1
$\begingroup$

How do you transform between the electromagnetic tensors $F_{\mu\nu}$ and $F^{\mu\nu}$? $$ F_{\mu \nu}= \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix},\\ \ F^{\mu \nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix} $$ In other words, what do you do to $F_{\mu\nu}$ to get $F^{\mu\nu}$?

$\endgroup$

2 Answers 2

2
$\begingroup$

Index raising and lowering is defined through the metric, in this case the flat space metric (Minkowski)

$$ g^{\mu\nu} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$

We raise an index by aplying the metric to a tensor, like this $A^\mu=g^{\mu\nu}A_\nu$. Now, if you want to raise two index you need to operate with the metric twice.

$$F^{\mu\nu}=g^{\mu\alpha}g^{\beta\nu}F_{\alpha\beta}$$

In a more formal language lowering and raising indices is a way to construct isomorphisms between covariant and contravariant tensorial spaces. We use the metric tensor because it help us to map basis vectors $e_i$ to dual basis vector $\beta^i$.

$\endgroup$
1
$\begingroup$

I see now, as with transforming one-forms to/from vectors, you apply the metric. Because you want to make two subscripts superscripts, apply it twice. So, with $$ g^{\mu\nu} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $$ $$ F^{\mu\nu} = g^{\mu\nu}F_{\mu\nu}g^{\mu\nu} $$

$\endgroup$
4
  • $\begingroup$ Though a more rigorous explanation would be appreciated $\endgroup$ Commented Feb 8, 2015 at 2:55
  • 6
    $\begingroup$ You should write $F^{\mu\nu}=g^{\mu\rho}F_{\rho\sigma}g^{\nu\sigma}$. An index can only appear twice in a factor. $\endgroup$
    – Ryan Unger
    Commented Feb 8, 2015 at 3:01
  • $\begingroup$ What "more rigorous explanation" are you looking for? Index raising/lowering is defined with the metric. $\endgroup$
    – user195162
    Commented Jun 25, 2019 at 18:19
  • $\begingroup$ @CactusHouse Is this more rigorous? $\endgroup$
    – Filippo
    Commented Nov 2, 2020 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.