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 $\nabla_a$ 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 $$A=\frac{8\pi G}{c^4}$$.

  • $\begingroup$ At least one way to get this constant is to consider the Newtonian limit. In fact, $G$ is defined in this limit, so it is pretty much the only way. $\endgroup$ – Peter Kravchuk Apr 30 '13 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.