I Null energy condition constrains the metric

[ergospherical]

Another GR question... in the thick of revision season. I would appreciate a sketch of how to approach the problem.

You basically are given a metric, involving a positive function $A(z)$, $$g = A(z)^2(-dt^2 + dx^2 + dy^2) + dz^2$$The game is to figure out somehow that the null-energy condition is satisfied if$$\frac{d^2}{dz^2} \log A(z) \leq 0$$I obtained the Riemann components using the tetrad formalism with the basis $(A dt, A dx, A dy, dz)$. You get non-zero connection 1-forms ${\omega^0}_3 = A^{-1} A' e^0$, then ${\omega^1}_3 = A^{-1} A' e^1$ and ${\omega^2}_3 = A^{-1} A' e^2$. You then get the curvature 2-forms from ${\Theta^{\mu}}_{\nu} = {d\omega^{\mu}}_{\nu} + {\omega^{\mu}}_{\rho} \wedge {\omega^{\rho}}_{\nu}$, and I read off the Riemann components via ${\Theta^{\mu}}_{\nu} = \tfrac{1}{2} {R^{\mu}}_{\nu \rho \sigma} e^{\rho} \wedge e^{\sigma}$ as:\begin{align*}
{R^{0}}_{303} &= {R^{1}}_{313} = {R^{2}}_{323} = \frac{A''}{A} \\
{R^{0}}_{202} &= {R^{0}}_{101} = {R^{1}}_{212} = - \left( \frac{A'}{A} \right)^2
\end{align*}
OK... so assuming these are correct, you need to figure out how to ensure that $T_{ab} k^a k^b \geq 0$ for any null vector $k$. In principle you can use these components to construct the energy-momentum tensor in the tetrad basis via $T_{\mu \nu} = (8\pi)^{-1} (R_{\mu \nu} - \tfrac{1}{2} R \eta_{\mu \nu})$, but this will take a bit more work.

Then, due to symmetry in x-y, I can rotate the x-y components of the null vector so that it takes the components $k^{\mu} = (k^0, k^1, 0, k^3)$ again in the tetrad basis, and there is the null constraint $\eta_{\mu \nu} k^{\mu} k^{\nu} = -(k^0)^2 + (k^1)^2 + (k^3)^2 = 0$ that I can use to consider only two independent components.

Is the way forward to go ahead with computing all of the matrix elements $T_{\mu \nu}$, and then forming a system of equations? It seems like there must be a better approach.

 

[PeterDonis]

the null constraint $\eta_{\mu \nu} k^{\mu} k^{\nu} = -(k^0)^2 + (k^1)^2 + (k^3)^2 = 0$

No, the null constraint is $g_{\mu \nu} k^\mu k^\nu = 0$, which means $- A^2 (k^0)^2 + A^2 (k^1)^2 + (k^3)^2 = 0$.

Is the way forward to go ahead with computing all of the matrix elements $T_{\mu \nu}$, and then forming a system of equations?

Maxima should be able to crank out the Einstein tensor of the metric. I would expect you will find that it does not have many nonzero components, which will simplify things.

 

[ergospherical]

No, the null constraint is $g_{\mu \nu} k^\mu k^\nu = 0$, which means $- A^2 (k^0)^2 + A^2 (k^1)^2 + (k^3)^2 = 0$.

The $k^{\mu}$ are here in the tetrad basis, so we use the Lorentz metric $\eta$.

 

[PeterDonis]

The $k^{\mu}$ are here in the tetrad basis, so we use the Lorentz metric $\eta$.

So that means the factors of $A$ appear in the components. They have to appear somewhere.

 

[ergospherical]

Well, indeed, but I did specify that I'm working entirely in the tetrad basis. (This choice obviously simplifies things because I already have the Riemann components in the tetrad basis).

 

[PeterDonis]

\begin{align*}
{R^{0}}_{303} &= {R^{1}}_{313} = {R^{2}}_{323} = \frac{A''}{A} \\
{R^{0}}_{202} &= {R^{0}}_{101} = {R^{1}}_{212} = - \left( \frac{A'}{A} \right)^2
\end{align*}

These look a lot like the terms you get when you expand out $d^2 / dz^2 ( \log A(z) )$. That should be helpful.