General Relativity

[Perpetual Student]

In my attempts to get a better handle on the mathematics of GR, new struggles have surfaced. So, here is what I hope is a straightforward question for the physicist here. In the Einstein field equations, why is there a "scalar curvature factor" multiplied by the metric tensor as well as the curvature tensor? This apparent redundancy obviously has some purpose, which is not well described (at least, for me) in the materials I have looked at? I notice that these are sometimes combined in a so called Einstein tensor. Any responses would be appreciated.

 

[sol invictus]

In my attempts to get a better handle on the mathematics of GR, new struggles have surfaced. So, here is what I hope is a straightforward question for the physicist here. In the Einstein field equations, why is there a "scalar curvature factor" multiplied by the metric tensor as well as the curvature tensor? This apparent redundancy obviously has some purpose, which is not well described (at least, for me) in the materials I have looked at? I notice that these are sometimes combined in a so called Einstein tensor. Any responses would be appreciated.

Well, a somewhat unsatisfactory answer is that without that term, Einstein's equations wouldn't correctly describe gravity.

A more specific answer is that without that term, energy would not be locally conserved. Mathematically, conservation of any quantity means that the time derivative of the density at a point is equal to the divergence of the flow away from that point (basically, if the density of some stuff is decreasing somewhere it must be because some net amount of stuff is flowing away from that point).

If you take the appropriate derivatives of Einstein's equations (take the covariant derivative of both sides, contracted with one index), you see that they guarantee conservation of stress-energy - the left-hand side is zero due to a geometrical identity (called the "Bianchi identity"), and therefore the right-hand side must vanish as well.

If you eliminated the term you asked about, the derivative of the left-hand side wouldn't be zero, and stress-energy wouldn't be conserved.

 

[Perpetual Student]

OK, thanks, I'll struggle with the above for a bit.
In the meantime, All I can find on the Internet relating to the Einstein field equations is the same tensor equation using the sum of the curvature tensor, the scalar curvature times the metric tensor, Λ times the metric tensor = the stress energy tensor with G, Newton's gravitational constant, etc.
But, Ι inevitably see reference made to a set of 10 partial differential equations, but those 10 equations are never shown. Does anyone have a link or other source to those 10 equations which might include some descriptive text? I assume these 10 equations are all included in the concise form of these tensor equations.
I am hoping the 10 equations might be a bit more transparent (with a little descriptive help) since the tensor equations are so densely packed with components.

 

[Vorpal]

But, Ι inevitably see reference made to a set of 10 partial differential equations, but those 10 equations are never shown. Does anyone have a link or other source to those 10 equations which might include some descriptive text? I assume these 10 equations are all included in the concise form of these tensor equations.

Yes, they are: you have a symmetric rank-2 tensor in four dimensions, so its number of independent components is 10.

I am hoping the 10 equations might be a bit more transparent (with a little descriptive help) since the tensor equations are so densely packed with components.

All that would do is pick out the individual components of the tensor in some frame.

The Riemann curvature tensor describes are the gravitational tidal forces, i.e., how nearby geodesics deviate from one another. The Ricci curvature tensor can be interpreted as describing how the volume of a close ball of test particles behaves.

With that in mind, if you contract Einstein's equation, you will see the that the Ricci scalar R = R^α_α is -(8πG)T, i.e., proportional to the trace of the stress-energy tensor. So an equivalent formulation of Einstein's equation is:
[latex]$ R_{\mu\nu} = 8\pi G\left(T_{\mu\nu} - \frac{1}{2}Tg_{\mu\nu}\right) $[/latex]
In other words, what the term you're asking about introduces is a correction of energy density and the some pressure terms (in general, spatial momentum in the corresponding spatial direction) through the trace of the stress-energy tensor. Pressure gravitates.

