Definition of the Einstein Tensor

  • Thread starter Dreak
  • Start date
  • #1
52
0
Hello.


The Einstein has following definition (in my course):

Gμε = Rμε - 1/2Rgμε.

But why don't we just:

gμεGμε = gμεRμε - 1/2Rgμεgμε.

<=>

G = R- 1/2 . R . 4 = R- 2R = -R? Is this wrong or.?

Also, what is the meaning of the ricci scalair and tensor?
 
Last edited:

Answers and Replies

  • #2
140
16
Well, no, it's not wrong. That's how you calculate the trace of the Einstein tensor in 4-dimensional spacetime. I'm not sure I understand what your problem is, though.
 
  • #3
5,432
292
The Ricci tensor encapsulates the curvature that can be attributed to the presence of a non-zero EMT on the rhs of the field equations. Clearly ##G_{\mu\nu}=R_{\mu\nu}=0## in the vacuum case.
 
  • #4
52
0
Well, no, it's not wrong. That's how you calculate the trace of the Einstein tensor in 4-dimensional spacetime. I'm not sure I understand what your problem is, though.

When I posted the topic, I realised I made a mistake and that I made a mistake in something I had in mind :). But thanks for the help.

The Ricci tensor encapsulates the curvature that can be attributed to the presence of a non-zero EMT on the rhs of the field equations. Clearly ##G_{\mu\nu}=R_{\mu\nu}=0## in the vacuum case.


So... The rieman tensors tells how space is curved and the Ricci tensor tells how the EMT curves the space?
 
  • #5
87
4
So... The rieman tensors tells how space is curved and the Ricci tensor tells how the EMT curves the space?

The Riemann tensor can be decomposed into the Ricci tensor and the Weyl tensor. The Ricci describes the curvature originating in EMT (immediate presence of energy / momentum), while the Weyl tensor describes curvature coming from far away (distant sources) - for example in the form of gravitational waves, or tidal forces.

The Weyl tensor allows that gravity is propagating through vacuum.
 
  • #6
3,507
26
Hello.


The Einstein has following definition (in my course):

Gμε = Rμε - 1/2Rgμε.

But why don't we just:

gμεGμε = gμεRμε - 1/2Rgμεgμε.

<=>

G = R- 1/2 . R . 4 = R- 2R = -R? Is this wrong or.?

Also, what is the meaning of the ricci scalair and tensor?


So... The rieman tensors tells how space is curved and the Ricci tensor tells how the EMT curves the space?

There is a way to explain the different curvatures in terms of more classic operators like the Laplacian, the Hessian, grad, div...once one understands connections and how they relate to them, too.
The main curvature of course is the one obtained thru the Riemann tensor, the Riemann tensor is obtained from the second derivatives of the metric tensor, this latter has the important peculiarity that it has no gradient (its covariant derivative is vanishing), but if we take the connection as a sort of substitute of the gradient, we can think of the Riemann tensor as the Hessian of the metric tensor. And following the parallelism, we can relate the Ricci tensor (trace of the Riemann tensor) to the generalized Laplacian of the metric tensor.
So the different curvatures like the Ricci, Einstein, Weyl... can be thought of as different manipulations of the trace(well its generalization to tensors, that is, tensor contractions) of the Riemann curvature tensor.
To answer the OP title the Einstein tensor can be defined as the Ricci tensor minus a expression wich divergence be equal to Ricci tensor's divergence(by virtue of the symmetries of the Riemann tensor manifest in the second Bianchi identity), since we want to obtain a divergence-free tensor for physical (conservation) reasons.

This is a bit heuristically obtained so I hope I didn't make any obvious mistake in the above.
 
