3
  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.

Qmechanic
  • 201,751
Hamed
  • 345
  • @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 06:14
  • 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 07:18
  • About your computation you should include your additional term in the electric field component. – user40276 Feb 17 '16 at 07:27

1 Answers1

1
  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 $E_i$.

See also e.g. my Phys.SE answer here.

Qmechanic
  • 201,751