The Casimir effect causes a limited number of wave functions, causing fewer particles between plates than outside and this could be considered as a negative mass. Then, if we have a technology good enough so that the total energy of the system is negative, will the system have acceleration towards the opposite direction of a force applied to this system?
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1$\begingroup$ There are no on mass shell particles between the plates, they are virtual particles , just a mathematical function. see this answer of mine for vrtual particles physics.stackexchange.com/questions/185110/… $\endgroup$– anna vOct 9, 2020 at 17:40
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It's a good question, and the simple answer is that we don't know.
I would repeat Anna's comment that virtual particles are a calculational device and do not really exist, so you need to be careful when talking about there being fewer virtual particles between the plates. However it is definitely true that the vacuum energy density is lower between the plates than it is outside them. That is, after all, what causes the Casimir force.
The problem is that we don't know how vacuum energy gravitates, or indeed if it does. If we do a naive calculation of how much gravitational force we expect from the vacuum energy we get a ridiculous number, so clearly the naive calculation doesn't work. We have to conclude we don't understand how the vacuum energy behaves, and therefore we don't understand how the reduced vacuum energy between the plates behaves.
It is possible the region between the plates does behave as a negative mass, though the end result would simply be that the total mass of the two plates plus the space between them was slightly less than the mass of the two plates alone. And we see this sort of thing routinely as the mass of any bound system is slightly less than the mass of its parts. This is what causes the mass deficit in nuclei.
answered Oct 11, 2020 at 5:35
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$\begingroup$ So as we don't know how the vacuum energy behaves, so that we also don't know the possibility of negative total mass system. Am I right? $\endgroup$ Oct 11, 2020 at 12:36
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$\begingroup$ > "mass of any bound system is slightly less than the mass of its parts" -- not any bound system; only those where the energy increase due to repulsion is low enough, such as alpha particle. If energy increase due to repulsion is high enough, like in uranium 235, bound system mass is higher than sum of masses of its decayed products. $\endgroup$ Dec 18, 2021 at 16:50