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Recently, I was wondering if the homogeneous wave equation is Lorentz invariant. It can be easily proved that the operator $\Box$ is invariant under Lorentz transformations, and so I concluded that, if acting upon a scalar field $\phi(x)$, then $\Box\phi(x)=0$ is also invariant.

If one considers for example the fields $\vec{E}$ or $\vec{B}$, I don't think this result holds, since such fields are not Lorentz invariant. However, the following doubt struck me. The wave equations for the fields are derived from Maxwell's equations, which are Lorentz invariant. One could then think that the resulting homogeneous wave equation is invariant, but I'm pretty sure there is something wrong with my reasoning, could you help me pointing it out?

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    $\begingroup$ What if you use the 4-vector $\:\left(\mathbf{A}, \phi\right)\:$ instead of the two 3-vectors $\:\mathbf{E}, \mathbf{B}$ ? $\endgroup$ – Frobenius Nov 26 '16 at 10:55
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Firstly, wave equations in a relativistic field theory need not be Lorentz invariant. They are Lorentz covariant i.e., unchanged in form under a Lorentz transformation. The wave equation for the electromagnetic field, in a relativistic classical field theory is written in terms of the four-potential $A^\mu$ given by $$\Box A_\mu=j_\mu$$ in Lorentz gauge.

Secondly, wave equations written in terms of $\textbf{E}$, $\textbf{B}$ need not be even be relativistically covariant because they are not four-vectors i.e., not vectors in spacetime.

Maxwell's equations when written in a covariant form, the wave equation derived from it are Lorentz covariant.

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