# How is the magnetic field related to Special Relativity

1. Jan 25, 2016

### arydberg

Purcell has a paper in which he derives the magnetic field by assuming a charged particle travels alongside a conducting wire at the same velocity as the conduction electrons in the wire. The conduction electrons and the test charge are moving together but the positive charges are moving backwards with respect to the electrons and the test charge. The positive charges undergo a length contraction as explained by special relativity. The unbalanced charges then produce the magnetic field. Is it possible that there is more to this. That there are things we don't understand yet. Is it possible to totally explain the magnetic field with special relativity?

2. Jan 25, 2016

### Orodruin

Staff Emeritus
The electric field cannot exist without the magnetic field in special relativity. They are both parts of the very same anti-symmetric rank two tensor. This is very well understood.

3. Jan 25, 2016

### bcrowell

Staff Emeritus
The presentation you're referring to was published in a textbook (Purcell, Electricity and Magnetism), not a paper. The physics dates back to Einstein's 1905 paper on relativity, 60 years before Purcell did his presentation.

4. Jan 25, 2016

### pervect

Staff Emeritus
The way I would describe the situation is this. Special relativity is perfectly compatible with Maxwell's equations, for instance Maxwell's equations predict that light is a wave that moves at a constant speed. Maxwell's equations, of course, include the idea of a magnetic field.

If you try to modify Maxwell's equations to try to totally excise the magnetic field, the resulting theory will no longer be compatible with special relativity. (It will also, of course, be incomatible with observation, as we've observed the magnetic field). So one needs the magnetic field or something like it to have a theory that's compatible with special relativity. The basic concepts involved are called "covariance" or sometimes "general covariance". The key issue here is what "or something like it" might mean. At this point, though, I'm going to follow Bohr's dictum. “Never express yourself more clearly than you are able to think."

5. Jan 25, 2016

### arydberg

So if the speed of light was infinite would the magnetic field exist?

6. Jan 25, 2016

### arydberg

Sorry my mistake.

7. Jan 25, 2016

### Staff: Mentor

There's no way of knowing, because in the universe we live in the speed of light is not infinite and we don't have any other universes to compare with. We can say that in our universe the finite speed of light means that there must be a magnetic field (and vice versa!).... But that's as far as it goes.

8. Jan 26, 2016

### pervect

Staff Emeritus
It's rather problematical whether light as we know it would exist without a magnetic field, considering that the wave equation that explains light requires the magnetic field to exist as we know it. In the wave equation as we know it, a changing electric field generates a changing magnetic field, which regenerates the changing electric field.

People without a mathematical background talk about "setting the speed of light to infinity" all the time, as if we were expected for the reader to know what was meant by saying that phrase. Unfortunately, it's unclear what this phrase might actually mean in mathematical terms, i.e. in writing a replacement set of equations for Maxwell's equations, so serious discussions tend to bog down at this point.

9. Jan 26, 2016

### andresB

As a nitpick, there is this interpretation

"... neither Maxwell's equations nor their solutions indicate an existence of causal links between electric and magnetic fields. Therefore, we must conclude that an electromagnetic field is a dual entity always having an electric and a magnetic component simultaneously created by their common sources: time-variable electric charges and currents."

https://en.wikipedia.org/wiki/Jefimenko's_equations (and reference in there of course)

10. Jan 26, 2016

### bcrowell

Staff Emeritus
The Galilean limit of electromagnetism has been studied, and it's not speculative or unknowable.

Marc De Montigny, Germain Rousseaux, "On the electrodynamics of moving bodies at low velocities," http://arxiv.org/abs/physics/0512200

11. Jan 27, 2016

### pervect

Staff Emeritus
An interesting paper, thank you. Reading this paper in the context of the original question, if by "speed of light is infinite", we meant that we are considering problems for which L << cT, so that in the terminology of the paper we would have Gallilean kinematics (and not Caroll kinematics, which is mentioned in the paper though I'm not really familiar with it), my reading of the paper is that we can say nothing about the magnetic field. I see three possibilities for the magnetic field, we could have a negligible one (the electric limit), we could have a dominant one (the magnetic limit), or neither could be dominant, they could be of similar magnitude. It seems to me that only in the last case (when both are present and neither can be ignored) would we have electromagnetic radiation.

[add]Additionally, in the case where the charge density $c \rho$ is of the same order as the current desnity $j$, it seems to me we would not have a valid Gallilean approximation available at all - we'd have to transform $(c\rho, j)$ via the Lorentz transform to get accurate results, not a Gallilean transform. So while we could transform (ct, x) via a Gallilean transform, we would need to reintroduce the Lorentz transform to handle the charge-current 4 vector.

Since I haven't thought this over for a long period of time, I'd appreciate any comments or corrections from people who have.

Last edited: Jan 27, 2016
12. Jan 27, 2016

### pervect

Staff Emeritus
One other comment. The DeMontigy paper mentions the following:

In the Gallilean formalism, the effective charge density seems rather mysterious, a mathematical trick. Purcell's example could be interpreted as pointing out that the idea of an effective charge density is not so mysterious after all - it's just a consequence of Lorentz contraction.

13. Jan 28, 2016

### vanhees71

This is, of course, misleading, although completely correct. There's nothing "dual" (except the mathematical symmetry known as "duality transformation" of classical electrodynamics, which becomes even more convincing if one introduces magnetic charges in addition to the usual electric charges into the theory). According to classical (and quantum) electrodynamics there's an electromagnetic field, which can be split into electric an magnetic components with respect to a reference frame.

More precisely the electromagnetic field is a massless vector field and thus (because observations exclude the possibility of having a continuous intrinsic polarization-like structure) is necessarily a gauge field. The equations of motion thus deal with redundant field-degrees of freedom. In the classical theory you can describe everything in terms of the gauge invariant field-strength tensor, and you have the field equations of motion (Heaviside-Lorentz units with $c=1$)
$$\partial_{\mu} F^{\mu \nu} = j^{\nu}$$
together with constraints
$$\partial_{\mu} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}=0.$$
That's first of all manifestly Lorentz covariant, and it's logically not that "relativity is compatible with Maxwell electromagnetics" but the other way around, i.e., Maxwell electromagnetics is a relativistic field theory.

Now to understand the Jefimenko equations, it's clear that the sources of the electromagnetic field are charge-current densities, and thus you have a causal description in terms of a retarded Green's function only within this interpretation of the Maxwell equations. You can, of course, rewrite these most natural equations of expressing the electromagnetic field in terms of their sources in other ways to make them look as if some field components are "causing" or are "sources" of other field components. However, those equations are non-local and the causality structure becomes completely hidden.

I don't comment anymore on Purcell's book. Better leave it out of this discussion (see the longer thread(s) about this topic from some time ago in this forum).