Resultant time dilation from both gravity and motion

[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<\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<\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?

 

[starthaus]

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)".

No, this is not what I'm saying. What I am saying is something very basic and totally different. Yet, you seem clearly unable to grasp it.

You make it seem like there is a preferred coordinate system where validity of circular orbits is decided.

Not at all. It is very basic, really but you are so insistent, I'll explain it (maybe JesseM) will also get this one). The Earth rotates about the NS axis. \theta represents the angle from the N pole, therefore \frac{d\theta}{dt} is a very bad choice to represent the Earth rotation. By contrast, \frac{d\phi}{dt} is the correct choice.

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

LOL

 

[starthaus]

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 }

OK.

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 \text{d}\Omega_o/c\text{d}t_o)^2 }

This is wrong. Since you are putting in results by hand again, try deriving it from the basics and you'll find out why.

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 \text{d}\Omega_o/c\text{d}t_o)^2 }}

No.

 

[starthaus]

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}.

You sure did, they must agree. You just botched the SR Doppler effect. Now, where did you make the blunder?You and kev are piling up them errors.

 

[espen180]

Not at all. It is very basic, really but you are so insistent, I'll explain it (maybe JesseM) will also get this one). The Earth rotates about the NS axis. \theta represents the angle from the N pole, therefore \frac{d\theta}{dt} is a very bad choice to represent the Earth rotation. By contrast, \frac{d\phi}{dt} is the correct choice.

This is exactly why I opted to contract the angles into a single one which is in the direction on tangential motion. Then there is no need to worry about which angle to rotate around.

 

[starthaus]

This is exactly why I opted to contract the angles into a single one which is in the direction on tangential motion. Then there is no need to worry about which angle to rotate around.

Oh, but you do. Do you think that a mathematical sleigh of hand fixes your misunderstanding of basics physics? What axis does the Earth rotate about?

 

[espen180]

Oh, but you do. Do you think that a mathematical sleigh of hand fixes your misunderstanding of basics physics? What axis does the Earth rotate about?

I thought we were working with the Schwartzschild geometry, a non-rotating body.

Still, what difference does Earth's rotation axis make for orbits around the Earth? It's not like the Earth exhibits a non-negligible amount of rotational frame-dragging.

 

[starthaus]

I thought we were working with the Schwartzschild geometry, a non-rotating body.

So, what is \frac{d\phi}{dt} again?

Still, what difference does Earth's rotation axis make for orbits around the Earth?

You are making the same mistakes as JesseM, we are talking about the delay experienced by clocks on the Erath surface due to Earth rotation. What do you think I have been trying to explain to you starting with post 6?

It's not like the Earth exhibits a non-negligible amount of rotational frame-dragging.

This is not what we are talking about.

 

[DrGreg]

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?

I'm not sure the context you have in mind. In my original post, the whole point was to calculate w from everything else that was in the same equation. That's probably not what you meant. So what did you mean? (In other words, if you want w in terms of something else, what is the "something else"?)

 

[starthaus]

I'm not sure the context you have in mind. In my original post, the whole point was to calculate w from everything else that was in the same equation. That's probably not what you meant. So what did you mean? (In other words, if you want w in terms of something else, what is the "something else"?)

In this thread I am calculating \frac{d\tau}{dt} as a function of coordinate speed \frac{dr}{dt} from the Schwarzschild metric:

ds^2=(1-r_s/r)(cdt)^2-...

So, it would appear that your w is equal to \frac{dr/dt}{1-r_s/r}. I asked this question before in the thread, in post 10, probably it got missed in the tremendous noise. Is this correct? How would you arrive to w's value in your derivation? You do not appear to use the same approach I am using, this is why I am interested.

 

[DrGreg]

In this thread I am calculating \frac{d\tau}{dt} as a function of coordinate speed \frac{dr}{dt} from the Schwarzschild metric:

ds^2=(1-r_s/r)(cdt)^2-...

