There is an explanation in the Wikipedia. Unfortunately the article is quite verbose and doesn't clearly explain why both positive and negative charges vary density even if only one is moving.

It is also not possible to figure out with respect to which frame of reference they shrink/expand. Can this be made more clear or is there a source that already does so?

Actually, I resorted to wikipedia after Feynmann because he said something like

model of a wire with a current of positive charges, separated by an average distance, L. The wire has to be electrically neutral in the lab frame, so there must be a bunch of negative charges, at rest, separated by the same average distance. Therefore there's no electrostatic force on a test charge Q outside the wire.

I have got this quotation from the last answer. I do not understand why nobody, who discusses the relativistic density growth, cannot see the problem. How, after saying that

  1. motion increases density and
  2. positive charges are moving

can they say that we have negative charges, standing still in the same frame, having the same density? I see that when positive charges are stopped, their density drops and we'll get the negative charge prevailing. No current = object is charged negatively! Do you see that? This means that all objects in the Universe must be negatively charged in order to be neutral when positive charge starts moving in them. But, we know that objects are normally neutral. So, when current appears in the loop, you must first explain where the extra charge is coming from and why the loop remains neutral. But, teachers do not do that. They have a conspiracy to avoid discussing this simplest case and jump immediately to the case of test charge moving at speed of current charges. So, is it right that all objects in your Universe are electrically charged when no current circulates in them?

  • $\begingroup$ I am not sure I agree with what the title suggests. It does not seem good to me to give more rational weight to the electric field than to the magnetic field. The magnetic field does not derive from the electric one it is just that they are two faces of the same poligon (that's an image). It just so happens that when you change your point of view you can sometimes see just one of them or both exactly like when looking at a cube from different angles. $\endgroup$ – gatsu May 2 '13 at 10:57
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    $\begingroup$ I don't think so. The quantity $\vec{E}^2-\vec{B}^2$ is invariant upon changes of frame of reference and therefore if you go from a situation in which you have a steady current in a wire in a lab frame of reference (and therefeore have a pure magnetostatic field in this frame of reference), then there is no way that you get only an electric field when going in the frame of reference that is comoving with the current. $\endgroup$ – gatsu May 2 '13 at 11:24
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    $\begingroup$ And by the way, we don't have only electric charge in Nature, all particles carry also a magnetic moment as well that generates a magnetic field. Even the neutron that is neutral has a magnetic moment. Also, in special relativity, charge density is only one coordinate of a current four-vector, so I do not see why one component should be privileged whith respect to the others. $\endgroup$ – gatsu May 2 '13 at 11:31
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    $\begingroup$ I don't know why people are attacking the derivation requested. Electromagnetism in its entirety can be derived from Coulomb's Law plus SR, full stop. Classical magnetism is nothing more than the effect of transforming frames with nothing but electrostatics at work, and it is this connection that the OP wants elucidated. For a full exposition, I heartily recommend Purcell's book, which is being re-released after going out of print for a time. It's one of the few undegrad texts that derives rather than assumes Maxwell's equations. $\endgroup$ – user10851 May 2 '13 at 21:08
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    $\begingroup$ @ChrisWhite: I don't think the statement that "Electromagnetism can be derived from SR plus colulomb's law" is valid. If that were the case, you could have a potential derived from a scalar field, which with the right coupling, would give you coulomb's law and be perfectly covariant under special relativity. $\endgroup$ – Jerry Schirmer Oct 19 '13 at 16:13

follow the link at WP:
The simplest, and the full derivation of Magnetism as a Relativistic side effect of ElectroStatics by Hans de Vries. His paper linked in that page is very clear.
Motion of charge is needed to perceive a magnetic field.
In the Maxwell equations the 'Ampère's circuital law' says :
no current and no electric field variation(a) in time -> curl of magnetic field = 0
and Gauss's law for magnetism : divergence of of magnetic field = 0
(a) An electric field variation is created by charges in motion.

How can we presume that the magnetic field, that only exists in presence of motion and is observer dependent, has the same fundamental existence as the electostactic field that exists in all circunstances ?
'Motion' by itself can not be an entity creator. Luckily special relativity brings order.

edit add:
I will recall an example from a common life experience: rainfall. You are not in motion and rain is falling in a non windy day. It falls uniformly downwards and you get equally wetted from all sides.
The moment you start running you will get more wetted in your front than in your backside, i.e. more droplets per unit time will hit your front than your back. Your motion created the illusion/perception that the rainfall is not uniform.
Now, using relativity, we can reverse the situation: it is a windy day and you are stopped ...
In the same way a test charge in motion irt a uniformly charged wire will perceive it as having more charge from ahead than from behind.
This fundamental question about what is real versus what is perceived make me wonder why so many theorists are trying to find magnetic monopoles. Inglorious and Insane task. There is more to Physics than equations. KISS (keep it simple, stupid).

The fundamental question is: can motion create anything? NO.

  • $\begingroup$ I do not understand why he speaks "We can even ignore the factor $\gamma$ in the formula for $t$. The difference at low speeds can be neglected for our purpose. The typical average drift velocity of electrons in domestic electromotors is in the order of a millimeter per second" after he explains that the speed of electrons is irrelevant and hides it behind the current I? He states that it a typical error to take the speed of electrons into account in this computation and his purpose is to fix it. Which $v$ is he talking about? Why electrons in the motor? Why $\gamma$ is related with that? $\endgroup$ – Val May 3 '13 at 14:18
  • $\begingroup$ @Val electron motion is slow motion Drift velocity. $\gamma$ =1 for non relativistics speeds. $v$ is the relative speed of the test charge in relation to the wire. 'motor' is irrelevant here. I edited my answer with the rainfall example. $\endgroup$ – Helder Velez May 3 '13 at 20:40

There's also the discussion by Daniel V. Schroeder, called Magnetism, Radiation, and Relativity. (Webpage on the Weber State University website)

The subtitle is 'Purcell simplified'.

