Resultant time dilation from both gravity and motion

[espen180]

So what? his derivation is wrong just the same.

You are missing the point. Your original claim that kev is using the wrong metric is false. Now that we have established that there is nothing wrong with the definitions, please point to the spesific place the error occurs, and preferably propose the correct result is its place.

 

[JesseM]

Not relevant.

You seemed to think it was relevant before when you said "Do you now understand what my objection is to your citing the inappropriate material for answering Dmitry7's OP?" The problem is that rather than sticking to a single criticism, you keep changing your line of attack, never really admitting that you made any mistakes in your previous attacks, as if you somehow believe that as long as you can show kev was wrong in some way, you have "won", even if the way you finally decide he is wrong had not even occurred to you at the moment you started attacking his post. Your latest criticism in post #41 about angles isn't any better than your previous attacks. The Schwarzschild metric is spherically symmetric, so although the fact that \theta only ranges from 0 to \pi means you can't have a full orbit with constant r and \phi, the time dilation equation is only talking about the instantaneous rate a clock is ticking relative to a clock at infinity over an infinitesimally short section of its orbit. It is certainly possible to have a circular orbit which for one half of the orbit has \theta varying from 0 to \pi while r has a constant value of R and \phi has a constant value of \pi/2 (so for any infinitesimal section of an orbiting object's worldline whose endpoints lie on this half of the orbit, dr and d\phi would be 0), while the other half of the orbit also has \theta varying from 0 to \pi and r having a constant value of R, but now with \phi having a constant value of -\pi/2 (so for any infinitesimal section of an orbiting object's worldline whose endpoints lie on this half of the orbit, dr and d\phi would still be 0). kev's derivation would work just fine in this case.

 

[starthaus]

You seemed to think it was relevant before when you said "Do you now understand what my objection is to your citing the inappropriate material for answering Dmitry7's OP?" The problem is that rather than sticking to a single criticism, you keep changing your line of attack, never really admitting that you made any mistakes in your previous attacks, as if you somehow believe that as long as you can show kev was wrong in some way, you have "won", even if the way you finally decide he is wrong had not even occurred to you at the moment you started attacking his post. Your latest criticism in post #41 about angles isn't any better than your previous attacks.

His derivation is a hack and you've been doing your darnest to defend it. Why is it so difficult for you to admit that it is wrong?

The Schwarzschild metric is spherically symmetric, so although the fact that \theta only ranges from 0 to \pi means you can't have a full orbit with constant r and \phi, the time dilation equation is only talking about the instantaneous rate a clock is ticking relative to a clock at infinity over an infinitesimally short section of its orbit. It is certainly possible to have a circular orbit which for one half of the orbit has \theta varying from 0 to \pi while r has a constant value of R and \phi has a constant value of \pi/2 (so for any infinitesimal section of an orbiting object's worldline whose endpoints lie on this half of the orbit, dr and d\phi would be 0), while the other half of the orbit also has \theta varying from 0 to \pi and r having a constant value of R, but now with \phi having a constant value of -\pi/2 (so for any infinitesimal section of an orbiting object's worldline whose endpoints lie on this half of the orbit, dr and d\phi would still be 0). kev's derivation would work just fine in this case.

Please read here. Sorry, but no matter how hard you may try, \omega is not \frac{d\theta}{dt}
We are talking about rigotous derivations, not about hacks, right?

 

[espen180]

Please read here.
We are talking about rigotous derivations, not about hacks, right?

Everything in that post has been adressed above.

Regarding "hacks", I would like to hear your definition of one, and why you think using algebra is "against the rules" if they don't conform to your rules (also state those rules, please).

 

[starthaus]

Everything in that post has been adressed above.

Regarding "hacks", I would like to hear your definition of one, and why you think using algebra is "against the rules" if they don't conform to your rules (also state those rules, please).

You mean using algebra badly? Like in truncating the metric by missing non-null terms? You claimed that you knew the metric by heart.
Like in using the wrong definition of angular speed?

 

[JesseM]

His derivation is a hack and you've been doing your darnest to defend it. Why is it so difficult for you to admit that it is wrong?

But currently your only basis for saying it's wrong is the argument in post #41. Regardless of whether that argument is valid, can you not admit that all your previous unrelated arguments which had nothing to do with \phi vs. \theta were on the wrong track?

Please read here.
We are talking about rigotous derivations, not about hacks, right?

Yes, I already did read that post, it's the same argument as the one I was responding to when I referred to "Your latest criticism in post #41 about angles". It is perfectly "rigorous" to consider a circular orbit which has constant r=R and constant \phi=\pi/2 for one half, and constant r=R and constant \phi=-\pi/2 for the other half, so that for any infinitesimal section of the object's worldline on either half, dr = d\phi = 0; do you deny that such an orbit should be physically possible in the Schwarzschild spacetime? It may be true that for the purposes of a derivation, it might be a bit more "elegant" to consider a different orbit where r and \theta remain constant for the whole orbit, but there's nothing physically wrong or non-rigorous about the way kev did it.

 

[starthaus]

Yes, I already did read that post, that's exactly what I was responding to above. It is perfectly "rigorous" to consider a circular orbit which has constant r=R and constant \phi=\pi/2 for one half, and constant r=R and constant \phi=-\pi/2 for the other half, so that for any infinitesimal section of the object's worldline on either half, dr = d\phi = 0; do you deny that such an orbit should be physically possible in the Schwarzschild spacetime? It may be true that for the purposes of a derivation,

What about the missing terms in \phi? What about the v=r\frac{d\theta}{dt}. Wouldn't it be easier for you to admit that you are backing the wrong formulas rather than patching in all kinds of special pleads? I gave you the correct general formula, it does not agree with kev's formula. I gave you the general derivation, it does not agree with the pervect/kev derivation. Can you at least decide which is right and which is wrong?

it might be a bit more "elegant" to consider a different orbit where r and \theta remain constant for the whole orbit, but there's nothing physically wrong or non-rigorous about the way kev did it.

Isn't this the problem that needs to be solved? Isn't this the problem I solved at post 2?

 

[espen180]

In #2 you gave the metric, which of course is at the heart of all the results suggested in the thread. Personally I am unsure where the problem is. For my part, I choose dr/dt=0 and theta=pi/2 at the beginning of the derivation, but the argument here was that keeping these zero is not neccesary. The calculation is just as valid, for example at the apogee of the particle's trajectory, is what I think JesseM meant.

The other disputes have been focused on individual pieces of the derivation, like how to treat the metric or how to define certain variables.

 

[JesseM]

What about the missing terms in \phi?

It's true that kev did not write out the full metric, but given that he was assuming an orbit where for any infinitesimal segment you'd have d\phi = 0, those extra terms would disappear anyway so this wouldn't affect his final results. And kev never claimed he was starting from the full metric, he said in post #8 that he was "Starting with this equation given by pervect", and pervect had already eliminated terms that went to zero.