So, it would appear that your w is equal to \frac{dr/dt}{1-r_s/r}. I asked this question before in the thread, in post 10, probably it got missed in the tremendous noise. Is this correct? How would you arrive to w's value in your derivation? You do not appear to use the same approach I am using, this is why I am interested.

First of all I had better come clean about a detail that I glossed over. In my derivation in the other thread I referred to w as speed measured in...

the locally-Minkowski coordinates of a free-falling and momentarily-at-rest local observer at that event

...whereas in post #8 of this thread I referred to what I am now calling w as...

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

In case somebody complains, I should point out that the two speeds must be the same. As far as relative velocity is concerned, it doesn't matter whether the observer is accelerating or not, the relative velocity (in this sense) will be the same. (Of course you cannot use that argument for other quantities such as acceleration.)

If you want to express w in terms of Schwarzschild coordinates, you could construct "locally-rescaled Schwarzschild coordinates" at the event of interest (that is multiply each coordinate by a constant such that the metric equals the Minkowski metric at that event only) and then w will be the coordinate velocity in those coordinates, which you can then rescale back into Schwarzschild coordinate velocity.

So, for radial motion only (where \theta and \phi are constant and can be ignored), change coordinates to

T = t \sqrt{1-r_s/r_0}
R = r / \sqrt{1-r_s/r_0}​

where r₀ is the value of r where you want to make the measurement, so that the metric becomes

ds^2=c^2\,dT^2-dR^2​

at that point only. Then, along the worldline being measured,

w = \frac{dR}{dT} = \frac{dR/dr}{dT/dt}\cdot \frac{dr}{dt} = \frac{dr/dt}{1-r_s/r_0}​

So, yes, you are correct about w in this case.

Note that if you simply want to calculate dt/d\tau in terms of dr/dt you don't really need to involve w at all, you just plug dr = (dr/dt)\,dt into the metric and it all falls out.

 

[starthaus]

First of all I had better come clean about a detail that I glossed over. In my derivation in the other thread I referred to w as speed measured in... ...whereas in post #8 of this thread I referred to what I am now calling w as...In case somebody complains, I should point out that the two speeds must be the same. As far as relative velocity is concerned, it doesn't matter whether the observer is accelerating or not, the relative velocity (in this sense) will be the same. (Of course you cannot use that argument for other quantities such as acceleration.)

If you want to express w in terms of Schwarzschild coordinates, you could construct "locally-rescaled Schwarzschild coordinates" at the event of interest (that is multiply each coordinate by a constant such that the metric equals the Minkowski metric at that event only) and then w will be the coordinate velocity in those coordinates, which you can then rescale back into Schwarzschild coordinate velocity.

So, for radial motion only (where \theta and \phi are constant and can be ignored), change coordinates to

T = t \sqrt{1-r_s/r_0}
R = r / \sqrt{1-r_s/r_0}​

So, R and T are simply r and t rescaled to make 1-r_s/r "go away" from the Schwrazschild metric. I am having trouble assigning any physical properties to R and T and, consequently to w. To me, they are just rescaled versions of r,t,dr/dt.

where r₀ is the value of r where you want to make the measurement, so that the metric becomes

ds^2=c^2\,dT^2-dR^2​

at that point only. Then, along the worldline being measured,

w = \frac{dR}{dT} = \frac{dR/dr}{dT/dt}\cdot \frac{dr}{dt} = \frac{dr/dt}{1-r_s/r_0}​

So, yes, you are correct about w in this case.

Thank you

My "w" simply falls out the metric (see post 6).

Note that if you simply want to calculate dt/d\tau in terms of dr/dt you don't really need to involve w at all, you just plug dr = (dr/dt)\,dt into the metric and it all falls out.

Yes, our methods are identical, I am just skipping the coordinate rescaling step.

 

[yuiop]

...
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 }

OK.

...
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 \text{d}\Omega_o/c\text{d}t_o)^2 }

This is wrong. Since you are putting in results by hand again, try deriving it from the basics and you'll find out why.

