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- Thread starter snoopies622
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Although you are right that it shows up in the derivation of the SE tensor for a fluid, since there you want to convert the rest frame picture into a general frame picture.

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Well, there is a local inertial frame in which the metric is flat around one point. Alternatively, you can always look at the components of the tensor in an orthonormal frame.

MTW has a good discussion.

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Yes. That is correct. A very useful thing to realize in GR.

Here's a resource you might find useful.

http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec12.html [Broken]

Here's a resource you might find useful.

http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec12.html [Broken]

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Thanks, Ibrits (or is it lbrits?). I'll check out that 'Gravitation' book, too.

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I think its called "local cartesian coordinate"

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From what I understand, gamma does play a direct role in general relativity.

Both gravitational and kinematic time dilation contribute toward the relativistic precession of an orbiting mass (ex: Mercury around the Sun), where kinematic time dilation is what arises from special relativity.

In terms of "orbits per orbit" of relativistic precession, where r is the average orbit radius (semi-major axis), a common formula is:

[tex]\delta = \frac{3GM}{rc^2}[/tex]

Two-thirds of this is from gravitational time dilation (due to the gravitational field of the source mass), and one-third from kinematic time dilation (due to the current velocity of the orbiting mass in relation to the source mass).

The value [tex]\delta 2\pi[/tex] gives the relativistic precession in terms of radians per orbit.

Both gravitational and kinematic time dilation contribute toward the relativistic precession of an orbiting mass (ex: Mercury around the Sun), where kinematic time dilation is what arises from special relativity.

In terms of "orbits per orbit" of relativistic precession, where r is the average orbit radius (semi-major axis), a common formula is:

[tex]\delta = \frac{3GM}{rc^2}[/tex]

Two-thirds of this is from gravitational time dilation (due to the gravitational field of the source mass), and one-third from kinematic time dilation (due to the current velocity of the orbiting mass in relation to the source mass).

The value [tex]\delta 2\pi[/tex] gives the relativistic precession in terms of radians per orbit.

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Yep. I believe that's a fair summary.

One other point I left out was that when a mass is radially in-falling (no transverse acceleration), kinematic and gravitational time dilation are actually the same thing.

One other point I left out was that when a mass is radially in-falling (no transverse acceleration), kinematic and gravitational time dilation are actually the same thing.

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Thanks, shalayka. I will look further into this. That 3GM/(rc^2) term looks familiar.

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It might also help to see it from the perspective of:

[tex]

3 \frac{GM}{rc^2} \frac{1}{1 - e^2} = 3 \frac{v^2}{c^2} \frac{1}{1 - e^2},

[/tex]

[tex]

v = \sqrt{\frac{GM}{r}}.

[/tex]

... where e is eccentricity, and r is the average orbit radius (semi-major axis).

From what we know about special relativity, kinematic time dilation, in relation to an orbiting body, is given by:

[tex]

\tau = t \sqrt{1 - \frac{v^2}{c^2}} = t \sqrt{1 - \frac{GM}{rc^2}},

[/tex]

From general relativity, gravitational time dilation is given by:

[tex]

\tau = t \sqrt{1 - \frac{2GM}{rc^2}}.

[/tex]

As you can see, both forms of time dilation are quite similar when considering it from the perspective of an orbiting body.

For fun, we can obtain the Newtonian acceleration in a backwards sort of way, by differentiating the gravitational time dilation formula with respect to r:

[tex]

\tau \approx t \approx 1,

[/tex]

[tex]

\frac{\partial \tau}{\partial r} \approx \frac{GM}{r^2 c^2},

[/tex]

[tex]

a = \frac{\partial \tau}{\partial r} {c^2} = \frac{GM}{r^2},

[/tex]

[tex]

v = \sqrt{a r} = \sqrt{\frac{GM}{r}}.

[/tex]

[tex]

3 \frac{GM}{rc^2} \frac{1}{1 - e^2} = 3 \frac{v^2}{c^2} \frac{1}{1 - e^2},

[/tex]

[tex]

v = \sqrt{\frac{GM}{r}}.

[/tex]

... where e is eccentricity, and r is the average orbit radius (semi-major axis).

From what we know about special relativity, kinematic time dilation, in relation to an orbiting body, is given by:

[tex]

\tau = t \sqrt{1 - \frac{v^2}{c^2}} = t \sqrt{1 - \frac{GM}{rc^2}},

[/tex]

From general relativity, gravitational time dilation is given by:

[tex]

\tau = t \sqrt{1 - \frac{2GM}{rc^2}}.

[/tex]

As you can see, both forms of time dilation are quite similar when considering it from the perspective of an orbiting body.

For fun, we can obtain the Newtonian acceleration in a backwards sort of way, by differentiating the gravitational time dilation formula with respect to r:

[tex]

\tau \approx t \approx 1,

[/tex]

[tex]

\frac{\partial \tau}{\partial r} \approx \frac{GM}{r^2 c^2},

[/tex]

[tex]

a = \frac{\partial \tau}{\partial r} {c^2} = \frac{GM}{r^2},

[/tex]

[tex]

v = \sqrt{a r} = \sqrt{\frac{GM}{r}}.