What about the v=r\frac{d\theta}{dt}.

What about it? That would appear to be an equation for Schwarzschild coordinate velocity (as opposed to kev's 'local velocity') for an object in circular orbit with varying \theta coordinate, as with the type of orbit I described--again, do you agree that the type of orbit I described is a physically valid one? If you agree there would be a valid physical orbit with that type of coordinate description (with \phi having one constant value for half the orbit and a different constant value for the other half), do you disagree that the above equation would be the correct coordinate velocity for an object in this orbit?

Wouldn't it be easier for you to admit that you are backing the wrong formulas rather than patching in all kinds of special pleads?

I think you don't understand what http://www.nizkor.org/features/fallacies/special-pleading.html is, the fact that I and others respond to each of your various arguments with counterarguments, resulting in you continually abandoning your previous arguments in favor of new arguments you have invented on the spot, does not qualify as "special pleading". Yes or no, do you acknowledge that the arguments you made against kev's derivation prior to the new argument you've made in the posts here and here were flawed?

I gave you the correct general formula, it does not agree with kev's formula.

kev's formula is not intended to be a "general" one for arbitrary motion, it deals specifically with the case of an object in circular orbit. And since the OP was asking about whether total time dilation was a sum of gravitational and velocity-based time dilation, I thought it would be interesting to point out that for this specific case, total time dilation was actually a product of the two (whereas your more general formula does not relate in any obvious way to the formulas for gravitational and velocity-based time dilation)

I gave you the general derivation, it does not agree with the pervect/kev derivation.

Do you deny that the general formula would reduce to the specific formulas found by pervect/kev in the specific case they were considering, namely an infinitesimal section of a circular orbit where the radial coordinate and one of the two angular coordinates are constant?

Can you at least decide which is right and which is wrong?

If a general formula reduces to a more specific formula under the specific conditions assumed in the derivation of the specific formula, I'd say that both are right.

 

[starthaus]

whereas your more general formula does not relate in any obvious way to the formulas for gravitational and velocity-based time dilation

Are you even reading what I am writing? Can you re-read posts 2,6,39,47?

 

[starthaus]

A more general equation is:

\frac{\text{d}\tau}{\text{d}t}= \sqrt{\frac{1-r_s/r}{1-r_s/r_o}} \sqrt{1- \left (\frac{1-r_s/r_o}{1-r_s/r} \right )^2 \left (\frac{\text{d}r}{c\text{d}t} \right )^2 - \left (\frac{1-r_s/r_o}{1-r_s/r} \right ) \left(\frac{r \text{d}\theta}{c\text{d}t}\right)^2 - \left (\frac{1-r_s/r_o}{1-r_s/r} \right) \left(\frac{r \sin \theta \text{d}\phi}{c\text{d}t}\right)^2 }

where r_o is the Schwarzschild radial coordinate of the stationary observer and r is the Schwarzschild radial coordinate of the test particle and dr and dt are understood to be measurements made by the stationary observer at r_o in this particular equation.

For r_o = r the time dilation ratio is:

\frac{\text{d}\tau}{\text{d}t} = \sqrt{1-\frac{v'^2}{c^2}}

No, you are missing a lot of terms.

 

[JesseM]

Are you even reading what I am writing? Can you re-read posts 2,6,39,47?

Post 47 was by espen180, and as to the others, yes you derived equations for special cases that were closer to a product of gravitational time dilation and something else, but since you didn't use the concept of local velocity the "something else" (i.e. the second part of the product that didn't look like the gravitational time dilation equation) didn't look very much like the SR velocity-based time dilation equation. Since espen180's original post was asking about the total time dilation in relation to the gravitational time dilation and velocity-based time dilation formulas, I thought kev and pervect's equations were relevant to the OP. Again, I'm not saying your equations are wrong in any way, but you haven't made a convincing case that kev's are wrong either--if you still maintain that, please answer the questions in my previous post.

 

[starthaus]

Post 47 was by espen180, and as to the others, yes you derived equations for special cases that were closer to a product of gravitational time dilation and something else, but since you didn't use the concept of local velocity the "something else" (i.e. the second part of the product that didn't look like the gravitational time dilation equation) didn't look very much like the SR velocity-based time dilation equation.

Because it isn't. Both effects are GR effects, in fact, there is only one effect. There is no such thing as an SR effect. The effect falls out the Schwarzschild metric.I have already explained this here

Since espen180's original post was asking about the total time dilation in relation to the gravitational time dilation and velocity-based time dilation formulas, I thought kev and pervect's equations were relevant to the OP. Again, I'm not saying your equations are wrong in any way, but you haven't made a convincing case that kev's are wrong either--if you still maintain that, please answer the questions in my previous post.

I have answered your questions, I would like you to answer mine.

 

[JesseM]

Because it isn't. Both effects are GR effects. There is no such thing as an SR effect.

This is a strawman, I didn't say anything about "SR effect", I just said that the formula for total time dilation of an orbiting object calculated using GR (not SR) broke down into the product of two formulas that look like the formulas for gravitational time dilation for a stationary object and velocity-based time dilation for a moving object in SR.

I have answered your questions, I would like you to answer mine.

What questions of yours have I not answered? You didn't even ask any questions in this post! Anyway, there are several questions in that previous post #59 that I have asked variations on in the past and you have not answered, such as "again, do you agree that the type of orbit I described is a physically valid one?" (referring to the type of orbit I mentioned earlier in post 56) and "Yes or no, do you acknowledge that the arguments you made against kev's derivation prior to the new argument you've made in the posts here and here were flawed?" and "Do you deny that the general formula would reduce to the specific formulas found by pervect/kev in the specific case they were considering, namely an infinitesimal section of a circular orbit where the radial coordinate and one of the two angular coordinates are constant?" If you are interested in good-faith debate here, please answer the questions I ask you (and I will do likewise of course) rather than just picking one part of my post to criticize and ignoring everything else.

 

[Passionflower]

Because it isn't. Both effects are GR effects, in fact, there is only one effect. There is no such thing as an SR effect. The effect falls out the Schwarzschild metric.

That does not exclude the possibility to identify them as separate things.

For instance in the Gullstrand–Painlevé chart you can readily identify the Lorentz factor for relative motions.

But in non-stationary spacetimes this will obviously be a very daunting task.

 

[starthaus]

but since you didn't use the concept of local velocity

Because, contrary to kev's claims, the term \frac{dr/dt}{\sqrt{1-r_s/r}} does not represent "local velocity". It represents nothing. We had a very lengthy discussion in another thread, if you want the correct formula for local velocity you can find it in my blog, in the file "General Euler-Lagrange derivation for proper and coordinate acceleration".

the "something else" (i.e. the second part of the product that didn't look like the gravitational time dilation equation) didn't look very much like the SR velocity-based time dilation equation.

