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In an arbitrary number of spacetime dimensions $D$, Maxwell's equations are \begin{align*} \mathrm{d}F &= 0, \\ \mathrm{d}(\star F) &= -J. \end{align*}

How many independent equations does this represent for arbitary $D$?

Qmechanic
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tparker
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1 Answers1

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I think the second equation is missing a $*$. The first one has n(n-1)(n-2)/6 independent equations (it's setting a 3 form equal to 0 and there's one equation for $dx_idx_jdx_k$ for each $i, j,k$ distinct and order doesn't matter). The second equation is one of 1 forms so it has $n$ independent equations. This yields a total of $(n^3-3n^2+8n)/6$ independent equations. For $n=4$, this yields 8 as expected.

Toyesh Jayaswal
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  • I think the second equation is missing a ∗. Wikipedia leaves out the second Hodge star on the inhomogeneous equation and considers the current density to be a 3-form. So how does that approach — with two 3-form equations — give the same counting for arbitrary dimension as one 3-form and one 1-form? – Ghoster Dec 22 '23 at 07:16
  • it's not a 3 form - the dual of a 2 form is an n-2 form, d of this is an n-1 form ("pseudo vector"), the dimension of this space is also n. Under this convention, j is generally a pseudovector, not a 3 form. – Toyesh Jayaswal Dec 22 '23 at 07:22
  • The second bullet point explicitly says “$J$ … is the current 3-form”. – Ghoster Dec 22 '23 at 07:23
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    Yes, it is a 3 form in 4 dimensions, but looking at the definition of J_a *dx_a (which generalizes), it's an n-1 form generally – Toyesh Jayaswal Dec 22 '23 at 07:25
  • OK, I understand now. I upvoted. – Ghoster Dec 22 '23 at 07:30