(I haven't checked whether the discussion on Wikipedia and the one by Schroeder are any different from each other. Hopefully they are, in such a way that each one helps to understand any unclear points in the other)

EDIT (in response to your comment)

My apologies for reading your question only superficially (Also, the Schroeder article is linked to on the wikipedia article page, so I should have assumed you had already read that.)

About that quote from a Feynman text:

model of a wire with a current of positive charges, separated by an average distance, L. The wire has to be electrically neutral in the lab frame, so there must be a bunch of negative charges, at rest, separated by the same average distance. Therefore there's no electrostatic force on a test charge Q outside the wire.

I think you do have a point.

The underlying problem is, I think, that the available explanations do not consider relativity of simultaneity. The explanations seem to get away with that, as the sought after result is produced, but it may be that the explanations contain errors that fortuitously drop away against each other, keeping the errors hidden.

It may well be that the explanation just cannot be pushed as far into detail as you demand. In my opinion your demand is valid, but the explanation may not be up to it.

  • $\begingroup$ Thanks for the like. It has the same trouble at the others (I have fixed my question to incorporate it). Additionally, I do not understand his radiation point. The says that "there's an inner region of field lines pointing away from the particle, and an outer region of field lines pointing away from the imaginary point where it would be [if not bounced]." How does the field far away know the position that particle would have but never did? Is it inertial so that it moves with the speed of the particle that radiated it? This completely breaks my picture of relativity and electromagnetism. $\endgroup$ – Val May 4 '13 at 11:05

I think the wiki article's example is a little confusing because it looks at the case of the charge being comoving with the current first and with charge neutrality in that frame. I think this needlessly complicates things, as it requires you to think about magnetic fields first. Instead, I'll consider a slightly different situation.

Case: Let there be a four-current-density of positive charges moving toward the +x direction, $j_+ = \rho \gamma (e_t + \beta e_x)$.

Let there be stationary negative charges forming another four-current density $j_- = -\rho \gamma e_t$.

Note that the negative charges are spaced out further (by a factor of $\gamma$) than the moving positive charges are (edit: in the positive charges' frame; so as a point of fact, both charges are equally spaced in our current reference frame), but overall the total current density is $j = j_+ + j_- = \rho \gamma \beta e_x$, a purely spatial current.

A stationary test charge will experience no force here (and just to be clear, no four-force either), as it is not moving and there is no net charge from the current distribution.

Let's consider going into the frame in which the negative charges move to the left and the positive charges are stationary. In this primed frame, $j_+' = \rho e_t$ and $j_-' = - \rho \gamma^2 (e_t - \beta e_x)$. The total current density is $j' =- \rho (\gamma^2 - 1) e_t + \rho \gamma^2 \beta e_x$. The same test charge from before now moves with velocity $u = \gamma (e_t - \beta e_x)$.

As the current distribution has net negative charge, a positive test charge must feel some attractive electric force to the current distribution. But we already concluded that the total force was zero! And this should not change under a Lorentz transformation, which is no more exotic than a simple change of coordinates. Hence, we infer the existence of another force, equal and opposite to the electric one, that repels the moving test charge. This is the magnetic force.

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    $\begingroup$ I doubt the OP knows what a four-current-density is $\endgroup$ – Larry Harson May 2 '13 at 21:19
  • $\begingroup$ @LarryHarson, your doubt is confirmed by my question physics.stackexchange.com/questions/63008/…, asked simultaneously with this one :) $\endgroup$ – Val May 2 '13 at 21:47
  • $\begingroup$ If charges are in motion then by do relativistically scale them by just adding a constant term $\beta e_x$? What the $\beta e_x$ term is? If charges are stationary then why do you scale them by γ? How is it possible that increased, due to motion positive charge, is balanced exactly by the stationary negative? $\endgroup$ – Val May 2 '13 at 22:07
  • $\begingroup$ If you're unfamiliar with them, you may need to consult wiki for information on Lorentz transformations. I'll just say that two terms are required, just as two terms are required to rotate vectors--Lorentz boosts are very similar to rotations. You ask, "How is it possible that increased..." I do not understand this question. $\endgroup$ – Muphrid May 2 '13 at 22:16
  • $\begingroup$ You say that positive charge becomes more dense but net charge is zero. How it is possible? $\endgroup$ – Val May 2 '13 at 22:41

I suggest you try to get "Principles of Electrodynamics" by Melvin Schwartz. It's a short book that explains Electromagnetism very neatly using Special Relativity.


I think that the answer here, regarding the electrons in the closed loop, answers this question also.

Basically, it explains why the charges moving along all 4 directions of a rectangular frame are not seen relativistically contracted in stationary (lab) frame. This implies that if we move along with the bottom wire electrons, we see that the distances between them have expanded as current is developed. The distances in the opposite, upper wire have contracted purely relativistically because there is no force to compensate the contraction. The vertical electron density should stays intact seen from either bottom wire electrons or stationary frame. This means that any electron, moving along the bottom wire electrons, will see more electron density from above than from underneath. If it is moving half-way between upper and bottom wires, it is attracted to the bottom and repelled from the top. This is ampere law and interpreted as magnetic field in the stationary (lab) frame where densities are unchanged.


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