Physics 106b – Problem Set 8 – Due Jan 13, 2006Version 2 – Jan 9, 2006We cover special relativity in this set, Chapter 6 of the lecture notes and Chapter 12 of Hand andFinch. Problems 1 through 4 are required, problem 5 is extra credit and equal in weight to the firstfour problems.Changes since version 1: Correction to sign error in expression for |aµ|2in Problem 2.1. Hand and Finch 12-4. This is a classic problem that is not too different from the discussionof length contraction in the lecture notes. Doing the problem, though, will hopefully helpyou deepen your understanding.2. Hand and Finch 12-6 (relativistic acceleration). This problem is useful for calculating the mo-tion of a charged particle in various accelerators (cyclotron, synchrotron, linear accelerator).In addition to what is asked, do the following:• Show that|aµ|2= −γ6γ2⊥d~βdt2whereγ2⊥=1 − β2⊥−1=1 −β2−~β ·d~βdtd~βdt2−1is the γ factor due to the lab-frame velocity perpendicular to the lab-frame acceleration.(It’s less complicated than it looks.)• Show that the above formula reduces to the two that you found for circular motion andparallel acceleration.Notes:• Equation 12.116 erroneously implies that the three-acceleration ~a is given by the spacecomponents of aµ. The three-acceleration ~a is alwaysd~βdt. Just as the space componentsof the four velocity uµare not just the three-velocity~β but are rather γ~β, the spacecomponents of the four-acceleration are not just the three-acceleration ~a. There wouldbe no ambiguity if H&F simply dropped the (a0,~a) part of Equation 12.116.• As in the definition of four-velocity, you will find it necessary to conve rtddτtoddttoobtain the desired explicit form. In our demonstration that uµ= γ1,~β, we use dτ =p|xµ|2and took the derivative with respect to t. We implicitly assumed there that1~β was constant. Since we are no longer m aking that assumption, the original relationfor τ(t) may no longer hold. What we can be sure of, though, is that the infinitesimalversion of the relation holds:(dτ)2= (dt)2− (d~x · d~x)2dτ = dtp1 − β2=dtγwhere~β may now be a function of time. You will want to use the differential relation.1• There is a minor error in the problem: it should ask you to demonstrate that “therelation between the laboratory acceleration of a particle undergoing circular motion atconstant s peed and the acceleration in the instantaneous rest frame of the particle isarest= γ2alab.” That this is a typo is confirmed by Goldstein derivation 7.7 and thegeneric formula for |aµ|2given above.3. Hand and Finch 12-17 (Compton scattering). Notes:• In class we have not used the notation kµkµor pµpµ. In general, for any four-vector,the expression aµaµis the invariant norm |aµ|2= (a0)2− (a1)2− (a2)2− (a3)2. Theformal meaning of lowered indices is discussed in the lecture notes, but you don’t needto know that for this problem.• The kµdefined in this problem is different from the kµdefined in class by a factor ~. Youcan basically forget about the kµdefined in class when doing this problem – just take itas given that kµis the four-momentum of the photon, with the time component beingthe energy and the space c omponents being the spatial momentum, and that |kµ|2= 0because the photon rest mass vanishes.• A “backscattered” photon is one with outgoing angle θ0= π.4. Relativistic harmonic oscillator (clearer version of Hand and Finch 12-21). The Lagrangianfor the relativistic simple harmonic oscillator is (c = 1 as usual):L = −mp1 − β2− V (x) V (x) =12k x2Let a be the amplitude of the motion. Use the usual Euler-Lagrange procedure to obtain theequation of motion for the particle, realizing that β = ˙x. Integrate the equation of motion toobtain the following expression for the energy:E = γ m +12k x2which, when evaluated at x = a (when ˙x = 0), can be writtenE = m +12k a21Two finer points: 1) with the differential relation, we could of course have derived the four-velocity expressionwithout the assumption of constant velocity; hence, it holds even when the velocity is not constant. 2) If we integratethe differential relation, we findτ (t) =Zt0d˜tq1 −ˆβ(˜t)˜2which may in general be different fromp|xµ|2=√t2− ~x · ~x, hence the distinction between using the differential andintegral relations.2One can calculate the period of motion via the integralτ = 4Zt=τ/4t=0dt = 4Zx=ax=0dx˙xShow that this integral isτ =2 aκZu=1u=01 + 2 κ21 − u2p1 + κ2(1 − u2)du√1 − u2where u = x/a and the dimensionless parameter κ2=k a24 m. We see that the period is nowamplitude-dependent via κ2, which is half the the ratio of the maximum p otential energy tothe particle rest mass. Show that this integral reduces to the nonrelativistic period τNR=2 πpm/k when calculated to zeroth order in κ2. Show that, to first order in κ2, the period isτ = τNR1 +34κ2= τNR1 +38V (a)mHints:• For help with integrating the equation of motion, refer back to the relativistic calcu-lation of the relation between work and kinetic energy under the topic The Energy-Momentum Four-Vector in Section 6.1.4 of the lecture notes.• To do the period integral, you will need to rewrite ˙x in terms of x using the energy.5. Hand and Finch 12-18. The expression given for ∂µis incorrect, it should be ∂µ=1c∂∂t, −~∇.You’ll have to read Section 6.1.3 of the lecture notes to make sense of the tensor