# The necessity of the B field

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?

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This topic got a fair amount of discussion here: physics.stackexchange.com/questions/3618/… –  Art Brown Oct 4 '12 at 0:13
Possible duplicate: physics.stackexchange.com/q/38151/2451 –  Qmechanic Oct 4 '12 at 6:52
Many thanks for pointing those out, apologies for the repetition –  Jallen Oct 4 '12 at 11:32

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.

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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 $F_{\mu\nu}$.

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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:

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

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.

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Covariance of the first equation just means $ma' = qE'$ for $u'= 0$. The second equation implies the first, and not the other way around. –  Larry Harson Jun 23 '13 at 16:12
@LarryHarson: Under the assumption of covariance, the special case at low velocities implies the general thing: once you know the behavior at slow speeds, you find the behavior at high speeds by boosting. How do you find the correct covariant behavior quickly? You find a convariant expression which reduces to the correct nonrelativistic limit, in this case it's uniquely determined, from the tensor form of F, which follows from the invariance of charge under boosts. This is how you derive the second equation from the special case of the first, and it is how Einstein did so. –  Ron Maimon Jun 24 '13 at 12:53
If you already know how $E$ is supposed to transform, then yes, covariance means the first equation implies the second. But this is too easy. Are you talking about Einstein's method in his 1916 GR paper? –  Larry Harson Jun 24 '13 at 16:20
@LarryHarson: I was talking about the 1905 paper, where he just knows that E and B exist. But the derivation can be made using only the fact that q is independent of frame. If E were a gradient of a scalar, so that there were no B, then q would have to transform like mass under boost. If E were the gradient of a metric tensor, like in GR, again, q would have to transform. This how Purcell gets the general form, although I don't like it so much. There are only a discrete set of tensorial covariant expressions that match the low energy limit, and it's not wrong to select one based on experience. –  Ron Maimon Jun 25 '13 at 3:26

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: $$m \frac{d^2x^{\alpha}}{d\tau^2} = q F^{\alpha\beta} \frac{dx_{\beta}}{d\tau}$$ 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.

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