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How to calculate cosmological constant?

  1. Apr 1, 2015 #1

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    I reading book, part about gravity. It is Landau-Lifschiz, "Theory of field". And I trying to understand how to calculate cosmological constant. Re-read part with introduction of cosmological constant several times, still not understood.
    Let's imagine function of space curvature, without presence of mass, is known. In such case it is easy to calculate Cristoffel symbols, Richi tensor etc. But how to find value of cosmological constant from them?
    I have impression what cosmological constant = 1/(R1*R2), where R1 and R2 are main radiuses of curvature. Is it correct?
     
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  3. Apr 1, 2015 #2

    PeterDonis

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    I'm not sure I understand what this function is supposed to be. Is it the metric? Is it the stress-energy tensor? Is it something else?
     
  4. Apr 1, 2015 #3

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    I saying about geometry, so metric.
     
  5. Apr 1, 2015 #4

    PeterDonis

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    Then you compute the Einstein tensor of that metric, and divide it by ##8 \pi G / c^4## to convert it to a stress-energy tensor, in accordance with the Einstein Field Equation. If that stress-energy tensor has a part that is a constant multiple of the metric, then that is the cosmological constant. (More precisely, it's minus the cosmological constant, since it's on the RHS of the EFE.)
     
  6. Apr 4, 2015 #5

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    If tensor divide by something, it still would be tensor.
    If calculate cosmological constant from metric, it is equal to
    [itex]\Omega=\frac{1}{2} g^{lm} g^{ik} R_{iklm}[/itex]
    However, I think the equation is too complex. May it be found from curvature of space, radiuses of curvature?
     
    Last edited by a moderator: Apr 4, 2015
  7. Apr 4, 2015 #6

    Nugatory

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    I don't understand how you could come up with that expression from the metric (which metric?)... But if you take advantage of the symmetries ##g_{ij}=g_{ji}## and ##R_{ijkl}=-R_{jikl}## that expression simplifies down to zero.
     
  8. Apr 4, 2015 #7

    PeterDonis

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    Yes, just with different units--stress-energy units in this case, instead of curvature units.

    That's not what I said. I said, if the stress-energy tensor is a constant multiple of the metric, i.e., ##\Lambda g_{ab}## with ##\Lambda## a constant. You don't have to "calculate" a constant from the metric or anything else.
     
  9. Apr 4, 2015 #8

    wabbit

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    Is it possible to identify a CC from the metric alone ? I mean, if the stress energy tensor is not otherwise constrained, you can always add a term ## \Lambda g_{ab} ## to both sides, is there a condition that identifies a preferred value of ## \Lambda ##?
     
  10. Apr 4, 2015 #9

    PeterDonis

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    Yes. You do what I described in post #4. If you know the metric, you know its Einstein tensor (just compute it). If the Einstein tensor is a multiple of the metric, then that multiple is the cosmological constant. (The bit in post #4 about dividing by ##8 \pi G / c^4## is just to change the units, as I noted in post #7. You can tell whether you have a multiple of the metric without doing that.)

    For example, take the metric for de Sitter spacetime (there are actually several different ones, depending on which chart you choose). Compute the Einstein tensor of that metric. You will find that it is a multiple of the metric--i.e., as matrices, the Einstein tensor is just some ##\Lambda## times the metric tensor. That multiple ##\Lambda## is the cosmological constant (with a minus sign).

    Yes, but that changes the Einstein tensor, which changes the metric--i.e., the metric that gives rise to this modified Einstein tensor is different from the metric that gave rise to the original Einstein tensor (before you added ##\Lambda g_{ab}## ).
     
    Last edited: Apr 4, 2015
  11. Apr 4, 2015 #10

    wabbit

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    What I meant is, the metric ## g ## gives you the Einstein tensor ## G ##, and the EFE says ## G+\Lambda g = 8\pi T ## . If ## (\Lambda, T) ## is a solution, so is ## (\Lambda+x,T+\frac{x}{8\pi} g) ##. My question was, is there a condition that fixes x ?

