Solving Electromagnetic Tensor & B-Field

[StuartY]

Hello, first off, I'm not sure if I put this question in the right place so sorry about that.
Given B_i = 1/2 ε_ijk F_jk how would you find F in terms of B? I think you multiply through by another Levi-Civita, but then I don't know what to do after that. Any help would much appreciated.

 

[DuckAmuck]

That's correct. When you multiple two levi-civitas, and end up with delta functions. Then you take advantage of F being antisymmetric.

B_i = \frac{1}{2} \epsilon_{ijk} F^{jk}
\epsilon^{imn} B_i = \frac{1}{2} \epsilon^{imn} \epsilon_{ijk} F^{jk}
\epsilon^{imn} B_i = \frac{1}{2} (\delta^m_j \delta^n_k - \delta^n_j \delta^m_k) F^{jk}
\epsilon^{imn} B_i = \frac{1}{2} (F^{mn} - F^{nm})
F^{nm} = - F^{mn}

after some relabeling:
\epsilon^{ijk} B_i = F^{jk}

 

[Orodruin]

That's correct. When you multiple two levi-civitas, you end up with a variety of results depending on how many indices are shared. In the case you want here, 2 are shared: \epsilon^{jkl} \epsilon_{ijk} = 2 \delta^{l}_i

B_i = \frac{1}{2} \epsilon_{ijk} F^{jk}
\epsilon^{jkl} B_i = \frac{1}{2} \epsilon^{jkl} \epsilon_{ijk} F^{jk}
\epsilon^{jkl} B_i = \frac{1}{2} 2 \delta^{l}_i F^{jk}

This is wrong, you can never have three of the same index when using the Einstein summation convention. You are using the same letter for some free indices as for some summation indices and then you confuse which is what. The thing to use is the standard epsilon-delta relation with only one index summed over and compute $\epsilon_{mni} B_i$. I will leave the actual computation for the OP.

 

[DuckAmuck]

This is wrong, you can never have three of the same index when using the Einstein summation convention. You are using the same letter for some free indices as for some summation indices and then you confuse which is what. The thing to use is the standard epsilon-delta relation with only one index summed over and compute $\epsilon_{mni} B_i$. I will leave the actual computation for the OP.

My bad. I guess I just remember the answer, and forgot how to do math. I fixed it.

 

[vanhees71]

Just use the identity
$$\epsilon_{ijk} \epsilon_{ilm}=\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}.$$
The Hodge dual of a Hodge dual leads back to the original (up to signs), e.g.,
$$B_j=\epsilon_{jkl} F_{kl}=2(^{\dagger} F)_{j}$$
leads to
$$\epsilon_{abj}B_j=\epsilon_{jab} B_j=\epsilon_{jab} \epsilon_{jkl} F_{kl}=(\delta_{ak} \delta_{bl}-\delta_{al} \delta_{bk})) F_{kl}=F_{ab}-F_{ba}=2F_{ab}.$$
So indeed you have
$$^{\dagger \dagger} F=F.$$
Note that all this is in usual 3D Carrtesian notation. If you want to have this covariantly in Minkowski space you must be careful with the signs of the 4D Levi-Civita tensor. It also differs from one textbook to the other. So make sure you know the convention used in your textbook!