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For a photon, $p^i{p}_i=0$ and the null displacement along the ray is parallel to $p^i$. Consequently $p_{j}d{x}^j=0$, and the first term of the momentum law vanishes. Write $d{x}^i=p^{i}d\lambda$ with an appropriate path parameter. ThenFor a static potential, a photon emitted and received by laboratories at rest in the background frame has energy proportional to $p^0$. Using $E=h_{\mathrm{P}}\nu$, the scalar-force gravitational redshift isRestoring $c$, the leading fractional shift is $\Delta \nu /\nu =-\Delta {\Phi }_N/c^2$, where ${\Phi }_N$ is the dimensional Newtonian gravitational potential. Near Earth's surface, an upward displacement $H$ has $\Delta {\Phi }_N=gH>0$, hence $\Delta \nu /\nu =-gH/c^2$: the receiver sees a redshift.

Thus yes: the predicted weak-field frequency shift agrees with the Pound-Rebka experiment. The calculation concerns stationary emitter and receiver, with identical local transition-energy standards; it does not add a Doppler shift due to their relative motion. Agreement with this gravitational redshift measurement alone does not establish the tensor theory of general relativity.