- #316
PeterDonis
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Q-reeus said:How do you get that? If gross gas flow is v wrt stationary capacitor plates, we have E.v = 0, and there is zero energy transfer from capacitor E field to gas momentum/energy.
Before ionization, yes. But the energy transfer you were talking about (from the KE of gas flow to the plates) happens *after* ionization, because it's only after ionization that the gas is moving towards the plates at all! (Before ionization the gas flow is entirely parallel to the plates.) Where does the KE that the ions pick up, and deposit in the plates, come from? From the E field of the capacitor. Where else?
Q-reeus said:Sure. In flowing gas rest frame, there is a B = vxE. Upon dissociation, ions accelerate initially just along applied E axis. As soon as finite velocity v' perpendicular to v is obtained, a magnetic Lorentz force F = v'xB acts against the gross flow velocity v.
You're mixing frames. In the gas rest frame, there is no "gross flow velocity" v. There is only the velocity v', perpendicular to v, that is caused by the E field of the capacitor when ionization occurs. I assume what you are trying to say is that this velocity v' causes a Lorentz force that is directed parallel to the plates. But since this Lorentz force is parallel to the plates, it has no effect on the momentum transfer normal to the plates.
Q-reeus said:Recall the ions begin as ionically bonded molecules - they need to be dissociated under the action of E, just as vacuum 'virtual particle pairs' do in order to become real electron/positron pairs.
No, these are different processes. The ionically bonded molecules are bound states; the E field has to supply the binding energy to convert them into free states (more precisely, to move an electron from each molecule from a bound state to a free state, leaving behind an ion, which still has other electrons in bound states but can be viewed overall as a free particle).
The virtual pairs in vacuum are not bound states; they are free particle states with an average energy of zero (one particle of each pair has positive energy, one has negative energy). The E field has to supply the rest energy of a pair to make it real; that doesn't involve breaking any bonds or dissociating anything or driving particles from one state to another. It's just pumping energy into a state that starts out with zero energy.
Q-reeus said:I have shown above and previously the complete decoupling between any transverse motion and energetics of pair creation (by appropriate analogy with ionic molecular dissociation). Can we accept that and move the argument along?
No, because your analogy fails.
Q-reeus said:The main gist is it's ok to roughly model vacuum breakdown, energetics wise, in analogy to gas breakdown.
No, it isn't. The link you provided is behind a paywall, so I can't read it. The abstract doesn't tell me anything useful except that it's talking about quantum field theory.
Q-reeus said:one might as well say dielectric breakdown has nothing to do with the dielectric!
I didn't say anything of the kind. I did not say the breakdown has nothing to do with the vacuum; I said that since there is an obvious ponderable medium present, the field source, it makes no sense to analyze the phenomenon as though that object wasn't there.
Q-reeus said:Dielectric/vacuum *plus* acting E -> breakdown.
No, vacuum plus acting E plus *source* of E -> breakdown. You can't have the E without a source for it. Therefore the properties of the source are relevant for analyzing the phenomenon. Again, why is this so hard to understand?
Q-reeus said:Opaque dielectric screens intervene between an observer and some source of static E. One source is a pair of charged capacitor plates at rest wrt observer. The other is a 'conveyor belt' capacitor in rapid relative transverse motion, but where the LT's yield identical field strength E as for the stationary source case.
But different B--zero in one case, nonzero in the other. Easy to detect. But of course...
Q-reeus said:the B field present in the second case can either be ignored or exactly canceled via stationary source of opposing B.
In other words, you're destroying the evidence that distinguishes one case from the other, then claiming that they're the same. I expect this of politicians, but not here on PF.
(Oh, and before you object that the B field is irrelevant to pair creation, see further comment below.)
Q-reeus said:"Virtual particle filled Vac the vacuum knows it has the same properties in any inertial frame, and that by the LT's will have an easier time of it breaking down by choosing that frame with the largest applied E, consistent with a minimum duration for effecting breakdown.
So you think that, if the pairs are moving relative to the source, they will have *less* energy relative to the source? Have you actually read what I wrote? I did not say anything about whether the rest frame of the source has the largest E, smallest E, or whatever. What I said was, the created pairs have minimum energy in the rest frame of the source; therefore, for the source to supply minimum energy to the pairs, they must be created in the rest frame of the source.
The paper you linked to makes the same assumption; I've already made that clear by pointing out their equation #21, which says that the time rate of change of the electric field is proportional to the current. By Maxwell's equations, that can only be true if the B field is zero; and the only frame in which the EM field of any source can be a pure E field, with no B, is the rest frame of the source. So the paper you linked to agrees with me that the pairs are created in the rest frame of the source. Do you think the paper is wrong?
Q-reeus said:My point though is, the concept of a breakdown field Ecrit implies a purely intensive property, whereas linkage to the source frame implies something quite different - there is Ecrit + 'something else'.
Ecrit must be defined relative to a frame; the E field alone is not Lorentz covariant (only the full EM tensor is). Which frame do you think Ecrit is defined relative to, if it isn't the source's rest frame? (This question, btw, illustrates why you can't just ignore the B field. Yes, it has no direct effect on pair production because the B field can't pump any energy into the pairs, but it is certainly relevant in forming a properly Lorentz covariant description of what is going on. Ignoring the B field is basically pretending that the E field is covariant when it isn't.)
Q-reeus said:The sole something else I will admit belongs in this picture is minimum duration. I gave a link in #257 that mentioned temporal influence on breakdown, but a better article can be found here: http://www4.rcf.bnl.gov/~swhite/erice_proc/adrian2.ps
I'll take a look.