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F(R) gravity

  1. Jul 12, 2012 #1
    hey guys , is there anybody here to give me a nice reference for f(R) gavity ?


    Regards


    thecoop
     
  2. jcsd
  3. Jul 12, 2012 #2
    What is f(R) gravity?
     
  4. Jul 12, 2012 #3

    jtbell

    User Avatar

    Staff: Mentor

    There's a Wikipedia page about it, for what it's worth. It has references to what appear to be legitimate publications.

    http://en.wikipedia.org/wiki/F(R)_gravity

    It appears to be a serious research area, although probably a bit out of the mainstream. I'd never heard of it, either.
     
  5. Jul 12, 2012 #4
    It would be a theory with a modified Hilbert-Einstein action, where the density is not proportiomal to the Ricci scalar R, but to some non-linear function of it.

    The modified field equations will contain the Ricci curvature in a non-linear way, and might disallow the possibility of a black hole.

    By dimensional analysis, one can construct a dimensionless quantity from c, [itex]\hbar[/itex], G, and R:
    [tex]
    \frac{G \, R}{\hbar \, c^{3}}
    [/tex]
    Also, we can construct a combination with the dimension of an action density:
    [tex]
    \left[ \frac{S}{x^4} \right] = \mathrm{T}^{-1} \, \mathrm{L}^{-2} \, \mathrm{M} \stackrel{\mathrm{n.u.}}{\rightarrow} \mathrm{L}^{-4}
    [/tex]
    Because [itex]\left[ G \right] = \mathrm{T}^{-2} \, \mathrm{L}^{3} \, \mathrm{M}^{-1} \stackrel{\mathrm{n.u.}}{\rightarrow} \mathrm{L}^{2}[/itex], we ought to have [itex]\mathcal{L} \propto G^{-2}[/itex]. Returning the proper powers of [itex]\hbar[/itex], and c, we get:
    [tex]
    \mathcal{L} \propto \frac{c^6}{\hbar \, G^2}
    [/tex]
    Thus, we can write:
    [tex]
    S_{g} = \frac{c^6}{\hbar G^2} \, \int{f\left( \frac{\hbar \, G \, R}{c^3} \right) \, \sqrt{-g} \, d^4 x}
    [/tex]

    If you require the Planck's constant to drop out of the expression, you ought to have:
    [tex]
    f(x) = A \, x
    [/tex]
    i.e. the Lagrangian density is proportional to the scalar curvature. This is the ordinary Hilbert-Einstein action. The numerical constant A is fixed by Newton's Law of universal Gravitation.

    We might be tempted to choose a different functional form. For example, let us see if we can get G, the Universal Gravitational constant, to drop out from the expression. In a sense, this would make the action purely quantum effect. It is easy to see that this happens if:
    [tex]
    f_{2} = B \, x^2
    [/tex]
    A quadratic function dominates a linear function for large values of the argument, i.e. for:
    [tex]
    R \gg \frac{A}{B} \, L^{-2}_{P}
    [/tex]
    where [itex]L_{P} = \sqrt{G \, \hbar/c^{3}} \sim 10^{-35} \, \mathrm{m}[/itex] is the Planck length.

    There is, however, a problem with choosing a non-linear function. Namely, [itex]R[/itex] contains second derivatives of the metric tensor, but as a 4-gradient. Therefore, they get integrated out when the function is linear. But, one cannot do the same for a nonlinear function, and the equations that we would get would contain derivatives higher than two of the metric tensor.
     
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