# 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?

-

2
– Art Brown Oct 4 '12 at 0:13
– Qmechanic♦ Oct 4 '12 at 6:52
Many thanks for pointing those out, apologies for the repetition – Jallen Oct 4 '12 at 11:32

## 4 Answers

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.

-

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.

-

Of course you also need another assumption to pick out spin 1 rather than spin 0: the transformation law for E and B, the charge is independent of the velocity, the theory is renormalizable and natural. The scalar alternative is the Higgs coupling and it occurs in nature too. – Ron Maimon Oct 4 '12 at 16:08
@Ron: Interesting.  I hadn't considered the spin connection (my understanding of qm is abysmal) and will have to think on it.  I always appreciate your insights (even when, too often, they go over my head).  Thanks! – Art Brown Oct 4 '12 at 16:44
@Ron: So my takeaway from your comment is that there are even more ways than the one I mentioned in which the Maxwell generalization of coulomb+SR is not unique.  I studied Purcell as a kid and fell in love with this idea; it was quite a shock to realize t'aint so.  Frankly, I think this treatment is corrupting the youth.  ;-) – Art Brown Oct 4 '12 at 19:43
it's so, using Purcell's explicit additional assumption--- "The charge is independent of the velocity". This is a discrete choice, and it is equivalent to saying "Electromagnetism is spin 1". For fundamental theories, there are only 3 options--- scalar, spin-1, gravitational, and each one has a different transformation law for the charge, but Purcell gives you the physical point of view regarding spin 1 in a reasonable way (although I never liked the idea of spending so much physical and intuitive effort on something that only takes 3 lines once you learn covariant formalism). – Ron Maimon Oct 4 '12 at 23:40
@Ron: It's certainly required that charge be velocity-independent for standard em, but I don't think that assumption rules out a scalar potential alternative.  Compare four-force $f^\alpha = q F^{\alpha \beta} U_\beta$ with the scalar version $f^\alpha = g[\partial^\alpha \phi - U^\alpha U_\beta \partial^\beta \phi ]$; in both cases the charge is invariant.  Instead, the discriminant is that, for an external time-invariant electric field, the resultant force be velocity-independent.  That characteristic rules out the scalar potential case. – Art Brown Nov 29 '12 at 4:50

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.

-

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 $F_{\mu\nu}$.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.919478714466095, "perplexity_flag": "middle"}