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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".

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  • $\begingroup$ 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
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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.

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

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