In general relativity the four-velocity of a timelike, massive particle following a path $x^{\mu}(\tau)$ is defined as the derivative $$u^{\mu} = \frac{dx^{\mu}}{d \tau}$$ where $\tau$ is the proper time of the curve, i. e. $$\tau = \int_{\lambda_0}^{\lambda_1} \sqrt{-g_{\mu \nu} \frac{dx^{\mu}}{d \lambda} \frac{dx^{\nu}}{d \lambda}}$$ where $\lambda$ is an arbitrary parametrization of the curve. Among all the possible affine parameters the preferred one is the proper time for which we define the four-velocity. Similarly for a space-like curve the preferred affine parameter is the "length" of the curve, however we wouldn't call the derivative of the coordinates with respect to such parameter a four-velocity.

On the other hand, for null-like geodesics, with no preferred choice of the affine parameter, corresponding to a massless particle/photon, it still make sense to define the four-momentum as $$p^{\mu} = \frac{dx^{\mu}}{d \lambda}$$ but how can we fix the multiplicative constant which can arise from $$\lambda \rightarrow a \lambda$$ which gives $$p'^{\mu} = \frac{dx^{\mu}}{d (a \lambda)} = \frac{1}{a} p^{\mu}~?$$ It sees to me that it cannot be done in this context and one should impose the correct four-momentum "by hand".

  • $\begingroup$ In the integral defining proper time, the measure $d\lambda$ is missing. I tried to edit the question but my edit was not accepted by the peer. Can someone please explain to me, why in the second equation one wouldn't expect the measure $d\lambda$ outside the root? $\endgroup$ – VacuuM Aug 10 '18 at 8:29
  • $\begingroup$ $c^2d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu}\implies d\tau=\frac{1}{c}\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda\implies\tau=\int_{\lambda_0}^{\lambda}\frac{1}{c}\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda$? $\endgroup$ – VacuuM Aug 10 '18 at 9:23

how can we fix the multiplicative constant

The choice is not made on geometric, but on physical grounds:

Contraction of the momentum with a time-like vector has to yield the photon energy as measured by an observer moving with corresponding 4-velocity.

  • $\begingroup$ I suspected this but I was not sure. Thank you. $\endgroup$ – MrRobot Aug 6 '17 at 9:23

Two things to realize

(1) The GR trappings are irrelevant here. This is an SR issue.

(2) You're effectively trying to define momentum in SR as $p=m\gamma v$, but that doesn't work in general. It only works for massive particles. A more general definition is that momentum is the spacelike part of the energy-momentum vector, which implies $m^2=E^2-p^2$. As described in the answer by Christoph, ultimately all quantities, e.g., the energy-momentum vector, have to be defined in terms of a measurement process.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.