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Question about Einstein inserting the cosmological constant

  1. Aug 25, 2012 #1
    In Gravitation by MTW in section 17.3, they state
    (I assume it was a typo putting the cosmological constant in the center portion)

    Changing the symbol of the Einstein tensor and the cosmological constant to L and Ω so as not to confuse myself, I can see how with [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}+\Omega g_{\alpha\beta}[/tex] if I first set [tex]\Omega=0[/tex] I then get
    [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}[/tex] which is of course nothing more than the EFE. But if instead I take [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}+\Omega g_{\alpha\beta}[/tex] but let [tex]\Omega\neq 0[/tex] I get [tex]L_{\alpha\beta}+\Omega g_{\alpha\beta}=0[/tex] so that [tex]L_{\alpha\beta}=-\Omega g_{\alpha\beta}[/tex] when spacetime is flat. If I then set [tex]\Omega=0[/tex] I then end up again with the Einstein tensor [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}[/tex] by pulling a -1 out of the coefficients since [tex]\Omega g_{\alpha\beta}=L_{\alpha\beta}+G_{\alpha\beta}[/tex].
    So what I am not understanding is if Einstein required a multiple of the metric in the field equation but it caused him so much anguish to not have flat spacetime, why didn't he just go with a [tex]\Omega g_{\alpha\beta}-L_{\alpha\beta}=G_{\alpha\beta}[/tex] substitution to do so?
     
  2. jcsd
  3. Aug 25, 2012 #2

    PeterDonis

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    Staff: Mentor

    No, it wasn't. See below.

    That's not what Einstein did; at least, not if I'm understand you correctly. Also, you have left out the RHS of the EFE; perhaps it will help clarify things if we put it back in, and also if we always expand out the Einstein tensor (the original one) to avoid confusion.

    The original EFE was (I'll use a and b for the indexes since I'm too lazy to type the LaTeX for Greek letters all the time :redface:):

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

    There are no solutions to the above equation that describe a static universe containing a non-zero density of matter (i.e., with [itex]T_{ab}[/itex] nonzero). Einstein's proposal to allow such a solution was to modify the equation to:

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

    where [itex]\Lambda[/itex] is a constant that is the same everywhere in the spacetime, the "cosmological constant". This modified equation *does* allow solutions with nonzero [itex]T_{ab}[/itex] that are static (although it turns out that they are unstable equilibrium solutions, like a pencil balanced on its point--which is now neither here nor there, since the actual universe is expanding, not static, but it's interesting to note that Einstein apparently never considered it).

    Now, however, let's consider what happens if [itex]T_{ab} = 0[/itex]. (Btw, I think MTW's language is rather misleading when they describe this condition as describing a "flat" spacetime; the more usual terminology would be that [itex]T_{ab} = 0[/itex] indicates a "vacuum" spacetime. A "flat" spacetime would be when the Riemann curvature tensor is zero, which would mean that [itex]R_{ab} - 1/2 g_{ab} R = 0[/itex]. See further comments below.)

    If [itex]T_{ab} = 0[/itex], the original equation becomes:

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

    which does indeed describe a flat spacetime (zero curvature). However, the modified equation becomes:

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

    which does *not* describe a spacetime with zero curvature (and hence it's rather misleading to call this a "flat" spacetime, as MTW apparently do). To see that the curvature is not zero, move the cosmological term to the RHS, to obtain:

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

    where now we can consider the RHS to be a stress-energy tensor associated with "dark energy", or some term like that (MTW note later on in the same section that this is how the cosmological constant is currently viewed by many physicists). But however we interpret the RHS, it clearly shows that the spacetime is curved, not flat, since the LHS, which describes the curvature, is not zero.

    Now for those comments in MTW:

    You can see why this is, because the modified "G" includes a term, the cosmological term, which does not describe curvature. This may be one thing that is confusing you (and, as I said above, MTW's terminology is itself confusing): what describes curvature here is specifically the combination [itex]R_{ab} - 1/2 g_{ab} R[/itex], and nothing else. So if we want an equation that says "curvature = stress-energy", the modified equation, with the cosmological term, doesn't meet this requirement (unless we move the cosmological term to the RHS and interpret it as a "dark energy" stress-energy tensor--but nobody at the time Einstein was considering all this would have even thought about such a possibility).

    Again, this is basically saying that we want an equation saying "curvature = stress-energy", and the modified equation doesn't do that. But again, MTW's terminology is confusing here, because strictly speaking, "flat spacetime" means [itex]R_{ab} - 1/2 g_{ab} R = 0[/itex]; "empty" spacetime, in the sense of "vacuum" spacetime, is what [itex]T_{ab} = 0[/itex] describes.

    I hope the above helps to clarify things.

    [Edit: To further add to the confusion from MTW's language, even the condition [itex]R_{ab} - 1/2 g_{ab} R = 0[/itex] does not, strictly speaking, describe a "flat" spacetime. It only describes a "Ricci flat" spacetime. A truly, entirely flat spacetime is, of course, Ricci flat, but the converse is not true; there are spacetimes, such as the Schwarzschild spacetime of a black hole, that are Ricci flat but not flat. MTW appear to ignore this point entirely in the discussion quoted, perhaps because it's not necessary to understand the issue Einstein was grappling with; but I wanted to clarify it to avoid possible misinterpretations of what I said above. In the above, wherever I say "flat", the strictly correct term is "Ricci flat".]
     
    Last edited: Aug 25, 2012
  4. Aug 25, 2012 #3
    Ah, now that I take a second look at it, I do see you are correct.


    Ok, will take a closer look in the text to see how they refer to flatness for the Schwarzchild metric. Appreciate your answering.
     
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