A clock in freefall

Question

Time dilation calculated using Schwarzschild metric for a non rotating spherical body is:
$$t_0=t_fsqrt{1-frac{2GM}{rc^2}}$$

For such a non rotating spherical body, what would the time dilation of a clock in vacuum free-falling from infinity be?  (If the answer is non-trivial; a high level outline of the calculation would suffice / be appreciated)

Edit:
I am currently working on an iOS app that is trying to model the mechanism underpinning relativity.  So, far the mechanism that I have created is shockingly simple and shockingly good at conforming to Relativity.  However, I am trying to break it.  I am trying to find any possible areas where the two may diverge.  I have noted that using my model a clock in freefall will experience no time dilation, i.e. $t_0=t_f$ and I want to make sure Relativity agrees.

I have noted the gravitational component of time dilation above.  Since my clock is moving one might also expect a kinematic time dilation.  I can calculate the velocity of my clock:
$$E_k=frac{1}{2}mv^2$$
$$E_p=frac{-GMm}{r}$$
$$v=sqrt{frac{2GM}{r}}$$
Plugging this velocity into the kinematic time dilation equation:
$$Delta t'=frac{Delta t}{sqrt{1-frac{v^2}{c^2}}}$$
$$Delta t'=frac{Delta t}{sqrt{1-frac{2GM}{rc^2}}}$$

At this point one might make the observation that the kinematic dilation is the inverse of the gravitational dilation and therefore conclude that:
$$t_0=t_f$$

John Rennie · Accepted Answer

This is how to calculate the time dilation for an object moving at velocity $v$ in a radial direction towards or away from the black hole.

Because the object is moving radially $dtheta = dphi = 0$ and the Schwarzschild metric simplifies to:

$$ c^2dtau^2 = c^2left(1-frac{r_s}{r}right)dt^2 - frac{dr^2}{1-r_s/r} tag{1}$$

$dtau$ is the proper time, and this corresponds to the time shown on the falling objects clock. $dt$ and $dr$ and the time and radial displacement measured by the distant observer. The time dilation is $dtau/dt$, and to calculate this we have to note that if the velocity measured by the Schwarzschild observer is $v$ then $dr = vdt$. Substituting this into equation (1) we get:

$$ c^2dtau^2 = c^2left(1-frac{r_s}{r}right)dt^2 - frac{v^2dt^2}{1-r_s/r} $$

And rearranging this gives:

$$ left(frac{dtau}{dt}right)^2 = 1 - frac{r_s}{r} - frac{v^2}{c^2}frac{1}{1-r_s/r} tag{2} $$

I've left $v$ in the equation. To eliminate $v$ you need to use the expression relating $v$ to $r$ for an object free-falling from infinity:

$$ frac{v}{c} = - left( 1 - frac{r_s}{r} right) left( frac{r_s}{r} right)^{1/2} $$

I'll leave the working as an exercise for the reader. The rather surprising result after we've done the substitution is:

$$ frac{dtau}{dt} = 1 - frac{r_s}{r} tag{3} $$

ProfRob · Answer

Simpler:
A constant of motion for an inertial observer in the Schwarzschild metric:
$$ left(1 - frac{r_s}{r}right)frac{dt}{dtau} = frac{E}{mc^2} .$$
For an observer starting at rest at infinity then $E/mc^2=1$, so
$$ dtau = left(1 - frac{r_s}{r}right) dt$$
Thus, on a very fundamental level, a clock in freefall experiences time dilation, whether it starts from rest or not.

A clock in freefall

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