# A clock in freefall

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: $$t_0=t_f$$

• This is not homework. What I really want to know is if in this case $t_0=t_f$. – aepryus Aug 10 '14 at 12:46
• 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)? – ACuriousMind Aug 10 '14 at 12:50
• @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 $t_0=t_f$. – aepryus Aug 10 '14 at 12:54
• @ACuriousMind I am wondering if a clock in freefall experiences time dilation at all, relative to a clock at infinity. – aepryus Aug 10 '14 at 12:57
• 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. – John Rennie 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:

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

• Actually, I think I'll give it a go. – aepryus Aug 10 '14 at 15:31
• Perhaps based on the above analysis this isn't necessary? – aepryus Aug 10 '14 at 23:43
• 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. – aepryus Aug 11 '14 at 6:03
• @aepryus: I realised the calculation was easier than I thought, so I've completely rewritten my answer to do it. – John Rennie Aug 11 '14 at 15:39
• @aepryus: are you numerically solving Einstein's equation? That has traditionally required supercomputers, unless you have some symmetry to reduce the order of the equation. If you're just solving for time dialation factors and orbits of observers, it's at worst a system of four ODEs. – Jerry Schirmer Aug 11 '14 at 16:19