A Maxwell's equation in differential forms formalism

  • #1
123
0

Homework Statement


This is not actually a homework but a personal work. Here it is:
Using the differential forms:

[itex]F=\tfrac{1}{2!}{{F}_{\mu \nu }}d{{x}^{\mu }}\wedge d{{x}^{\nu }}[/itex] and [itex]J=\tfrac{1}{3!}{{J}^{\mu }}{{\varepsilon }_{\mu \alpha \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}[/itex]

proof that the Maxwell equation: [itex]{{F}^{\mu \nu }}_{,\mu }={{\mu }_{0}}{{J}^{\nu }}[/itex] can be written as:

[itex]d*F={{\mu }_{0}}J[/itex]

Homework Equations



[itex]*F\equiv \tfrac{1}{4}{{F}^{\mu \nu }}{{\varepsilon }_{\mu \nu \rho \sigma }}d{{x}^{\rho }}\wedge d{{x}^{\sigma }}[/itex]

[itex]d*F=\tfrac{1}{4}{{F}^{\mu \nu }}_{,a}{{\varepsilon }_{\mu \nu \rho \sigma }}d{{x}^{\alpha }}\wedge d{{x}^{\rho }}\wedge d{{x}^{\sigma }}[/itex]

The Attempt at a Solution



I have tried several ways to do this but they either fail (not giving the desired result) or get to a point where I cannot proceed any further.

Here are some attempts:
A) I multiplied both sides of the given equation with [itex]{{\varepsilon }_{\nu \alpha \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}[/itex] in order to make the current 3-form to appear in the RHS and then I tried to manipulate the LHS of the equation in order to make the [itex]d*F[/itex] to appear. This is the process:

[itex]{{F}^{\mu \nu }}_{,\mu }={{\mu }_{0}}{{J}^{\nu }}\Rightarrow {{F}^{\mu \nu }}_{,\mu }{{\varepsilon }_{\nu \alpha \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}=3!J[/itex]

As you can see, only one of the indices of Levi-Civita Symbol (LCS) contracts with the one indices of the field tensor (in contrast with the [itex]d*F[/itex] expression where two indices contract with the two indices of the field tensor) and also in the [itex]d*F[/itex] expression the derivative contracts with a basis 1-form while in the above equation the derivative contracts with one indice of the field tensor. In order to bypass the problem, I tried to decouple those disturbing contractions using Kronecker deltas:

[itex]{{F}^{\mu \nu }}_{,\mu }{{\varepsilon }_{\nu \alpha \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}={{\delta }_{\mu }}^{\rho }{{\delta }_{\alpha }}^{\sigma }{{F}^{\mu \nu }}_{,\rho }{{\varepsilon }_{\nu \sigma \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}[/itex]

I don’t know how to proceed further, in order to get the desired contractions. I tried to represent the deltas with LCSs, like this:

[itex]{{\varepsilon }_{\rho \sigma \mu \alpha }}{{\varepsilon }^{\rho \sigma \mu \nu }}=3!{{\delta }_{\alpha }}^{\nu }[/itex]

but then the equation gets more complex. I suppose that I have to use some property of the LCSs in order to manipulate the 5 LCS that appear.

B) I tried to connect [itex]*d*F[/itex] with [itex]*J[/itex] , a process that involves simpler manipulations, but I come up with an equation that has an opposite sign of the desired. Here is what I did:

[itex]*d*F=\tfrac{1}{24}{{\varepsilon }^{\mu \nu \beta \gamma }}{{F}_{\mu \nu }}^{,\alpha }{{\varepsilon }_{\alpha \beta \gamma \delta }}d{{x}^{\delta }}=\tfrac{1}{6}{{F}_{\mu \nu }}^{,\mu }d{{x}^{\nu }}\Rightarrow 6*d*F={{F}_{\mu \nu }}^{,\mu }d{{x}^{\nu }}[/itex]

[itex]*{{J}_{}}=\tfrac{1}{3!}\tfrac{1}{3!}{{J}_{\mu }}{{\varepsilon }^{\mu \alpha \beta \gamma }}{{\varepsilon }_{\alpha \beta \gamma \delta }}d{{x}^{\delta }}=-\tfrac{1}{3!}{{J}_{\nu }}d{{x}^{\nu }}\Rightarrow -6*{J}={{J}_{\nu }}d{{x}^{\nu }}[/itex]

(I used the identity [itex]{{\varepsilon }_{\mu \nu \alpha \beta }}{{\varepsilon }^{\mu \nu \rho \sigma }}=2\left( {{\delta }_{\alpha }}^{\rho }{{\delta }_{\beta }}^{\sigma }-{{\delta }_{\beta }}^{\rho }{{\delta }_{\alpha }}^{\sigma } \right) [/itex] in the first equation and [itex]{{\varepsilon }_{\rho \sigma \mu \alpha }}{{\varepsilon }^{\rho \sigma \mu \nu }}=3!{{\delta }_{\alpha }}^{\nu }[/itex] in the second)

So, according to the given equation, the above 1-forms are related like this:

[itex]{{F}_{\mu \nu }}^{,\mu }={{\mu }_{0}}{{J}_{\nu }}\Rightarrow *d*F=-{{\mu }_{0}}*{J}\Rightarrow d*F=-{{\mu }_{0}}{J}[/itex]

That minus sign shouldn’t be there!!!

Please check this last process for mistakes; it is the one that gets me closer to the result. If it correct, then apparently what I want to proof is the result with minus sign. But in the literature this equation has always a positive sign. I checked the definitions of the forms and the Hodge duality that are used in literature (which are the same as I those presented above), but the result they get has a positive sign (for reference check here http://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field , in the “Differential forms approach” section)
 

Answers and Replies

  • #2
1,024
32
it is better to go with one form current.
j=jadxa
with the definition of F,we have
*F=1/4(εabcd Fcd dxaΔdxb)
we can employ riemann normal coordinates in case of curved spacetime,
g=η,zero christoffel symbols this all means dε=0
d*F=1/4 d(εabcd Fcd dxaΔdxb)=1/4 ε[ab cde]Fcd dxeΔdxaΔdxb
this is three form with components
(d*F)eab=3/2 ε[ab cde]Fcd
then component of *d*F are
(*d*F)p=1/6 εpeab(3/2 ε[ab cde]Fcd)=1/4 εabpe εabcdeFcd=-δ[c|p ge|d]eFcd
=-δcpgedeFcd=-∂d Fpd=∂d Fdp
which from maxwell eqn
aFab=-jb gives what is required.your minus sign is ok.they are just using different faraday tensor ie. the equation used is ∇aFba=jb .that's all.see wiki page on electromagnetic tensor.
 
Last edited:
  • #3
123
0
Thank's!
 

Related Threads on A Maxwell's equation in differential forms formalism

Replies
1
Views
2K
Replies
6
Views
5K
Replies
0
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
0
Views
309
  • Last Post
Replies
11
Views
2K
Replies
1
Views
1K
Replies
0
Views
464
Replies
1
Views
4K
Replies
7
Views
1K
Top