Because it doesn't. Because it has absolutely nothing to do with any "SR velocity-based time dilation".

What questions of yours have I not answered?

One very basic one: that the derivation and the result I gave are the rigorous, complete ansers. Yes or no?

You didn't even ask any questions in this post!

Sure I did, you need to look at the sentences ending with question marks.

Anyway, there are several questions in that previous post #59 that I have asked variations on in the past and you have not answered, such as "again, do you agree that the type of orbit I described is a physically valid one?" (referring to the type of orbit I mentioned earlier in post 56) and "Yes or no, do you acknowledge that the arguments you made against kev's derivation prior to the new argument you've made in the posts here and here were flawed?" and "Do you deny that the general formula would reduce to the specific formulas found by pervect/kev in the specific case they were considering, namely an infinitesimal section of a circular orbit where the radial coordinate and one of the two angular coordinates are constant?" If you are interested in good-faith debate here, please answer the questions I ask you (and I will do likewise of course) rather than just picking one part of my post to criticize and ignoring everything else.

If you are happy with hacks and with formulas that are correct for restrictive conditions , like only in the equatorial plane, the answer is yes. Happy?

 

[JesseM]

Because, contrary to kev's claims, the term \frac{dr/dt}{\sqrt{1-r_s/r}} does not represent "local velocity". It represents nothing. We had a very lengthy discussion in another thread, if you want the correct formula for local velocity you can find it in my blog.

What thread did you discuss it? Do you also dispute DrGreg's derivations which he linked to in post #8?

Because it doesn't. Because it has absolutely nothing to do with any "SR velocity-based time dilation".

"Has to do with" is a rather ill-defined phrase. I'd say that if it uses the same equation as "SR velocity-based time dilation", then it "has to do with" it in at least some limited sense.

One very basic one: that the derivation and the result I gave are the rigorous, complete ansers. Yes or no?

Your derivation seems rigorous but you'll have to define what you mean by "complete". Do you just mean that it's the most general case, and that all more specific answers would be derivable from it? If so then I agree. But if you're saying that from a pedagogical point of view it's "complete" in the sense that there's no point in discussing any more specific cases, I disagree, the more specific cases may be more helpful for gaining physical intuitions than the more general equation.

What questions of yours have I not answered? You didn't even ask any questions in this post!

Sure I did, you need to look at the sentences ending with question marks.

When I said "in this post" I meant the post I was responding to (the one where you said 'I have answered your questions, I would like you to answer mine'). There were no sentences ending with question marks in that post. If there were questions in other posts that you think I didn't address and you'd still like answers to, just point them out.

If you are happy with hacks and with formulas that are correct for restrictive conditions , like only in the equatorial plane, the answer is yes. Happy?

Not completely, because your comments about "hacks" and "restrictive conditions" still seem to imply you think there is something better about your own suggestion here that we should set d\theta to 0 rather than d\phi, when in fact this would be every bit as restrictive in terms of the set of circular orbits that would meet this condition--do you disagree that both are equally restrictive? Anyway, as I think espen180 pointed out earlier, because of the spherical symmetry of the Schwarzschild spacetime, for any circular orbit you can always do a simple coordinate transformation into a coordinate system that still has the same metric but where the circular orbit now meets this condition, so in fact kev's final equation should apply to arbitrary circular orbits. If you're familiar with the phrase without loss of generality in proofs, kev could have said "without loss of generality, assume we're dealing with a circular orbit where d\phi = 0" and a physicist would understand the implied argument about why the final results should apply to all circular orbits.

 

[starthaus]

What thread did you discuss it?

Here

Do you also dispute DrGreg's derivations which he linked to in post #8?

DrGreg's time dilation formula is a subset of mine, so "no". The point is that the quantity in discussion (v) is not what you and kev claim it is. The correct formula can be found in my blog.

"Has to do with" is a rather ill-defined phrase. I'd say that if it uses the same equation as "SR velocity-based time dilation", then it "has to do with" it in at least some limited sense.

That's what you said. And I explained to you that it has nothing to do with any "SR-based time dilation. ".

Your derivation seems rigorous

It either is or it isn't. Can you answer with yes or no, please?

but you'll have to define what you mean by "complete". Do you just mean that it's the most general case, and that all more specific answers would be derivable from it? If so then I agree.

Good. This is what I meant.

your comments about "hacks" and "restrictive conditions" still seem to imply you think there is something better about your own suggestion here that we should set d\theta to 0 rather than d\phi,

Absolutely.d \theta=0 means no motion along the meridian whereas d \phi=0 means no rotation, contradicting the problem statement. Why are we even discussing this?

when in fact this would be every bit as restrictive in terms of the set of circular orbits that would meet this condition--do you disagree that both are equally restrictive?

Of course I do, how many posts do we need to waste on this obvious issue?

Anyway, as I think espen180 pointed out earlier, because of the spherical symmetry of the Schwarzschild spacetime, for any circular orbit you can always do a simple coordinate transformation into a coordinate system that still has the same metric but where the circular orbit now meets this condition,

That's not the point. \phi and \theta are not intechangeable. They have different meanings , both mathematically and physically. They have different domains of definition, \omega is \frac{d\phi}{dt} (and not \frac{d\theta}{dt}). If you insist on interchanging them, you would need to exchange their domains of definition (\theta would need to be in the interval [0,2\pi]), you would also need to rewrite the Schwarzschild metric. This is not what kev did in his hack.

so in fact kev's final equation should apply to arbitrary circular orbits. If you're familiar with the phrase without loss of generality in proofs, kev could have said "without loss of generality, assume we're dealing with a circular orbit where d\phi = 0" and a physicist would understand the implied argument about why the final results should apply to all circular orbits.

This is incorrect, since \phi and \theta are not intechangeable.

 

[yuiop]

-The correct answer to Dmitry7's question is:

\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}\sqrt{\frac{1-(r_1sin\theta_1\omega/c\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega/c\sqrt{1-r_s/r_2})^2}}

-The correct answer to espen180's question is :


\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{r\omega sin(\theta)}{c \sqrt{1-r_s/r}})^2}

I hope that this clarifies things once and for all.

This is not the correct answer to the question posed by espen180 in the OP of this thread. Here it the original question again to remind you:

When a frame is moving in relation to an observer in his rest frame at infinity, and the frame is in a gravitational well, is the resultant time dilation simply the sum of the motional and gravitational dilation, e.g.

t=\tau\left(\gamma^{-1}+\gamma_g^{-1}\right)=\tau\left(\sqrt{1-\frac{v^2}{c^2}}+\sqrt{1-\frac{GM}{c^2r}}\right)

Where \tau is proper time and t is measured by the observer?

If, not what is the correct expression?

