# The signature of the metric and the definition of the electromagnetic tensor

Let $$ \eta_{\mu\nu}={\rm diag}(+1,-1,-1,-1) \qquad \bar\eta_{\mu\nu}={\rm diag}(-1,+1,+1,+1) $$ with corresponding Lorentz force laws (in units where mass equals charge) $$ \ddot x^\mu=\eta_{\nu\lambda}F^{\mu\nu}\dot x^\lambda \qquad \ddot{\bar x}^\mu=\bar\eta_{\nu\lambda}\bar F^{\mu\nu}\dot{\bar x}^\lambda $$

As the trajectories $x^\mu, \bar x^\mu$ should agree (and so will all its derivatives) for all initial conditions, we can equate the terms $$ \tag{1} \eta_{\nu\lambda}F^{\mu\nu} = \bar\eta_{\nu\lambda}\bar F^{\mu\nu} $$ Contracting with the inverse $\eta^{\lambda\sigma}$ of $\eta_{\nu\lambda}$ finally yields $$ F^{\mu\sigma} = -\bar F^{\mu\sigma} $$ as $$ \bar\eta_{\nu\lambda}\eta^{\lambda\sigma} = -\delta_\nu^{\sigma} $$ This means the signs of the components of the electromagnetic tensor $F^{\mu\nu}$ do indeed depend on the metric convention. This also applies to $F_{\mu\nu}$, whereas the tensor of mixed rank $F^\mu{}_\nu$ is independant of this choice (which is just (1)).

We will work in unit with $c=1$. In both sign conventions for the metric $\eta_{\mu\nu}$ we define the field strength as

$$\tag{1} A^{\mu}~=~(\Phi,{\bf A}). $$

$$\tag{2} F_{\mu\nu}~:=~ \partial_{\mu} A_{\mu} -\partial_{\nu} A_{\mu}, \qquad \mu,\nu~\in~\{0,1,2,3\}. $$

$$\tag{3} E_i~:=~- \partial_i\Phi -\partial_0 A^i, \qquad i~\in~\{1,2,3\}. $$

[The relation (3) can be partially remembered by the fact that in electrostatics, one demands that ${\bf E}~=~-{\bf \nabla}\Phi$. It turns out that the rest of eq. (3) is then fixed by consistency.] Tensors are raised and lowered with the metric tensor $\eta_{\mu\nu}$.

It is then straightforward to check that this implies that in signature

$$\tag{4} (+,-,-,-)\qquad \text{resp.} \qquad(-,+,+,+), $$

the $4$-potential $A_{\mu}$ with lower index is

$$\tag{5} A_{\mu}~=~(\Phi,-{\bf A}) \qquad \text{resp.} \qquad A_{\mu}~=~(-\Phi,{\bf A}),$$

and the electric field ${\bf E}$ is

$$\tag{6} E_i~=~F_{0i} \qquad \text{resp.} \qquad E_i~=~F_{i0}. $$

See also this related Phys.SE post.