Hodge operators and Maxwell's equations

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  • Thread starter Silviu
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Hello! I am reading this paper and on page 18 it states that "in (2 + 1)D electrodynamics, p−form Maxwell equations in the Fourier domain Σ are written as: ##dE=i \omega B ##, ##dB=0##, ##dH=-i\omega D + J##, ##dD = Q## where H is a 0-form (magnetizing field), D (electric displacement field), J(electric current density) and E (electric field) are 1-forms, while B (magnetic field) and Q (electric charge density) are 2-forms." Can someone explain to me how does he associates these physical quantities with tensors? How, for example, can the magnetic field be a 2-form, as the magnetic field on its own is not even relativistically invariant (if I would force myself to make it a one-form), let alone a 2-form? Also what is the Fourier domain Σ? Thank you!
 

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  • #2
haushofer
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Well, for one thing, a p-form is dual to a D-p form, because they have the same amount of independent components. But I don't see directly how to fit the 2 (?) components of B in a one-form.
 
  • #3
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If H is a 0 form, and you pull back on it, it should map back to a 2 form, no? I.E take the cartesian basis ## \omega = \left\{ dx, dy \right\} ##. If I take the hodge dual of a 0 form it'll look like this ##\star 1 = dx \wedge dy ## But it's late here, and I'm confusing myself with my own responses, so I'll post the place where I learned about maxwell as differential forms, and try to answer better in the morning.

He has 3 sections on it, none of them are very long:
(1): http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/maxwell1
(2): http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/maxwell2
(3): http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/maxwell3
 

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