[/tex]

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Fascinating. This is from a general geodesic equation using the Schwarzschild metric?

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Yep. The gravitational time dilation formula appears in the standard Schwarzschild solution's line element / metric, in the form of:

[tex]

{\rm d}s^2 = {\rm d}t^2 \left(\sqrt{1 - \frac{2GM}{rc^2}}\right)^2 - \frac{{\rm d}r^2}{\left(\sqrt{1 - \frac{2GM}{rc^2}}\right)^2} - ...

[/tex]

The fact that Einstein managed to work backwards from Newtonian velocity equation to get the gravitational time dilation equation just blows my mind!

[tex]

{\rm d}s^2 = {\rm d}t^2 \left(\sqrt{1 - \frac{2GM}{rc^2}}\right)^2 - \frac{{\rm d}r^2}{\left(\sqrt{1 - \frac{2GM}{rc^2}}\right)^2} - ...

[/tex]

The fact that Einstein managed to work backwards from Newtonian velocity equation to get the gravitational time dilation equation just blows my mind!

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yes. It does show up in gr. In gr, gamma (ratio of coordinate tmie to proper time) is a function of both speed and the gravitational potential (especially when g_0t =0).

Pete

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Thanks. What is "g_0t"? Aren't zero and t as metric tensor subscripts used to mean the same thing?

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Oops! You're right! Sorry about that. I meant to write g_0k = 0, k = 1,2,3. I can send you a PM which shows the calculation if you'd like? Its far too difficult for me to do that in PM since all I have to do is send you a URL (which we're not allowed to post in open forum).Thanks. What is "g_0t"? Aren't zero and t as metric tensor subscripts used to mean the same thing?

Pete

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Correction: People are not allowed to post URLe to pages under their personal websites. The powers that be never made sense to me on this point so if you want to know the real reason you'd be better off asking a moderator why this is policy.

Pete

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Since there is no difference between what I'd write on my web page and what I'd post here then I don't see the problem. The reason I created my web site was because it was easier to provide derivations and it let me have the material readily handy in case I wanted to bring it up again. So if I write an equation here on on my web page and then provide the link, what's the difference .. besides the increased amount of work required by myself?

Pee

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Probably because in ("standard") general relativity, the gravitational mass of an object does not increase with relative velocity (inertial mass is not equal to relativistic mass). It's only an illusion based on the kinematic time dilation of the gravitated object (special relativity says that the gravitated object may be equally considered to be the moving object), since the final deflection angle is always the same (as judged by a third object at a large distance). It just appears to the gravitated object that the total deflection occurred over a shorter period of time, since its rate of time is less than that if it were at rest.

This follows from the notion that stronger gravitation leads to a greater deflection angle over the same period of time. Since the gravitated object cannot perceive that its rate of time is not equal (and thus, not measuring deflection over a differing period of time), it can only assume that gravitation is stronger. However, it's not actually the case.

This follows from the notion that stronger gravitation leads to a greater deflection angle over the same period of time. Since the gravitated object cannot perceive that its rate of time is not equal (and thus, not measuring deflection over a differing period of time), it can only assume that gravitation is stronger. However, it's not actually the case.

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"Since the gravitated object cannot perceive that its rate of time is not equal..."

... that is, not equal to the rate of time of the third object that is at rest relative to the gravitating object, residing at a large distance away.

Just wanted to clarify, and ran past the edit time limit of 30 minutes.

To favour a perspective different than that given by standard general relativity is not necessarily "wrong" in your specific case, but it could end up confusing those learning physics. I think that's what the moderators are trying to avoid, by disallowing you to post these links in the forum.

P.S. Thanks for the links in the private message! :) I will read them more thoroughly tonight.

... that is, not equal to the rate of time of the third object that is at rest relative to the gravitating object, residing at a large distance away.

Just wanted to clarify, and ran past the edit time limit of 30 minutes.

To favour a perspective different than that given by standard general relativity is not necessarily "wrong" in your specific case, but it could end up confusing those learning physics. I think that's what the moderators are trying to avoid, by disallowing you to post these links in the forum.

P.S. Thanks for the links in the private message! :) I will read them more thoroughly tonight.

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Where did you get that idea from?Probably because in ("standard") general relativity, the gravitational mass of an object does not increase with relative velocity (inertial mass is not equal to relativistic mass).

Its not an illusion but is aIt's only an illusion based on the kinematic time dilation ...

The fact is that the active gravitational mass is the source of gravity. The active gravitational mass of a body is a function of its mass, which in term is a function of velocity as well as pressure/stress. Also the passive gravitational mass has the same value as the inertial mass (which is the same as relativistic mass). Its not that hard to show these things. If you're familiar with the math I can send you URLs with derivations on them so you can see exactly what all this means. I also have an article from the American Journal of Physics which addresses the active gravitational mass of a moving point mass.

Pete

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