❓What is the final expression for the tensor product when computed in terms of the magnetic field com

What is the final expression for the tensor product when computed in terms of the magnetic field components?

The final expression for the tensor product $\mathbf{F}{ij} \mathbf{F}{jk}$ in terms of the magnetic field components $B_i$ is: $\mathbf{F}{ij} \mathbf{F}{jk} = B^2 \delta_{ik} - B_i B_k$

Where:

• $\mathbf{F}{ij}$ is the magnetic field tensor, defined as $\mathbf{F}{ij} = \epsilon_{ijk} B_k$.

• $B_i$ and $B_k$ are the components of the magnetic field vector $\mathbf{B}$.

• $B^2$ is the magnitude squared of the magnetic field, $B^2 = B_m B_m$ (summation over $m$).

• $\delta_{ik}$ is the Kronecker delta, which is 1 if $i=k$ and 0 if $i \neq k$.

Derivation

The result is derived using the definition of the magnetic field tensor and the vector identity for the product of two Levi-Civita symbols.

1. Substitute the definition $\mathbf{F}{ij} = \epsilon{ijm} B_m$ (using $m$ as a dummy index for the first $\mathbf{F}$) and $\mathbf{F}{jk} = \epsilon{jkl} B_l$ (using $l$ for the second $\mathbf{F}$):

   $\mathbf{F}{ij} \mathbf{F}{jk} = (\epsilon_{ijm} B_m) (\epsilon_{jkl} B_l) = B_m B_l (\epsilon_{ijm} \epsilon_{jkl})$

2. Rearrange the Levi-Civita symbols to group the repeated summation index $j$:

   $\epsilon_{jkl} = -\epsilon_{kjl}$

   $\mathbf{F}{ij} \mathbf{F}{jk} = - B_m B_l (\epsilon_{ijm} \epsilon_{kjl})$

3. Use the $\epsilon-\delta$ identity: $\epsilon_{abc} \epsilon_{dbc} = \delta_{ad} \delta_{cb} \delta_{cb} - \delta_{ab} \delta_{dc} \delta_{cb}$. A simpler form for the expression $\epsilon_{ijm} \epsilon_{kjl}$ (with summation over $j$) is:

   $\epsilon_{ijm} \epsilon_{kjl} = \delta_{ik} \delta_{ml} - \delta_{il} \delta_{mk}$

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