Null energy condition constrains the metric

In summary, the null energy condition (NEC) imposes restrictions on the spacetime metric in general relativity. It states that for any null vector, the energy density as measured by an observer following that vector must be non-negative. This condition helps to determine the allowable shapes and dynamics of spacetime, influencing the formation of structures like black holes and the expansion of the universe. Violations of the NEC can lead to exotic phenomena, such as traversable wormholes or warp drives, which challenge our understanding of physics.
  • #1
ergospherical
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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.
 
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  • #2
ergospherical said:
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##.

ergospherical said:
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.
 
  • #3
PeterDonis said:
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##.
 
  • #4
ergospherical said:
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.
 
  • #5
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).
 
  • #6
ergospherical said:
\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.
 

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