Espen is asking about the resultant time dilation due to motional and gravitational dilation. He asks about the contribution due to motion but does not specify that the motion should be orbital. You supposedly "correct answer" is for the limiting case of orbital motion only. This might seem a bit picky, but you have set the standard here:

If you are happy with hacks and with formulas that are correct for restrictive conditions , like only in the equatorial plane, the answer is yes. Happy?

You have put a restrictive condition of only considering orbital (horizontal) motion. The complete general and correct answer to the OP was given by DrGreg in #8 and I quote him again here:

I believe that the equation

<br /> \frac{dt}{d\tau} = \frac{1}{\sqrt{1-v^2/c^2}\sqrt{1 - 2GM/rc^2}}<br />​

always applies (for radial, tangential or any other motion) where v is speed relative to a local hovering observer using local proper distance and local proper time.

I derived this in posts #9 and #7 of the thread "Speed in general relativity"[/color] (and repeated in post #46).

DrGreg's conclusions agree with the conclusions of pervect and myself.


You seem to think that you equation given in #7:

\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{1}{c^2}\frac{dr/dt}{1-r_s/r})^2}

and my equation:

\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-\frac{(dr&#039;/dt&#039;)^2}{c^2}

are in disagreement, with yours right and mine wrong and fail to understand that they are numerically identical.

[EDIT] Well they would be numerically identical, when you correct the error in your equation. Your equation in #7 should read:

\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{1}{c}\frac{dr/dt}{(1-r_s/r)})^2}

 

[starthaus]

This is not the correct answer to the question posed by espen180 in the OP of this thread.

:lol:

DrGreg's conclusions agree with the conclusions of pervect and myself.

How do you get the "general solution" from the truncated metric you've been using? Run this by us again, please.

Espen is asking about the resultant time dilation due to motional and gravitational dilation. He asks about the contribution due to motion but does not specify that the motion should be orbital. You supposedly "correct answer" is for the limiting case of orbital motion only. This might seem a bit picky, but you have set the standard here:

If you don't understand the use of Schwarzschild solution in deriving the answer, then ask and I'll try to help you.

 

[starthaus]

[EDIT] Well they would be numerically identical, when you correct the error in your equation. Your equation in #7 should read:

\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{1}{c}\frac{dr/dt}{(1-r_s/r)})^2}

Congratulations, you found a missing closed/open parens after \frac{1}{c^2}. You are good at finding typos. :-)
The complete equation is obviously correct:

\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{1}{c^2}\frac{dr/dt}{1-r_s/r})^2}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{v/c}{1-r_s/r})^2}

 

[yuiop]

This is incorrect, since \phi and \theta are not intechangeable.

Let us say an exam question asks what is the time dilation ratio d\tau/dt of a particle moving in Schwarzschild coordinates with velocity components dr/dt = 0, r d\theta/dt = 0 and r d\phi/dt = 0.5 and numerical values for G, M, c and r are given.

Student A states that is not possible to give a numerical solution because the value of \theta has not been given.

Student B realizes the spherical symmetry of Schwarzschild coordinates allows him to re-orientate the axes of the coordinates and interchange d\phi and d\theta so that he can obtain a correct numerical solution, even though the value of \theta is not known.

Who deserves the most points for their answer? Student A who is plugging numbers into a formula or student B who is using understanding of the physical situation and using a bit of ingenuity to actually obtain the required numerical answer?

 

[starthaus]

Let us say an exam question asks what is the time dilation ratio d\tau/dt of a particle moving in Schwarzschild coordinates with velocity components dr/dt = 0, d\theta/dt = 0 and d\phi/dt = 0.5 and numerical values for G, M, c and r are given.

Student A states that is not possible to give a numerical solution because the value of \theta has not been given.

And he's right.

Student B realizes the spherical symmetry of Schwarzschild coordinates allows him to re-orientate the axes of the coordinates and interchange d\phi and d\theta so that he can obtain a correct numerical solution, even though the value of \theta is not known.

Try doing that and I'll show you where the mistake is.

Who deserves the most points for their answer? Student A who is plugging numbers into a formula or student B who is using understanding of the physical situation and using a bit of ingenuity to actually obtain the required numerical answer?

Neither, student C who knows how to use the general (not the truncated) Schwarzschild solution such that he/she produces the correct general solution that has \theta as a parameter. How would you do this , kev, without resorting to silly hacks?

 

[yuiop]

Why not just contract the angle differentials into \text{d}\theta^2+\sin^2\theta\text{d}\phi^2=\text{d}\Omega^2 and avoid the problem alltogether?

Because \phi and \theta are independent coordinates. So your hack is illegal.

LOL. In which countries is this illegal?

I have seen this equation given by espen used in enough references to be fairly certain it is legitimate and pretty much standard procedure.

 

[starthaus]

LOL. In which countries is this illegal?

I have seen this equation given by espen used in enough references to be fairly certain it is legitimate and pretty much standard procedure.

calculus is not your strong suit.

 

[JesseM]

DrGreg's time dilation formula is a subset of mine, so "no". The point is that the quantity in discussion (v) is not what you and kev claim it is.

In post #8 DrGreg said that his v stood for "speed relative to a local hovering observer using local proper distance and local proper time". Do you disagree with this definition? If not, what do you think kev "claims" that is different? (I haven't claimed anything about it myself, so the phrase 'what you and kev claim' is pure imagination on your part)

That's what you said. And I explained to you that it has nothing to do with any "SR-based time dilation. ".

I don't think you understood my point. I was saying that "has to do with" is a semantically ambiguous phrase, and that under one reasonable definition of "has to do with", the fact that the term in the total GR time dilation equation for circular orbits looks just like the SR time dilation equation would by definition mean it "has to do with" the SR time dilation equation, since "has to do with" can be defined in a broad way that does not imply any deeper connection besides a superficial similarity in equations. You certainly never "explained" why it doesn't have to do with SR time dilation, since you never gave any meaningful definition of the vague phrase "has to do with".

It either is or it isn't. Can you answer with yes or no, please?

