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Einstein tensor with the cosmological constant present.

  1. Dec 3, 2012 #1
    I can fairly well grasp the trace relationship between the Einstein tensor and the Ricci tensor, and see that Ricci tensor is a multiple of the metric. If the cosmological constant is included I don't get why the Einstein tensor shouldn't become a multiple of the metric (leaving out physical considerations) to still achieve a Ricci flat manifold. Any text suggestions to help me explore this appreciated.

    [tex]G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
    [tex]g^{\mu\nu} G_{\mu\nu}=g^{\mu\nu} R_{\mu\nu}-\frac{1}{2}R g^{\mu\nu} g_{\mu\nu}=0[/tex]
    [tex]G=R-2R=0[/tex]
    [tex]G=-R=0[/tex]

    or


    [tex]G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
    [tex]G_{\mu\nu}-\Lambda g_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
    [tex]\Lambda g_{\mu\nu}-\Lambda g_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
    [tex]\Lambda g^{\mu\nu} g_{\mu\nu}-\Lambda g^{\mu\nu} g_{\mu\nu}=g^{\mu\nu} R_{\mu\nu}-\frac{1}{2}R g^{\mu\nu}g_{\mu\nu}=0[/tex]
    [tex]4\Lambda - 4\Lambda =R- 2R=-R=0[/tex]
     
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  3. Dec 3, 2012 #2

    Mentz114

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    As far as I know the cosmological constant adds curvature so that if R=0, then [itex]G^a_a=g^a_a\Lambda[/itex]. This is your equation
    [tex]
    G_{\mu\nu}-\Lambda g_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0
    [/tex]
    with R=0
     
  4. Dec 3, 2012 #3

    PeterDonis

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    That's one way of writing it (although there is usually a minus sign on the RHS when written in this form; see below). However, it doesn't result in what you and Messenger have been writing. The complete Einstein Field Equation, including the cosmological constant, is:

    [tex]G_{ab} + \Lambda g_{ab} = R_{ab} - \frac{1}{2} g_{ab} R + \Lambda g_{ab} = 8 \pi T_{ab}[/tex]

    If we have a spacetime with no matter or radiation present, then [itex]T_{ab} = 0[/itex] and we have

    [tex]G_{ab} + \Lambda g_{ab} = R_{ab} - \frac{1}{2} g_{ab} R + \Lambda g_{ab} = 0[/tex]

    In other words, the presence of a cosmological constant doesn't change the definition of the Einstein tensor; it's still [itex]G_{ab} = R_{ab} - 1/2 g_{ab} R[/itex]. What the cosmological constant does is give more possible solutions for [itex]G_{ab}[/itex] when [itex]T_{ab} = 0[/itex]. One way of interpreting this is that the cosmological constant provides another form of "stress-energy" which isn't ordinary matter or radiation. (Sometimes the term "dark energy" is used for this, although that term is also used to refer to other ways for non-zero stress-energy to be present without any normal matter or radiation.) In other words, we rearrange the second equation above to

    [tex]G_{ab} = R_{ab} - \frac{1}{2} g_{ab} R = - \Lambda g_{ab}[/tex]

    So the cosmological constant now looks like a form of "stress-energy tensor" that is proportional to the metric; we just write [itex](T_{ab})_{\Lambda} = - ( \Lambda / 8 \pi ) g_{ab}[/itex].
     
  5. Dec 3, 2012 #4
    My thoughts on this were that R=0 implies Minkowski space [-1,1,1,1] (such that [tex]g^{\mu\nu}g_{\mu\nu}=4[/tex], so that [tex]\Lambda g_{\mu\nu}[/tex] is a rank 0 tensor implying zero curvature but that would also imply the Einstein tensor has vanished. If the Einstein tensor also became a four dimensional rank 0 tensor then it would seem that I could still get a Ricci flat manifold with a cosmological constant. I don't quite grasp how the cosmological constant can be a rank 0 tensor representing at the same time a Euclidean scalar field and curvature when the Einstein tensor vanishes.
     
