Physics, General RelativityHomeworkDue Wednesday,thSeptemberJacob Lewis BourjailyProblem 1a)We are to use the spacetime diagram of an observerOto describe an ‘experiment’ specified by theproblem 1.5 in Schutz’ text.We have shown the spacetime diagram in Figure 1 below.txFigure 1.A spacetime diagram representing the experiment which was required to bedescribed in Problem 1.a.b)The experimenter observes that the two particles arrive back at the same point in spacetime afterleaving from equidistant sources.The experimenter argues that this implies that they were released‘simultaneously;’ comment.In his frame, his reasoning is just, and implies thathist-coordinates of the two eventshave the same value. However, there is no absolute simultaneity in spacetime, so adifferent observer would be free to say that inher frame, the two events were notsimultaneous.c)A second observerOmoves with speedv= 3c/4 in thenegativex-direction relative toO. We areasked to draw the corresponding spacetime diagram of the experiment in this frame and comment onsimultaneity.Calculating the transformation by hand (so the image is accurate), the experiment ob-served in frameOis shown in Figure 2. Notice that observerOdoes not see the twoemission events as occurring simultaneously.txFigure 2.A spacetime diagram representing the experiment in two different frames.The worldlines in blue represent those recorded by observerOand those in green rep-resent the event as recorded by an observer in frameO. Notice that there is obvious‘length contraction’ in the negativex-direction and time dilation as well.1
2JACOB LEWIS BOURJAILYd)We are to show that the invariant interval between the two emission events is invariant.We can proceed directly.It is necessary to know that in frameOthe events havecoordinatesp1= (5/2,-2) andp2= (5/2,2) while in frameOthey have coordinatesp1=γ(1,-1/8) andp2=γ(4,31/8) whereγ2=167.Δs2= (p1-p2)2= 16;Δs2= (p1-p2)2=γ2(-9 + 16) = 16.We see that the invariant interval is indeed invariant in this pointless example.Problem 2.a)We are to show that rapidity is additive upon successive boosts in the same direction.We may as well introduce the notation used in the problem: letv= tanhαandw=tanhβ; this allows us to writeγ=1√1-tanh2α= coshαandvγ= sinhα, and similarexpressions apply forβ. We see that using this language, the boost transformationsare realized by the matrices1γ-vγ-vγγ¶7→coshα-sinha-sinhαcoshα¶,(a.1)and similarly for the boost with velocityw. Two successive boosts are then composedby2:coshβ-sinhβ-sinhβcoshβ¶coshα-sinhα-sinhαcoshα¶=coshαcoshβ+ sinhαsinhβ-sinhαcoshβ-coshαsinhβ-sinhαcoshβ-coshαsinhβcoshαcoshβ+ sinhαsinhβ¶=cosh(α+β)-sinh(α+β)-sinh(α+β)cosh(α+β)¶This matrix is itself a boost matrix, now parameterized by a rapidity parameter(α+β). Therefore, successive boosts are additive for rapidity.