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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: $$t_0=t_f$$

aepryus
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    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
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    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
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    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
  • I'm closing this because it reads like a homework-like question as is, but aepryus, I think you can edit it a little to make it not be homework-like. For instance, you might edit the question to focus more on what you really want to ask, i.e. whether $t_0 = t_f$, and also describe some of what you did on your own to try to figure it out. – David Z Aug 10 '14 at 16:58
  • Interesting question. I don't see a good reason to close. – Halfdan Faber Aug 10 '14 at 17:42

2 Answers2

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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:

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

Community
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John Rennie
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    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 06:03
  • @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. – Zo the Relativist Aug 11 '14 at 16:19
  • @JerrySchirmer: since you're around - is my equation (3) really correct? It seems suspiciously simple. The working looks fine though ... – John Rennie Aug 11 '14 at 16:20
  • @JohnRennie: your work looks right. THe fact that you require the object to fall from infinity rather than specifying something like the energy of the orbit or the maximum radius is certainly a simplification, so I'm not horribly surprised that you got a simple answer. And we have an exact closed form answer for $t(\tau)$ and $r(\tau)$ for radial orbits, so this is certainly the simplest case possible. – Zo the Relativist Aug 11 '14 at 16:26
  • So when we reach $r=r_s$ time stops ${\rm d}\tau =0$ and we are moving at the speed of light $v=c$. But since a massive object can never reach $c$, so we will never reach the radius $r_s$ and hence will never actually fall into a black hole ... ? – John Alexiou Aug 11 '14 at 17:32
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    @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. – John Rennie Aug 12 '14 at 04:41
  • I have a string of comments... First, thanks a lot for analysis - it is extremely helpful. Second, I too find this result "surprising". I'm still a little confused by the homework hub-bub, but is this a common homework question? Given the result, it seems this would be a famous problem, but, for what it's worth, have never heard of it before. – aepryus Aug 12 '14 at 05:20
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    @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 ... – John Rennie Aug 12 '14 at 05:27
  • ... a very important point. Beginners to SR tend to think of time dilstion as a simple thing with a simple formula. Then they get massively confused about the twin paradox. It's important to understand that time dilation comes from the metric, and I think you question neatly illustrates how the metric can be used to calculate it. – John Rennie Aug 12 '14 at 05:29
  • Third, I realize for most cases $r_s/r$ is going to be very small, but wonder if perhaps that term also should go away. I will try to reconcile your math with my simplistic analysis above, but wonder if this is perhaps some sort of artifact of the Schwarzschild coordinates. My coordinate transformations are a little rusty but does $dr=vdt$ hold in the Schwarzschild coordinates or could there be an additional term that would eliminate the last $r_s/r$? – aepryus Aug 12 '14 at 05:29
  • @aepryus: if you measure $r$ in a coordinate system and $t$ in the same coordinate system then by definition $v = dr/dt$ in that coordinate system. – John Rennie Aug 12 '14 at 05:31
  • Ok, I'll go through this and try to figure out where the discrepancy lies - And lastly, I apologize for the heresy, but I really have come up with an underlying model that cleanly derives all of the phenomena of relativity except for one, the twin paradox. It has given me much intuition about SR/GR, for example leading to this question. It resolves the twin paradox by making the v of the Lorentz transformations always relative to the frame co-moving with the underlying spacetime at any point. My previous question asking about experimental confirmation of the twin paradox hasn't been helpful. – aepryus Aug 12 '14 at 05:50
  • @aepryus: you draw the wrong conclusion from the comparison between the two time dilation formulae in your question. The two formulae are related only in that they are both derived from a metric, but it's a different metric in the two cases. You should have a go at deriving the SR time dilation equation from the Minkowski metric. The basic principle is the same as I've used in my answer. – John Rennie Aug 12 '14 at 06:10
  • @aepryus: post a question if you need a clue. If you do ask emphasise you're trying to understand how time dilation derives from a metric so it doesn't get flagged as homwork! – John Rennie Aug 12 '14 at 06:59
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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.

ProfRob
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  • 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 =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? – aepryus Nov 29 '22 at 05:25
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    @aepryus the sign of the velocity is unimportant (as is also true in SR). – ProfRob Nov 29 '22 at 07:36