Constructing the Einstein Field Equation

[MarkovMarakov]

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 R_{ab} is R_{ab}=AT_{ab}+Bg_{ab}+CRg_{ab}
where R is the Ricci scalar.

Why is this the most general form (involving up to the second derivative of the metric --- by definition of R_{ab})? 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 C=\frac{1}{2}.

This I understand.

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

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

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

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

But I don't understand how it leads to A=\frac{8\pi G}{c^4}.

 

[Takeshin]

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\phi= 4\pi G\mu .

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 : \Box \hbar_{\mu \nu} = -2K T_{\mu \nu} 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 \left|T^{00}\right| >> \left|T^{i0}\right| and \left|T^{00}\right| >> \left|T^{ij}\right|. You also have \left|T^{00}\right| = μc^2 . It leaves you with only one equation out of ten.

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

Now, you can let \hbar^{00} be equal to -4 \frac{\phi}{c^2} (if you calculate Christoffel symbols, you can prove that \phi 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)

 

[PeterDonis]

In my notes

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.

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

This can't be just the most general form of the Ricci tensor, because it has T_{ab} 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.)

I only know the energy-momentum tensor T_{ab} to be of the form
\begin{pmatrix}H&\pi_i\\\frac{s_i}{c}&T_{ij}\end{pmatrix} where H is the energy density, \pi_i is the momentum density, s_i is the energy flux.

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