1 2 − c Note that once we have found γ it does not matter that we got it form this specific event involving a light pulse. It can applied to Eqns. (5,6) valid for a generic event. Putting γ back into Eqn. (6) we obtain x ut x = − . (12) u2 1 2 − c If we now go to Eqn 5 and isolate t we find ux t c2 t = − (13) u2 1 2 − c upon remembering that 1 u2 1 = . (14) − γ2 c2 To summarize, if I am frame S and you are S’, and you are moving to the right (increasing x direction) at speed u and my coordinates for an event are (x, t)andyoursare(x,t) the Lorentz transformation tells us that x ut x = − (15) 1 u2/c2 − u t c2 x t = − (16) 1 u2/c2 − 4 If we consider two events labelled 1 and 2, then the coordinate differences obey ∆x u∆t ∆x = − (17) 1 u2/c2 − u ∆t c2 ∆x ∆t = − (18) 1 u2/c2 − where ∆x = x x etc., and differences are not necessarily small. 2 − 1 If you want to get my coordinates in terms of yours, you need to invert Eqns. (15) and 16. The answer is we get the same equations but with u u.Thesamegoesforthedifferences.Asaresult →− the differences will be related as follows; ∆x + u∆t ∆x = (19) 1 u2/c2 − ∆t + u ∆x ∆t = c2 (20) 1 u2/c2 − Now for the velocity transformation law . Let us follow a particle as it moves by an amount ∆x in time ∆t according to me and ∆x in time ∆t according to you. Let us agree that velocities are defined as follows once and for all: ∆x v = according to me (21) ∆t ∆x w = according to you (22) ∆t u =yourvelocityrelativetome (23) In the above all ∆s better be infinitesimals going to zero, as we are talking about instantaneous velocities and the derivative needs to be taken. 5 Suppose I see a particle with velocity v.Whatisthevelocityw according to you? To get this we take the ratio of the equations (37 -38) that give the primed coordinate differences in terms of unprimed ones: ∆x ∆x u∆t v u w = = − = − (24) ∆t ∆t u ∆x 1 uv − c2 − c2 where in the last step we have divided top an bottom by ∆t.Itis good to check that for small velocities (dropping the terms that go as 1/c2)wegetresultsagreeingwithcommonsense. Let us get used to going from your description to mine. Suppose you think a particle has velocity w.WhatwillIthinkitsvelocityis? Now we use Eqns.(19-20). Taking the ratios as before and recalling the definition of w we get w + u v = vw , (25) 1+ c2 which just has the sign of u reversed in Eqn. (24) as expected and w v. ↔ Suppose you see an object moving at w =3c/4andyouyourself are moving relative to me at u =3c/4. In the old days I would expect that object to be moving at 1.5c.Howeverthecorrectanswer is 3c 3c 4 + 4 24 v = 9 = c. (26) 1+16 25 It is interesting to see that if you choose to apply this to a light pulse seen by you (w = c)thespeedIwillfind(foranyrelativevelocityu between us) c + u v = = c. (27) 1+u/c 6 Time dilatation. Let me carry a clock with ticks ∆t = τ0 seconds apart. If the two events are two successive ticks, the spatial separation is ∆x =0sincetheclockisatrestforme.Youthinkthe time difference between ticks is u2 τ0 0 c2 τ0 ∆t = − = (28) 1 u2/c2 1 u2/c2 − − Thus you say my clock is slow. This is what I will hear from anyone moving relative to me. Let us rederive this using Eqns (19-20). Thus set u ∆t + c2 ∆x ∆t = τ0 = (29) 1 u2/c2 − and note that the other equation says ∆x + ut ∆x =0= (30) 1 u2/c2 − which means ∆x = u∆t.(ThisjustmeansyouthinkIandmy − clock are moving to the left at speed u)FeedingthisintoEqn.29to get 2 2 ∆t(1 u /c ) τ0 = − (31) 1 u2/c2 − which agrees with Eqn. 28. Length contraction Suppose you are carrying a rod of length L0.Thisisitsrestlength.Supposeyouaremovingrelativetome at velocity u.HowlongwillIsayitis?TofinditslengthIhave two of my co-moving assistants measure the location of its front and back ends at the same time.Thesetwomeasurements,whichcan be seen as two events, have ∆x = L,thelengthaccordingtome, and ∆t =0sincethemeasurementsofamovingrod’sendshaveto 7 be simultaneous. (What happens if I measure one end, take a break and them measure the other?) The separation in space between the two events is the rest length L0 of the rod according to you. Thus L 0 2 2 L0 = − or L = L0 1 u /c . (32) 1 u2/c2 − − that is I will say your meter stick is actually 1 u2/c2 meters long, − i.e., shortened. Likewise you will see my meter sticks are contracted. Suppose the contraction factor 1 u2/c2 = .5. I say your meter sticks are − .5m long. You say yours are fine and mine are .5m long. How do you understand a guy with a shortened meter stick accusing you of having a shortened meter stick? Should I not say your are 2m long? The answer is this : You will say that I measured the leading end of your meter stick first and then after a delay , during which time the stick slid by, I measured the other end. That is, you will say I did not measure the ends simultaneously. I will insist I did, and both are right since simultaneity is not absolute. Order of events and causality If event 1 causes event 2, no one should see the cause come after the effect for this leaves time for some one to prevent the cause from itself occurring. On the other hand if the two events are not causally related, a reversed order, while strange from our daily experience, will not lead to logical contradictions. According to the LT u ∆t c2 ∆x ∆t = − (33) 1 u2/c2 − Let ∆t = t t > 0, so that I think 2 occurred after 1. Then we 2 − 1 8 can find an observer for whom ∆t < 0provided u ∆x>∆t (34) c2 u c∆t > (35) c ∆x If c∆t>∆x the order can be reversed only if u/c > 1whichis impossible.