Gamma in GR

  • #1
snoopies622
791
22
Does the Lorentz factor show up much (or at all) in general relativity? For instance, is it appropriate to use it as part of the four-momentum vector when finding the components of the stress-energy tensor even in a non-flat spacetime? I was wondering because the derivations of it that I've seen always assume either a Euclidean or Minkowskian metric.
 

Answers and Replies

  • #2
lbrits
410
3
[tex]\gamma[/tex] is the ratio of (Lorentz) coordinate time to proper time, and so it only shows up when you are using local Lorentz coordinates. In other words, it almost never shows up in GR.

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.
 
  • #3
snoopies622
791
22
I'm a little confused (what else is new?): if there is fluid, the spacetime isn't flat. Why then is it valid to use the Lorentz factor at all?
 
  • #4
lbrits
410
3
I'm a little confused (what else is new?): if there is fluid, the spacetime isn't flat. Why then is it valid to use the Lorentz factor at all?

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.
 
  • #5
snoopies622
791
22
So, are you saying that since spacetime is a manifold, if we take a small enough section of it, we can treat it as flat, even in the presence of mass-energy? If so, then I think I understand.
 
  • #6
lbrits
410
3
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]
 
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  • #7
snoopies622
791
22
Thanks, Ibrits (or is it lbrits?). I'll check out that 'Gravitation' book, too.
 
  • #8
off-diagonal
30
0
So, are you saying that since spacetime is a manifold, if we take a small enough section of it, we can treat it as flat, even in the presence of mass-energy? If so, then I think I understand.


I think its called "local cartesian coordinate"
 
  • #9
shalayka
126
0
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.
 
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  • #10
snoopies622
791
22
So in that case the kinematic time dilation would involve the Lorentz factor, even though the metric is non-Minkowskian. Hmm.
 
  • #11
shalayka
126
0
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.
 
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  • #12
snoopies622
791
22
Thanks, shalayka. I will look further into this. That 3GM/(rc^2) term looks familiar.
 
  • #13
shalayka
126
0
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]
 
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  • #14
snoopies622
791
22
Fascinating. This is from a general geodesic equation using the Schwarzschild metric?
 
  • #15
shalayka
126
0
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!
 
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  • #16
pmb_phy
2,952
1
Does the Lorentz factor show up much (or at all) in general relativity? For instance, is it appropriate to use it as part of the four-momentum vector when finding the components of the stress-energy tensor even in a non-flat spacetime? I was wondering because the derivations of it that I've seen always assume either a Euclidean or Minkowskian metric.

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
 
  • #17
snoopies622
791
22
Thanks. What is "g_0t"? Aren't zero and t as metric tensor subscripts used to mean the same thing?
 
  • #18
pmb_phy
2,952
1
Thanks. What is "g_0t"? Aren't zero and t as metric tensor subscripts used to mean the same thing?
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).

Pete
 
  • #19
shalayka
126
0
That's weird that you can't post URLs. I am always interested in reading new material related to GR, so I would make as much use of it as snoopies.
 
  • #20
pmb_phy
2,952
1
That's weird that you can't post URLs. I am always interested in reading new material related to GR, so I would make as much use of it as snoopies.
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
 
  • #21
shalayka
126
0
I think it probably has to do with physics being inherently complex. The moderators most likely want to keep things as simple as possible by avoiding material that hasn't gone through the peer-review process. Not that things that have passed through the peer-review process cannot be subsequently proven incorrect, but still...
 
  • #22
pmb_phy
2,952
1
I think it probably has to do with physics being inherently complex. The moderators most likely want to keep things as simple as possible by avoiding material that hasn't gone through the peer-review process. Not that things that have passed through the peer-review process cannot be subsequently proven incorrect, but still...
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 website 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
 
  • #23
shalayka
126
0
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.
 
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  • #24
shalayka
126
0
"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.
 
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  • #25
pmb_phy
2,952
1
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).
Where did you get that idea from?
It's only an illusion based on the kinematic time dilation ...
Its not an illusion but is a result of things like 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
 
  • #26
shalayka
126
0
Ah yes, I was incorrect in saying that inertial mass does not increase with velocity. Of course that's true, since it's the mechanism behind momentum. I don't know what I was thinking.

Perhaps I am still confused about gravitation however.

