Rewriting Maxwell's Equations in Tensor Form

[Sycobob]

Homework Statement

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

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

Homework Equations

F_{\mu\nu} is the electromagnetic tensor

\Phi_{,\nu} \equiv \partial_{\nu}\Phi

F_{i0}= E_{i}

F_{ij}= \epsilon^{ijk}B_{k}

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 F_{[\alpha\beta,\gamma]} = 0. I sort of have a solution, but I feel like I'm missing a step.

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

which can be rewritten as:

\epsilon^{\mu\nu\rho\sigma}\partial_{\rho}F_{\mu \nu} = 0

which, up to a normalization constant, is just:

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

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 \epsilon^{013}.

 

[OnesieWithaZ]

Bump. : (

 

[Fredrik]

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}

 

[bres gres]

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

 

[haushofer]

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 ;)

 

[haushofer]

See e.g. https://en.m.wikipedia.org/wiki/Levi-Civita_symbol

 

[bres gres]

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 don't 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)