Smith
Joined: 23 Aug 2010, 13:12 Posts: 33

Re: Exercise [18.18]
Another way to treat the problem:
Rescale Penrose's velocity hyperbola, H+, by M0. Then with v = tanh (rho), the components of the transformed energymomentum vector can be read off as M0(cosh(rho),sinh(rho))= M0*gamma*(1,v).

deant
Joined: 12 Jul 2010, 07:44 Posts: 154

Re: Exercise [18.18]
Even simpler, Smith:
Noting that mass is (in general) defined to be proportional to the zero'th component of the p 4vector, just take the 4vector (m0c^2,0,0,0), apply a Lorentz transform (a boost by velocity v, in any direction), and look at just the zero'th component again: The result is immediately apparent, by comparing the original and transformed 0components.
(NB Vectors transform the same way coordinates do, since the tangent vector space can be regarded as the limit of a "very small" region around a point; tangent vectors therefore transform like "very small" coordinate displacements).
...of course, the simplicity of this solution is mitigated by the fact that Lorentz transforms haven't been explicitly described in the text up to this point (that I can recall), although the Lorentz group has been discussed. But they'll be familiar to anyone who's done even the most basic introductory course on Special Relativity, and the details are easily found on Wikipedia (for example).
I think Robin is right though, and his/her Exercise [18.18b] post (different solution to that given above) is probably what Penrose was after in this exercise, especially since it's only got a "needs a bit of thought" (intermediate) difficulty level.