Do you agree that an increase in gravitation caused by the Sun occurs where its kinematic time dilation is actual (as in, caused by its orbit around the galactic centre), but not where its kinematic time dilation is relative (as in, where the Sun appears to travel rectilinearly because the gravitated object is flying past it, like in my post above)? This I can can agree to, since they are completely different situations.
 
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  • #27
pmb_phy
2,952
1
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.
The point is that if its okay for me to speak of it in a post then why not in a web page that I can refer to so that (2) I don't have to keep repeating myself and (2) the derivation becomes that much clearer (pretty pictures etc :smile:)? It is tobe noted that nothing I've ever posted cannot be found in a modern textbook on special/general relativity or in the physics literature. Same goes with my web pages.
I think that's what the moderators are trying to avoid, by disallowing you to post these links in the forum.
Then they should not allow a lot of the material presented in modern relativity texts to be cited since the material is no different.
P.S. Thanks for the links in the private message! :) I will read them more thoroughly tonight.
I look forward to your comments/reply on the material.

Pete
 
  • #28
shalayka
126
0
I agree. I see now what you are trying to get at. Perhaps they are worried about broken links, or dispersal of information across too broad of a network (imagine Wikipedia being a conglomeration of multiple sites, and not just one), because I can't think of any other reason. Again, your point about referencing books is contrary to this, making the whole thing questionable... though books generally tend to be a more "permanent" record of information than personal web sites.

Thanks for being patient with me about this Pete. I'm glad that you set me straight about the message presented in your (seemingly excellent) work.

I suppose the only answer that the moderators actually have to give though is: "Because it's our forum, and we say so", as is their right.
 
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  • #29
pmb_phy
2,952
1
Again, your point about referencing books is contrary to this, ..
Why is that?

Pete
 
  • #30
shalayka
126
0
Sorry, what I meant was that it's odd to be able to reference books, but not web pages.

Another quick question: Am I correct in my assumption about the difference in gravitational fields caused by the kinetic energy in curvilinear vs. rectilinear motion? If not, please let me know. I'm not an expert by any means, but I'm trying. :)
 
  • #31
shalayka
126
0
... The reason I'm wondering about this is because where two bodies undergoing rectilinear motion, passing each other at high velocity, it seems non-intuitive that they could appear as black holes to each other (if the rest+kinetic energy vs. body volume ratio were adequate).
 
  • #32
pmb_phy
2,952
1
... The reason I'm wondering about this is because where two bodies undergoing rectilinear motion, passing each other at high velocity, it seems non-intuitive that they could appear as black holes to each other (if the rest+kinetic energy vs. body volume ratio were adequate).
Your intuition is right, it won't become a black hole. It seems that people associate high intensity gravitational fields with black holes and that isn't really the case. Its possible to have a naked singularity in which case the intensity of the gravitational field increases without bound and yet there is no event horizon associated with it. An infinitely long mass filament is just such an object.

If you look carefully at that paper I sent you a link to you will see that the author addresses this very question.

Pete
 
  • #33
pmb_phy
2,952
1
Correction: That last post was inaccurate:

Your intuition is right, it won't become a black hole. I don't understand why people would believe that it would. Granted that the intensity of the gravitational field will increase with an increase in velocity but an object which is not a black hold cannot become a black hole by a mere change in reference frames.

If you look carefully at that paper I sent you a link to you will see that the author addresses this very question.

Pete
 
  • #34
pmb_phy
2,952
1
... The reason I'm wondering about this is because where two bodies undergoing rectilinear motion, passing each other at high velocity, it seems non-intuitive that they could appear as black holes to each other (if the rest+kinetic energy vs. body volume ratio were adequate).
Recall that a black hole is an object whose mass resides entirely within a partcular surface. In the rest frame of the object this surface is a sphere as measured whose radius is less than or equal to the Schwarzschild radius. If that is true in one frame of reference then its true in all frames of reference.

How was the article? Did you find time to read it yet? I have a lot of articles on this whole topic of mass if you'd like to read any of them.

Pete
 
  • #35
shalayka
126
0
Sorry Pete, I got sidetracked in a big way this week, so I didn't end up looking at the articles. :( I would love to read the ones you have on mass as well though! This week will be very quiet for me, so I will definitely have ample time then.
 

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