= Solution
For a <photon>, $p^ip_i=0$ and the null displacement along the ray is parallel to $p^i$. Consequently $p_jdx^j=0$, and the first term of the momentum law vanishes. Write $dx^i=p^i d\lambda$ with an appropriate path parameter. Then
$$
dp^i=-(p^j\partial_j\Phi)p^i d\lambda=-p^i d\Phi,
\qquad\boxed{d(e^\Phi p^i)=0.}
$$
For a static potential, a <photon> emitted and received by laboratories at rest in the background frame has <energy> proportional to $p^0$. Using $E=h_{\rm P}\nu$, the <scalar-force gravitational redshift> is
$$
\boxed{\frac{\nu_r}{\nu_e}=e^{\Phi_e-\Phi_r}
=1-(\Phi_r-\Phi_e)+O((\Delta\Phi)^2).}
$$
Restoring $c$, 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 \b[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>.
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