The reason I said "seems" is that I looked at your answer in post #13 and got the gist of how you derived it, the approach seemed fine but I didn't check the details of the math, trusting you probably got it right. If you insist that I double-check your work in detail, fine:

<br /> (cd\tau)^2=(1-r_s/r)(cdt)^2-(1-r_s/r)^{-1}(dr)^2-(rd\theta)^2-(rd\phi sin\theta)^2<br />

d\tau /dt = (1/c)\sqrt{(1-r_s/r)c^2-(1-r_s/r)^{-1}(dr/dt)^2-(rd\theta/dt)^2-(rd\phi/dt sin\theta)^2}

Strictly speaking this equation is itself the most general answer for the time dilation for an object moving along an arbitrary worldline where dr/dt, d\theta/dt, and d\phi/dt might all be nonzero. But then if we introduce the condition that we are looking at a portion of an orbit where dr = d\theta = 0 as you did in post #6, and set \omega = d\phi/dt, then this reduces to:

d\tau /dt = (1/c)\sqrt{(1-r_s/r)c^2 - (r sin\theta \omega)^2}

Factoring out \sqrt{(1-r_s/r)c^2} gives:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin\theta \omega)^2 / (1-r_s/r)c^2}

or, to make it closer to the form you chose:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin\theta \omega/ c\sqrt{1-r_s/r})^2}

Then for the ratio of times for two clocks in circular paths with r₁ and r₂ and \theta_1 and \theta_2 and d\phi_1/dt = \omega_1 and d\phi_2/dt = \omega_2 we'd have:

<br /> \frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}\sqrt{\frac{1-(r_1sin\theta_1\omega_1/c\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega_2/c\sqrt{1-r_s/r_2})^2}}<br />

This is almost like the equation you got, but you do seem to have made the minor error of leaving out the "c" that appears in my equation (yours is not dimensionally correct, since \omega has units of 1/time), and also you just wrote \omega in both parts of the fraction, neglecting to account for the possibility that \omega_1 differs from \omega_2. An additional thing to note is that if you want to have a circular orbit where d\theta = 0 as opposed to an arbitrary circular path, you must pick \theta = 0, so that sin\theta = 0 and the whole thing reduces to \frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}; a circular path with d\theta = 0 but \theta not equal to zero would be one where the center of the path did not coincide with the center of the Schwarzschild coordinate system at r=0, like how a line of latitude on the Earth forms a circle whose center does not coincide with the center of the Earth (unless it's latitude 0, in which case the line of latitude would just be the equator), so this would not actually be a physically valid "orbit".

when in fact this would be every bit as restrictive in terms of the set of circular orbits that would meet this condition--do you disagree that both are equally restrictive?

Of course I do, how many posts do we need to waste on this obvious issue?

In what sense do you imagine that your suggestion of setting d\theta = 0 is less "restrictive" than setting d\phi = 0? Setting d\theta=0 is equivalent to picking a circle of constant latitude on a globe (latitude varies 180 degrees from the South Pole to the North Pole, just like the \theta coordinate varies over half a circle from 0 to \pi), like the first image on http://literacynet.org/sciencelincs/showcase/drifters/activity1b.html:

Whereas setting d\phi = 0 is equivalent to picking a circle of constant longitude on a globe (longitude varies 360 degrees from 180 west of the prime meridian to 180 east of it, just like \phi varies in a full circle from -\pi to \pi), like the second image:

It should be clear that all the infinite possible circles with d\phi = 0 have centers that coincide with the center of the coordinate system at r=0, whereas only a single circle with d\theta = 0 has a center that coincides with the center of the coordinate system (the one at \theta = 0). So if you're interested in circular orbits, d\theta = 0 seems more restrictive, not less restrictive! But either way you could also find an infinite number of circular orbits at different angles relative the coordinate system such that neither d\theta = 0 nor d\phi = 0 would apply.

Do you disagree with any of the above? If so, which part? If not, can you explain your reasoning behind calling d\theta = 0 the less "restrictive" condition?

That's not the point. \phi and \theta are not intechangeable. They have different meanings , both mathematically and physically. They have different domains of definition, \omega is \frac{d\phi}{dt} (and not \frac{d\theta}{dt}). If you insist on interchanging them, you would need to exchange their domains of definition (\theta would need to be in the interval [0,2\pi]), you would also need to rewrite the Schwarzschild metric.

No, you would not need to rewrite the Schwarzschild metric, that's exactly what I meant when I said "Anyway, as I think espen180 pointed out earlier, because of the spherical symmetry of the Schwarzschild spacetime, for any circular orbit you can always do a simple coordinate transformation into a coordinate system that still has the same metric but where the circular orbit now meets this condition". Just as the metric has exactly the same form in the different inertial coordinate systems used in flat spacetime, so it is also true that because of the spherical symmetry of the Schwarzschild metric, one can find a family of different spherical coordinate systems with the \theta = 0 and \phi = 0 axes oriented in different directions (all of them having the same r=0 point), but with the same Schwarzschild metric applying to all these coordinate systems. And for any circular orbit whose center is at r=0, there will be one member of this family of spherical coordinate systems where the orbit has a constant \theta coordinate, and another where the orbit can be divided into two halves that each have a constant \phi coordinate. Do you disagree?

 

[starthaus]

In post #8 DrGreg said that his v stood for "speed relative to a local hovering observer using local proper distance and local proper time". Do you disagree with this definition? If not, what do you think kev "claims" that is different? (I haven't claimed anything about it myself, so the phrase 'what you and kev claim' is pure imagination on your part)

I don't think you understood my point. I was saying that "has to do with" is a semantically ambiguous phrase, and that under one reasonable definition of "has to do with", the fact that the term in the total GR time dilation equation for circular orbits looks just like the SR time dilation equation would by definition mean it "has to do with" the SR time dilation equation, since "has to do with" can be defined in a broad way that does not imply any deeper connection besides a superficial similarity in equations. You certainly never "explained" why it doesn't have to do with SR time dilation, since you never gave any meaningful definition of the vague phrase "has to do with".

The reason I said "seems" is that I looked at your answer in post #13 and got the gist of how you derived it, the approach seemed fine but I didn't check the details of the math, trusting you probably got it right. If you insist that I double-check your work in detail, fine:

<br /> (cd\tau)^2=(1-r_s/r)(cdt)^2-(1-r_s/r)^{-1}(dr)^2-(rd\theta)^2-(rd\phi sin\theta)^2<br />

d\tau /dt = (1/c)\sqrt{(1-r_s/r)c^2-(1-r_s/r)^{-1}(dr/dt)^2-(rd\theta/dt)^2-(rd\phi/dt sin\theta)^2}

Strictly speaking this equation is itself the most general answer for the time dilation for an object moving along an arbitrary worldline where dr/dt, d\theta/dt, and d\phi/dt might all be nonzero. But then if we introduce the condition that we are looking at a portion of an orbit where dr = d\theta = 0 as you did in post #6, and set \omega = d\phi/dt, then this reduces to:

d\tau /dt = (1/c)\sqrt{(1-r_s/r)c^2 - (r sin\theta \omega)^2}

Factoring out \sqrt{(1-r_s/r)c^2} gives:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin\theta \omega)^2 / (1-r_s/r)c^2}

or, to make it closer to the form you chose:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin\theta \omega/ c\sqrt{1-r_s/r})^2}

Then for the ratio of times for two clocks in circular paths with r₁ and r₂ and \theta_1 and \theta_2 and d\phi_1/dt = \omega_1 and d\phi_2/dt = \omega_2 we'd have:

<br /> \frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}\sqrt{\frac{1-(r_1sin\theta_1\omega_1/c\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega_2/c\sqrt{1-r_s/r_2})^2}}<br />

This is almost like the equation you got, but you do seem to have made the minor error of leaving out the "c"

It's a typo, corrected in many subsequent posts. See 37 for example.

that appears in my equation (yours is not dimensionally correct, since \omega has units of 1/time), and also you just wrote \omega in both parts of the fraction, neglecting to account for the possibility that \omega_1 differs from \omega_2.

