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## Resultant time dilation from both gravity and motion

 Quote by JesseM 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?

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 Quote by JesseM 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?

Recognitions:
 Quote by starthaus 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".
 Quote by starthaus 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"?

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 Quote by JesseM 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

Recognitions:
 Quote by starthaus 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?

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 Quote by JesseM 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?

Recognitions:
 Quote by starthaus 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"?
 Quote by starthaus 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.
 Quote by starthaus 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.
 Quote by starthaus 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.

Recognitions:
 Quote by starthaus 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?
 Quote by starthaus 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...

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 Quote by JesseM 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.

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 Quote by DrGreg 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?

 Quote by starthaus 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.

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 Quote by espen180 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 realised 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.

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Gold Member
 Quote by starthaus $$v$$, 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 of that same thread.

 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 travelling 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' dont seem to agree on the value of $$\frac{\text{d}\tau_2}{\text{d}\tau_1}$$.

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 Quote by DrGreg 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 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 $$w$$?

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 Quote by espen180 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

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 Quote by kev Hi espen, You are right to question this and I apologise for any confusion caused. When trying to answer your question I realised 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.