Last edited:
  • #7
52
0
There is a way to explain the different curvatures in terms of more classic operators like the Laplacian, the Hessian, grad, div...once one understands connections and how they relate to them, too.
The main curvature of course is the one obtained thru the Riemann tensor, the Riemann tensor is obtained from the second derivatives of the metric tensor, this latter has the important peculiarity that it has no gradient (its covariant derivative is vanishing), but if we take the connection as a sort of substitute of the gradient, we can think of the Riemann tensor as the Hessian of the metric tensor. And following the parallelism, we can relate the Ricci tensor (trace of the Riemann tensor) to the generalized Laplacian of the metric tensor.
So the different curvatures like the Ricci, Einstein, Weyl... can be thought of as different manipulations of the trace(well its generalization to tensors, that is, tensor contractions) of the Riemann curvature tensor.
To answer the OP title the Einstein tensor can be defined as the Ricci tensor minus its divergence(by virtue of the symmetries of the Riemann tensor manifest in the second Bianchi identity), since we want to obtain a divergence-free tensor for physical (conservation) reasons.

This is a bit heuristically obtained so I hope I didn't make any obvious mistake in the above.

It made it a bit more clear; shame I mis some basic differential geometry :(.

One last thing

gμεR is not the same as Rμε. Or is it?

Somehow I have the feel it's not right, because you must use the same up and low indices to do this, but on the other hand i have:

gμεR = gμε x gμεRμε = 4Rμε

Something tells me the last stap isn't allowed because there are 2 lowers μ and ε and 1 upper μ
and ε?

I find it all a bit confusing unfortunately :(
 
  • #8
5,432
292
gμεR is not the same as Rμε. Or is it?
No way. Rμε = gμα gεβRαβ

This is correct,

gμεR = gμε gαβ Rαβ

You can't re-use dummy indexes, they are already tied up in the contraction.

Keep going, you'll soon be familiar with index manipulation.
 
Last edited:
  • #9
3,507
26
It made it a bit more clear; shame I mis some basic differential geometry :(

Well I thought maybe relating diff. geometry to generalizations of Calculus II concepts might help, but I don't know, it might confuse even more.
 
  • #10
52
0
No way. Rμε = gμα gεβRαβ

This is correct,

gμεR = gμε gαβ Rαβ

You can't re-use dummy indexes, they are already tied up in the contraction.

Keep going, you'll soon be familiar with index manipulation.

Ok, thanks for the conformation. I was writing it down on paper before I posted and I already thought the triple indices were something messy :).
It has already improved a lot during the day by using it continually and finding all my (stupid) mistakes, but your help (and from the others) have helped a lot too!
 
  • #11
WannabeNewton
Science Advisor
5,815
543
If you know about geodesic congruences, the Ricci tensor is essentially related to the expansion scalar of geodesic congruences. It controls how the volume of a geodesic ball carried along a given geodesic in the congruence change along said geodesic. The Weyl tensor contains information about other changes of the geodesic ball, for example how the ball shears into an ellipsoid as it gets carried along the geodesic.
 
  • #12
3,507
26
If you know about geodesic congruences, the Ricci tensor is essentially related to the expansion scalar of geodesic congruences. It controls how the volume of a geodesic ball carried along a given geodesic in the congruence change along said geodesic. The Weyl tensor contains information about other changes of the geodesic ball, for example how the ball shears into an ellipsoid as it gets carried along the geodesic.

And how does the Einstein tensor fit in the geodesic congruences explanation of curvatures?
 
  • #15
WannabeNewton
Science Advisor
5,815
543
It's not as easy to give ##G_{ab}## by itself as simple and elegant a geometric meaning as ##R_{ab}##. The main reason for taking up ##G_{ab}## as opposed to ##R_{ab}## for the gravitational field equations is of course that ##\nabla^{a}G_{ab} = 0##. Perhaps MTW has a good geometric description of ##G_{ab}##; I'll check (it can take millennia to traverse the pages of that book :p) and let ya know.
 
  • #16
3,507
26
It's not as easy to give ##G_{ab}## by itself as simple and elegant a geometric meaning as ##R_{ab}##.
Yeah, I know.

Perhaps MTW has a good geometric description of ##G_{ab}##; I'll check (it can take millennia to traverse the pages of that book :p) and let ya know.

:rofl:

I'll look it up too.
 
  • #17
52
0
That's a good reference i already knew, thanks. It's too bad the Einstein tensor is only mentioned once to say the obvious: "the `Einstein tensor', is just []a quantity constructed from the Ricci tensor".