 

[Farsight]

OK, thanks, I'll struggle with the above for a bit. In the meantime, All I can find on the Internet relating to the Einstein field equations is the same tensor equation using the sum of the curvature tensor, the scalar curvature times the metric tensor, Λ times the metric tensor = the stress energy tensor with G, Newton's gravitational constant, etc. But, Ι inevitably see reference made to a set of 10 partial differential equations, but those 10 equations are never shown. Does anyone have a link or other source to those 10 equations which might include some descriptive text? I assume these 10 equations are all included in the concise form of these tensor equations. I am hoping the 10 equations might be a bit more transparent (with a little descriptive help) since the tensor equations are so densely packed with components.

Try this:

http://www.maths.qmul.ac.uk/~agp/MTH6123/co_notes20_10.pdf

It's lecture 20 of a Mathematical Cosmology course given by Dr AG Polnarev of University of London. Note the mention of "shear stress". Interestingly you don't hear much mention of elasticity on a forum like this, but you can find it if you look for it. You can access all the lectures at http://www.maths.qmul.ac.uk/~agp/MTH6123/ and see his webpage at http://www.maths.qmul.ac.uk/~agp/ . He's got a more current course, but it looks like it derives the EFEs rather than giving historical info.

 

[Reality Check]

Try this:

http://www.maths.qmul.ac.uk/~agp/MTH6123/co_notes20_10.pdf

It's lecture 20 of a Mathematical Cosmology course given by Dr AG Polnarev of University of London. Note the mention of "shear stress". Interestingly you don't hear much mention of elasticity on a forum like this, but you can find it if you look for it. You can access all the lectures at http://www.maths.qmul.ac.uk/~agp/MTH6123/ and see his webpage at http://www.maths.qmul.ac.uk/~agp/ . He's got a more current course, but it looks like it derives the EFEs rather than giving historical info.

Shear stress is the standard term for the off-diagonal components of any stress tensor. GR has a stress-energy tensor with additional energy components.

Actually deriving the EFEs is more useful than just giving the historical info.

Another good source of GR education is Leonard Susskind's YouTube lecures on Modern Physics concentrating on General Relativity (recorded September 22, 2008 at Stanford University).

 

[Perpetual Student]

Thanks for all the explanations and information -- and all the great suggestions for further reading and lectures. I intend to review all of them over the next several weeks -- or months -- I do have lots of time.
I have been making some slow progress with some of the sources given earlier on tensors. I am taking the approach of first getting a handle on how tensors are used in physics before making any attempt to fathom the mathematics of GR. As I mentioned earlier, it's my ultimate hope to obtain a general feel for how Einstein's amazing tensor equation determines all of the remarkable things discussed here. If nothing else, it's forcing the neurons of this 72 year old brain to keep firing away -- which, I am told, may keep dementia at bay for a few more years.

 

[ynot]

Thanks for all the explanations and information -- and all the great suggestions for further reading and lectures. I intend to review all of them over the next several weeks -- or months -- I do have lots of time.
I have been making some slow progress with some of the sources given earlier on tensors. I am taking the approach of first getting a handle on how tensors are used in physics before making any attempt to fathom the mathematics of GR. As I mentioned earlier, it's my ultimate hope to obtain a general feel for how Einstein's amazing tensor equation determines all of the remarkable things discussed here. If nothing else, it's forcing the neurons of this 72 year old brain to keep firing away -- which, I am told, may keep dementia at bay for a few more years.

You're doing what I would like to have the time and patience to do. As I share many of your “what ifs” and “yeah buts” I look forward to knowing how you “get on” with Relativity after your study. Assuming you will let us know of course (no pressure ;-).

 

[Perpetual Student]

Tensors are not for the faint of heart.

 

[dasmiller]

Tensors are not for the faint of heart.

If they were, they'd be Relaxors.

 

[Perpetual Student]

I would appreciate any responses from the physicists here.
In my review of the properties and algebra of tensors, it has been made clear that a tensor equation holds true regardless of any transformation of coordinate system. This is simply a mathematical consequence of how tensors are defined and constructed. This question may be a bit premature since, I have not yet studied the specifics of the EFE, but it's very compelling at this point in my studies. Is the consensus opinion expressed here regarding the equivalence of all frames of reference under GR due to this aspect of tensors and the fact that Einstein chose the tensor equation below to express his theory?

