Would/should the mass of a particle measured between the two plates (typical casimir experiment setup) be different than measured outside the plates? If so would this be evidence of the Higgs field/mechanism?
Yes, I see what you mean. In general all quantum fields are subject to vacuum fluctuations and the Casimir force. But afaik the coupling of the Higgs boson (not the vev) is suppressed by the usual Fermi coupling constant GF~ 1/MW2. That means that the weak Casimir effect is weak compared to the electromagnetic one. I have to check the details.It was my understanding that the Higgs field was created from vaccuum fluctuations of the Higgs boson. Do you see the implicatios now? If the higgs boson exists it's power may be different between the plates.
The Casimir Effect and the Quantum VacuumKimball A. Milton said:Zero-point fluctuations in quantum fields give rise to observable forces between material bodies, the so-called Casimir forces. In these lectures I present the theory of the Casimir effect, primarily formulated in terms of Green's functions. There is an intimate relation between the Casimir effect and van der Waals forces. Applications to conductors and dielectric bodies of various shapes will be given for the cases of scalar, electromagnetic, and fermionic fields. The dimensional dependence of the effect will be described. Finally, we ask the question: Is there a connection between the Casimir effect and the phenomenon of sonoluminescence?
R.L. Jaffe said:In discussions of the cosmological constant, the Casimir effect is often invoked as decisive evidence that the zero point energies of quantum fields are "real''. On the contrary, Casimir effects can be formulated and Casimir forces can be computed without reference to zero point energies. They are relativistic, quantum forces between charges and currents. The Casimir force (per unit area) between parallel plates vanishes as [itex]\alpha[/itex], the fine structure constant, goes to zero, and the standard result, which appears to be independent of [itex]\alpha[/itex], corresponds to the [itex]\alpha\to\infty[/itex] limit.
No, not precisely. The source of Hawking radiation is a pair of virtual particles created from vacuum (no particle).Isn't that precisely how Hawking Radiation works?
I agree that using (visual) models is sometimes helpful, but one has to be aware of the fact that they are models.How do you conceptualize all of these things without being able to have some visualization of what's going on!
No, not precisely. The source of Hawking radiation is a pair of virtual particles created from vacuum (no particle).
I agree that using (visual) models is sometimes helpful, but one has to be aware of the fact that they are models.