Gravity's Effect on Time and the Gravitational Red Shift

We may consider how the proper time for a particle \bgroup\color{black}$ \tau$\egroup relates to the coordinate time \bgroup\color{black}$ t$\egroup as a function of distance from the mass at the origin. In the rest frame of the particle,

\bgroup\color{black}$\displaystyle -c^2d\tau^2=-\left(1 - \frac{r_s}{r} \right) c^2 dt^2 $\egroup

so the proper time interval at a radius \bgroup\color{black}$ r$\egroup is.

\bgroup\color{black}$\displaystyle d\tau=\sqrt{1-{r_s\over r}}dt $\egroup

Applying this for \bgroup\color{black}$ r\rightarrow\infty$\egroup we have.

\bgroup\color{black}$\displaystyle d\tau_\infty=dt $\egroup

So for light that escapes to infinity, the time to emit the light in the rest frame of the atom at radius \bgroup\color{black}$ r$\egroup is shorter than the time seen at infinity.
\bgroup\color{black}$ \displaystyle {d\tau\over d\tau_\infty}=\sqrt{1-{r_s\over r}} $\egroup
This causes a red shift in the observed frequency.

\bgroup\color{black}$\displaystyle {\nu_{obs}\over\nu_{emit}}=\sqrt{1-{r_s\over r}} $\egroup

Since the energy of a photon is \bgroup\color{black}$ E=h\nu$\egroup, the photon loses energy as it moves outward in the gravitational field.

Jim Branson 2012-10-21