[latex]R_\mu_\nu - \dfrac{1}{2}g_\mu_\nu R + g_\mu_\nu \Lambda = \dfrac{8\pi G}{c^4}T_\mu_\nu[/latex]

An interesting corollary to this question is, could their be a "non-tensor" formulation of the field equations? Can this last question be rejected as trivially invalid for some easily expressed reason?

 

[Vorpal]

Disclaimer: not a physicist.

Is the consensus opinion expressed here regarding the equivalence of all frames of reference under GR due to this aspect of tensors and the fact that Einstein chose the tensor equation below to express his theory?

To the first part: no! Merely having tensors is not enough. For example, it's possible to have a tensorial law of Newtonian gravitation in Minkowksi spacetime, at the cost of introducing a fixed "God-given" field that singles out a specific frame in which Newton's law of gravitation holds--but can still be written with tensors.

To the second part: sort of, but more the reverse. The equivalence of all frames of reference in GTR is actually a byproduct of an even stronger principle, the principle of general covariance, which is responsible for the field equation. General covariance requires all frames to be equivalent, but it actually goes deeper than that (to be fair, Einstein himself wasn't very clear in expressing it). An example to illustrate the difference: Nordström's theory of gravitation is...
[latex]\begin{eqnarray*} R = 24\pi T & & C_{\alpha\beta\mu\nu} = 0 \end{eqnarray*}[/latex]
where C_αβμν is the Weyl curvature tensor. Here, it's expressed in a purely tensorial manner, and it's also does not single out any special frame. But it has something GTR does not: immutable requirements on spacetime geometry; in this case, no matter what the matter distribution is, spacetime must be conformally flat (no Weyl curvature).

But GTR doesn't do any of that. There is no "prior geometry".

An interesting corollary to this question is, could their be a "non-tensor" formulation of the field equations? Can this last question be rejected as trivially invalid for some easily expressed reason?

The field equation is equivalent to the following statement: in every local Lorentz frame,
[latex]\[ R_{00} = 4\pi G\left(\rho + p_x + p_y + p_z\right) \][/latex]
Where p_x pressure in the x-direction, etc. In such a frame, the physical meaning of the Ricci tensor component R₀₀ is that it's the negative of relative acceleration of the volume of a small ball of test particles that are at rest (i.e., -V̈/V), under gravitational effects only.

ETA: Λ = 0 above. A nonzero Λ is equivalent to shifting the energy density and pressures everywhere.

 

[Farsight]

Disclaimer: I'm not a physicist either. I'm sometimes called an armchair physicist or an amateur physicist, but I'm just an IT guy with a computer science degree and a special interest.

Vorpal, should that be stress rather than pressure for the last three terms? Only I was reading this, and it says Pressure is a scalar (non-directional) quantity; stress is directional.

Perpetual Student: it might be worth your while asking the same question about elasticity.

 

[Vorpal]

Yep; I'm abusing the term. It's T_xx, the flow of x-momentum in the x-direction, or the normal stress in that direction. It would have better to parallel the usual definition of pressure as the average of those stresses, come to think of it. Oh well.

 

[Farsight]

Good man Vorpal. I only mentioned it because in a previous response here I referred to this.

... As I mentioned earlier, it's my ultimate hope to obtain a general feel for how Einstein's amazing tensor equation determines all of the remarkable things discussed here....

For a general feel, think in terms of "elastic space". Imagine you've got a large glass box full of transparent jelly. Now imagine you can insert a very long hypodermic needle and inject more jelly into the centre, then withdraw it and mend the hole in the glass. You've now got a spherical pressure gradient, the pressure diminishing as you move outwards from the centre of the jelly. That's your gravitational field. If you took a horizontal slice through it and plotted the pressure, you'd end up drawing something like the "upturned hat" on the wikipedia gravitational potential page. Only the hat would be the right way up.

