# Is the equal sharing of EM field energy between $\textbf{E}$ and $\textbf{B}$ fields a Lorentz invariant statement?

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.

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).

• So do you think the claim is true? – akhmeteli Nov 21 '16 at 9:25
• @akhmeteli See edited answer. – Emilio Pisanty Nov 21 '16 at 9:26
• 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. – akhmeteli Nov 21 '16 at 9:28
• 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. – akhmeteli Nov 21 '16 at 9:54
• @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. – Emilio Pisanty Nov 21 '16 at 10:53

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.