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Definition of the Einstein Tensor

  1. Aug 16, 2013 #1

    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: Aug 16, 2013
  2. jcsd
  3. Aug 16, 2013 #2
    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.
  4. Aug 16, 2013 #3


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    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.
  5. Aug 16, 2013 #4
    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.

    So... The rieman tensors tells how space is curved and the Ricci tensor tells how the EMT curves the space?
  6. Aug 16, 2013 #5
    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.
  7. Aug 16, 2013 #6

    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: Aug 16, 2013
  8. Aug 16, 2013 #7
    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 :(
  9. Aug 16, 2013 #8


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    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: Aug 16, 2013
  10. Aug 16, 2013 #9
    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.
  11. Aug 16, 2013 #10
    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!
  12. Aug 16, 2013 #11


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    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.
  13. Aug 16, 2013 #12
    And how does the Einstein tensor fit in the geodesic congruences explanation of curvatures?
  14. Aug 16, 2013 #13


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  15. Aug 16, 2013 #14
    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".
  16. Aug 16, 2013 #15


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    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.
  17. Aug 16, 2013 #16
    Yeah, I know.


    I'll look it up too.
  18. Aug 16, 2013 #17
    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)?
  19. Aug 16, 2013 #18


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    [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?
  20. Aug 16, 2013 #19


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    If you have access to it, check out chapter 12 of O'Neill's text "Semi-Riemannian Geometry With Applications to Relativity".
  21. Aug 16, 2013 #20


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    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).
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