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I have to show, how the contravariant components of the electromagnetic field tensor behave under Lorentz transformation.

I guess the answer should look something like this $$F'^{\mu\nu}=\frac{\partial x'^\alpha}{\partial x^\gamma}\frac{\partial x'^\beta}{\partial x^\delta}=\Lambda^\alpha{}_\gamma\Lambda^\beta{}_\delta F^{\gamma\delta}$$ That's not a big deal, one can see this immediately (although I don't really get the message of this exercise, since $F^{\mu\nu}$ just transforms as a 2nd rank tensor has to by definition)

However, the exercise wants me to derive this from the inhomogeneous Maxwell equations in covariant formulation:

$$\partial_\mu F^{\mu\nu}=\mu_0j^\nu$$

I know how $\partial_\mu$ transforms: $$\partial'_\mu=\frac{\partial x^\nu}{\partial x'^\nu}\frac{\partial}{\partial x^\nu}=\Lambda_\mu{}^\nu \partial_\nu=(\Lambda^{-1})^\nu{}_\mu \partial_\nu$$

as well as $j^\nu$:

$$j'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}j^\nu=\Lambda^\mu{}_\nu j^\nu$$

But how do I link these equations to the 3d one and arrive at the 1st ?

Any hints?

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  • $\begingroup$ Noiralef gives a proper hint, however also look up something called "quotient theorem". Its proof is essentially the answer to your question. $\endgroup$ – Bence Racskó Jan 19 '17 at 16:17
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What you're maybe missing is the requirement that the covariant equation should still be valid in the new coordinate system: $$ \partial'_\mu F^{\mu\nu} = \mu_0 j^{\prime\nu} $$ You can now plug in your three equations and solve for $F'$.

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