# Is there a “special” inertial frame determined by the value of E and B at a point in an EM field?

Given the facts that $$(E^2 - B^2)$$ and $$(E\cdot B)$$ are Lorentz invariants of the EM field, and that the energy density $$(E^2 + B^2)$$ is not invariant, it seems that at each point in an EM field there should be unique inertial frame in which the field energy is minimum. Can that minimum value, and that inertial frame, be considered Lorentz invariants?

• The energy density is proportional to $E^2+B^2$. I don’t see a general connection between this and the two invariants. – G. Smith Mar 23 at 4:34

Since $$E^2-B^2$$ is invariant under boosts, any boost must change both $$E^2$$ and $$B^2$$ by the same increment $$\epsilon$$, if it changes them at all. In other words, a boost either increases both $$E^2$$ and $$B^2$$ by the same amount or decreases them both by the same amount, if it changes them at all.

There are two cases two consider: $$E\cdot B=0$$ and $$E\cdot B\neq 0$$. Throughout this answer, only a single point is considered, and "energy density" means the energy density at that point.

First suppose $$E\cdot B=0$$. In this case, there is a frame in which either $$E$$ or $$B$$ is zero. (See the appendix for an outline of a proof.) This must be the frame in which the energy density is a minimum, because any boost away from that frame will increase both $$E^2$$ and $$B^2$$. (It can't decrease them, because then one of them would end up being negative, but $$E^2$$ and $$B^2$$ cannot be negative.)

Now consider the case $$E\cdot B\neq 0$$. In this case, there is a frame in which the electric and magnetic field vectors are parallel to each other. (See the appendix for an outline of a proof.) Denote these fields by $$E_0$$ and $$B_0$$. In such a frame, we have $$|E_0\cdot B_0| = \sqrt{E_0^2B_0^2}. \tag{1}$$ After a boost, we have $$E\cdot B = \sqrt{E^2B^2}\cos\theta = \sqrt{(E_0^2+\epsilon)(B_0^2+\epsilon)}\,\cos\theta, \tag{2}$$ and since $$E\cdot B$$ is invariant, this gives $$\sqrt{E_0^2B_0^2} = \sqrt{(E_0^2+\epsilon)(B_0^2+\epsilon)}\,\cos\theta. \tag{3}$$ This shows that we must have $$\epsilon\geq 0$$. Therefore, if we start in a frame in which the vectors $$E$$ and $$B$$ are parallel, a boost cannot decrease the value of $$E^2+B^2$$, so it cannot decrease the energy density at the given point. This shows that the energy at any given point is minimized in any frame where $$E$$ and $$B$$ are parallel to each other.

The frame that minimizes the energy density is not unique. For example, if we start in a frame where $$E$$ and $$B$$ are parallel to each other (or if one of them is zero) and then apply a parallel boost, the fields $$E$$ and $$B$$ are unchanged.

Summary: There is always a frame in which (at the given point) either $$E$$ and $$B$$ are parallel to each other or one of them is zero. The frame that does this is not unique. Any such frame minimizes the energy density (at that point).

Appendix

The claim is that if $$E\cdot B=0$$, then there is a frame in which either $$E$$ or $$B$$ is zero; and if $$E\cdot B\neq 0$$, then there is a frame in which they are parallel. Here's an outline of the proof, using the Clifford-algebra approach that was described in my answer to

Wigner Rotation of static E & M fields is dizzying

Let $$\gamma^0$$, $$\gamma^1$$, $$\gamma^2$$, $$\gamma^3$$ be mutually orthogonal basis vectors, with $$\gamma^0$$ being timelike and the others being spacelike. In Clifford algebra, vectors can be multiplied, and the product is associative. I'll use $$I$$ to denote the identity element of the algebra, and I'll use $$\Gamma$$ to denote the special element $$\Gamma\equiv \gamma^0\gamma^1\gamma^2\gamma^3. \tag{4}$$ Mutually orthogonal vectors anticommute with each other, and the product of parallel vectors is proportional to $$I$$.

