Using the Lorentz transformation

1. Using the Lorentz transformation The relativity of simultaneity Let us consider two events A and B, and look at them in both IRF and IRF’, using the Lorentz Transformation LT to find their coordinates in IRF once we have them in IRF’. IRF and IRF’ are the usual systems already drawn in lesson 11. I will add the remark that the origins of the two frames are chosen to be coincident or, to specify things to the detail, the “space” coordinates of both origins are null at times t=t’=0. Just to avoid pains with fixed translations in the formulas. Now, given the coordinates of A and B in IRF’ [tA’, xA’] we can easily evaluate their values in IRF: [tA, xA] = [γ(tA’+β xA’), γ(xA’+β tA’)] [tB, xB] = [γ(tB’+β xB’), γ(xB’+β tB’)] and, being Δt’= tB’ - tA’ and Δx’= xB’ - xA’ the following equations hold: Δt= γ(Δt’+β Δx’) Δx= γ(Δx’+β Δt’) Now, we choose A and B to be simultaneous in IRF’: Δt’= tB’ - tA’ = 0. Then, we have Δt= γβ Δx’ Δx= γΔx’ This results shows that two events which are simultaneous in one frame are not in another. They still are simultaneous only when they happen with the same value of the coordinate in relative motion in IRF’. To be clear: with the same value of “x”. The transverse coordinates – “y” and “z” in our case – can take any value. Advanced EM - Master in Physics - 2011-2012

2. Now, let us take an extra step: let the two events be the measurement of the length of a ruler at rest in IRF’ and parallel to the “x” axis. Then, in IRF’ Δx’=L, while in IRF Δt= βγL and Δx= γL Which means that the length seems to expand, instead of shrinking, as we have learnt from the example of the train they should do. We shall soon deal with that matter. Before being done with this example, one further remark: take two events in IRF’ which take place at the same x’, but at different times. Well, it is obvious that they will be at different x in IRF, since IRF’ is moving and an object moving with IRF’ will be –seen by IRF- at different places in different times. That to us is perfectly obvious. What is not obvious is the other way around, i.e. with the words “time” and “space” inverted: that events simultaneous, i.e. with the same time “t’” in IRF’ but different positions (I mean: “x”) are seen in IRF at different times. There was an obvious asymmetry in the handling of space and time in the so called “Galilean relativity”. And some symmetry – although not equality- we still expect in 4-space. It was due not to error, but to the fact that, due to the value extremely high of the speed of light wrt our size and our speeds the fact that time and space are congruent was not apparent – actually, it is still not apparent. The contraction of length We have seen in the previous example that Δx= γΔx’ Where γΔx’ is p.ex. the length of a ruler at rest in IRF’ as measured at tA’ = tB’ = 0, and Δx its Lorentz transformed: it is larger by a factor γ than that seen in a moving frame. And while in the example of the paradox train its length measured in a frame in which the train is moving is shorter than that in the train’s frame by a factor γ. Obviously there is some contradiction which needs to be resolved. Advanced EM - Master in Physics - 2011-2012

3. We are well used to measure any kind of length, sizes, distances etc. Usually we do it on “objects”, and anyway always on still objects. If we measure a ruler, in the language of the IRFs we lay it on a coordinate axis, we keep it still and we read the two extreme coordinates, xA’ and xB’. If, in the system in which our ruler is at rest we measure first one end of the ruler and later on the other one, it does not make a difference. But, if we make the measurement in a frame in which the ruler is moving, then, as the formulas of two pages above indicate, the two events which were simultaneous in IRF’ are now, as seen in IRF, at a time distance Δt= γβ Δx’ . And if the two ends of a moving object are measured at different times then the measurement does obviously not yield the right measurement of length. The first thing to do then is to decide on a procedure for measuring the length of objects in movement: CONVENTION: For a measurement of the dimensions of a given object, the coordinates should be measuredat the same instant. (which is what had happened in the case of the External Observer of the train Paradox). In this case instead the 2 events of measurement of one coordinate only have times equal in IRF’ but not in IRF. And, since a moving object is measured obviously the result depends on the time delay… which is Δt= βγ Δx’ , at which time we measure Δx= γΔx’. Going back to the measurement in IRF’, let us specify that the events measured in that case were lying on the x-axis, A was at the frame’s origin, B was at x=L, both had time=0. In such case the TDL gives the following coordinates for A and B in IRF: A = [0,0] B = [βγL, γL] Which confirm that in IRF the measurement of B is made a time βγL later than the measurement of A. Advanced EM - Master in Physics - 2011-2012

4. Given the fact that the movement between the two IRFs is at constant speed, it is not difficult for the observer in IRF, who wants to measure both points at the same time, to correct the length calculated from the LT at Δt= βγL in IRF by the distance βΔt= β2γL. The position of A in IRF must therefore be corrected for this distance, and it will be xA = β2γL. At this point the distance Δx= xB – xA becomes And we find again the length contraction that we found in the train’s example. We have not learnt anything new… just to be careful in an environment where time changes with the reference frame. The convention of measuring distances between events at the same time has the advantage that it is a clear, simple experimental procedure. But it has the disadvantage that it is a different pair of events for every IRF. Still on distances: the segmented bar A cylindrical bar of length L0 when in movement appears to have a length shorter by a factor γ. What on Earth has happened? Have the interatomic distances shrunk? Has the number of atoms decreased? They both seem to be rather unlikely situations. Let us try to understand what has happened. So, let us (imagine to) slice the bar in n disks. And now accelerate it by a quantity δβ. If we derive wrt β we obtain: Advanced EM - Master in Physics - 2011-2012

