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It is fairly easy using basic special relativity to arrive at the conclusion that the magnetic force effect on nearby charges of wires carrying currents on nearby charges is only due to the length contraction in certain inertial frames from the point of view of charges, which gives rise to a perceived change in charge density. However, we structure the (classical field) laws of electromagnetism in Maxwell's laws using a B field.

My question is, is it possible to formulate Maxwell's equations only in terms of Lorentz transformed E fields? If not, why not?

Jallen
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4 Answers4

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No. $\boldsymbol{B}$ is an independent entity.

Reference Jackson, Classical Electrodynamics, Section 12.2.

Jackson argues that the anti-symmetric tensor $F$ formulation of EM that we all know and love: $$m \frac{\mathrm d^2x^{\alpha}}{\mathrm d\tau^2} = q F^{\alpha\beta} \frac{\mathrm dx_{\beta}}{\mathrm d\tau}$$ is not the only possible covariant generalization of the rest frame force law: $$ma=qE$$
In fact, he constructs a counter-example based on a Lorentz scalar potential $\phi$. Non-relativistically, this potential gives a Coulomb and a magnetic-like force, just like we see, but there's no independent $\boldsymbol{B}$-field: in any frame, the force is given in terms of the 4-gradient of the scalar potential, $\partial_\mu \phi$, not the 6 components of $F$. If this Lorentz scalar were the way the world works, the answer to your question would be yes. But it's not.

Art Brown
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I think the key here is your requirement of a formulation "in terms of Lorentz transformed E-fields".

The E-field is a 3-vector field. A Lorentz covariant field must be a 4-vector or 4-tensor field (a Lorentz invariant field must be a Lorentz scalar field).

In fact, the E-field is, within SR, a component of a rank 2 electromagnetic 4-tensor field whilst the scalar and vector potentials are components of a electromagnetic 4-vector potential.

Since the E-field is only a part of a Lorentz covariant tensor, the notion of a "Lorentz transformed E-field" needs further clarification.

Alfred Centauri
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Yes--- electromagnetism is developed from this point of view, as suitable for an undergraduate course, in Purcell. The equation of motion is

$$ma = qE$$

where E and a are both in the rest frame. The covariant form of this equation is the usual equation of motion:

$$ m {\mathrm d^2x_{\nu}\over \mathrm d\tau^2} = q F_{\mu\nu} {\mathrm dx^\mu\over \mathrm d\tau}$$

As you can see by specialising to the rest frame, so the Newton's law in the rest frame, only using E, is indeed covariantly equivalent to both E and B in the usual frame.

Ron Maimon
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If you look at the covariant formalism of classical electrodynamics, you can see that you don't have to mention either the $\bar{E}$ or $\bar{B}$ field. You can do all your calculations with the four-potential $A_\mu$ and the electromangetic tensor $F_{\mu\nu}$.

Deep Blue
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