# Asymmetry in Lorentz force on a charged particle in two different inertial systems

One of the first examples of problems involving Maxwell's electromagnetism used in order to introducing special relativity is the problem of a charged particle $q$ in a stationary magnetic field, let's say $\vec{B} = B_0 \vec{e}_z$. If we are in a rest frame relatively to the particle, the charge is motionless and so there is no Lorentz force applied and the particle remain still.

If the observer is on a inertial system moving with a relative velocity $\vec{v}$ in relation to the first system, the particle would have a velocity of $-\vec{v}$ and should move in response to the Lorentz force $\vec{F}=\vec{v} \times \vec{B}$. As far as I know, this paradox is solved using Lorentz transformation between the two systems.

Hence there is my doubt: when speaking of kinematics (for example lengths contraction), it is clear that at low velocity the relativistic effect are neglegible and this is even more clear looking how Lorentz transformation tend to Galilean ones for $v<<c$. However, this classical limit seems not to be appliable in the electrodynamics case because even at very low speed the Lorentz force make a great difference between the two scenarios.

If the Lorentz transformation have the Galilean ones as a limit for low velocities, why this is not appliable in electrodynamics?

• How do you rule out the presence of an electrical field, in the Galielian-boosted reference frame? Special relativity tells you that such a field actually appears and counterbalance the magnetic field so that the particle appear in an inertial motion in both frame. On the other hand, I think it's quite difficult to obtain the expression for the fields under a Galileian boost...what I mean is that i don't see a radical different description from the two relativities, because I don't know the Galileian one. – giulio bullsaver Nov 17 '16 at 22:39

Our intuition of this is badly skewed, in two different ways. As usual, with SR, we never experience large velocities, so the effects tend to come as a surprise. For this question, there's an additional bit that's important, which is that our experience of magnetic and electric fields is strongly asymmetric, but we don't usually think about it.

You and I haven't encountered large concentrations of charge. When you have a "charged" object, it's really just a very small net charge. The vast majority of electrons and positive ions present, cancel each other out.

If you take such an object, and wave it in a magnetic field, you certainly don't observe any Lorentz force, basically because the mass is so large in proportion to the net charge. The Lorentz force is something mainly studied in considering synchrotron radiation, or currents, or other effects where the charge-to-mass ratio is somewhat similar to that of a fundamental particle.

Additionally, when you consider magnets (as in ferromagnets) or electromagnets, a large proportion of the material present is engaged in magnetic activity. This makes the overall effect stronger. For electric fields, this situation is unsustainable, because a strongly charged object will simply create sparks to discharge.

The effect of this is that while we think of electric and magnetic fields being comparable, they're really not. Magnetic fields are much weaker, in the sense that if we select a certain amount of material to be engaged in creating the field, we can get a stronger electric field than magnetic field.

In any case, we can also be somewhat more quantitative: Lorentz-transformed quantities go like the hyperbolic trig functions $\cosh$ and $\sinh$. In this case, the electric field should be $\vec E \sim \sinh \frac v c$, which is linear in $\frac v c$ for small velocities. The nonrelativistic Lorentz force, of course, is proportional to $v \times \vec B$. So, they have the same order of dependence on velocity, as you'd expect. (You can work this out in more detail to derive the Lorentz force from SR considerations. It's easiest if you've been introduced to the 4-potential and EM field strength tensor.)

• Great answer! It's much clearer now. I've been thinking about a possible answer on my own: whatever instrument is used to measure a magnetic field in the end is a measure of a force on a charged particle. So if every time we measure the magnetic field in the moving inertial system we will get a null result, is possible to assume that, in this system, the field is actually zero. So it's obvious the first particle will not move. Is this answer correct and compatible with yours? – skdys Nov 16 '16 at 7:46
• I think that answer is correct, yeah. One quick note: a magnetic field measurement doesn't have to involve a moving charge, in the sense of a particle whose position $\vec x$ has a non-zero time derivative $\frac{\mathrm{d} \vec x}{\mathrm{d} t}$. Electrons, for instance, have a magnetic dipole moment, due to their intrinsic angular momentum. – Scott Lawrence Nov 16 '16 at 15:05

Electric current in a coil is the same in all frames, if we use Galilean transformations.

If we use Lorentz transformations, electric current in a coil is larger in frames where the coil moves slower. (Because at large coil-speeds charges move around the coil like in slow motion)

I bet you didn't think of that while considering relativistic effects at slow speeds.