# Electromagnetic tensor and the B-field

1. Jan 12, 2016

### StuartY

Hello, first off, I'm not sure if I put this question in the right place so sorry about that.
Given Bi = 1/2 εijk Fjk 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.

2. Jan 12, 2016

### 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:
$$\epsilon^{ijk} B_i = F^{jk}$$

Last edited: Jan 12, 2016
3. Jan 12, 2016

### Orodruin

Staff Emeritus
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.

4. Jan 12, 2016

### DuckAmuck

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

5. Jan 14, 2016

### 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.,
$$B_j=\epsilon_{jkl} F_{kl}=2(^{\dagger} F)_{j}$$
$$\epsilon_{abj}B_j=\epsilon_{jab} B_j=\epsilon_{jab} \epsilon_{jkl} F_{kl}=(\delta_{ak} \delta_{bl}-\delta_{al} \delta_{bk})) F_{kl}=F_{ab}-F_{ba}=2F_{ab}.$$
$$^{\dagger \dagger} F=F.$$