# Four-dimensional generalization of the equation $\vec{D} = \epsilon \vec{E}$

In "The electrodynamics of moving dielectrics" (p. 262) Landau and Lifshitz have defined four-tensors and four-vector

$$F^{\mu \nu} = \begin{pmatrix} 0&-E_x&-E_y&-E_z\\ E_x&0&-B_z&B_y\\ E_y&B_z&0&-B_x\\ E_z&-B_y&B_x&0\\ \end{pmatrix}$$

$$H^{\mu \nu} = \begin{pmatrix} 0&-D_x&-D_y&-D_z\\ D_x&0&-H_z&H_y\\ D_y&H_z&0&-H_x\\ D_z&-H_y&H_x&0\\ \end{pmatrix}$$

$$u^{\mu} = \begin{pmatrix} \frac{1}{\sqrt{1-v^2/c^2}},&\frac{\vec{v}}{c\sqrt{1-v^2/c^2}} \end{pmatrix}$$

Then they write

From this four-vector and the four-tensors $$F^{\mu \nu}$$ and $$H^{\mu \nu}$$ we form combinations which become $$\vec{E}$$ and $$\vec{D}$$ in a medium at rest. These combinations are the four-vectors $$F^{\lambda \mu}u_{\mu}$$ and $$H^{\lambda \mu}u_{\mu}$$; for $$\vec{v} = 0$$ their time components are zero and their space components are $$\vec{E}$$ and $$\vec{D}$$ respectively. The four-dimensional generalization of the equation $$\vec{D} = \epsilon \vec{E}$$ is therefore evidently $$H^{\lambda \mu} u_{\mu} = \epsilon F^{\lambda \mu}u_{\mu}$$

How is that evident?

I understand that with the assumption $$\vec{v} = 0$$ that generalized equation gives us $$\vec{D} = \epsilon \vec{E}$$. But couldn't that just be a lucky coincidence?

Or is the reasoning something along these lines: "We found this equation, it gives good results, all the experiments work with that too, so it's probably right." ?

We know that $$\vec{D}=\epsilon\vec{E}$$ in a medium at rest, assuming Einstein's principle of relativity to hold not only for Newtonian mechanics but for Maxwell's electrodynamics as well. Landau and Lifshitz manage to express this in a manifestly covariant way (as one says) as $$H^{\lambda\mu}u_\mu=\epsilon F^{\lambda\mu}u_\mu$$. Equations of the same form will then automatically be valid in all other inertial frames, since both sides of the equation transform as contravariant four-vectors under the Lorentz group.