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Vanishing Ricci scalar always implies vacuum?

  1. Feb 28, 2009 #1
    I have been searching through the literature and popular textbooks for this simple answer.

    I know that in the absence of soures, i.e. matter fields the Ricci scalar is zero. This is synonymous with saying that the Ricci scalar vanishes in vacuum and that the resulting space is flat. However, this is simply a sufficient requirement.

    Is it also necessary? That is to say, if the Ricci tensor is found to be zero in a very complicated spacetime (I work in brane worlds) does that also mean that the space is vacuum? I am pretty sure that it is flat, since geometrically R=0 implies flatness...

    To me is seems that this is indeed the case, but my supervisor and another guy I know from mathematics says it might not be (he wasnt sure).

    Thank you in advance,

    /blake
     
  2. jcsd
  3. Feb 28, 2009 #2

    jambaugh

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    No...I don't think so. If you allow a cosmological constant the trace of which can cancel the trace of the non-zero stress-energy of a non-vacuum then the Ricci scalar will be zero even though the Ricci tensor is non-zero.

    I think you can also do this without invoking the cosmological constant. I seem to remember something singular about the contribution of electromagnetic fields to the scalar part of Einstein's equations but it has been quite a while since I looked at this.

    Remember we "trace" with the metric which is indefinite so you can get a null trace from all positive quantities of a diagonalized tensor.
     
  4. Apr 8, 2009 #3

    A/4

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    Bumping this thread. Can anyone elaborate on the significance of a vanishing Ricci scalar? What would the implications be of an external field that was included on the metric? For example, the Ricci scalar of the Reissner-Nordstrom metric vanishes everywhere, but Schwarzschild does not. Naively, the latter can be used to show the coordinate nature of the horizon singularity. What about the former case?
     
  5. Apr 8, 2009 #4

    StatusX

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    Vanishing Ricci tensor is equivalent to zero stress-energy tensor, ie, vacuum. But it's possible to have a non-zero Ricci tensor whose trace (which is just the Ricci scalar) is still zero. This would correspond to a stress energy tensor whose trace is zero. Such stress energy tensors are associated with scale invariance, and arise, eg, for electromagnetic fields. By the way, the Schwarzschild solution should have vanishing Ricci scalar. Maybe you're thinking of the invariant [itex]R^{\mu \nu \rho \sigma} R_{\mu \nu \rho \sigma}[/itex].
     
  6. Apr 8, 2009 #5
    Let me risk (I am not a physicist) a reply.
    My understanding about vacuum in GR is the following;
    In a region of space with no energy-momentum (pure nothingness), the energy-momentum
    tensor is identically zero.

    If we suppose that the cosmological constant is zero, by simple algebra (contacting the 2
    indices on the left hand-side of Einstein equation by multiplying by the metric tensor), it is easy to show that the curvature scalar is zero and that also the Ricci tensor is identically zero.

    Does that means that in this region of nothingness, space-time is flat?
    No (or not enough), since flatness for dimension > 3 requires the full riemann curvature tensor to be zero.
    The solutions in this case are called vacuum solutions in GR and A/4 talked about a couple
    of them.

    Well, in the original statement the question was about the resulting space being flat (not spacetime). Hmm..., I may have missed something important.
     
  7. Apr 8, 2009 #6
    Let me risk (I am not a physicist) a reply.
    My understanding about vacuum in GR is the following;
    In a region of space with no energy-momentum (pure nothingness), the energy-momentum
    tensor is identically zero.

    If we suppose that the cosmological constant is zero, by simple algebra (contacting the 2
    indices on the left hand-side of Einstein equation by multiplying by the metric tensor), it is easy to show that the curvature scalar is zero and that also the Ricci tensor is identically zero.

    Does that means that in this region of nothingness, space-time is flat?
    No (or not enough), since flatness for dimension > 3 requires the full riemann curvature tensor to be zero.
    The solutions in this case are called vacuum solutions in GR and A/4 talked about a couple
    of them.

    Well, in the original statement the question was about the resulting space being flat (not spacetime). Hmm..., I may have missed something important.
     
  8. Apr 8, 2009 #7

    A/4

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    Oops. Yes, my mistake.
     
  9. Apr 9, 2009 #8
    1°) Einstein's field equations are of the generic following form:
    Rab - k. gab = Tab
    Where R is the Ricci, g is the metric and T is the energy momentum tensor.

    Consequently, in M4(Real), if [R] = 0, you only get [T] = k. [g].
    The metric and the energy momentum tensors are proportional. One could also say: the metric "contains entirely" the energy momentum and conversely the former determines the metric. Now, for me, the relevant question related to your thread is: what exactly is vacuum?

    2°) In general the trace of [R] = R00 + R11 + R22 + R33

    If the trace of [R] vanishes, the above sum vanishes, = 0. Which obviously doesnot automaticaly imply a vanishing Ricci tensor.
     
  10. May 1, 2009 #9

    A/4

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    I'd like to continue this thread with a slightly new question. Suppose one writes down a metric similar to the Schwarzschild metric, but with an added f(r) term in the g_tt and g_rr components, e.g. 1-2m/r - f(r). The purpose of the exercise will be to interpret the resulting physical (or unphysical) nature of the spacetime, instead of solving the field equations and obtaining the metric. In general, the Ricci scalar will not vanish (unless f(r) ~ -1/r^2, in which case you recover the RN metric). This will give a Ricci scalar R = g(r), which can be singular at r=0 for proper choice of f(r).

    For example, consider f(r) = a/r^n (with a = const.). This gives a Ricci scalar of the form R ~ a(n-1)(n-2)/r^{n+2}. Technically this is no longer a vacuum solution if n > 2, but it can be if a cosmological (non)-constant term is introduced to counter it.

    Has this ever been discussed? How would one physically interpret this result?

    Edit: On further reflection, the problem is more intricate since the components of the Riemann/Ricci tensor will also change. So, perhaps the main question is: what are the physical properties/consequences of a spacetime with non-vanishing Ricci scalar?
     
    Last edited: May 1, 2009
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