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Linearized Gravity and the smalness of the pertubation

  1. Dec 9, 2013 #1
    The approach taken in linearized gravity seems to be to 'perturb' the 'Minkowski metric' such that

    $$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$

    where ##|h_{\mu \nu}| <<1##. As I've understood it, the goal is to get an approximate theory for gravity, i.e. for weak gravitational fields and thus for small curvature. However the perturbation above seems more to be about choice of basis. After all if we had ##g_{\mu \nu} = \eta_{\mu \nu}## everywhere, that would say nothing about the curvature of spacetime, but rather that we had chosen to use an orthonormal basis everywhere. In this light the perturbation above seems more like the demand that we are only allowing 'almost orthonormal bases' than a demand on the weakness of gravity.

    Would it not be more sensible to perturb about a ##g_{\mu \nu}## with vanishing second derivatives and demand that the second derivatives of the perturbation be small, since it is the second derivatives of the metric that contribute to the Riemann tensor?
    Has approaches like this been tried somewhere?
     
  2. jcsd
  3. Dec 9, 2013 #2

    WannabeNewton

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    Well the point is that given any physical system, we can form a dimensionless constant ##\hat{G}## out of the characteristic length scale of the system, its characteristic time scale, Newton's constant, and the speed of light. Linearized gravity is the regime containing first order corrections to the ##\hat{G} \rightarrow 0## regime. In such a case it's extremely intuitively obvious that you can apply perturbation theory to ##g_{\mu\nu}## and expand about a parameter ##\epsilon## up to accuracy ##O(\epsilon^2)## i.e. ##g_{\mu\nu} = \eta_{\mu\nu} + \epsilon h_{\mu\nu} + O(\epsilon^2)## because you're assuming that the solution ##g_{\mu\nu}## is very close to ##\eta_{\mu\nu}## in the space of solutions to Einstein's equation (by which I mean we have a one-parameter family of solutions ##g_{\mu\nu}(\lambda)## with ##g_{\mu\nu}(0) = \eta_{\mu\nu}## and we restrict ourselves to solutions such that ##\lambda << 1##).

    Furthermore your statement is incorrect. If ##g_{\mu\nu} = \eta_{\mu\nu}## everywhere then the space-time is trivially flat. I have no idea why you think this is just a statement about a choice to use an orthonormal basis. It seems you're confusing this with the fact that at a given event ##p\in M## in an arbitrary space-time ##(M,g_{\mu\nu})##, we can find a basis ##\{e_{a}\}## for ##T_p M## such that ##g_{\mu\nu}(e_a)^{\mu}(e_b)^{\nu} = \eta_{ab}##. This is obviously not the same thing as saying that ##g_{\mu\nu} = \eta_{\mu\nu}, \forall p\in M## because that would obviously trivially imply that ##M## is flat.

    If you want a more rigorous explanation of what's going on then check out section 7.5 of Wald "General Relativity"; it deals with the more general case of perturbing about a curved background ##g^0_{\mu\nu}## and should eradicate your doubts/concerns.
     
    Last edited: Dec 9, 2013
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