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How can we prove that the electromagnetic field strength $F_{\mu\nu}=\partial_\nu A_\mu-\partial_\mu A_\nu$ is a tensor of type (0,2), where $A$ is the four potential.

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    $\begingroup$ You want to consider the coordinate transformation of the tensor, and check whether it satisfies the one of a rank-2 tensor. $\endgroup$ – Drake Marquis Sep 3 '17 at 10:39
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    $\begingroup$ Technically, that is not the tensor, those are the tensor's components in the coordinate basis at each point in spacetime. $\endgroup$ – DanielC Sep 3 '17 at 11:11
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As Daniel has mentioned, these are the components of the tensor. And as Drake has mentioned, we can easily perform a coordinate transformation and see what is the outcome. So, for the coordinate transformation $x^\mu \mapsto x'^\mu (x^\nu)$ we have

$$ \begin{align} F'_{\mu\nu}=\partial '_\mu A'_\nu - \partial'_\mu A'_\nu &= \frac{\partial }{\partial x'^\mu} \left(\frac{\partial x^\beta}{\partial x'^\nu} A_\beta\right) - \frac{\partial}{\partial x'^\nu} \left(\frac{\partial x^\beta}{\partial x'^\mu} A_\beta\right) \\ &=\frac{\partial^2 x^\beta}{\partial x'^\mu\partial x'^\nu} A_\beta + \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} \partial_\alpha A_\beta -\frac{\partial^2 x^\beta}{\partial x'^\nu\partial x'^\mu} A_\beta - \frac{\partial x^\alpha}{\partial x'^\nu} \frac{\partial x^\beta}{\partial x'^\mu} \partial_\alpha A_\beta \\ &=\frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} \partial_\alpha A_\beta - \frac{\partial x^\beta}{\partial x'^\mu} \frac{\partial x^\alpha}{\partial x'^\nu} \partial_\beta A_\alpha \\ &= \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} \left(\partial_\alpha A_\beta -\partial_\beta A_\alpha \right)\\ &=\frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} F_{\alpha\beta}\,. \end{align} $$ And this is nothing but the transformation behavior of a $(0,2)-tensor$.

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An alternative approach to Immanuel's notes that, because tensors are closed under covariant derivatives (but in general not under partial ones), since $A_\mu$ is a rank-$(0,\,1)$ tensor $\nabla_\mu A_\nu=\partial_\mu A_\nu -\Gamma_{\mu\nu}^\rho A_\rho$ is a rank-$(0,\,2)$ tensor, as is its antisymmetric part $$\nabla_\mu A_\nu-\nabla_\nu A_\mu=F_{\mu\nu}+(\Gamma_{\nu\mu}^\rho-\Gamma_{\mu\nu}^\rho)A_\rho.$$The $A_\rho$ coefficient is a torsion tensor, so we're done.

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    $\begingroup$ Well, the conclusion is valid, even with the torsion tensor ('s components) in the second term of the RHS. $\endgroup$ – DanielC Sep 3 '17 at 21:39

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