The electric and magnetic fields are components of the Faraday tensor $$F_{ab}$$, which are the components of a bivector $$F = \sum_{a In the frame defined by the given basis, the electric and magnetic parts of $$F$$, which I'll denote $$F_E$$ and $$F_B$$ (so $$F=F_E+F_B$$), are the parts that do and do not involve a factor of the timelike basis vector $$\gamma^0$$, respectively. The quantity $$F$$ satisfies $$F^2=(F^2)_I + (F^2)_\Gamma \tag{6}$$ where subscripts $$I$$ and $$\Gamma$$ denote terms proportional to $$I$$ and $$\Gamma$$, respectively. Both parts, $$(F^2)_I$$ and $$(F^2)_\Gamma$$, are invariant under proper Lorentz transformations. The part $$(F^2)_I$$ is proportional to $$E^2-B^2$$, and the part $$(F^2)_\Gamma$$ is proportional to $$E\cdot B$$.

The starting point for the proof is that in four-dimensional spacetime, any bivector may be written in the form $$F = (\alpha+\beta\Gamma)uv \tag{7}$$ where $$\alpha,\beta$$ are scalars and where $$u$$ and $$v$$ are mutually orthogonal vectors. If neither vector is null (the null case wil be treated last), then we can always find a frame in which they are both proportional to basis vectors in a canonical basis. The factor $$\alpha+\beta\Gamma$$ is the same in all frames (only $$u$$ and $$v$$ are affected by Lorentz transformations, because $$\Gamma$$ is invariant), so in the latter frame, $$F_E$$ is proportional to the product of two basis vectors (one of which is timelike), and $$F_B$$ is proportional to the product of the other two. After reverting to the traditional formulation in which the electric and magnetic fields are represented by vectors in 3-d space rather than by bivectors in 4-d spacetime, this is equivalent to the condition that $$E$$ abd $$B$$ both be proportional to a single vector.

If one of the coefficients $$\alpha,\beta$$ in (7) is zero (so that $$E\cdot B=0$$), then this proves the existence of a frame in which one of them is zero. If both coefficients $$\alpha,\beta$$ are non-zero (so that $$E\cdot B\neq 0$$), then then this proves the existence of a frame in which they are parallel.

Finally, if one of the vectors $$u,v$$ in (7) is null, then $$E$$ and $$B$$ are already parallel to each other and have the same magnitude.

• By "factor", do you mean "increment? That is, delta(E^2) = delta(B^2)? In that case, it seems that at some boost (keeping E and B parallel) B would go to zero, E^2 would be at a minimum, and E^2 +B^2 would also be at a minimum. – S. McGrew Mar 23 at 4:57
• In general there is no frame in which E and B are parallel. – my2cts Mar 23 at 8:14
• @S.McGrew I updated the answer to address both cases ($E\cdot B=0$ and $E\cdot B\neq 0$), and I added an appendix that outlines the proof of the assertion about the existence of a frame in which $E$ and $B$ are parallel (or in which one of them is zero). – Chiral Anomaly Mar 23 at 14:12
• OK, it appears that you've given a simple proof that a) the field energy density is minimum in any frame where E is parallel to B, and b) the field energy density is the same for all frames in which E is parallel to B. Right? If so, that says that the minimum field energy is an invariant -- but that it does not correspond to a unique inertial frame because there are an infinite number of frames in which E is parallel to B. – S. McGrew Mar 23 at 14:21
• @S.McGrew Yes, that sums it up. Using the word "invariant" this way seems a little non-standard to me, but I think I get what you're saying. Given $E$ and $B$ in an arbitrary frame, we have enough information to determine the minimum possible value of $E^2+B^2$ among all frames. I'm assuming that's what you mean by "invariant" in this context. – Chiral Anomaly Mar 23 at 15:43