The Lorentz TransformationThis is a derivation of the Lorentz transformation of Special Relativity. Thebasic idea is to derive a relationship between the spacetime coordinates x, y, z, tas seen by observer O and the coordinates x′, y′, z′, t′seen by observer O′movingat a velocity V with respect to O along the positive x axis.xyx′y′OO′VThese observers are assumed to be inertial. In other words, they are moving ata constant velocity with respect to each other and in the absence of any externalforces or acc elerations (which is somewhat redundant). In particular, there isno rotational motion or gravitational field present.Our derivation is based on two assumptions:1. The Principle of Relativity: Physics is the same for all observers in allinertial coordinate systems.2. The speed of light c in a vacuum is the same for all observers independentlyof their relative motio n or the motion of the light source.We first show that this trans formation must be of the formt′= at + bx (1a)x′= dt + ex (1b)y′= yz′= zwhere we assume that the origins coincide at t = t′= 0. The above figure showsthe coordinate systems displaced simply for ease of visualization.The first thing to note is that the y and z coor dinates are the same for bothobservers. (This is only true in this case because the relative motion is alongthe x axis only. If the motion were in an arbitrary direction, then each spatialcoordinate of O′would depend o n all of the spatial coordinates of O. However,this is the case that is used in almost all situations, at least at an elementarylevel.) To see that this is necessary, suppose there is a yardstick at the origin o feach coordinate system aligned along ea ch of the y- and y′-axes, and supposethere is a pa intbrush a t the end of each yardstick pointed towards the other.1If O′’s yards tick along the y′-axis gets shorter as seen by O, then when theorigins pass each other O’s yardstick will get paint on it. But by the Principleof Relativity, O′should also see O’s yardstick ge t shorter and hence O′wouldget paint on his yardstick. Since this clea rly can’t happen, there can be nochange in a direction perpendicular to the direction of motion.The next thing to notice is that the transformation equations are linear.This is a result of space being homogeneous. To put this very loosely, “thingshere are the same as things there.” For example, if there is a yardstick lyingalong the x axis between x = 1 and x = 2, then the length of this yardstickas seen by O′should be the same as another yardstick lying between x = 2and x = 3. But if there were a nonlinear dependence, say ∆x′goes like ∆x2,then the first yardstick would have a length that goes like 22− 12= 3 while thesecond would have a length that goes like 32− 22= 5. Since this is also not theway the world works, equations (1) must be linear as shown. We now want tofigure out what the coe fficients a, b , d and e must be.First, let O look at the origin of O′(i.e., x′= 0). Since O′is moving at asp e e d V along the x-axis, the x coordinate corresponding to x′= 0 is x = V t.Using this in (1b) yields x′= 0 = dt + eV t or d = −eV . Similarly, O′looks atO (i.e., x = 0) and it has the coordinate x′= −V t′with respect to O′(sinceO moves in the negative x′direction as seen by O′). Then from (1b) we have−V t′= dt and from (1a) we have t′= at, and hence t′= −dt/V = at so that−aV = d = −eV and thus also a = e and d = −aV . Using these results inequations (1) now gives ust′= at +bax(2a)x′= a(x − V t ) (2b)y′= yz′= z.Now let a photon move along the x-axis (and hence also along the x′-axis)and pass both origins when they co incide at t = t′= 0. Then the x coo rdinateof the photon as seen by O is x = ct, and the x′coordinate as see n by O′is x′= ct′. Note that the value of c is the same for both observers. This isassumption (2). Using these in equations (2) yieldst′= at +bact= at1 +bcax′= a(ct − V t) = cat1 −Vcso thatcat1 −Vc= x′= ct′= cat1 +bca2and therefore −V/c = bc/a orba= −Vc2.So now equations (2) becomet′= at −Vc2x(3a)x′= a(x − V t ) (3b)y′= yz′= z.We still need to determine a. To do this, we will again use the Principle ofRelativity. L e t O look at a clock situated at O′. Then ∆x = V ∆t and from(3a), O and O′will measure time intervals related by∆t′= a∆t −Vc2∆x= a1 −V2c2∆t.Now let O′look at a clock at O (so ∆x = 0). Then ∆x′= −V ∆t′so (3b) y ie lds−V ∆t′= ∆x′= a(0 − V ∆t) = −aV ∆tand hence∆t =1a∆t′.By the Principle of Relativity, the relative factors in the time measurementsmust be the same in both cases. In other words, O sees O′’s time related to hisby the factor a(1 − V2/c2), and O′sees O’s time related to his by the factor1/a. This means that1a= a1 −V2c2ora =1q1 −V2c2and therefore the final Lorentz transformation equations aret′=t −Vc2xq1 −V2c2(4a)x′=x − V tq1 −V2c2(4b)y′= yz′= z.3It is very common to define the dimensionless variablesβ =Vcandγ =1q1 −V2c2=1p1 − β2.In terms of these variables, equations (4) becomet′= γt −βcx(5a)x′= γ(x − βct) (5b)y′= yz′= z.Since c is a universal co nstant, it is essentially a conversion factor between unitsof time and units of le ngth. Because of this, we may further change to unitswhere c = 1 (so time is measured in units of length) and in this case the Lorentztransformation equations b e comet′= γ(t − βx) (6a)x′= γ(x − βt) (6b)y′= yz′= z.Note that 0 ≤ β ≤ 1 so that 1 ≤ γ < ∞. We also see thatγ2=11 − β2so thatγ2− γ2β2= 1.Then recalling the hyperbolic trigonometric identitiescosh2θ − sinh2θ = 1and1 − tanh2θ = sech2θwe may define a parameter θ (sometimes called the rapidity) byβ = tanh θso thatγ = cosh θ4andγβ = sinh θ.In terms of θ, equations (6) becomet′= (cosh θ )t − (sinh θ)xx′= −(sinh θ)t + (cosh θ)xy′= yz′= zwhich looks very similar to a rotation in the xt-plane, except that now we havehyperbolic functions instead of the usual trigonometric ones. Note also thatboth sinh terms have the same