# Homogenuous Maxwell Equations in the Language of Differential Forms

1. I understand that if I define electric field to be $E=E_i dx^i$, magnetic field to be $B=B_1 dx^2 \wedge dx^3 + B_2 dx^3 \wedge dx^1 + B_3 dx^1 \wedge dx^2$, and field strength to be $F= dx^0 \wedge E + B$, I would get the two homogenuous Maxwell equations from $dF=0$. The last equation is a nontrivial equation.

2. However I've read somewhere else that if I define the vector potential to be the one form $A=A_\mu dx^{\mu}$, and the field strength to be $F=dA$ I would get the two homogenuous Maxwell equations.

My questions:

1. First, with this second definition the equation $dF=0$ is trivially true since $d^2=0$ and I don't understand how this would give me anything non-trivial.

2. Secondly, from the second definition if I write $F$ in components I would have [*]: $$F=dA=\partial_{\beta}A_{\mu} dx^{\beta} \wedge dx^{\mu}\\ = \partial_0 A_i dx^0 \wedge dx^i + \partial_i A_0 dx^i \wedge dx^0 + \partial_j A_i dx^j \wedge dx^i\\ = \partial_0 A_i dx^0 \wedge dx^i + x^0 \wedge E + B$$ which has the first term extra compared to the first definition, and I don't have any reason that this term is zero. So, what am I missing here? Which of the the two above approaches are correct?

[*] I use Greek letter super/subscripts for 4 space-time components and small english letters for 3 space componenets.

• @0celo7 But, doesn't that give the non-homogenuous equations? Which with the source term is $d\star F=J$. – Hamed Feb 17 '16 at 6:14
• If you're using magnetic charges and electric charges, maxwell equation turns into $dF = j_m$ and $d \star F = j_e$. Note that $F_A = dA$ only if you have a globally defined electromagnetic potential. In general what physicists do is to work in a contractible manifold and add magnetic charges (distributional 3-forms). This is equivalent to working in a manifold with the support of the magnetic charges removed. – user40276 Feb 17 '16 at 7:18
1. Yes, written in terms of the gauge potential $A_{\mu}$, the source-free Maxwell equations become trivially satisfied.
2. It seems OP is using the electrostatic definition of $E_i$. In full electromagnetism, besides the $\partial_i A_0$ term, there is also a $\partial_0 A_i$ term in the definition of $E_i$.