According to this Wikipedia article, Alfvén's theorem states that "in a fluid with infinite electric conductivity, the magnetic field is frozen into the fluid and has to move along with it". However, in this Wikipedia article it gives a number of ways particles can drift across the magnetic field due to a 'Nonuniform B'. Do these particles violate Alfvén's theorem? Does Alfvén's theorem only work for fluids and not for test particles?
1 Answers
Do these particles violate Alfvén's theorem?
Yes. In MHD there are no particles, only a fluid. One could, in principle, try to describe the system as being comprised of multiple magnetized fluids including a separate one for the cross-field drifting particles, however such models are not within the confines of ideal MHD. You can see some of the assumptions of ideal MHD at https://physics.stackexchange.com/a/452325/59023 and https://physics.stackexchange.com/a/551944/59023.
In a magnetized plasma with infinite conductivity, the result of this assumption is that the only electric field field which can exist is the convective electric field, i.e., that resulting from Lorentz transformation (see links above for derivation and discussion).
Does Alfvén's theorem only work for fluids and not for test particles?
Hans Alfvén only considered magnetized fluids, not kinetic plasmas. So yes, his theorem is only appropriate in the limit where you treat the kinetic plasma as a single, idealized, magnetized fluid.

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Thanks. I thought MHD was supposed to be accurate provided the three conditions outlined in this Wikipedia article are satisfied? If we have enough particles in the kinetic model, shouldn't we recover the results you get from MHD? I guess the Wikipedia article on guiding center drifts doesn't model the evolution of the background magnetic field due to the particles? If we did include that in our model would we recover the MHD results? – Peanutlex Sep 21 '21 at 16:00
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@Peanutlex - No, MHD has some rather strict limitations that are often not at all satisfied in most space plasma environments. Does this mean it is useless? No, not at all. Does this mean one needs to be very careful in the application and interpretation of the results? Yes, very much so... – honeste_vivere Sep 21 '21 at 17:46