# 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 $t_0=t_f$. Aug 10, 2014 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)? Aug 10, 2014 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 $t_0=t_f$. Aug 10, 2014 at 12:54
• @ACuriousMind I am wondering if a clock in freefall experiences time dilation at all, relative to a clock at infinity. Aug 10, 2014 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. Aug 10, 2014 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. Aug 10, 2014 at 15:31
• Perhaps based on the above analysis this isn't necessary? Aug 10, 2014 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. Aug 11, 2014 at 6:03
• @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. Aug 12, 2014 at 4:41
• @aepryus: the homework tag is meant to indicate that the working is straightforward for any reasonably experienced physicist. For example we get endless questions about blocks sliding down slopes etc, and solving these is just routine mechanics. The point is not that the question really was homework, but that the working is routine and therefore uninteresting. In this case I suspect the VTCers underestimated the subtlety of the question. Even though I've done this sort of thing before it took me a day to realise how it could be answered. I also think the question helps illustrate ... Aug 12, 2014 at 5:27

Simpler:

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.

• Just to confirm; based on this is the time dilation of a clock blasted from the surface at escape velocity (so that it reaches infinity at 𝑣=0) exactly the same as the falling clock's time dilation? I.e., is the time dilation the same for both clocks at any given r? Nov 29, 2022 at 5:25
• @aepryus the sign of the velocity is unimportant (as is also true in SR). Nov 29, 2022 at 7:36