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My book claims that in vacuum, the total energy density of the electromagnetic field is equally shared between electric and magnetic fields by explicitly showing that $$\frac{1}{2}\epsilon_0 \textbf{E}^2=\frac{1}{2}\frac{\textbf{B}^2}{\mu_0}.$$

$\bullet$ However, one can change to a different inertial frame by a Lorentz boost where $\textbf{E}$ or $\textbf{B}$ will mix up. Is this statement a Lorentz invariant assertion? If not, does it imply we are in a special Lorentz frame (not violating relativity, of course!) in which this claim is true?

$\bullet$ Is it possible to go a frame in which one of the fields ($\textbf{E}$ or $\textbf{B}$ ) of an EM wave is made to vanish? Then it is clear that the above claim is meaningless.

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The answer to the question in the title is yes. There are two fundamental Lorentz invariant scalars of the EM field,

$$\frac12F^{\mu\nu}F_{\mu\nu}=\mathbf{B}^2-\mathbf{E}^2$$ and $$\frac14F_{\mu\nu}\ {}^\ast F^{\mu\nu}=\frac14\epsilon^{\mu\nu\alpha\beta }F_{\mu\nu}F_{\alpha\beta}=\mathbf{B}\cdot\mathbf{E}$$

(modulo units and constants). The former reflects exactly the equal-energy-density property of electromagnetic radiation, and it guarantees that if $\mathbf B ^2 = \mathbf E^2$ in one frame, then that stays true in all frames of reference.

In general, for radiation, the equal sharing is at least morally true, and it's easy to verify for plane-wave radiation. On the other hand, there are plenty of fields for which energy is not equally shared, with the simplest example being the electrostatic field of a point charge.

Moreover, for homogeneous fields (or locally for inhomogeneous fields) any two configurations that share those two invariants are almost certainly connected by a Lorentz transformation (but check for a proof before you take my word for it).

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  • $\begingroup$ So do you think the claim is true? $\endgroup$ – akhmeteli Nov 21 '16 at 9:25
  • $\begingroup$ @akhmeteli See edited answer. $\endgroup$ – Emilio Pisanty Nov 21 '16 at 9:26
  • $\begingroup$ Neither do I think that the claim is true (please see my answer), but your example of the field of a point charge does not seem applicable as there are no charges in vacuum. $\endgroup$ – akhmeteli Nov 21 '16 at 9:28
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    $\begingroup$ I don't see why radiation without charges cannot be physical: for example, if an electron and positron annihilate, you have pure radiation, without charges. $\endgroup$ – akhmeteli Nov 21 '16 at 9:54
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    $\begingroup$ @akhmeteli That's mostly beside the point. I just want to contest the interpretation of the OP's claim of "in vacuum" as requiring the absence of charges everywhere. SRS's comment is correct. $\endgroup$ – Emilio Pisanty Nov 21 '16 at 10:53
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I strongly doubt this claim: you can choose initial fields (at some initial time point in the entire 3d space) arbitrarily (provided divergence of electric and magnetic field vanishes). The claim is true, however, for a plane (sinusoidal) electromagnetic wave, and this does not depend of the (inertial) frame of reference, as $E^2-B^2$ is a Lorentz invariant.

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