  6. Dec 3, 2012 #5
    I see where MTW:Gravitation show that equation as a "redefintion" of the Einstein tensor, but they state that the theory is no longer geometric when written that way. If there is no proof for that, doesn't that bother you? They write it as:
    [tex]"G_{\alpha\beta}"=R_{ab} - \frac{1}{2} g_{ab} R + \Lambda g_{ab}=G_{ab} + \Lambda g_{ab}[/tex]
     
  7. Dec 3, 2012 #6

    PeterDonis

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    Yes, you could call it a "redefinition" of the Einstein tensor, if you want to insist on only putting ordinary matter and radiation on the RHS of the EFE. MTW's discussion, which mentions drawbacks to redefining the Einstein tensor this way, assumes, IIRC, that there is no feasible alternative to that; when MTW was published in 1973, the alternate view that I gave in my last post, of treating the cosmological constant term as a kind of "stress-energy" and putting it on the RHS of the EFE, so that the LHS remains the standard Einstein tensor, was, AFAIK, not very well-developed. Today it's much better developed.

    What do you mean by "no proof"? The EFE including cosmological constant is actually easier, in a sense, to "prove" than the standard EFE, at least if you derive it from a Lagrangian. The Einstein-Hilbert Lagrangian, from which the standard EFE is derived by variation with respect to the metric (after integrating over all spacetime), is

    [tex]\frac{1}{16 \pi} R \sqrt{-g}[/tex]

    Hilbert and Einstein claimed that this was the right Lagrangian because it was the most general one using no greater than second derivatives of the metric. But they hadn't realized that that was, strictly speaking, false: the most general such Lagrangian also includes a constant term:

    [tex]\frac{1}{16 \pi} \left( R + 2 \Lambda \right) \sqrt{-g}[/tex]

    Varying this with respect to the metric gives the EFE with a cosmological constant, as I wrote it down. So there is certainly "proof" of that version. So no, I'm not bothered.

    (Technical note: I left out the Lagrangian due to ordinary matter and radiation in the above, so strictly speaking, varying what I wrote above gives the vacuum EFE, with or without a cosmological constant. Adding the matter action term doesn't change anything I said, it just adds the ordinary stress-energy tensor [itex]T_{ab}[/itex] on the RHS of the EFE.)
     
  8. Dec 3, 2012 #7

    PeterDonis

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    No, it doesn't. Any vacuum spacetime has R = 0. For example, Schwarzschild spacetime has R = 0, and it certainly isn't Minkowski space.
     
  9. Dec 3, 2012 #8
    Peter,
    Thanks for some thought provoking comments, I will be back later with more questions after researching more in depth what you have stated.
     
  10. Dec 3, 2012 #9

    PeterDonis

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  11. Dec 3, 2012 #10
    From http://eagle.phys.utk.edu/guidry/astro490/lectures/lecture490_ch9.pdf

    it states that g22 and g33 are the same as flat spacetime, and that g00 and g11 are subject to the boundary conditions that spacetime becomes flat at infinity. Isn't the [tex]1-\frac{2M}{r}[/tex] the same derivation for Newtonian gravity, as in a small perturbation away from flat spacetime? Isn't this solution for Ruv=0 Schwarzschild solution only applicable since it is so near Minkowski space?

    This is also what I don't understand. The cosmological constant for Minkowski space would seem to be added to the "1" with the perturbation becoming the Newtonian phi. How does one exactly derive the cosmological constant into the formula for Newtonian gravity? I have seen the result that Hobbs et. al. arrived at General Relatvity: An Introduction for Physicists but not the actual logic in the derivation.
     
  12. Dec 4, 2012 #11

    PeterDonis

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    No. The correspondence with Newtonian gravity helps to confirm that the "M" that appears in the metric is the externally measured mass of the body; but there's nothing in the derivation that requires g_00 or g_11 to be close to their Minkowski values.

