In Wikipedia the components of the EM Field Tensor are listed as $$F^{\mu\nu}=\left( \begin{array}{cccc} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \\ \end{array} \right)$$ in an inertial frame with units where $c=1$ and with the $(+---)$ sign convention.

What is the correct expression for the opposite sign convention: $(-+++)$? And more importantly, what is the general rule for expressing the coordinates of tensors under the opposite sign convention? Do I switch the sign of all components, or only some, and if so which components are switched? Does it differ for covariant or contravariant tensors?


1 Answer 1


The signature-independent definition is $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu.$$ The definitions of $\partial_\mu$ and $A_\mu$, as well as $\mathbf{E}$ and $\mathbf{B}$ in terms of $A_\mu$, are also signature independent.

The raised version of the field tensor you cited is produced by raising both indices, $$F^{\mu\nu} = \eta^{\mu\rho} \eta^{\nu\sigma} F_{\rho\sigma}.$$ Changing the metric sign convention, in this case, does nothing besides swap $\eta \to -\eta$. Since there are two factors of $\eta$, nothing changes.

There is no general rule for an arbitrary tensor, because the rule depends on what is not changed by swapping the signature, which in turn depends on the conventions you're using for the physical quantities you're describing. For example, the expression you wrote down is in terms of $E_i$ and $B_i$. Whether these are equal to $E^i$ and $B^i$ (independent of metric signature) or $\pm E^i$ and $\pm B^i$ (dependent on metric signature) depends on the source. If it's the latter, then the field tensor would pick up a minus sign. Just looking for a canonical source that says "the" answer is useless, because the answer you'll find in the next canonical source will be different.

  • $\begingroup$ Thanks! This is excellent $\endgroup$
    – Dale
    Commented Sep 14, 2019 at 17:51
  • $\begingroup$ In my electrodynamic course we use the tensor $F_{\mu\nu} = - \partial_\mu A_\nu + \partial_\nu A_\mu$ but it works too. We just have minus signs where you have a plus sign. $\endgroup$ Commented Aug 5, 2022 at 8:58
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    $\begingroup$ "The signature-independent definition..." - The first equation is $F:=\mathrm d(A^\flat)$, where $A$ is the electromagnetic four-potential, so it is signature-dependent, isn't it? At least that's the essence of Qmechanics's answer to essentially the same question. $\endgroup$
    – Filippo
    Commented Oct 19, 2022 at 9:07

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