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: $$t_0=t_f$$

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    $\begingroup$ This is not homework. What I really want to know is if in this case $t_0=t_f$. $\endgroup$
    – aepryus
    Aug 10 '14 at 12:46
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    $\begingroup$ Have a look at our homework policy. It is a specific question, where the value would lie in understanding the method by which one arrives at the solution, and thus homework-like. Also, are you just asking if a clock at infinity experiences time dilation compared to a clock at infinity (since that's what $t_f$ is from the Wiki article)? $\endgroup$
    – ACuriousMind
    Aug 10 '14 at 12:50
  • $\begingroup$ @ACuriousMind I'm not interested in the solution, but rather a discussion of the relevant physics impacting the situation. At some level, I really just want a yes or no to the question: Is $t_0=t_f$. $\endgroup$
    – aepryus
    Aug 10 '14 at 12:54
  • $\begingroup$ @ACuriousMind I am wondering if a clock in freefall experiences time dilation at all, relative to a clock at infinity. $\endgroup$
    – aepryus
    Aug 10 '14 at 12:57
  • $\begingroup$ I don't think this is a homework question. I think there is an interesting underlying concept of comparing coordinate to proper time. Have a +1 from me for the question, and if you feel up to integrating $r(t)$ numerically and posting the answer here I'll +1 that as well. $\endgroup$ Aug 10 '14 at 15:10

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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    $\begingroup$ Actually, I think I'll give it a go. $\endgroup$
    – aepryus
    Aug 10 '14 at 15:31
  • $\begingroup$ Perhaps based on the above analysis this isn't necessary? $\endgroup$
    – aepryus
    Aug 10 '14 at 23:43
  • $\begingroup$ Well, aside from this particular question. I do need to start wading back into the deep end with PDEs and both analytic and numeric solutions. This might be a reasonable place to start. $\endgroup$
    – aepryus
    Aug 11 '14 at 6:03
  • $\begingroup$ @aepryus: I realised the calculation was easier than I thought, so I've completely rewritten my answer to do it. $\endgroup$ Aug 11 '14 at 15:39
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    $\begingroup$ @ja72: in the above $v$ is the velocity measured by the Schwarzschild observer, and $v \rightarrow 0$ as $r \rightarrow r_s$. So you will indeed never cross the event horizon. This is a well known result and discussed in many, many questions on this site. $\endgroup$ Aug 12 '14 at 4:41


A constant of motion for an inertial observer in the Schwarzschild metric: $$ \left(1 - \frac{r_s}{r}\right)\frac{dt}{d\tau} = \frac{E}{mc^2}\ .$$

For an observer starting at rest at infinity then $E/mc^2=1$, so $$ d\tau = \left(1 - \frac{r_s}{r}\right) dt$$

Thus, on a very fundamental level, a clock in freefall experiences time dilation, whether it starts from rest or not.


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