# Why are magnetic fields a consequence of special relativity? [duplicate]

I'm very confused about why it is a consequence of special relativity.

Let's forget about anything quantitative at all.

Special relativity gives you length contraction -- so, when you're moving at a certain speed, distances along your direction of motion are compressed. Amongst many other things, this means that volumes will shrink, which also means that densities will increase.

Now, electromagnetism tells us that the electric force is proportional to the charge density. So, naïvely, we'd expect the electric force on a test particle external to the charge distribution to be higher in a boosted reference frame. This, however, contradicts the central assumption of special relativity that the net force on an object doesn't depend on the speed of the reference frame.

So, you need some new force that isn't present in the stationary reference frame. Well, in the boosted frame, the compressed charges are moving, so there is a current, so you could perhaps cancel out your excess force with some force that depends on the current distribution of spacetime. If you work this out, it turns out that magentism exactly does the trick, and if you factor in both electricity and magnetism, then the net force on the particle does not depend on whether you are in a stationary or moving reference frame.

• Jerry, its fine if we can explain the induction of magnetic fields by the acceleration of electric charged particles. But wouldn't it be a simplification not to mention that this electrons carry a intrinsic magnetic field and wouldn't it be better to mention that this field exist unchanged and remarkable in all frames of reference? – HolgerFiedler Feb 27 '17 at 5:07
• @stuffu The force HAS to not care whether the charge distribution is stationary or moving at a constant velocity -- that is just an axiom of special relativity. If the force was higher in the boosted frame, you could measure the higher force, and know you were boosted, after all. The reason why is that you have a magnetic force in the boosted frame, and you have none in the non-boosted frame, and the magnetic force cancels the excess electric force. – Jerry Schirmer Feb 27 '17 at 12:27
• @HolgerFiedler: this answer has nothing to do with the fundamental electric charges and magnetic moments. It's not a "where does magnetic force" come from thing, so much as a "If you have special relativity and electricity, then you necessarily would HAVE to have magnetism" thing. – Jerry Schirmer Feb 27 '17 at 12:28
• @JerrySchirmer No, seriously. When you go into a boosted frame from a frame where charges were sitting still, you will be able to see the charges moving, and you will be able to measure that the force between the charges is smaller than what it was in the previous frame. Charges measure a force between them, you measure a different force between them. (Maybe that is hard to believe?) Quantitatively: charges measure a Coulomb force, you measure a Coulomb force minus a magnetic force. – stuffu Feb 27 '17 at 12:43
• While I don't contest that under special relativity a theory with electrostatic force also requires a magnetic force in order to obey relativistic invariance, I feel that the explanation here is too simple to be entirely convincing. All that is said here should apply to gravity in the same way as it does to electrostatic force, after all both are $1/r^2$ forces generated by a density. So why does gravity not also have a need for an accompanying gravito-magnetic field? – Marc van Leeuwen Feb 27 '17 at 16:10

To summarise rapidly in a different way some of the ideas presented in the first link given by ACuriousMind: no the magnetic field does not derive from electrostatics + special relativity.

There is a very simple way to make sure of it. In the standard example of Purcell's you have a frame in which an electrically neutral wire carrying a stationary current $I$ generates a constant magnetic field $B$. In this setup there is only a magnetostatic field $\vec{B}$ and zero electric field $\vec{E} = \vec{0}$.

In this frame, a free charge is moving parallel to the wire with velocity $\vec{v}$ and will experience a Lorentz force $q \vec{v} \times \vec{B}$.

The standard argument then goes-on by applying a boost to put oneself in a new inertial frame co-moving with the mobile charge and therefore both the instantaneous velocity of that charge and the associated Lorentz force vanish in that new frame; abracadabra the magnetic force has disappeared!

Does that imply that the magnetic field does not exist in that frame? No.

It is easy to make sure of that because:

• a) there is still a current in the wire so there should be an associated magnetic field and,

• b) the quantity $E^2-c^2B^2$ is a Lorentz invariant so that \begin{equation} -c^2 B^2 = E'^2 - c^2 B'^2 \end{equation}where the primed quantities are in the co-moving frame. From the above equation it follows that the right-hand-side ought to be negative; which means that there has to be a magnetic field in the co-moving frame no matter what the force on some cherry-picked mobile charge may be.

To show that in some circumstances there is no magnetic field, one would need to show that there is no magnetic force on any moving charge; which is not the case at all in the usual textbook "derivation".

Now, to be fair, the magnetic force (or rather its expression) does disappear when placing oneself in a frame co-moving with a mobile charge but that is because the electromagnetic force is a 4-vector and applying a boost to it simply gives us a relationship between the force as observed in one frame (just the Lorentz force) and the force as observed in the co-moving frame (pure electrostatics). Note that this would only be a relationship and one cannot infer from it that electrostatics is somehow more fundamental than magnetism, for that would give a preference to frames in which forces on some particles are expressible only in term of electric forces; which in itself kind of contradicts the whole story about principle of relativity.

• I think the cleaner argument here is that "If you only have electric forces and Lorentz invariance, you must end up with Magnetic forces". It's not an argument that one is more "fundamental" than the other, so much as you need both electricity and mangetism in order to have a consistent theory. – Jerry Schirmer Feb 27 '17 at 17:04
• @JerrySchirmer: sure but note that the original question is about magnetic fields not magnetic forces. And again the reason is because one can think at the 4-vector level when it comes to force components so the whole picture about fields is of course not fully captured since focusing on 1 particle does not say anything about all the possible other moving charges. – gatsu Feb 27 '17 at 17:17
• Sure, but a magnetic force presupposes a magnetic field, and vice versa. The most obvious way to know a field is there to measure forces on a test particle, at least. But sure, I get the point of your answer, and have even upvoted. – Jerry Schirmer Feb 27 '17 at 17:31

If we have a moving test charge, and we want to know what force is accelerating it, a simple way to calculate that is this: First we transform to a frame where the velocity of the charge is zero. Special relativity tells us how we correctly transform to that frame.

Then in this new frame every charge is exerting a Coulomb force on the test charge. We calculate the net force that the test charge feels from all the other charges, lets call that force F.

Finally we find out what that force F looks like in the original frame, by transforming to that frame, again special relativity tells us how that is done. As we are interested about the force, we transform the force.

Now if there is a person saying that a magnetic field is causing a force on that test particle, then the magnetic field must be such that when calculating the force on the particle, the result is the same as what we calculated using special relativity. So special relativity dictates what the magnetic field is like.

One special case I want to mention: A test charge and another charge are moving side by side at the same velocity. When we transform to the test charge's frame, we calculate that the force affecting the test charge is F. When we transform back to the original frame, we learn that the force affecting the test charge in this frame is F/gamma, where gamma is the same gamma as in relativistic length transformation, or relativistic time transformation.