Consider... [the formula given by special relativity for the magnitude of the ]<math>P \equiv m_0 \sqrt{g_{ij}\frac{dx^i}{d\tau}\frac{dx^j}{d\tau}}</… - Steven Weinberg

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Consider... [the formula given by special relativity for the magnitude of the ]<math>P \equiv m_0 \sqrt{g_{ij}\frac{dx^i}{d\tau}\frac{dx^j}{d\tau}}</math>...where <math>d\tau^2 = dt^2 - g_{ij} dx^i dx^j</math>. [This holds because in] a locally inertial Cartesian coordinate system, for which <math>g_{ij} = \delta_{ij}</math>, we have <math>d\tau = dt\sqrt{1 - \mathbf {v}^2}</math> where <math>v^i = \frac{dx^i}{dt}</math>... [The <math>P</math>] is evidently invariant under arbitrary changes in the spatial coordinates, so we can evaluate it... in Robertson-Walker coordinates. ...[T]o save work ...adopt a spatial coordinate system in which the particle position is near the origin <math>x^i = 0</math>, where <math>\tilde{g}_{ij} = \delta_{ij} + \mathit0(\mathbf{x})</math>, and we can therefore ignore the purely spatial components of <math>\Gamma_{jk}^i</math> of the . General relativity gives [the momentum]... with a metric <math>g_{ij} = a^2(t)\delta_{ij}</math>...<math>P(t) \propto 1/a(t)</math>... for any non-zero mass, however small... Hence, although for photons both <math>m_0</math> and <math>d\tau</math> vanish... [the momentum relation] is still valid.