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Rewriting Maxwell's Equations in Tensor Form

  • Thread starter Sycobob
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  • #1
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Homework Statement



From Sean Carroll's notes on general relativity (chapter 1, pg. 20):

Show that [itex]F_{[\alpha\beta,\gamma]} = 0[/itex] is equivalent to half of the Maxwell equations.

Homework Equations



[itex]F_{\mu\nu}[/itex] is the electromagnetic tensor

[itex]\Phi_{,\nu} \equiv \partial_{\nu}\Phi[/itex]

[itex]F_{i0}= E_{i}[/itex]

[itex]F_{ij}= \epsilon^{ijk}B_{k}[/itex]

The Attempt at a Solution



I'm specifically looking to turn Maxwell's (homogeneous) equations into tensor form, not just show that they fall out of [itex]F_{[\alpha\beta,\gamma]} = 0[/itex]. I sort of have a solution, but I feel like I'm missing a step.

[itex]
\begin{eqnarray*}
\nabla×\textbf{E} + \partial_{t}\textbf{B} = 0 \\
\nabla\cdot \textbf{B} = 0
\\
\\
\epsilon^{ijk}\partial_{j}E_{k} + \partial_{0}B^{i} = 0 \\
\partial_{i}B^{i} = 0
\\
\\
\epsilon^{ijk}\partial_{j}F_{k0} + \frac{1}{2}\epsilon^{ijk}\partial_{0}F_{jk} = 0 \\
\frac{1}{2}\epsilon^{ijk}\partial_{i}F_{jk} = 0
\end{eqnarray*}
[/itex]

which can be rewritten as:

[itex]\epsilon^{\mu\nu\rho\sigma}\partial_{\rho}F_{\mu \nu} = 0[/itex]

which, up to a normalization constant, is just:

[itex]F_{[\alpha\beta,\gamma]} = 0[/itex]

My question is about going from step 3 to step 4. I sort of pulled it out of my hat and checked that it was correct (term by term). I'm looking for some kind of justification for this step, or a nudge in the right direction if I'm approaching this all wrong.

Also, I'm still getting the hang of tensor notation, and I feel like equation 4 doesn't make sense. Only 3 indices are contracted, leaving the right side a vector, not a scalar. On the other hand, trying to use the Levi-Civita tensor with 3 indices here seems wrong too, as the indices run from 0 to 4 leaving you with stuff like [itex]\epsilon^{013}[/itex].
 
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Answers and Replies

  • #2
Bump. : (
 
  • #3
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
407
The first equation in your step 3 is actually three equations, one for each value of i. And the second equation is just one. That's why the final result is a set of four equations.

I think the final step will be easier to understand if you make sure that you understand every step of the following rewrites: (Here A is anything with three indices from 0 to 3).

\begin{align}
&\varepsilon^{ijk} A_{ijk} =\varepsilon^{ijk0} A_{ijk} =\varepsilon^{\mu\nu\rho 0} A_{\mu\nu\rho}\\
&\varepsilon^{ijk} A_{0jk} =\varepsilon^{0jki} A_{0jk} =\varepsilon^{0\nu\rho i} A_{0\nu\rho} =\varepsilon^{\mu\nu\rho i} A_{\mu\nu\rho}
\end{align}
 
  • #4
18
1
The first equation in your step 3 is actually three equations, one for each value of i. And the second equation is just one. That's why the final result is a set of four equations.

I think the final step will be easier to understand if you make sure that you understand every step of the following rewrites: (Here A is anything with three indices from 0 to 3).

\begin{align}
&\varepsilon^{ijk} A_{ijk} =\varepsilon^{ijk0} A_{ijk} =\varepsilon^{\mu\nu\rho 0} A_{\mu\nu\rho}\\
&\varepsilon^{ijk} A_{0jk} =\varepsilon^{0jki} A_{0jk} =\varepsilon^{0\nu\rho i} A_{0\nu\rho} =\varepsilon^{\mu\nu\rho i} A_{\mu\nu\rho}
\end{align}
need help here

how did you obtain these rules>
thank you
 
  • #5
haushofer
Science Advisor
Insights Author
2,299
674
How many independent components does a 1-form have in 4 dimensions? And a 3-form? This is a first step to check whether your result makes sense (hint: it does :P )

Hit your equation then with an (4-comp.) epsilon symbol with the free index contracted. Find/derive the identities of contracted epsilon symbols in terms of kronecker deltas and apply those.

Hope this helps ;)
 
  • #7
18
1
How many independent components does a 1-form have in 4 dimensions? And a 3-form? This is a first step to check whether your result makes sense (hint: it does :P )

Hit your equation then with an (4-comp.) epsilon symbol with the free index contracted. Find/derive the identities of contracted epsilon symbols in terms of kronecker deltas and apply those.

Hope this helps ;)
thank for your help
i have trouble in "εijk A0jk=ε0jki A0jk "
i get stuck in this equation
i dont know how can they put a zero into the εijk
such that εijk ---> ε0jki
i think A0 is "fixed variable" and you cannot do anything for it
right? if you put a zero there,then A0 will become dummy variable (so we can contract an upper and lower index together)
 

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