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I've often seen the Casimir effect cited as a source of negative energy/exotic matter with regards to ideas like the Alcubierre drive. The articles then go on to note that the energy required by the Alcubierre drive is orders of magnitude more than that produced in theory by the Casimir effect.

Fair enough. What I haven't seen is numbers, or formulae: as far as I can see, nobody has put up the actual resulting stress-energy tensor online, nor an expression for "effective negative mass" produced. I can see in principle how to calculate it, but I don't yet have the education to do it myself, so I'm turning to StackExchange.

So: How do you calculate the curvature of spacetime due to a parallel-plate Casimir setup; and, if you were to replace the setup with a suitable amount of exotic matter, what is the mass (and optionally, geometry) of the matter required?

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    $\begingroup$ I recommend looking at "Lectures on quantum energy inequalities" by Christopher Fewster arXiv:1208.5399 for an introduction. And it does have the computations of stress-energy tensor for a simple two-plate Casimir effect. $\endgroup$ – user23660 Aug 12 '13 at 3:58
  • $\begingroup$ I'm afraid the coverage consists of one page of material that assumes a bit more material than I can handle. Thanks for the reference, though - I'll come back to it in a few years when I've learned more. $\endgroup$ – linkhyrule5 Aug 12 '13 at 19:20
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Following Brown, Maclay, "Vacuum Stress between Conducting Plates: An Image Solution" (DOI:10.1103/PhysRev.184.1272) consider two ideally conducting parallel conducting plates separated by distance $a$ along the $z$-axis.

Simple symmetry argument allows us to obtain the possible structure of vacuum stress-energy tensor $\langle T_{\mu\nu}\rangle_0$. The plates remain invariant under rotation in $xy$-plane as well as under boosts along any direction in $xy$-plane. Therefore tensor $\langle T_{\mu\nu}\rangle_0$ must have the corresponding $SO(2,1)$ symmetry. In addition it must be traceless. The only possible structure that satisfies these requirements is $$\langle T_{\mu\nu}\rangle_0 = \left(\frac14 g_{\mu\nu} - \hat{z}_\mu\hat{z}_\nu\right)\cdot f(z),$$ where $\hat{z}_\mu$ is space-like unit vector along the $z$-axis and $f(z)$ is unknown yet function.

Furthermore, conservation of energy-momentum means that $f(z)$ must be constant, except at the plates, so this function takes two different values $C_1$ and $C_0$ outside and inside the plates (reflection symmetry means that values of $f(z)$ on both sides outside must be equal). Considering limits $a\to 0$ and $a \to \infty$ (which both correspond to a single mirror situation) we must conclude that $C_1 = 0$.

The remaining unknown constant $C_0$ must be computed through some sort of regularization. Brown, Maclay use explicit Green funciton construction, one can use $\zeta$-function regularization or simply compute total energy as $E= \sum_i \hbar \omega_i /2 $. The result is (inside the plates): $$ \langle T_{\mu\nu}\rangle_0 = \mathrm{diag}(-1,1,1,-3) \frac{\hbar c}{a^4}\frac{\pi^2}{720},$$ and zero outside. So here it is: the negative energy... and also pressure causing Casimir force.

This result should be the stress-energy tensor standing in the rhs of Einstein equations: $$ R_{\mu\nu} - \frac12 g_{\mu\nu} R = \frac{8\pi G }{c^4} \langle T_{\mu\nu}\rangle_0 .$$ So that is how you calculate geometry: by solving Einstein equations with stress-energy tensor including $\langle T_{\mu\nu}\rangle_0 $ (and matter, which will be present).

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  • $\begingroup$ Good explanation, but I'm not accepting it yet - I'm hoping for just a little more detail. I should probably confess that I haven't taken a formal GR course; I've read enough of MTW to follow your work, but I can't carry it through to answer my original question. $\endgroup$ – linkhyrule5 Aug 12 '13 at 21:37
  • $\begingroup$ Aaand it looks like nobody else is going to answer, so fair enough. Thanks - I'll come back to this when I learn enough GR to carry it through. $\endgroup$ – linkhyrule5 Aug 14 '13 at 0:41
  • $\begingroup$ The calculation of the Casimir effect stress-energy tensor has nothing to do with GR: it's a quantum field theory calculation. $\endgroup$ – John Baez Sep 25 '16 at 2:17

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