    (I do not argue with your statement that if the Einstein tensor is precisely a multiple of the metric, then there is a natural choice with T=0, but this is about the general case - is there a natural choice there too ? )
     
    Last edited: Apr 4, 2015
  12. Apr 4, 2015 #11

    PeterDonis

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    Experimental data does, because experimental data tells you the stress-energy tensor. Your transformation changes the stress-energy tensor from ##T## to ##T + g x / 8 \pi##.

    To put it another way, ##\Lambda## has a physical meaning: it tells you the degree to which the Einstein tensor fails to vanish if ##T = 0##, i.e., if there is no matter or energy present (other than ##\Lambda## itself). That can be measured experimentally, though of course it isn't easy (the only measurement of it we have currently is the data that tells us the expansion of the universe is accelerating). And once you have that value for ##T = 0##, you use the same value in regions of spacetime where ##T \neq 0##.
     
    Last edited: Apr 4, 2015
  13. Apr 4, 2015 #12

    wabbit

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    OK so going back to OP original question, I suppose we can say

    - the CC is unrelated to radii of curvature,

    - you can determine it from the curvature, through Einstein's tensor, using the EFE, provided you have a little bit of information about the stress-energy tensor and its sources. If for instance the sources are known/assumed to be matter and radiation, it is enough to measure G anywhere far away from any source and use the fact that T=0 there : the Einstein tensor as a whole is proportional to the metric at any such point, and ##\Lambda=-\frac{G_{00}}{g_{00}} ##.
     
  14. Apr 4, 2015 #13

    PeterDonis

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    I'm not sure I would say that, since, as you note, you can determine the CC from the curvature. You just need to measure the curvature in a region where there are no sources present.
     
  15. Apr 4, 2015 #14

    wabbit

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    Ah yes you're right, thanks for the correction.

    This would be a much better, coordinate-free way to express the CC than using ## \Lambda=-\frac{G_{00}}{g_{00}} ## at such a point.

    I am not sure what the exact result is, is it as simple as CC=scalar curvature ?

    And even if it's not exactly practical, I like this recipe for measuring the CC : go to outer space, find a really quiet place away from any matter, electric or magnetic field and such, and measure the sum of the angles, and surface, of a triangle. A little easy formula and here's your CC : )
     
    Last edited: Apr 4, 2015
  16. Apr 4, 2015 #15

    PeterDonis

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    Almost. The Ricci scalar ##R## is equal to ##4 \Lambda## for de Sitter spacetime.

    One triangle won't be enough, and they won't all be ordinary spatial triangles; remember that you're measuring spacetime curvature, not space curvature.
     
  17. Apr 5, 2015 #16

    wabbit

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    OK that's simple enough for me thanks : )
    Spatial curvature should be OK though since we're at a point where T=0, no ? In suitable coordinates I suppose we must have for Ricci curvature, ## R=diag(-\Lambda,\Lambda,\Lambda,\Lambda) ## or something close to that so I thought one triangle would be enough. If not, one tetrahedron then ?
     
    Last edited: Apr 5, 2015
  18. Apr 5, 2015 #17

    PeterDonis

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    No. Space curvature is coordinate-dependent. Spacetime curvature is not.

    For example, consider de Sitter spacetime in the flat slicing, i.e., each spatial slice has zero curvature. The spacetime curvature is not zero: as I said before, the Ricci scalar is ##R = 4 \Lambda##. But in this slicing, since the space curvature is zero, the only way to measure ##R## is to look at things that happen over time--for example, to see that initially "comoving" test particles diverge: they accelerate away from each other.
     
  19. Apr 5, 2015 #18

    wabbit

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    Thanks, I see. No way around that de Sitter case, we do need both... OK I am now aiming a laser beam from one fixed point towards the 4 vertices of my tetrahedron and measuring the aiming angles required to hit the targets - that ought do do it this time - unless those zero-length paths break it somehow, then I'll throw billiard balls : )
     
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