[Physics] A clock in freefall

general-relativitytime-dilation

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

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:
$$t_0=t_f$$

Best 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 $d\theta = d\phi = 0$ and the Schwarzschild metric simplifies to:

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

$d\tau$ 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 $d\tau/dt$, 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:

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

And rearranging this gives:

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

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:

$$ \frac{d\tau}{dt} = 1 - \frac{r_s}{r} \tag{3} $$

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