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What if G is a function of R?

  1. Aug 21, 2013 #1
    In Einsteins Field equations

    [itex] R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} + \Lambda\,g_{\mu \nu} = {8 \pi G \over c^4} T_{\mu \nu} [/itex]

    if you make an ansatz that the gravitational constant G is not constant, but a function of either the Ricci scalar R or the Ricci tensor [itex] R_{\mu \nu} [/itex], would you still require the cosmological constant λ to balance out the unsteady nature of the universe? The procedure would be either to solve for G's function or make a guess as to what it is in order to get a steady-state universe without λ.

    [itex] R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} = {8 \pi G(R) \over c^4} T_{\mu \nu} [/itex]

    Actually the more I think of this, the more dumb I think I am, because (and correct me if I'm wrong because I probably am!) we can express the field equations with the cosmological constant this way

    [itex] R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} = {8 \pi \over c^4} \left(G-\frac{c^4}{8 \pi}\Lambda\,g_{\mu \nu}T^{\mu \nu}\right)T_{\mu \nu} [/itex]

    Which seems to imply that the cosmological constant may be looked at as modifying G, such that G varies as

    [itex]G=G_{0}-\frac{c^4}{8 \pi}\Lambda\,g_{\mu \nu}T^{\mu \nu}[/itex]

    in Einstein's original derivation of

    [itex] R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} = {8 \pi G \over c^4} T_{\mu \nu} [/itex]

    Where perhaps [itex]G_{0}[/itex] is the gravitational constant in flat space.
     
    Last edited: Aug 21, 2013
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  3. Aug 21, 2013 #2

    Nabeshin

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    If you make G(R), than from say the Einstein Hilbert action it's easy to see that what you end up with is just f(R) gravity.
     
  4. Aug 21, 2013 #3
    Thank you. I'm now learning about f(R) gravity. I'm not sure I like the approach that f(R) gravity uses though, because as it appears, the idea is to replace the Ricci scalar with a function of the Ricci scalar in the field equations. It doesn't seem to me that this is valid, since it can be derived that

    [itex]\nabla R_{\mu \nu} = \frac{1}{2} \nabla g_{\mu \nu} R[/itex]

    the covariant derivative of the Ricci curvature tensor equals 1/2 the covariant derivative of the metric tensor times the Ricci scalar... not some function of the scalar, but only the scalar. thereby allowing

    [itex]\nabla \left( R_{\mu \nu} - \frac{1}{2} g_{\mu \nu}R \right) = 0 = \nabla \left( \frac{8 \pi G}{c^4} T_{\mu \nu} \right) [/itex]

    So it seems to me that messing with the Ricci scalar is off limits, because if it's a function, then the left hand side won't always be zero like the right hand side will...

    I figure that messing with G isn't necessarily off limits, because it's not a derived mathematical object. Sure it's great to first order, but who knows what G really is, or what mechanism(s) may produce its value? Therefore, perhaps it's wrong to assume it's constant. I do see the connection with f(R) gravity that you make though if I mess with G as a function of R since it would make the RHS not always zero, so modifying G will depart from canonical GR, and I guess thats what I'm suggesting we do... question the assumption that G is constant in all metrics (or for all [itex] T_{\mu \nu}[/itex] if you like).

    I see your point. This is already being studied, though in a slightly less mathematically rigorous way in my opinion. Of course, my goal is the same as those who study f(R) gravity--to try and explain nature without the need to invent dark matter and dark energy.
     
  5. Aug 21, 2013 #4

    WannabeNewton

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    Check out Brans-Dicke gravity; there ##G## is also taken to be variable although in a different fashion.
     
  6. Aug 21, 2013 #5

    Nabeshin

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    ? The Einstein equations obviously don't hold in f(R) gravity. The "Left hand side" Einstein tensor is a complicated function containing an arbitrary function of the Ricci scalar. See the wiki page, for example, for the full equations.

    Well, there have been extensive studies on the variability of G. The bounds are (I would say) fairly stringent. Since G isn't dimensionless it requires some care to discuss it's variability (Dicke has some great papers on this from the 60's). Present limits for a cosmic variability in G have something like [itex] \dot{G}/G \lesssim 10^{-13} \text{yr}^{-1}[/itex]. There are limits on [itex]\nabla G [/itex] as well, and they're similarly restrictive, I believe.
     
  7. Aug 22, 2013 #6
    yes.

    At any rate, R is not arbitrary. I'm thinking that replacing R with some arbitrary function is not the best way to go. I was thinking that instead of f(R), we do G(R). It seems to me that replacing G with G(R) lends us more credibility in a theory since it's exploiting our lack of knowledge of G rather than knowingly ignoring mathematical truths such as the left hand side of the equation you're referring to being equal to zero. R is a derived quantity based on the curvature metric whereas G is just a fudge factor that is indirectly measured and thought of as one of the many mysteries of the way nature behaves. My point is, we dont know why G is the value it is, but we do know why R is the value that it is-R depends on the geometry in question. You can't just replace R with something else; it seems a bit cavalier. It seems less cavalier (though still a bit) to replace G with G(R).

    Thanks for your response. I have some more learning to do regarding variability of G, and how those numbers were determined. Do you know if anyone has measured G on very small scales? What if there is much more variability on small scales?
     
  8. Aug 22, 2013 #7

    Nabeshin

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    I think you are misunderstanding. It's not as though we're changing what the Ricci scalar is. The manifold structure of spacetime remains perfectly in tact, exactly as it is in GR. We're simply endowing the metric with different dynamics. There's no real reason why you should just pick R (other than it's the simplest nontrivial thing), when in principle you can have ANY object made up from tensors like Riemann or Weyl.

    Like I said, if you wanted to do such a programme anyways, it's formally equivalent to f(R) gravity. If you want to lump it into G and call it G(R), fine, but you're doing exactly the same things. In words, making G depend on R means the strength of gravity depends on the spacetime curvature, which you could equally envisage as constant coupling but additional dynamics. It makes no difference which viewpoint you take.

    As wbn mentioned above, people HAVE thought a lot about variable Newton constants, but most serious discussion of it has them varying with the matter content rather than the geometry. In these cases, the Newton constant has its own dynamics which, although linked to the metric, is clearly separate. So in this case there is something physical going on in the interpretation: it literally is the G constant changing with position and time.


    This is probably the domain in which we understand gravity the least. We've only really verified gravity, and the inverse square law, down to something like mm or micrometers. Compared with the cosmological scales gravity is known to, this is not far at all. So there is some wiggle room here.
     
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