If nothing else, it's forcing the neurons of this 72 year old brain to keep firing away -- which, I am told, may keep dementia at bay for a few more years.

IMHO the trick to understand what the maths is telling you is to understand what the terms mean and relate them to the real world. You know that a gravitational field is in space. If you see 4π you know we're talking about a sphere. If you see p_x you know we're talking spatial pressure/stress in the x direction, and so on.

 

[Perpetual Student]

The only "trick" that I know to understanding any mathematics is to actually understand the mathematics.

 

[sol invictus]

In my review of the properties and algebra of tensors, it has been made clear that a tensor equation holds true regardless of any transformation of coordinate system. This is simply a mathematical consequence of how tensors are defined and constructed. This question may be a bit premature since, I have not yet studied the specifics of the EFE, but it's very compelling at this point in my studies. Is the consensus opinion expressed here regarding the equivalence of all frames of reference under GR due to this aspect of tensors and the fact that Einstein chose the tensor equation below to express his theory?

[latex]R_\mu_\nu - \dfrac{1}{2}g_\mu_\nu R + g_\mu_\nu \Lambda = \dfrac{8\pi G}{c^4}T_\mu_\nu[/latex]

It's certainly true that the equation as written indicates that GR is generally covariant, because the equation above is true and looks exactly the same in all coordinate systems. For that to be possible requires tensors, or something like them.

An interesting corollary to this question is, could their be a "non-tensor" formulation of the field equations? Can this last question be rejected as trivially invalid for some easily expressed reason?

You can pick a coordinate system, and then write out the components of Einstein's equations in those coordinates.

 

[sol invictus]

The equivalence of all frames of reference in GTR is actually a byproduct of an even stronger principle, the principle of general covariance, which is responsible for the field equation. General covariance requires all frames to be equivalent, but it actually goes deeper than that (to be fair, Einstein himself wasn't very clear in expressing it). An example to illustrate the difference: Nordström's theory of gravitation is...
[latex]\begin{eqnarray*} R = 24\pi T & & C_{\alpha\beta\mu\nu} = 0 \end{eqnarray*}[/latex]
where C_αβμν is the Weyl curvature tensor. Here, it's expressed in a purely tensorial manner, and it's also does not single out any special frame. But it has something GTR does not: immutable requirements on spacetime geometry; in this case, no matter what the matter distribution is, spacetime must be conformally flat (no Weyl curvature).

But GTR doesn't do any of that. There is no "prior geometry".

I don't understand the distinction you're making here. It sounds like you're saying (to translate to my language) that GR allows gravity waves, while Nordstrom's theory doesn't. Said another way, that specifying the stress-energy tensor in a region doesn't uniquely specify the metric.

That's true, but I don't see what it has to do with general covariance or the existence of a "prior geometry". It just means that there are some additional degrees of freedom besides matter. If you were to specify the appropriate boundary conditions on those degrees of freedom - meaning in the case of GR that you specify the metric and some derivatives on a Cauchy surface - you do fully specify the solution, including the geometry. I don't see how that's different from (say) including a scalar field in your theory, and then having to specify its initial conditions.

To me, any theory specified by an action that's an integral of a scalar density over spacetime is generally covariant. Do you disagree with that?

 

[Perpetual Student]

...
You can pick a coordinate system, and then write out the components of Einstein's equations in those coordinates.

If one were to do that, would it become immediately obvious that such an equation would violate some known experimental result?

 

[sol invictus]

If one were to do that, would it become immediately obvious that such an equation would violate some known experimental result?

The equations would be correct only in that one coordinate system. So they wouldn't correctly predict experimental results unless that coordinate system happened to coincide precisely with the one relevant to the experiment.

So the answer to your question is "yes" - either that, or you'd be smart enough to realize that the equations simply don't apply except in one special frame.