OK, there is a typo in the second equation where I missed the "o" subscript for r in the Omega term. I think it is obvious what was intended from the method.

The second equation should be:

\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 }

... 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}.

I will check it out.

 

[starthaus]

OK, there is a typo in the second equation where I missed the "o" subscript for r in the Omega term. I think it is obvious what was intended from the method.

The second equation should be:

\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 }

Still wrong, still put in by hand.

 

[yuiop]

Still wrong, still put in by hand.

If the first equation is right, how can the second equation be wrong?

All I have done is change the names of the variables.

 

[starthaus]

If the first equation is right, how can the second equation be wrong?

All I have done is change the names of the variables.

You did more than that. Look it over carefully.

 

[yuiop]

You did more than that. Look it over carefully.

Ah, guessing games again. It was not my intention to do more than changing the names of the variables, so if that is not the case you must be talking about a typo I can not spot.

 

[JesseM]

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?

"So what" is that since \theta = \pi/2 is the only case of the equation below (from my post 76, and basically the same as your own equation) that actually corresponds to a circular orbit rather than a non-orbiting circular path:

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

...then for a circular orbit with d\theta = 0 it must be true that sin(\theta) = 1 and therefore the equation reduces to:

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

If you then make the substitution v = r \omega / \sqrt{1 - r_s/r}, which was exactly the substitution kev made in post #8 (he defined u = r \left(\frac{d\phi}{dt}\right), equivalent to u = r\omega, and then he defined u = v \sqrt{1-\frac{r_s}{r}}, equivalent to v = r\omega / \sqrt{1 - r_s/r}), then this equation becomes:

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

So the equation is correct for the special case of a circular orbit where d\theta = 0. Do you disagree? If not, then the symmetry argument I already mentioned shows why this would hold for any circular orbit, even one where d\theta was not equal to 0 in our original coordinate system (since you could always rotate into a new coordinate system where d\theta was equal to 0 on the orbit, and the metric would be exactly the same in this new coordinate system since the Schwarzschild metric is invariant under rotations)

Actually now that I've looked back at kev's post #8 more carefully I have no idea why in post 28 you criticized him by saying:

You need to make

d\theta=dr=0

-v is equal to:

r\frac{d\phi}{dt} and not r\frac{d\theta}{dt}

...since it appears to me he did make d\theta = 0, and he did define the velocity in terms of \frac{d\phi}{dt} rather than \frac{d\theta}{dt}! I guess I shouldn't have taken your word for it that he did it differently there (even though you could still get exactly the same final result by assuming a circular orbit where d\phi = 0 along a short segment, do you disagree? If you do disagree, I can demonstrate)

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 point is that what doesn't? You didn't answer my question about whether you disagree that there are valid circular orbits in Schwarzschild spacetime which, in a given coordinate system, would have a description like the one above. If you do disagree then I think you need to do some thinking about how spherical coordinates work, in particular what the coordinate description would look like for an "upright" circle whose plane was at a right angle to the "horizontal" \theta = \pi/2 plane.

The domain for \theta is [0,\pi]. Do you dispute that?

No, of course not, why do you imagine I would? In post #56 I gave the example of a complete circle where the coordinate description would be such that one half of the circle would have a constant r=R and \phi = \pi/2 while the other half would have constant r=R and \phi = -\pi/2, I thought it was fairly obvious that the points covered by each half would then be defined by varying \theta from 0 to \pi. Again, do you disagree that this would be a valid coordinate description for the set of points on a single continuous circle, one whose center is at r=0 and whose plane is at a right angle to the \theta = \pi/2 plane, and where d\phi = 0 along any infinitesimal segment of this circle? If not you should see why, despite the fact that kev actually made d\theta = 0 rather than d\phi = 0, it would have been perfectly valid for him to do the reverse, either way there'd be a valid circular orbit meeting this condition, there'd be nothing non-rigorous or "hack"-y about such a starting assumption.

 

[JesseM]

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?