Unfortunatly, that's what is written in my course :(.
And that it somehow tells 'spacetime how to bend near matter'.


Ehm guys, another one last question (I somehow have a lot of last questions ^^).

If you got the EMT and you contract it; you get T = Tμμ.
But what is the meaning of this 'T' compared to the tensor? It's the trace but does it have a specific meaning?
Also; let's say I have a certain tensor

Aμε. How do I do the summation here? And what would be the difference with Aμε (so first upper, than lower and vice-versa)?
 
  • #18
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,037
1,352
[Along the lines of the original title/question....]
I have not found a satisfying answer to this question of mine:
What is the dimensionally- and signature-independent meaning of the Einstein tensor?
How would a pure-Riemannian or pseudo-Riemannian differential geometer interpret the Einstein tensor?
 
  • #19
WannabeNewton
Science Advisor
5,815
543
If you have access to it, check out chapter 12 of O'Neill's text "Semi-Riemannian Geometry With Applications to Relativity".
 
  • #20
vanhees71
Science Advisor
Insights Author
Gold Member
18,848
9,717
Unfortunatly, that's what is written in my course :(.
And that it somehow tells 'spacetime how to bend near matter'.


Ehm guys, another one last question (I somehow have a lot of last questions ^^).

If you got the EMT and you contract it; you get T = Tμμ.
But what is the meaning of this 'T' compared to the tensor? It's the trace but does it have a specific meaning?
Also; let's say I have a certain tensor

Aμε. How do I do the summation here? And what would be the difference with Aμε (so first upper, than lower and vice-versa)?

The only thing I can say about it is that it is vanishing when the theory is invariant under rescaling (dilatation invariance), i.e., if the theory doesn't contain any dimensionful couplings or masses.

Scaling invariance, however, is pretty fragile when the field theory is quantized. E.g., QCD with massless quarks is scale invariant and the classical field theory thus shows dilatation symmetry with a vanishing trace of the EMT. In the quantized theory the scale invariance is anomalously broken, and the trace of the energy-momentum tensor does not vanish (as can be seen in lattice-QCD calculations).
 
  • #21
3,507
26
Ehm guys, another one last question (I somehow have a lot of last questions ^^).

If you got the EMT and you contract it; you get T = Tμμ.
But what is the meaning of this 'T' compared to the tensor? It's the trace but does it have a specific meaning?
As commented by vanhees71 the EMT of pure radiation has vanishing trace :T=0.

Also; let's say I have a certain tensor

Aμε. How do I do the summation here?
It is just the tensor in mixed form instead of covariant or contravariant, you use the metric tensor to lower or raise the indices.
And what would be the difference with Aμε (so first upper, than lower and vice-versa)?
In this case all the second order tensors we are dealing with are symmetric so it doesn't make any difference. It does in the general case of course.
 
  • #22
52
0
Aha yes yes! It's all falling into place now!
I took huge leaps in understanding my course today. Thanks a lot guys, couldn't have don't it without you!
 
  • #23
29
1
[Along the lines of the original title/question....]
I have not found a satisfying answer to this question of mine:
What is the dimensionally- and signature-independent meaning of the Einstein tensor?
How would a pure-Riemannian or pseudo-Riemannian differential geometer interpret the Einstein tensor?

I provide an answer this question in my page General Relativity : the Einstein field equation. I just wrote this page in the last days; see also the cosmology page linked there for a more intuitive, less formal description. This explanation is quite brief and assumes some familiarity with tensor calculus; I might further expand it someday.
 
  • #24
11
0
In a d=4 space-time, gravitation is encoded in the metric gμε which is a 4 by 4 symmetric matrix, with therefore 6 degrees of freedom which have to be found by the Einstein's equation. That's great, because Gμε= 8 π G Tμε represent 6 equations due to the symmetry. If you consider only the trace, you will have to solve only one equation for 6 unknown.
 

Related Threads on Definition of the Einstein Tensor

Replies
2
Views
914
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
15
Views
4K
  • Last Post
Replies
23
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
912
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
918
Top