    Are you thinking of de Sitter spacetime?

    http://en.wikipedia.org/wiki/De_Sitter_space

    The static coordinates on this spacetime do indeed make the cosmological constant look like a Newtonian "potential". However, that doesn't mean you have to be able to "derive" de Sitter spacetime as a perturbation on Minkowski spacetime. Once again, there's nothing that requires the "correction" term in the de Sitter metric in static coordinates to be small compared to 1.

    I don't see why you would have to. GR is partly motivated by a desire to modify Newtonian gravity to take account of relativity; but that doesn't mean GR is derived starting with formulas from Newtonian gravity. You can derive the Einstein Field Equation, and solve it, without knowing or caring about Newtonian gravity.
     
  13. Dec 6, 2012 #12
    Peter,
    If for [tex]\frac{1}{16 \pi} R \sqrt{-g}[/tex] [tex]R \rightarrow 0 [/tex]
    does that mean for
    [tex]\frac{1}{16 \pi} \left( R + 2 \Lambda \right) \sqrt{-g}[/tex] [tex]R \rightarrow -2\Lambda[/tex]?
     
  14. Dec 6, 2012 #13

    PeterDonis

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    The definition of [itex]R[/itex] is independent of whether [itex]\Lambda[/itex] is zero or nonzero. So my answer would be "no". However, I'm not entirely sure what you mean by [itex]R \rightarrow 0[/itex]. Can you clarify what kind of scenario you're imagining?
     
  15. Dec 6, 2012 #14
    Based upon:
    http://en.wikipedia.org/wiki/Scalar_curvature

    so to let it also "vanish" with a constant present in Euclidean space (Minkowski), wouldn't R have to become equal to [itex]-2\Lambda[/itex]?
     
  16. Dec 6, 2012 #15

    PeterDonis

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    If [itex]\Lambda[/itex] is nonzero, but nothing else is present (i.e., the stress-energy tensor [itex]T_{ab}[/itex] is zero). it isn't Minkowski space. It's either de Sitter or anti-de Sitter space, depending on whether [itex]\Lambda[/itex] is positive or negative. Both of these spaces have nonzero scalar curvature. See:

    http://en.wikipedia.org/wiki/De_Sitter_space

    http://en.wikipedia.org/wiki/Anti_de_Sitter_space
     
  17. Dec 6, 2012 #16
    I need to read up more on the derivation of the LaGrangian that you showed. I would think there would be some correlation between R and [itex]\Lambda[/itex] for them to show up together in it, similar to potential and kinetic energy.
     
  18. Dec 6, 2012 #17

    Mentz114

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    In this paper, Padmanabhan proposes an action for the gravitational field equations such that the cosmological constant arises as a constant of integration.

    It's here http://uk.arxiv.org/abs/gr-qc/0609012v2
     
  19. Dec 6, 2012 #18

    PeterDonis

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    I don't have a handy online reference, but MTW discusses it. See below for a quick summary.

    No, there isn't. They have nothing to do with each other, other than appearing in the same Lagrangian.

    The idea behind the Lagrangian is to construct the most general Lorentz scalar that is composed of no higher than second derivatives of the metric. There are only two such scalars possible: the scalar curvature [itex]R[/itex], composed of second derivatives of the metric, and a constant [itex]\Lambda[/itex], which is a "zeroth derivative" of the metric. So the most general Lagrangian is just the sum of those two terms (the factor of 2 in front of [itex]\Lambda[/itex] is there so that the field equation will just contain [itex]\Lambda[/itex] instead of [itex]\Lambda / 2[/itex]). But there's no other connection between them; they are completely independent of each other as far as GR is concerned.

    The [itex]\sqrt{-g}[/itex] is there to make a Lorentz invariant integration measure (since we're going to integrate the Lagrangian over all spacetime and then vary it with respect to the metric to obtain the field equation). The factor of [itex]1 / 16 \pi[/itex] is for convenience, to make certain formulas look the way physicists were used to having them look.
     
  20. Dec 6, 2012 #19
  21. Dec 6, 2012 #20
    Gives me quite a bit to look into and think about. I really do appreciate you taking your time on this board Peter. Thanks.
     
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