I have a problem with the Maxwell equations in (2+1) dimensions using differential form. Following J. Baez "Gauge Fields, Knots and Gravity" page 93 (or any other book), the equations are

\begin{align}dF &=0\\ *d*F &=J \end{align}

with $J=-\rho dt+j$ which is a 1-form, and $F$ is the electromagnetic tensor field (2-form). These equations give the correct Maxwell equations in (3+1) dimension but in (2+1) there is a different equation with a minus sign

\begin{align}\text{div} \;E=-\rho. \end{align}

To detail the things \begin{align} F &=B dx\wedge dy-E_x dt\wedge dx-E_y dt \wedge dy\\ *F &=B dt+E_x dy-E_y dx\\ d*F &= -(\partial_t E_y-\partial_x B) dt\wedge dx+(\partial_t E_x-\partial_y B) dt\wedge dy+(\partial_x E_x+\partial_y E_y) dx\wedge dy\\ *d*F &= (\partial_x E_x+\partial_y E_y) dt+(\partial_t E_x-\partial_y B) dx+(\partial_t E_y-\partial_x B) dy \end{align}

which implies

\begin{align}\text{div} \;E=-\rho. \end{align}

WHY, there is a negative sign?


There's something wrong with your sign permutations in the Hodge star operator calculation. If

$F = B + E \wedge dt$,

then, in 2D,

$F = B dx \wedge dy + E_x dx \wedge dt + E_y dy \wedge dt$,

as you wrote yourself. Now, let us take our initial Hodge star as $\star dx \wedge dy = dt$. This means that $\star dt \wedge dx = dy$ and $\star dy \wedge dt = dx$, by cyclic permutation (right hand rule, and here's your mistake, I think). This means that

$\star F = B dt - E_x dy + E_y dx$,

which disagrees in sign to what you got. Indeed, it is easy to see that

$(\star d \star F)(\partial_t) = - (\partial_x E_x + \partial_y E_y) $ ,

which fixes the sign implying $\nabla \cdot E = \rho$.

  • $\begingroup$ The formula for the hodge operator I have is $*dx\wedge dt=\epsilon^{10}_{~~~~~2}dy=-\epsilon_{102}dy=dy$. This is why, I'm getting $+E_x dy$ $\endgroup$ – anubis Nov 4 '15 at 21:03
  • $\begingroup$ Well, if you want $\star dx \wedge dt = dy$, then you must necessarily have $\star dy \wedge dx = dt$ (sign change) and $\star dt \wedge dy = dx$, which fixes your sign yet again. You're using wrong Hodge duals in analyzing $\star d \star F$. $\endgroup$ – QuantumBrick Nov 4 '15 at 21:13
  • $\begingroup$ Thx. If you have the Nakahara (Geometry, Topology and physics), could you please tell me if his formula 7.172 is incorrect, because this is the one I'm using $\endgroup$ – anubis Nov 4 '15 at 22:20
  • $\begingroup$ No, Nakahara is not incorrect. It's just complicating something very simple. The Wikipedia article for the Hodge dual is much better than Nakahara's, in my opinion. I think you are probably doing something wrong with your signs when using Nakahara's formula, that's all $\endgroup$ – QuantumBrick Nov 5 '15 at 2:46
  • $\begingroup$ @anubis If I didn't answer your question please tell me and I will elaborate even more. If I did, and I think this is the case, please set my post as an answer so that this matter can be closed. $\endgroup$ – QuantumBrick Nov 5 '15 at 12:27

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