# Gravitation #3.14

1. Aug 14, 2010

### Living_Dog

How does one show that dF = 0 is the geometric version of Maxwell's equations??

2. Aug 14, 2010

### Petr Mugver

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.

3. Aug 14, 2010

### Living_Dog

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...huh? I know that dF=0 is Maxwell's equations. I asked:
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How does one show that dF = 0 is the geometric version of Maxwell's equations??
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E.g. if I wanted to show it was frame-independent, then I would perform a Lorentz boost and show how the same equation appears, but with primes.
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I'm sorry, but I don't know how to ask this question more clearly. I guess it's b/c I don't understand it. But then again, that's why I posted it.

4. Aug 16, 2010

### Living_Dog

I misunderstood the question. They were probably asking to show that Maxwell's equations can be obtained from this geometric version. I can do that having read section 4.5 of the text.