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Constructing the Einstein Field Equation

  1. Apr 30, 2013 #1
    This question is on the construction of the Einstein Field Equation.

    In my notes, it is said that

    >The most general form of the Ricci tensor [itex]R_{ab}[/itex] is [tex]R_{ab}=AT_{ab}+Bg_{ab}+CRg_{ab}[/tex]
    where [itex]R[/itex] is the Ricci scalar.

    Why is this the most general form (involving up to the second derivative of the metric --- by definition of [itex]R_{ab}[/itex])? I suppose there are symmetry and degrees of freedom arguments. But I can only see why the LHS is a *possible* form. I can't se why it is the most general form...

    >Taking the covariant derivative $\nabla_a[/itex] of the expression above gives [tex]C=\frac{1}{2}[/tex].

    This I understand.

    >Compare the resulting expression with the Poisson equation gives [tex]A=\frac{8\pi G}{c^4}[/tex].

    This I *don't* understand --- perhaps I am being silly again... but still.

    I assume the "Poisson equation" referred to here is [tex]E^i{}_i=4\pi\rho G[/tex]
    where [itex]E^i{}_i[/itex] is the tidal tensor and may be expressed as [tex]E^i{}_i=R^i{}_{aib}T^aT^b[/tex] where [itex]T^a[/itex] is the tangent vector of the geodesic.
    So contracting [tex]R_{ab}=AT_{ab}+Bg_{ab}+CRg_{ab}[/tex] with [itex]T^aT^b[/itex] gives [tex]4\pi\rho G=AT_{ab}T^aT^b[/tex].

    But then what? Or perhaps I have already made a mistake? I only know the energy-momentum tensor [itex]T_{ab}[/itex] to be of the form
    [tex]\begin{pmatrix}H&\pi_i\\\frac{s_i}{c}&T_{ij}\end{pmatrix}[/tex] where [itex]H[/itex] is the energy density, [itex]\pi_i[/itex] is the momentum density, [itex]s_i[/itex] is the energy flux.

    But I don't understand how it leads to [tex]A=\frac{8\pi G}{c^4}[/tex].
     
  2. jcsd
  3. Apr 30, 2013 #2
    I'm far from a specialist in the field, but the easiest way I know is taking the newtonian limit of the EFE and comparing it with Poisson law Δ[itex]\phi= 4\pi G\mu[/itex] .

    I don't precisely remember all the demonstrations, but "newtonian limit" implies :

    - Being in the weak field approximation, where you can reduce the EFE to an equation with the weak metric tensor : [itex]\Box \hbar_{\mu \nu}[/itex] = -[itex]2K T_{\mu \nu}[/itex] where K is the proportionnality constant between Einstein tensor and Stress-Energy tensor in the EFE.

    - Neglect the time derivative in the d'Alembert operator above so that you can reduce the equation to a laplace operator equation. I think it implies that the gravitation sources are quasi-stationnary.

    -Being in the perfect fluid approximations, so that [itex]\left|T^{00}\right|[/itex] >> [itex]\left|T^{i0}\right|[/itex] and [itex]\left|T^{00}\right|[/itex] >> [itex]\left|T^{ij}\right|[/itex]. You also have [itex]\left|T^{00}\right| = μc^2[/itex] . It leaves you with only one equation out of ten.

    Combining these approximations gives you the following equation : Δ[itex]\hbar^{00} = -2κμc^2[/itex]

    Now, you can let [itex]\hbar^{00}[/itex] be equal to -[itex]4 \frac{\phi}{c^2}[/itex] (if you calculate Christoffel symbols, you can prove that [itex]\phi[/itex] is indeed the gravitationnal potential), equal your equation to Poisson law and find the correct value for K.

    I am deeply sorry these are only hints, as I don't have the full demonstration here.

    (And I hope the LaTeX code will work, I'm not familiar with this forum, I only extrapolated from my standard LaTeX knowledge)
     
    Last edited: Apr 30, 2013
  4. Apr 30, 2013 #3

    PeterDonis

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    Staff: Mentor

    Are these notes available on the web? Or are they taken from a publicly available source? If so, can you give a reference? I think having the context will help in responding to your question.

    This can't be just the most general form of the Ricci tensor, because it has [itex]T_{ab}[/itex] in it, which is not derived from the metric or the curvature or the Ricci tensor; it's something separate. What do the notes actually say here? (This is why having a link to the source helps.)

    What you are calling the energy density [itex]H[/itex] is also (with an appropriate choice of coordinate chart and appropriate restrictions on the source, as Takeshin said) the [itex]\rho[/itex] that appears in the Poisson equation (note that you also have to choose units appropriately).
     
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