How do we express Maxwell's equations in terms of the field tensor?

[meteorologist1]

Hi, could someone show me how to express

\frac{\partial G^{\mu\nu}}{\partial x^\nu} = 0
which are Maxwell's equations, G is the dual tensor,

in terms of the field tensor F:
\frac{\partial F_{\mu\nu}}{\partial x^\lambda} + \frac{\partial F_{\nu\lambda}}{\partial x^\mu} + \frac{\partial F_{\lambda\mu}}{\partial x^\nu} = 0

Thanks.

 

[dextercioby]

(*F)^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu}{}_{\rho\lambda}F^{\rho\lambda}

Daniel.

 

[dextercioby]

And the other way around

T^{\lambda\mu\nu}=:\partial^{[\lambda}F^{\mu\nu]}

which is a totally antisymmetric tensor of rank 3.Its Hodge dual is a pseudovector

(*T)^{\sigma}=:\frac{1}{6}\epsilon^{\sigma}{}_{\lambda\mu\nu} T^{\lambda\mu\nu}=\frac{1}{6}\epsilon^{\sigma}{}_{\lambda\mu\nu} \partial^{[\lambda}F^{\mu\nu]}


Daniel.

 

[meteorologist1]

Thanks, but I'm having trouble understanding the notation. I don't really know much about relativity. We are just doing a chapter on special relativity in our E&M class, and all we have learned about field and dual tensors are that

<br /> F^{\mu\nu} = \left(<br /> \begin{array}{cccc}<br /> 0 &amp; E_x/c &amp; E_y/c &amp; E_z/c\\<br /> -E_x/c &amp; 0 &amp; B_z &amp; -B_y\\<br /> -E_y/c &amp; -B_z &amp; 0 &amp; B_x\\<br /> -E_z/c &amp; B_y &amp; -B_x &amp; 0<br /> \end{array}<br /> \right)<br />
<br /> G^{\mu\nu} = \left(<br /> \begin{array}{cccc}<br /> 0 &amp; B_x &amp; B_y &amp; B_z\\<br /> -B_x &amp; 0 &amp; -E_z/c &amp; E_y/c\\<br /> -B_y &amp; E_z/c &amp; 0 &amp; -E_x/c\\<br /> -B_z &amp; -E_y/c &amp; E_x/c &amp; 0<br /> \end{array}<br /> \right)<br />

and we wrote out Maxwell's equations using this. Could you please explain what you did without using the notation you used? Thanks again and sorry for the trouble.

 

[dextercioby]

The set of tensor equations has a logics.It all comes from differential geometry.

That "G^(\mu\nu)" notation is highly misfortunate,as people denote the matter field tensor by it.U basically do not need the dual for writing Maxwell's equations in Minkowski space.

So what did i do?Played around with tensors and duals.Again,knowedge of p-forms is required.So for non mathematics oriented EDyn class/course,i'd say that the equations using only the em-tensor are enough.

Daniel.

 

[pervect]

Thanks, but I'm having trouble understanding the notation. I don't really know much about relativity.

The main thing you need to understand to calculate the dual from Dexter's expression

<br /> G^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu}{}_{\rho \lambda}F^{\rho\lambda}

is the value of the Levi-civita tensor (and how to raise and lower indices in tensors).

The definition from the Wikipedia should help you there, note that you want the 4-d version of the Levi-civita tensor for GR.

The Levi-Civita symbol can be generalized to higher dimensions:

<br /> \epsilon_{ijkl\dots} = \left\{ \begin{matrix} +1 &amp; \mbox{if }(i,j,k,l,\dots) \mbox{ is an even permutation of } (1,2,3,4,\dots) \\ -1 &amp; \mbox{if }(i,j,k,l,\dots) \mbox{ is an odd permutation of } (1,2,3,4,\dots) \\ 0 &amp; \mbox{if any two labels are the same} \end{matrix} \right. <br />

taken from

http://en.wikipedia.org/wiki/Levi-Civita_symbol

Caution: the above definition, given by the component values, only works directly as stated in a right-handed orthonormal coordiante system. One can generalize from the components of the tensor in this coordinate system to any other in the usual manner.

Possibly the star operator confused you, too, the star operator is just the Hodges dual, which is how you get from F to G.

Clifford algebra's really make the Hodge dual operator much clearer. They're not nearly as scary as they sound, and are quite worthwhile.

http://www.mrao.cam.ac.uk/~clifford/introduction/intro/intro.html

 

[meteorologist1]

Ok, thanks for your help.

Since \frac{\partial G^{\mu\nu}}{\partial x^\nu} = 0
produces divB=0 and Faraday's Law, I ended up trying to get them from
\frac{\partial F_{\mu\nu}}{\partial x^\lambda} + \frac{\partial F_{\nu\lambda}}{\partial x^\mu} + \frac{\partial F_{\lambda\mu}}{\partial x^\nu} = 0 (*)

What was confusing me was how to write out the components of (*) since it didn't say anything about if mu, nu, and lambda can be equal or not. I looked in a lot of texts and finally found that the four components of (*) are
\frac{\partial F_{23}}{\partial x^1} + \frac{\partial F_{31}}{\partial x^2} + \frac{\partial F_{12}}{\partial x^3} = 0

\frac{\partial F_{23}}{\partial x^0} + \frac{\partial F_{30}}{\partial x^2} + \frac{\partial F_{02}}{\partial x^3} = 0

\frac{\partial F_{13}}{\partial x^0} + \frac{\partial F_{30}}{\partial x^1} + \frac{\partial F_{01}}{\partial x^3} = 0

\frac{\partial F_{12}}{\partial x^0} + \frac{\partial F_{20}}{\partial x^1} + \frac{\partial F_{01}}{\partial x^2} = 0

The rule is that the zeroth component lacks index 0, first component lacks index 1, and so on. And I did get divB=0 and Faraday's Law from these four equations.