If you insist to have the objects rotating a different angular speeds, yes. But the post is about different clocks at different altitudes and different latitudes. Remember that I was answering Dmitry67's question. So, the clocks share the same \omega

 

[yuiop]

And he's right.

OK, I will concede this one. They are only interchangeable if \theta = \pi/2 which is what espen and pervect specified. It is often a practical convenience to orientate the axes so that this condition is met, a bit like orientating the x axes of two inertial frames with each other and parallel with the relative motion of the two frames in SR even though this is not the most general situation.

 

[JesseM]

It's a typo, corrected in many subsequent posts. See 37 for example.

If you insist to have the objects rotating a different angular speeds, yes. But the post is about different clocks at different altitudes and different latitudes. Remember that I was answering Dmitry67's question. So, the clocks share the same \omega

Fair enough, now can you please answer the questions I asked in that post? Specifically the sentences ending in question marks.

 

[starthaus]

No, you would not need to rewrite the Schwarzschild metric, that's exactly what I meant when I said "Anyway, as I think espen180 pointed out earlier, because of the spherical symmetry of the Schwarzschild spacetime, for any circular orbit you can always do a simple coordinate transformation into a coordinate system that still has the same metric but where the circular orbit now meets this condition". Just as the metric has exactly the same form in the different inertial coordinate systems used in flat spacetime, so it is also true that because of the spherical symmetry of the Schwarzschild metric, one can find a family of different spherical coordinate systems with the \theta = 0 and \phi = 0 axes oriented in different directions (all of them having the same r=0 point), but with the same Schwarzschild metric applying to all these coordinate systems.

I think you explained it to yourself. \theta gives you the latitude of the plane of the circle defined by 0&lt;\phi&lt;2\pi, \theta=constant. So, the angular speed is \frac{d\phi}{dt}, not \frac{d\theta}{dt} that you keep trying to justify. You wrote it yourself , remember? \omega_1=\frac{d\phi_1}{dt} and \omega_2=\frac{d\phi_2}{dt} .

And for any circular orbit whose center is at r=0, there will be one member of this family of spherical coordinate systems where the orbit has a constant \theta coordinate, and another where the orbit can be divided into two halves that each have a constant \phi coordinate. Do you disagree?

You are trying to justify the inadvertent replacement of \phi (the correct coordinate) with \theta (the incorrect coordinate).
If you insist on doing that, at least do it correctly, the new metric should look like this:

ds^2=(1-r_s/r)(cdt)^2-\frac{dr^2}{1-r_s/r}-(rd\phi)^2-(rsin(\phi)d\theta)^2

So, you now have to make dr=d\phi=0 in order to get the time dilation and you are now obviously stuck with the term in sin(\phi)

 

[starthaus]

OK, I will concede this one. They are only interchangeable if \theta = \pi/2 which is what espen and pervect specified.

They aren't interchangeable unless you are in the business of producing hacks.There is a reason for the presence of sin(\theta) in the metric.

 

[JesseM]

I think you explained it to yourself. \theta gives you the latitude of the plane of the circle defined by 0&lt;\phi&lt;2\pi, \theta=constant. So, the angular speed is \frac{d\phi}{dt}, not \frac{d\theta}{dt} that you keep trying to justify.

That would only be the angular speed of a circle that happened to coincide with the equator (which would actually be \theta = \pi/2 rather than \theta = 0 as I incorrectly stated earlier). Do you disagree that any great circle on a sphere would correspond to a valid circular orbit, including a circle which could be divided into two halves of constant longitude (i.e. constant \phi), or plenty of circles where neither longitude nor latitude were constant?

If you insist on doing that, at least do it correctly, the new metric should look like this

No, you're just not getting it. I specifically said that despite the fact that the coordinate systems are different, the metric would have exactly the same form in each one:

Just as the metric has exactly the same form in the different inertial coordinate systems used in flat spacetime, so it is also true that because of the spherical symmetry of the Schwarzschild metric, one can find a family of different spherical coordinate systems with the \theta = 0 and \phi = 0 axes oriented in different directions (all of them having the same r=0 point), but with the same Schwarzschild metric applying to all these coordinate systems.

Do you understand the analogy with "different inertial coordinate systems used in flat spacetime"? If we have two coordinate systems in flat spacetime related by the Lorentz transformation, you'd agree that even though their coordinate axes point in different directions, they would both still have the same metric d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2), right? If you can understand that, you should be able to understand how two spherical coordinate systems with their axes pointed in different directions can nevertheless have the same metric too. I believe that's essentially what the "spherical symmetry" of the Schwarzschild metric means, just like the Lorentz symmetry of Minkowski spacetime can be taken to mean that the metric is unchanged in the different inertial coordinate systems related by the Lorentz transformation.

 

[JesseM]

They aren't interchangeable unless you are in the business of producing hacks.There is a reason for the presence of sin(\theta) in the metric.

And the only valid circular orbit (i.e. a circular path whose center coincides with the center of the Schwarzschild coordinate system at r=0) where d\theta = 0 is one in the "equatorial" plane where \theta = \pi /2, in which case sin(\theta) = 1. Do you disagree?

 

[starthaus]

In post #8 DrGreg said that his v stood for "speed relative to a local hovering observer using local proper distance and local proper time". Do you disagree with this definition?

This is a nit but if you want the coordinate speed, both at arbitrary r and at the "hovering point" r_0, you can get it from the file I pointed out to you.

If not, what do you think kev "claims" that is different?

You are banging on a nit, I call \frac{dr/dt}{\sqrt{1-r_s/r}} a nothing, you insist on calling it "speed relative to a local hovering observer using local proper distance and local proper time". How you name it does not affect the final result and that result is unique, it falls out the Schwarzschild metric. Do you dipute that?

I think I have answered all your sentences that end with "?". :-)

 

[starthaus]

That would only be the angular speed of a circle that happened to coincide with the equator (which would actually be \theta = \pi/2 rather than \theta = 0 as I incorrectly stated earlier).

Yes, so what?

Do you disagree that any great circle on a sphere would correspond to a valid circular orbit, including a circle which could be divided into two halves of constant longitude (i.e. constant \phi), or plenty of circles where neither longitude nor latitude were constant?

The point is that it doesn't. The domain for \theta is [0,\pi]. Do you dispute that?

 

[starthaus]

And the only valid circular orbit (i.e. a circular path whose center coincides with the center of the Schwarzschild coordinate system at r=0) where d\theta = 0 is one in the "equatorial" plane where \theta = \pi /2, in which case sin(\theta) = 1. Do you disagree?

