# Free fall orbit time dilation

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stevendaryl
Staff Emeritus
I am happy with them being brought together though, else there might be something to do with the elliptical orbit that has not been made clear. As you seemed to be able to conclude that B's clock would have gone faster, but in the bit you had quoted of what I had written, I had not mentioned which was in lower orbit, or their respective velocities. Any affects of bringing together tends to insignificant the longer A and B orbit anyway. If B's clock was in the lower orbit then time dilation due to gravity would slow B's clock relative to A's, and presumably that affect is invariant across frames of reference. So would there not just be the kinematic time dilation left, and would not A think, using the metric of A's clock, that B's clock had ticked less than would have been expected (taking gravitational time dilation into account) if it had been in A's rest frame, and B, using the metric of B's clock, think that A's clock had ticked less than would have been expected (taking gravitational time dilation into account) if A had been in B's rest frame?
To a pretty good accuracy, the elapsed time $\delta \tau$ on a clock in orbit is given by:

$\delta \tau = (1 - \frac{GM}{c^2 r} - \frac{1}{2} \frac{v^2}{c^2}) dt$

where $t$ is coordinate time (using the most natural coordinate system). For a perfectly circular orbit, we have, approximately:

$\frac{v^2}{c^2} = \frac{GM}{c^2 r}$

So $\delta \tau = (1 - \frac{3}{2} \frac{GM}{c^2 r}) dt$

So higher orbits correspond to smaller $\frac{1}{r}$, which correspond to more elapsed time. You can think of the factor of $\frac{3}{2}$ as being a sum of a factor of $1$ due to gravitational time dilation and a factor of $\frac{1}{2}$ due to velocity-dependent time dilation. I don't know if that answers the question, or not.

Dale
Mentor
As I mentioned earlier to Vanhees earlier

Could you possibly just use that equation (which would seem to show no change to the x, y, z coords) using a single coordinate (t=0, x = 1, y = 1, z = 1) at rest with the observer standing on Earth to illustrate how it shows proper acceleration?
I could, but it would be more productive if you did it yourself. The metric and the Christoffel symbols for a Schwarzschild spacetime are given in the text.

To a pretty good accuracy, the elapsed time $\delta \tau$ on a clock in orbit is given by:

$\delta \tau = (1 - \frac{GM}{c^2 r} - \frac{1}{2} \frac{v^2}{c^2}) dt$

where $t$ is coordinate time (using the most natural coordinate system). For a perfectly circular orbit, we have, approximately:

$\frac{v^2}{c^2} = \frac{GM}{c^2 r}$

So $\delta \tau = (1 - \frac{3}{2} \frac{GM}{c^2 r}) dt$

So higher orbits correspond to smaller $\frac{1}{r}$, which correspond to more elapsed time. You can think of the factor of $\frac{3}{2}$ as being a sum of a factor of $1$ due to gravitational time dilation and a factor of $\frac{1}{2}$ due to velocity-dependent time dilation. I don't know if that answers the question, or not.
Would you mind answering the question that I gave?

Any affects of bringing together tends to insignificant the longer A and B orbit anyway. If B's clock was in the lower orbit then time dilation due to gravity would slow B's clock relative to A's, and presumably that affect is invariant across frames of reference. So would there not just be the kinematic time dilation left, and would not A think, using the metric of A's clock, that B's clock had ticked less than would have been expected (taking gravitational time dilation into account) if it had been in A's rest frame, and B, using the metric of B's clock, think that A's clock had ticked less than would have been expected (taking gravitational time dilation into account) if A had been in B's rest frame?
Also with the equation you gave, is GM the gravitational mass? And r the radius from the mass? Also I did not understand the step you made where you said that for a perfectly circular orbit you have approximately

$\frac{v^2}{c^2} = \frac{GM}{c^2 r}$

Because would that not depend on the velocity they were orbiting at relative to the gravitational mass being orbited. With that equality the equation could be written as

$\delta \tau = (1 - \frac{3}{2}\frac{v^2}{c^2}) dt$

But consider where the mass was large, and the velocity of the orbiting object low. Where is the gravitational mass taken into account in the version I just gave, and where is the velocity taken into account in the version you gave?

Could you perhaps give a simple worked example, because my maths is quite basic, I have to often look up the symbols, and I am not currently clear how the equation you gave works.

I could, but it would be more productive if you did it yourself. The metric and the Christoffel symbols for a Schwarzschild spacetime are given in the text.
I think it would be easier for me (and other readers at this basic level) if you just gave the simple worked example. Lots of people I am sure find worked examples an aid in understanding. And I cannot understand how with the equation you gave there would be any change in the x, y, z coordinates in the example I gave. Regarding working through it myself, I am not clear on what equation 1.13 means for example. There are two new symbols introduced, and I am not clear what they mean. Is T and exponent, or a superscript? Is η another matrix?

