# Is there only one convention to define the electromagnetic field tensor?

I know that the electromagnetic field tensor depends on which metric is used. For example wikipedia uses the $$(+---)$$ sign convention, but in the Griffiths we have the $$(-+++)$$ sign convention.

That's why on wikipedia, to define $$F_{\mu \nu}$$ independently of the metric tensor, they define $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu.$$

However, in my electrodynamic course we define $$F_{\mu\nu} = - \partial_\mu A_\nu + \partial_\nu A_\mu$$ so all the signs are flipped. For example using the $$(-+++)$$ sign convention (and $$c=1$$) we get:

$$F_{\mu\nu}=\left( \begin{array}{cccc} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \\ \end{array} \right)$$

This matrix is the same as on wikipedia, and the opposite as the one in the Griffiths (all the signs flipped).

I haven't seen anyone else use the definition $$F_{\mu\nu} = - \partial_\mu A_\nu + \partial_\nu A_\mu$$ for the electromagnetic field tensor, so I am wondering if there is something wrong in my course, or if there is no problem by defining the tensor in this way.

• In some older textbooks, the semicolon ; was used to denote the covariant derivative operation so that your course's definition would appear as $F_{\mu \nu}=A_{\mu;\nu}-A_{\nu;\mu}$. This convention, together with the chosen signature-convention, must be accounted for. Aug 5, 2022 at 14:16

As long as you keep track of that sign in your formulas, there is no issue. All formulas in which the field tensor appears will simply have a sign difference. The fact most references use the same convention is probably the same we write $$2\pi$$ instead of $$\tau$$: someone choose that convention a lot of time ago and it would get messy if we used different conventions.
• Hopefully those who write $(+---)$ will also eventually perceive the convenience of a single convention and accept to use the superior choice $(-+++)$. Aug 5, 2022 at 10:05
• The $A_\mu$ are also differnt in the two metrics. If I remember correctly, it is $A^\mu=(\phi, {\bf A})$ where ${\bf E}= -\nabla \phi- \dot {\bf A}$, and the $A_\mu=g_{\mu\nu} A^\nu$. Aug 5, 2022 at 11:58