So what? There is an infinity of other circles that do not share \theta = \pi /2,.
None of these are captured by the solution that uses the truncated metric. Why is this so difficult for you to understand?

 

[starthaus]

Do you understand the analogy with "different inertial coordinate systems used in flat spacetime"? If we have two coordinate systems in flat spacetime related by the Lorentz transformation, you'd agree that even though their coordinate axes point in different directions, they would both still have the same metric d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2), right? If you can understand that, you should be able to understand how two spherical coordinate systems with their axes pointed in different directions can nevertheless have the same metric too. I believe that's essentially what the "spherical symmetry" of the Schwarzschild metric means, just like the Lorentz symmetry of Minkowski spacetime can be taken to mean that the metric is unchanged in the different inertial coordinate systems related by the Lorentz transformation.

Yes, I understand it very well, this is why I gave you the counter-example that shows what happens to the metric when you exchange the roles of \theta and \phi.
The point is that kev used a truncated metric. Do you dispute that?
Using a truncated metric, he got a particular solution, that isn't valid in the general case. Do you dispute that?
I posted the general solutions for both orbital and radial motion using the full metric. Do you dispute that?
Both solutions are correct. Do you dispute that?
Heck, I even posted a superset of the solution , using the Kerr metric. Do you have any complains about it?

 

[JesseM]

Yes, I understand it very well, this is why I gave you the counter-example that shows what happens to the metric when you exchange the roles of \theta and \phi.

You obviously don't understand at all, since doing a coordinate transformation which changes the direction the \theta and \phi axes point in, and then finding the new metric in this coordinate system, is NOT equivalent to switching the places of the \theta and \phi coordinates in the metric equation. My whole point is that the "spherical symmetry" of the Schwarzschild metric means you can reorient the \theta and \phi axes in arbitrary directions and the metric will always remain unchanged, just like the metric remains unchanged under a Lorentz transformation with an arbitrary choice of velocity. The wikipedia article on spherically symmetric spacetimes supports this by saying a spherically symmetric spacetime is often described as one whose metric is "invariant under rotations".

The point is that kev used a truncated metric.

I don't know what you mean by "truncated metric". The metric gives the proper time along an arbitrary path, and kev was considering a circular orbit, so he could set terms like dr/dt and d\phi/dt to 0, dropping some terms. You did exactly the same thing in your derivation, only with dr = d\theta = 0. Is that all you mean by "truncated"?

 

[starthaus]

I don't know what you mean by "truncated metric".

Missing terms right off the bat.

I understand very well, please stop talking down to me. I asked you a set of questions, would you please answer them as a set (all of them in one post)? Thank you

 

[JesseM]

So what? There is an infinity of other circles that do not share \theta = \pi /2,.
None of these are captured by the solution that uses the truncated metric. Why is this so difficult for you to understand?

There are no circular orbits where \theta is constant (so d\theta = 0) and has a value other than \theta = \pi / 2. There are other circular paths where the value of \theta is some other constant, but they are like circles of constant latitude on a globe (aside from the equator), the center of the path does not coincide with the center of the coordinate system and thus they are not valid free-fall orbits. Do you disagree?

 

[starthaus]

There are no circular orbits where \theta is constant (so d\theta = 0) and has a value other than \theta = \pi / 2. There are other circular paths where the value of \theta is some other constant, but they are like circles of constant latitude on a globe (aside from the equator),

Correct, these are precisely the circles covered by the solution I gave in post 6. You covered them just the same in the reconstruction of my sollution (see the rsin(\theta)?)

the center of the path does not coincide with the center of the coordinate system and thus they are not valid free-fall orbits. Do you disagree?

Who's talking about free-fall orbits? How many times do I need to tell you that the solution evolved from answering Dmitry67's question about the rate of ticking clocks a different latitudes? What do you think the different \theta's in the formula represent?

Could you please answer all my questions, in one post and without turning every point into your question?

 

[JesseM]

Missing terms right off the bat.

But kev explicitly said that he was starting from an equation pervect derived, where certain terms had already been eliminated. Do you think pervect's derivation was "missing terms right off the bat"?

I understand very well

A person who doesn't understand something will sometimes also fail to understand that they don't understand it. It's pretty clear that you didn't understand what I was saying if you thought that changing the metric equation by switching the roles of \theta and \phi had anything to do with what I was talking about, since I said very clearly that the metric should be invariant under rotations.

please stop talking down to me

When you keep repeating the same mistaken ideas about what I'm saying even though my words clearly show otherwise, I'm going to highlight the fact that you're not understanding me, if that seems like "talking down", well, better that than being over-polite and allowing you to persist in your mistaken understanding.

I asked you a set of questions, would you please answer them? Thank you

You asked a bunch of questions over a series of posts less than half an hour old and it's obvious I'm in the process of answering them, so hold your horses please.

 

[JesseM]

You are banging on a nit, I call \frac{dr/dt}{\sqrt{1-r_s/r}} a nothing, you insist on calling it "speed relative to a local hovering observer using local proper distance and local proper time".

I don't care what you choose to call it, but you seemed to imply that kev was actually incorrect in his description of what it meant when you said "The point is that the quantity in discussion (v) is not what you and kev claim it is". Are you actually saying there was any error in what kev "claims" about this quantity, or is it just that you prefer not to describe it at all?

How you name it does not affect the final result and that result is unique, it falls out the Schwarzschild metric.

Final result for what? An object in circular orbit, or some more general case?

Going to bed now, will continue tomorrow...

 

[starthaus]

Do you think pervect's derivation was "missing terms right off the bat"?

Yes, obviously. pervect not only truncated the metric, he also got the g_{tt} wrong. Do you disagree?

A person who doesn't understand something will sometimes also fail to understand that they don't understand it. It's pretty clear that you didn't understand what I was saying if you thought that changing the metric equation by switching the roles of \theta and \phi had anything to do with what I was talking about, since I said very clearly

You may not realize but what you are talking about is not a rotation of coordinates. What you are talking about is exchanging the rotational motion in the plane \theta=constant with a pseudo-rotation in the plane \phi=constant while all along refusing to admit that you can't complete such a motion since \theta&lt;\pi.
(Basically you are trying to convey the idea that a half circle is a full circle. )

When you keep repeating the same mistaken ideas about what I'm saying even though my words clearly show otherwise, I'm going to highlight the fact that you're not understanding me, if that seems like "talking down", well, better that than being over-polite and allowing you to persist in your mistaken understanding.

Goes both ways, I think that you refuse to understand something very basic and that you put up this strwaman in order not to admit that the solution you have been defending is incorrect and incomplete. We will not get any resolution on this item so I propose that we table this subject. OK?

You asked a bunch of questions over a series of posts less than half an hour old and it's obvious I'm in the process of answering them, so hold your horses please.

I'll wait. Please do not ask any more questions before answering all my questions. I would really appreciate that.