5. For a change of speed δβ the bar seem to shorten by δL. The first and last disks are distant (n-1) Δx’, where Δx’ is the thickness of the disks. Now the question is: How does the process of accelerating the bar take place? It is not sufficient to tap the bar at its base – I have to give n taps, one for each cylinder, all taps simultaneous in IRF’, otherwise the change of speed δβ proceeds from the base to the front with the velocity of sound in the metal (very slow wrt the speed of light). This way, I have managed to accelerate the whole bar simultaneously in IRF’. That is, all cylinders change velocity by δβsimultaneously. Since they are at different values of x’, they will obviously be accelerated at different times in IRF! The time difference between the first and the last is δt= βγ L0 [neglecting the factor (n-1)/n]; and all the individual taps will be delayed wrt to the preceding one by βγ L0/n. The distance between the first (which gets the first tap) and the last one will therefore be changed by a quantity δL= δβ δt= -βγ L0δβ: The same value we obtained at the beginning by derivating L. We have so understood that such length contraction seen in IRF is due to the action, diluted in time, of the acceleration of the various bar segments. It follows that integrating the found δL over the bar length we integrate the (dL/δβ) δβ, and obtain the overall Lorentz contraction! Advanced EM - Master in Physics - 2011-2012

6. A Paradox Two rockets, A and B, of length L, meet –going in opposite directions at relativistic velocity- in space. A B When observer A in rocket A sees the tip of his rocket next to the end of rocket B he shoots at B, with a gun located at the back of its rocket. The question: does he hit B or does he not? N.B. (exercise 3-7 in Taylor-Wheeler) Advanced EM - Master in Physics - 2011-2012

7. Transformation of angles Angle of a velocity We have seen how a velocity transforms. Let an object have velocity in IRF’ . Let us find its components in IRF, that sees IRF’ as usual flee in the forward direction in “x” with velocity β. In IRF the object’s velocity has components determined via the LT. Therefore tg(θ) < tg(θ’) (N.B.: if βand v’x have equal directions) Advanced EM - Master in Physics - 2011-2012

8. Y’ B Bar θ A X’ The coordinates of A and B are in IRF’ : A= [0,0,0,0] B=[0,cosθ’,sin θ’,0] IRF : A= [0,0,0,0] B=[βγcos θ’, γcosθ’,sin θ’,0] Then, because of the usual reason that the two ends of the bar have been measured at different times. If both measurements are reduced to the same time t= βγcos θ’ , then x’A = β2γcos θ’ and Angle of a rigid object (a bar, a ruler?) Advanced EM - Master in Physics - 2011-2012

9. A car travels overnight at relativistic speed on a straight road, we shall put there the “x” axis. It has the headlight on, a light beam with span ±ψ (in its own IRF’). What does the light beam look like in IRF (the road)? Now… a light beam is not a bar. It a something that moves, with its own velocity: ”c”. So, here we have to use the law of composition of velocities. In these formulas I have used the fact that the speed of light is 1, and its components are of course sinψ’ and cos ψ’ . Since in IRF’ the light is not a narrow, straight beam like a laser, but a beam with an angular span, it has nearly uniform intensity for angles < ψ’ . That angle in IRF’ converts in IRF (road) in a limit The angular aperture of the light beam seen on the road is much more narrow than in the frame of the car. The headlight effect If the light were a 4π lamp, sending light all around, the light emitted forward in IRF’ (cos ψ>0) in IRF is contained within a maximum angle ψM such that Advanced EM - Master in Physics - 2011-2012

10. The explosion In interstellar space an enormous body explodes in a large number of fragments. Question: How does a fragment see the motions of all others? Obviously, as uniform motions away in radial directions. Well, let each fragment have a sort of beacon that emits a light pulse in all directions with a precise period τ; and let τbe the same for all fragments. How can an observer on fragment A find how fragment B moves? A sees a light pulse arriving from fragment B every time interval Δt. This, beside knowing the emission period τ, is the only information he has to trace the relative velocity β. Well, let us define two reference frames: IRF’ is A’s frame, IRF B’s. B emits light flashes with a period τ, in IRF. The LI between two consecutive emissions is τ2.Such period –N.B.the period of emission- seen by an observer in IRF’ is τ’ = γτ, of course. But, τ‘ is not equal to Δt. It is the period with which B emits the flashes, seen by the frame of clocks on IRF’ that goes at velocity β wrt B. To calculate Δt we must add to it the time it takes light to travel from B to A – and which is not equal for all pulses. The time difference between two next flashes is obviously equal to βτ’= βγτ. Then Advanced EM - Master in Physics - 2011-2012

11. Rewrite the last formula: Or, inverting this formula in order to calculate the observed frequency as a function of the (known a priori) emission period and the relative velocity β: These formulas look very much like the relativistic Doppler effect. Advanced EM - Master in Physics - 2011-2012