Living_Dog
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How does one show that dF = 0 is the geometric version of Maxwell's equations??
Petr Mugver said:I guess you mean that dF is a 2-form in four dimensions, so it has six independent fields (the electromagnetic fields):
dF=E_idx^idt+\frac{1}{2}\epsilon_{ijk}B_idx^jdx^k
Now the homogeneous Maxwell equation read dF=0. For the other two equations, introduce a 3-form for the 4-current
J=J_1dx^1dx^2dt+J_2dx^3dx^4dt+J_3dx^1dx^2dt+\rho dx^1dx^2dx^3
So the inhomogeneous Maxwell equations are dF = -4 \pi J. Note that, since d^2 = 0, J satisfies the continuity equation dJ = 0.
Petr Mugver said:I guess you mean that dF is a 2-form in four dimensions, so it has six independent fields (the electromagnetic fields):
dF=E_idx^idt+\frac{1}{2}\epsilon_{ijk}B_idx^jdx^k
Now the homogeneous Maxwell equation read dF=0. For the other two equations, introduce a 3-form for the 4-current
J=J_1dx^1dx^2dt+J_2dx^3dx^4dt+J_3dx^1dx^2dt+\rho dx^1dx^2dx^3
So the inhomogeneous Maxwell equations are dF = -4 \pi J. Note that, since d^2 = 0, J satisfies the continuity equation dJ = 0.