# In general relativity, do light/time behave as in free space if the net force of gravity is zero?

If photons/particles are on lines equidistant from two bodies of mass, does time look the same to them as it would without the bodies of mass? Or is time slowed regardless?

(Given massive bodies $a$ and $b$, and particle $p$, where $p$ is located at a point equidistant from the bodies at time $t_0$, where $\vec{r}_{a->p} = -\vec{r}_{b->p}$, and gravitational acceleration $a_{g} = 0$, does general relativity predict an effect on the passage of time (at $t_0$) as observed by particle $p$? What about photon $p$?)

Summary: There is an effect, because it is potential that matters, not acceleration. And we have to be careful with photons.

Just so we're all on the same page, it's important to note that there is no absolute passage of time. When we talk about time "slowing down" in relativity, special or general, what we mean is the passage of one observer/frame/particle's time from the perspective of another.

Fortunately there is a natural "observer at infinity" in the case of a universe with just the two masses $a$ and $b$, so the question "How does a clock traveling between $a$ and $b$ appear to an observer at infinity?" is well defined.

Consider the Minkowski coordinates $t$, $x$, $y$, and $z$. For a stationary observer ($\mathrm{d}x = \mathrm{d}y = \mathrm{d}z = 0$), proper time $\tau$ obeys $$\mathrm{d}\tau^2 = \left(1 + \frac{2\Phi}{c^2}\right) \mathrm{d}t^2,$$ where $\Phi$ is the Newtonian potential, zeroed at infinity.1

At infinity, $\Phi = 0$ and a stationary observer's proper time $\tau$ progresses at the same rate as the coordinate $t$: $$\mathrm{d}\tau_\infty = \mathrm{d}t.$$

Now consider a timelike observer $p$ between two masses $a$ and $b$, each with mass $m$. Suppose the distance from $p$ to either mass is $r$. The Newtonian potential at $p$ is $\Phi = -2Gm/r$. Then proper time for such an observer is related to $t$ via $$\mathrm{d}\tau_p = \left(1 - \frac{4Gm}{c^2r}\right)^{1/2} \mathrm{d}t.$$

Putting our results together, we have the relation $$\frac{\mathrm{d}\tau_p}{\mathrm{d}\tau_\infty} = \left(1 - \frac{4Gm}{c^2r}\right)^{1/2} < 1.$$ If the observer at infinity watches $p$ for a time $\Delta\tau_\infty$ according to the observer at infinity's own watch, he will see $p$'s watch progress $\Delta\tau_p < \Delta\tau_\infty$ -- this is gravitational time dilation.

All that matters is that $p$ is in a gravitational potential well. This is not unexpected, since gravitational time dilation is just another way of looking at gravitational redshift. Even if $p$ is at a local extremum of the potential, a photon emitted from $p$ must do some net work to climb out of the well in order to escape to infinity, becoming redshifted in the process.

Asking the original question for photons, this analysis breaks down. Fundamentally, one can never transform into the rest frame of a photon, and one can never ask about its proper time. We could ask about a photon's trajectory by restoring the other terms in the metric and setting the whole thing equal to zero, $$0 = \mathrm{d}\tau^2 = \left(1 + \frac{2\Phi}{c^2}\right) \mathrm{d}t^2 + \text{(terms involving \mathrm{d}x, \mathrm{d}y, \mathrm{d}z)},$$ but as soon as it starts moving (and it always must move) $\Phi$ will change, making the problem rather more difficult and in need of further specification as to what exact quantity you really want to calculate.

1In fact this is an approximation, but it is very good unless $p$ is within a few Schwarzschild radii of the masses, which can only even happen if they are black holes or neutron stars. See also the gravitational redshift article, though be warned it may have sign errors.