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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.

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    $\begingroup$ 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. $\endgroup$
    – robphy
    Commented Aug 5, 2022 at 14:16

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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.

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    $\begingroup$ Hopefully those who write $(+---)$ will also eventually perceive the convenience of a single convention and accept to use the superior choice $(-+++)$. $\endgroup$ Commented Aug 5, 2022 at 10:05
  • $\begingroup$ 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 $A_\mu=g_{\mu\nu} A^\nu$. $\endgroup$
    – mike stone
    Commented Aug 5, 2022 at 11:58

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