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).