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Maxwell's equations differential forms

  1. Feb 24, 2013 #1

    WannabeNewton

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    1. The problem statement, all variables and given/known data
    I have to take the curved space - time homogenous and inhomogeneous maxwell equations, [itex]\triangledown ^{a}F_{ab} = -4\pi j_{b}[/itex] and [itex]\triangledown _{[a}F_{bc]} = 0[/itex], and show they can be put in terms of differential forms as [itex]dF = 0[/itex] and [itex] d*F = 4\pi *j[/itex] (here [itex]*[/itex] is the hodge dual defined for any p - form [itex]\alpha [/itex] as [itex](*\alpha )_{b_1...b_{n-p}} = \frac{1}{p!}\alpha ^{a_1...a_p}\epsilon _{a_1...a_pb_1...b_{n-p}}[/itex] where [itex]n[/itex] is the dimension of the manifold and [itex]\epsilon [/itex] is the natural volume element for the manifold i.e. a totally anti - symmetric nowhere vanishing continuous tensor field).

    3. The attempt at a solution
    Since the exterior derivative [itex]d[/itex] is independent of the choice of derivative operator, I chose for convenience the unique metric compatible derivative operator [itex]\triangledown _{a}[/itex] because [itex]\triangledown _{c}\epsilon _{a_1...a_p} = 0[/itex] identically and this simplifies the calculation. The homogenous ones are trivial since [itex](dF)_{ba_1a_2} = 3\triangledown_{[b}F_{a_1a_2]} = 0[/itex]. The inhomogeneous ones are really starting to annoy me :frown:. This is 4 dimensional space - time so we have that [itex]*F_{b_1b_2} = \frac{1}{2}F^{a_1a_2}\epsilon _{a_1a_2b_1b_2}[/itex] hence [itex](d*F)_{cb_1b_2} = 3\triangledown _{[c}*F_{b_1b_2]} = \triangledown _{c}*F_{b_1b_2} - \triangledown _{b_1}*F_{cb_2} + \triangledown _{b_2}*F_{cb_1}[/itex] (where I have assumed the anti - symmetry of the hodge dual of the EM field strength tensor to combine terms). Plugging in the equation for the hodge dual gives us [itex](d*F)_{cb_1b_2} = \frac{1}{2}(\epsilon _{a_1a_2b_1b_2}\triangledown _{c}F^{a_1a_2} - \epsilon _{a_1a_2cb_2}\triangledown _{b_1}F^{a_1a_2} + \epsilon _{a_1a_2cb_1}\triangledown _{b_2}F^{a_1a_2})[/itex]. On the other hand, [itex]4\pi (*j)_{b_1b_2b_3} = 4\pi j^{a_1}\epsilon _{a_1b_1b_2b_3} = \triangledown _{r}F^{a_1r}\epsilon _{a_1b_1b_2b_3}[/itex]. Has my work been correct so far? This is where I'm stuck because the two different equations (the one for the hodge dual of the 4 - current density and the one for the exterior derivative of the hodge dual of the EM tensor) are being summed over totally different indices, as you can see, so I cannot begin to even see how to resolve the two so they can be equal.
     
    Last edited: Feb 24, 2013
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  3. Feb 24, 2013 #2

    WannabeNewton

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    Ah ok I got it. It was the form of the final pair of equations written above that threw me off with their unsettling nature and the stagnation stemmed from the equation we started with. In fact, if we had instead started with [itex]*(d*F) = -4\pi j[/itex], instead of what the book asked to start with, then the solution would have been much more obvious and saved me some tine :). Anyways, we take [itex](d*F)_{cb_1b_2} = (\epsilon _{a_1a_2b_1b_2}\triangledown _{c}F^{a_1a_2} - \epsilon _{a_1a_2cb_2}\triangledown _{b_1}F^{a_1a_2} + \epsilon _{a_1a_2cb_1}\triangledown _{b_2}F^{a_1a_2})[/itex] and multiply both sides by [itex]\epsilon ^{b_1b_2b_3c}[/itex] to get [itex]\epsilon ^{b_1b_2b_3c}(d*F)_{b_1b_2b_3} = (\delta ^{[b_3 c]}_{a_1a_2}\triangledown _{b_3}F^{a_1a_2} - \delta ^{[b_2 c]}_{a_1a_2}\triangledown _{b_2}F^{a_1a_2} + \delta ^{[b_1 c]}_{a_1a_2}\triangledown _{b_1}F^{a_1a_2}) = \triangledown _{b_3}F^{b_3c} - \triangledown _{b_2}F^{b_2c} + \triangledown _{b_1}F^{b_1c} = \triangledown _{b_1}F^{b_1c} = -4\pi j^{c}[/itex] therefore [itex]\epsilon _{b_1b_2b_3c}(d*F)^{b_1b_2b_3} = *(d*F)_c = -4\pi j_{c}[/itex] so [itex]**(d*F)_{a_1a_2a_3} = (-1)^{1 + 2(4 - 2)}(d*F)_{a_1a_2a_3} = -(d*F)_{a_1a_2a_3} = -4\pi *j_{a_1a_2a_3}[/itex] thus [itex](d*F)= 4\pi *j[/itex]
     
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