# Can 'negative energy' be created by the Casimir Effect?

In several papers about Time machines I have read that the Casimir effect can be used to create negative energy, so the Alcubierre-device or wormholes could be produced.

How can negative energy be created?

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Maybe you can link the papers you refer to? – Bernhard Dec 30 '12 at 13:52
i thought it was a Book by Michio Kaku ' PHysics of the impossible' at least one of the paper i remember – Jose Javier Garcia Dec 30 '12 at 15:17
Reading a little I found this [wiki article] (en.wikipedia.org/wiki/Exotic_matter) In the second subsection "negative mass" it speaks of the cashmir effect. – KDecker Jul 3 '13 at 16:54
Alcubierre devices violate causality. They are highly unlikely to ever be buildable. This, when combined with the negative energy issue (and as seen below, great care should be taken in referring to Casimir energy densities as "negative") should be sufficient argument against these as a practical device. – Jerry Schirmer Jul 3 '13 at 17:31

The energy density between the Casimir conductors can indeed be positive or negative. The calculation of Casimir energies is often done by noting that the plates impose boundary conditions on the field modes that can exist between them. Therefore in the presence of the plates, a more restricted set of modes is allowed than would be the case if the plates were absent. In the vacuum, each mode $k$ contributes and energy $\frac{1}{2}\omega_k$ to the energy, so plates vs no plates gives different vacuum energy values.

Both cases, plates, and no-plates, of course give infinite answers so some regulation method must be introduced in order to give a finite answer for the difference in energies. The calculation is "reasonably" straightforward for the case of parallel plates, and leads to $$\mathcal{E} \propto -\frac{1}{L^4}$$ where $L$ is the plate separation. This gives a negative pressure (i.e. attraction) between the plates. The calculation is illustrated in Elizalde & Romeo "Essentials of the Casimir Effect and its Computation" Am J Phys $\bf{59}$ 8 (1991), available here.

If, instead of flat plates, the calculation is done for the two halves of a conducting spherical shell, this time the pressure is positive (repulsive), so the behaviour you get depends strongly on the geometry of your plates.

More interesting from your viewpoint is the interaction between the Casimir energy and the gravitational field. To work this out, you need the energy momentum tensor for the Casimir region. Given this, you can compute (semiclassically) the gravitational effect by taking its expectation value $\langle T_{\mu \nu}\rangle$.

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Hi twistor59: The link says HTTP Error 404 - File or directory not found. – Qmechanic Dec 30 '12 at 23:01
@Qmechanic Oops, sorry, link should be fixed now – twistor59 Dec 31 '12 at 7:46
Abstract page is here: dx.doi.org/10.1119/1.16749 – Qmechanic Jul 3 '13 at 16:07

No. It really only creates the illusion of negative energy. It's negative relative to the space around it; which only proves that there is a small amount of positive energy in what looks like flat empty space ("zero point energy").

But it is not possible to extract negative energy from this kind of system and use it to construct an Alcubierre warp drive, which is what NASA currently claims to be trying to do.

Unless NASA already has a source of exotic matter they are not allowed to talk about, they already know that harnessing the Casimir effect in this way is impossible, and are just desperate to get some media attention before their already pitiful budget is slashed again.

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As the posts above say, I believe it would just be a phantom due to a local drop in the energy density compared to the vacuum. What I'm curious about is if this might be a gauge issue. I know that some literature (http://iopscience.iop.org/0305-4470/20/10/026/pdf/0305-4470_20_10_026.pdf) has previously claimed that Casimir energies are gauge variant, and so if the calculations for Alcubierre and the canonical flat-plate Casimir effects are performed in different gauges they might be noncomensurate anyway.

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## protected by Qmechanic♦Nov 28 '13 at 10:34

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