Yes, I understand that's the question you were addressing in post #6, and (minor algebra errors aside) I don't dispute that your answer there is a good one to Dmitry67's post. But this whole debate got started when you disputed my answer to espen108's OP in post #3, in which I cited kev's result; you seem insistent that this is wrong somehow, although you seem to constantly change your mind about why it is wrong. kev's result is valid for all circular orbits, and it's relevant to espen108's OP because in that case the answer is a product of two equations that look just like GR time dilation and SR time dilation.

What do you think the different \theta's in the formula represent?

In your formula, you are assuming two circular paths which each have a different constant \theta coordinate. This makes sense as a response to Dmitry67's question about clocks at different latitudes.

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

I reply to posts individually, so if you write one big post I'll respond with one big post, if you write a lot of little posts I'll have an equal number of responses. Up until recently we were going back and forth with big posts, but then for some reason you decided to start breaking up your responses (your posts 77, 80, 84 all respond to my post 76, while your posts 85 and 87 respond to my post 82) which resulted in our back-and-forth being spread out over many more posts.

 

[JesseM]

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

He made a minor error with g_{tt} which kev corrected in his response, but other than that I still don't know what you mean by "truncated", in his post he started from the abstract form of the metric which included all the terms:

c^2 d\tau^2 = g_{tt} dt^2 + g_{rr} dr^2 + g_{\theta\theta} d\theta^2 + g_{\phi\phi} d\phi^2

Are you calling this "truncated" just because he didn't actually write out the equations for each "g"? There's nothing non-rigorous about this, anyone can look up what they'd all be in the Schwarzschild metric. Or do you think there was something wrong with pervect's next step of specifying that he was talking about a circular orbit where dr/dt and d\theta/dt would be zero, and eliminating the appropriate terms given these conditions?

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 and 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. )

Since nothing I said faintly resembles what you are saying, you need to actually explain where you got the idea that this is "what I am talking about", rather than just asserting it. I take it you think something I said implies this somehow? If so, what specific quote? Do you deny that in spherical coordinates, it is possible to find a valid continuous circle whose center is at r=0 and whose coordinates match the description I gave at the end of post 117? If you don't deny that, are you denying that it would be possible to rotate the original coordinate system into a new coordinate system where the same complete circle would now lie entirely in the \theta = \pi / 2 plane? If you deny either of these I suggest the problem lies with your understanding of how spherical coordinates work.

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?

Sure, I was only responding to your complaint about "talking down".

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.

I believe I've now responded to all your replies to me, but I can't agree to the request not to ask any further questions, as that would make it impossible for me to pin you down on a lot of your ambiguous arguments (like the one above where you tell me 'what I am saying' when your summary bears no resemblance to what I actually said and you don't explain how you think it was implied by what I said).

 

[yuiop]

Let's start with this equation for the time dilation ratio:

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

The above equation is obtained directly from the Schwarzschild metric and I think we are all in agreement about its validity.

Now define local velocities u_x, u_y, u_z as measured by a stationary observer at r using his proper length (dr') and proper time (dt'):

u_x = dr&#039;/dt&#039; = \frac{dr/dt}{(1-r_s/r)}

u_y = r d\theta/dt&#039; = \frac{r d\theta/dt}{\sqrt{1-r_s/r} }

u_z = r \sin\theta d\phi/dt&#039; = \frac{r \sin \theta d\phi/dt}{\sqrt{1-r_s/r} }

Substitute these local velocities into the time dilation equation:

d\tau/dt = \sqrt{1-r_s/r} \sqrt{1- u_x^2/c^2 - u_y^2/c^2 - u_y^2/c^2 }

Now define the local 3 velocity as:

w = \sqrt{ u_x^2 + u_y^2 + u_y^2 }

and substitute this value in:

d\tau/dt = \sqrt{1-r_s/r} \sqrt{1- w^2/c^2 }

This is the result obtained in more detail and more rigorously by DrGreg in #8 and valid for all vertical/horizontal or radial/orbital motion of a test particle at r, just as DrGreg claimed.

I do not think there is anything Starthaus can dispute there.