- #1
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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.
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.