 

[starthaus]

v is speed relative to a local hovering observer using local proper distance and local proper time.
.

v, in your time dilation formula is a scalar. Isn't the above in contradiction with your defining v as four-speed here?

 

[espen180]

You may not realize but what you are talking about is not a rotation of coordinates. What you are talking about is exchanging the rotational motion in the plane \theta=constant with a pseudo-rotation in the plane \phi=constant while all along refusing to admit that you can't complete such a motion since \theta&lt;\pi.
(Basically you are trying to convey the idea that a half circle is a full circle. )

You keep nagging on this point. There is nothing wrong with moving along a circular orbit around \theta rather than \phi. Since the metic is spherically symmetric, ALL circular paths coinciding with the center of the coordinate system is a valid circular orbit. What you are saying seems similar to "You can have a circular orbit about the equator of the Earth, but not perpendicular to the equator (concidering Earth as a perfect nonrotating sphere)". You make it seem like there is a preferred coordinate system where validity of circular orbits is decided.

It's as if you are just looking at the maths, but completely ignoring the physics.

 

[yuiop]

Let's first combine \text{d}\theta^2+\sin^2\theta \text{d}\phi^2=\text{d}\Omega^2 and so simplify the equation to

\frac{\text{d}\tau}{\text{d}t}= \sqrt{\frac{1-r_s/r}{1-r_s/r_o}} \sqrt{1- \left (\frac{1-r_s/r_o}{1-r_s/r} \right )^2 \left (\frac{\text{d}r}{c\text{d}t} \right )^2 - \left (\frac{1-r_s/r_o}{1-r_s/r} \right ) \left(\frac{r \text{d}\Omega}{c\text{d}t}\right)^2 }

Working backwards to get back to the metric gives me

c^2\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=c^2\frac{1-r_s/r}{1-r_s/r_o}-\left(\frac{1-r_s/r}{1-r_s/r_o}\right)^{-1}\left (\frac{\text{d}r}{\text{d}t} \right )^2-r^2\left (\frac{\text{d}\Omega}{\text{d}t} \right )^2

c^2\text{d}\tau^2=c^2\frac{1-r_s/r}{1-r_s/r_o}\text{d}t^2-\left(\frac{1-r_s/r}{1-r_s/r_o}\right)^{-1} \text{d}r^2-r^2\text{d}\Omega^2

I was hoping that doing this would lead me to an explanation as to where the \frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}} came from, but it seems it did not.

I do observe that in modeling this metric the metric coefficients are found by taking the ratio of the coefficients of the particle wrt an observer at infinity to the coefficients of the observer at r_0 to the same observer at infinity, but could I have an explanation of why that works?

Hi espen,

You are right to question this and I apologise for any confusion caused. When trying to answer your question I realized that basically I got this bit completely wrong. This is how it should be done:

The proper time of a moving clock at r relative to the reference clock at infinity in Schwarzschild coordinates is (using your notation):

\text{d}\tau = \text{d}t \sqrt{1-r_s/r} \sqrt{1- (1-r_s/r)^{-2}(\text{d}r/c\text{d}t)^2 - (1-r_s/r)^{-1}(r \text{d}\Omega/c\text{d}t)^2 }

The proper time of an observers clock d\tau_0 relative to the reference clock at infinity with motion dr_o/dt_o and d\Omega_o/dt_o at radius r_o, is given by:

\text{d}\tau_o = \text{d}t \sqrt{1-r_s/r_o} \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r_o \text{d}\Omega_o/c\text{d}t_o)^2 }

The ratio of the proper time of the two clocks is then:

\frac{\text{d}\tau}{\text{d}\tau_o} = \frac{\sqrt{1-r_s/r\;}}{\sqrt{1-r_s/r_o}} \frac{\sqrt{1- (1-r_s/r\;)^{-2}(\text{d}r\;/c\text{d}t\;)^2 - (1-r_s/r\;)^{-1}(r\: \text{d}\Omega\;/c\text{d}t\;)^2 }}{ \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r_o \text{d}\Omega_o/c\text{d}t_o)^2 }}

This is the completely general case of the ratio of two moving clocks in the Schwarzschild metric.

If the observer is stationary at r_o the equation reduces to:

\frac{\text{d}\tau}{\text{d}\tau_o}= \sqrt{\frac{1-r_s/r}{1-r_s/r_o}} \sqrt{1- \frac{(\text{d}r/\text{d}t)^2}{c^2(1-r_s/r)^2} - \frac{(r \text{d}\Omega/\text{d}t)^2}{c^2(1-r_s/r)} }

<EDIT> THe above has been edited to correct a typo.

 

[DrGreg]

v, in your time dilation formula is a scalar. Isn't the above in contradiction with your defining v as four-speed here?

With hindsight, it was a poor choice of notation, as the same symbol v was being used in different, incompatible ways. That's why I rewrote the argument in different notation in post #46[/color] of that same thread.

 

[espen180]

Thanks kev. So

\frac{\text{d}\tau}{\text{d}\tau_o}

here is the ratio of the observer's clock and the observed clock(for lack of a better term) as seen by a second observer at infinity, correct?

It is probably just my intuitive understanding that is failing me, but is this ratio the same ratio as observed by the first observer at r_0?

I think an SR example can explain my confusion:

Define the frame S, in which there is an observer at rest. In S, there are frames S' and S'' with observers at rest going at speeds v_1=\beta_1 c and v_2=\beta_2 c respectively in S. Now the rest observer can measure

t=\gamma_1\tau_1=\gamma_2\tau_2

The rest observer in S measures the ratio of the proper times of the rest observers in S' and S'' to be
\frac{\text{d}\tau_1}{\text{d}\tau_2}=\frac{\gamma_2}{\gamma_1}=\sqrt{\frac{1-\beta_1^2}{1-\beta_2^2}}

In S', the observer in S'' is traveling at the speed

v_2^{\prime}=\frac{\beta_2-\beta_1}{1-\beta_1\beta_2}c

and since \tau_1=\gamma_2^{\prime}\tau_2 the observer in S' measures

\left(\frac{\text{d}\tau_2}{\text{d}\tau_1}\right)^{\prime}=\sqrt{1-\left(\frac{\beta_2-\beta_1}{1-\beta_1\beta_2}\right)^2}=\frac{\sqrt{1+\beta_1^2\beta_2^2-\beta_2^2-\beta_1^2}}{1-\beta_1\beta_2}

Unless I made an error underway (which is very possible, this was a messy calculation), obervers S and S' don't seem to agree on the value of \frac{\text{d}\tau_2}{\text{d}\tau_1}.

 

[starthaus]

With hindsight, it was a poor choice of notation, as the same symbol v was being used in different, incompatible ways. That's why I rewrote the argument in different notation in post #46[/color] of that same thread.

Thank you for the honest answer. This brings me to a follow-up question. Can you please show how you calculate the value for w?