Resultant time dilation from both gravity and motion

[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:

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

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

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:

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

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:

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

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

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:

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

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 $$0<\phi<2\pi, \theta=constant$$. 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 $$0<\phi<2\pi, \theta=constant$$. 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 $$\theta<\pi$$.
(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 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 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?