# Can the magnetic field *always* be transformed away?

In the book, "Einstein's General Theory of Relativity..." by Øyvind Grøn and Sigbjorn Hervik, the following statement is made: "The Lagrangian density of an electromagnetic field is the energy-scalar representing the energy-density of the field in a local frame moving so that the magnetic field vanishes...".

However, it is my understanding that often it is not possible to make the magnetic field vanish by transforming to a moving frame -- for example, when the electric and magnetic fields are parallel in some frame, or when the fields are perpendicular but |E| < |B| in some frame. Was this an error on the author's part, or is my understanding wrong?

## 2 Answers

Written in terms of the electric and magnetic fields, the Lagrangian density for the electromagnetic field becomes $$\dfrac{1}{2}(E^2 - B^2)$$. So if there is a frame in which the magnetic field vanishes, the Lagrangian is simply $$\dfrac{1}{2}E^2$$ in that frame, and this, in turn, coincides with the energy density. However, your objection is right: you cannot in general guarantee the existence of a frame in which the magnetic field is zero, so the author's choice of words was indeed unfortunate and misleading.

• Would an elliptical current-carrying wire be a system in which the magnetic field can never be transformed away? – probably_someone Jan 31 at 1:02
• That's troubling, because it suggests that everything that follows in the book may be untrustworthy! It's not obvious to me that the Lagrangian given there really is "the energy-density of the field in a local frame moving so that the magnetic field vanishes...". – S. McGrew Jan 31 at 1:06
• that lagrangian density is also valid if the current is zero, that is, in free space, an assumption sometimes made – Wolphram jonny Jan 31 at 1:20
• @probably_someone even an infinitely long straight, current-carrying wire (which generates a magnetic field, but whose electric field is zero) is a system in which the magnetic field cannot be transformed away. – Bruno De Souza Leão Jan 31 at 21:24

$$E\cdot B$$ is an invariant, up to sign, under Lorentz transformations. Hence, if in a reference frame it does not vanish (and you can construct many field configurations of this type in a given reference frame), then $$B$$ cannot vanish in every reference frame. You are right. There are situations where the density Lagrangian coincides with the density energy in no reference frames.