Even in 1.1 I am not clear why it is not written as

${(∆s)^ 2} = {(∆x)^ 2} + {(∆y)^ 2}$

Reading through myself I have to try to work out what the symbols are referring to. A simple worked example makes it clear, and then I (and others) can use the equations to work out other things ourselves, once we are shown how they are used.

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Ibix
But consider where the mass was large, and the velocity of the orbiting object low.
That's not an orbiting object. That's a falling object.

That's not an orbiting object. That's a falling object.
I am not sure what "that" refers to, something in the equation, or something in the example I gave. In the example I gave the satellites are in free fall orbiting a mass.

Ibix
I am not sure what "that" refers to, something in the equation, or something in the example I gave. In the example I gave the satellites are in free fall orbiting a mass.
If they're moving slowly near a large mass, they're not orbiting, they're falling. If you want to consider objects in circular orbits then they have to be moving at orbital speed - the formula that Steven quoted.

If they're moving slowly near a large mass, they're not orbiting, they're falling. If you want to consider objects in circular orbits then they have to be moving at orbital speed - the formula that Steven quoted.
Well slow is a pretty relative term (given the speed of light), and they can be in free fall and orbiting can they not? Can you explain how the orbital speed is worked out using that equation, and perhaps explain how it relates to this question:

Any affects of bringing together tends to insignificant the longer A and B orbit anyway. If B's clock was in the lower orbit then time dilation due to gravity would slow B's clock relative to A's, and presumably that affect is invariant across frames of reference. So would there not just be the kinematic time dilation left, and would not A think, using the metric of A's clock, that B's clock had ticked less than would have been expected (taking gravitational time dilation into account) if it had been in A's rest frame, and B, using the metric of B's clock, think that A's clock had ticked less than would have been expected (taking gravitational time dilation into account) if A had been in B's rest frame?

Ibix
You could always look up orbital velocity on wikipedia. The derivation is about two lines long...

You could always look up orbital velocity on wikipedia. The derivation is about two lines long...
I didn't see the relevance of orbital velocity to the question (which Steven was replying to), (I thought his equation was to do with elapsed time on a clock):

Any affects of bringing together tends to insignificant the longer A and B orbit anyway. If B's clock was in the lower orbit then time dilation due to gravity would slow B's clock relative to A's, and presumably that affect is invariant across frames of reference. So would there not just be the kinematic time dilation left, and would not A think, using the metric of A's clock, that B's clock had ticked less than would have been expected (taking gravitational time dilation into account) if it had been in A's rest frame, and B, using the metric of B's clock, think that A's clock had ticked less than would have been expected (taking gravitational time dilation into account) if A had been in B's rest frame?

Could you explain its relevance, and if not perhaps give your own answer?

Ibix
If you don't understand why the velocity is relevant you need to read stevendarryl's last sentence. If you don't understand why orbital velocity is relevant you need to read your own posts where you talk about the satellites being in orbit.
If you do both those things you can answer your own question from Steven's post #76.

If you don't understand why the velocity is relevant you need to read stevendarryl's last sentence. If you don't understand why orbital velocity is relevant you need to read your own posts where you talk about the satellites being in orbit.
If you do both those things you can answer your own question from Steven's post #76.
I was not saying I did not know why the velocity the satellite was orbiting was relevant (I assumed it would be involved in the kinematic time dilation calculation). I just did not understand why the minimum velocity it could orbit before it would no longer be in free fall orbit, but just falling towards the mass was relevant. It could have just been assumed it was above that.

I explained in post #78 why I did not understand Steven's answer, and you seemed to bring up the minimum velocity it could orbit before it would free fall into the mass, which I did not understand the relevance of. The question below

I didn't see the relevance of orbital velocity to the question (which Steven was replying to), (I thought his equation was to do with elapsed time on a clock):

Any affects of bringing together tends to insignificant the longer A and B orbit anyway. If B's clock was in the lower orbit then time dilation due to gravity would slow B's clock relative to A's, and presumably that affect is invariant across frames of reference. So would there not just be the kinematic time dilation left, and would not A think, using the metric of A's clock, that B's clock had ticked less than would have been expected (taking gravitational time dilation into account) if it had been in A's rest frame, and B, using the metric of B's clock, think that A's clock had ticked less than would have been expected (taking gravitational time dilation into account) if A had been in B's rest frame?
Is simply a "yes" or "no" answer, and it has been asked on this thread before (it can be seen in a different form in #1). Could you possibly give a "yes" or "no" response to it, and perhaps some explanation if "no" (given that the kinematic time dilation would as I understand it be like that in flat spacetime)?

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PeterDonis
Mentor
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