I've read the definition of the electromagnetic field tensor to be \begin{equation}F^{\mu\nu}\equiv\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}\tag{*}\end{equation} in Introduction to Electrodynamics by David Griffiths, or as $$F_{\mu\nu}\equiv\begin{pmatrix}0&-E_x&-E_y&-E_z\\E_x&0&B_z&-B_y\\E_y&-B_z&0&B_x\\E_z&B_y&-B_x&0\end{pmatrix}$$ on the Lecture Notes on GR by Sean Carroll, which I know to be consistent via ${F_{\mu\nu}=\eta_{\mu\alpha}F^{\alpha\beta}\eta_{\beta\nu}}$ where the metric $\eta_{\rho\sigma}$ has a $(-+++)$ signature.

However on Wikipedia and other sources (sorry I can't remember) they use a $(+---)$ signature and they define the EM tensor to be the negative of ${(*)}$.

These are my thoughts about it: The antisymmetry $F^{\mu\nu}=-F^{\nu\mu}$ may point out that it's just an unfortunate mix of index letters and that for the sources notation to be consistent, either the first two or Wikipedia should change $\mu\nu$ to $\nu\mu$. If not the case, the properties seem to be the same; at first I thought the inner product would pop out a minus sign of difference, but it of course didn't happen, and as for other entities I've worked with, e. g. the 4-velocity, though the metric signature can change, the contravariant vector is the same in either case. However again, I've read the stress-energy tensor does change sign depending on the signature.

So is the signature of the metric involved in the definition of ${F^{\mu\nu}}$ or any tensor whatsoever? If so, how can I know what signature is involved? or if not, what's the matter with the minus sign difference on the definitions?


2 Answers 2


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)).

  • $\begingroup$ Great! Just one thing, you used ${u^\lambda=\bar u^\lambda}$ which is known by definition of the 4-position, but you also used ${\dot{u}^\mu=\dot{\bar{u}}^\mu}$, (as one supposedly doesn't know beforehand that ${F^{\mu\sigma}=-\bar F^{\mu\sigma}}$) right? So how is this justified? Can one always define contravariant vectors to be the same whatever the metric signature is? $\endgroup$
    – user24999
    Nov 20, 2013 at 17:48
  • 1
    $\begingroup$ @PedroFigueroa: same as velocities, accelerations (as well as any higher derivatives of position) agree - we're dealing with the same trajectory; I'll clarify $\endgroup$
    – Christoph
    Nov 20, 2013 at 19:28

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

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

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

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

[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.

  • $\begingroup$ So you postulate a vector field $A$ on $M$ and then you define $F:=\mathrm d(A^\flat)$? $\endgroup$
    – Filippo
    Oct 19, 2022 at 8:51
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    $\begingroup$ Hi Filippo. Thanks for your feedback. The priority is admittedly not elegance, but agreement with clunky historic sign conventions in the E&M literature. $\endgroup$
    – Qmechanic
    Oct 19, 2022 at 8:59

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