Finding the components of B in the Electromagnetic field tensor

Question

I have a bit of confusion about how magnetic field components are written in function of the field tensor.

I have the metric $(-,+,+,+)$ so my field tensor is this

$$ F^{\mu \nu} \, = \,\left(\begin{matrix}0 & E_1 & E_{2} & E_{3} \\ -E_{1} & 0 & B^{3} & -B^{2} \\ -E_2 & -B^{3} & 0 & B^1 \\ -E_3 & B^{2} & -B^{1} & 0\end{matrix}\right)\tag{3} $$

I normalize the Levi-Civita symbol in this way: $\epsilon^{123}=1$

In many textbooks and in the instructor's notes the components of B are defined like that:

$$B^k = \frac{1}{2} \epsilon^{ijk} F^{ij}$$

$i,j,k$ run from $1$ to $3$.

So, for example, to find $B^3$, which is clearly $F^{12}$, I set $k=3$ and sum over the permutation of the Levi-Civita symbol obtaining something like that

$$\frac{1}{2}( \epsilon^{123} F^{12} + \epsilon^{213} F^{21}) = \frac{1}{2}( F^{12} - F^{21}) = F^{12} = B^3$$

But I don't really know why I'm summing, or if there's something that tells me to sum over the permutations of $i$ and $j$, I'm summing just cause in this way I obtain the right result. But I think it would be simpler to write

$$ B^k=\epsilon^{ijk} F^{ij} $$

In this way no summation is required, you simply compute the value of $\epsilon^{ijk}$.

For example, $k=3$, $i$ and $j$ are now free so I write $\epsilon^{ijk}$ giving a value to $i$ and the other to $j$, picking the simpler choice I have $B^3= \epsilon^{123} F^{12}= 1 F^{12}$. The other choice is $B^3= \epsilon^{213} F^{21}= -F^{21}= F^{12}$

So my questions are:

why that way of writing $B^k$ and not the other simpler one?

Why summing over permutation? I mean what tells me that I have to sum over the permutations and not just pick a value for $i$ and $j$ and compute the value of the Levi-Civita tensor?

Answer 1

That's not how index notation works. You say that in $$ B^k = \epsilon^{ijk}F^{ij}\tag{1}$$ you are "free to choose" the $i$ and $j$, but then you only make the two choices (1,2) and (2,1) for which that gives the correct results. What about $i=1,j=3$, or $i=2,j=2$? You chose those specific $i$s and $j$s because you knew what result you wanted, not because it's somehow a choice dictated by the notation (1).

To avoid such ambiguous choices, all indices in index notation must occur either on both sides of the equation or be summed over. The general convention is to sum over repeated indices, although in your relativistic setting the index position is actually relevant and the equation should be written $$ B^k = \epsilon^{ijk} F_{ij}. \tag{2}$$ Here you are now summing over all possible values for $i$ and $j$, not only over their permutations - but those indices $ijk$ which are not permutations of $123$ simply yield zero because the Levi-Civita symbol is zero for those indices.