Magnetic force between 2 moving charges

Question

If 2 positive charges are moving in the same direction with a velocity v relative to the ground frame, will they experience a magnetic force? I’ve read several times that the velocity in the Lorentz force law is relative to the ground frame, not to the object creating the magnetic field, so I feel like there should be a magnetic force.

Moreover, assuming there is a force, I tried computing the velocity at which the electric and magnetic force cancel each other, and got the speed of light. Is this just a coincidence, or is there any proper reasoning behind this?

Answer 1

In any inertial frame of reference, the force experienced by one of the charges, X, is given by the Lorentz force law $$\mathbf F=q(\mathbf E + \mathbf v \times \mathbf B)$$ in which $\mathbf E$ and $\mathbf B$ are fields due to the other charge, Y. This equation is Lorentz-covariant, provided that we define $\mathbf F$ by $$\mathbf F=\frac{d\mathbf p}{dt}$$ in which $\mathbf p$ is the relativistic momentum of X (when the only force acting on X is the Lorentz force due to fields generated by Y).

Now let's suppose that X and Y are stationary in the S' frame, in which X is displaced from Y in the y direction. If we release the charges from rest, then in the S' frame, the force on X is $$F'_y=\frac{dp'_y}{dt'}$$ In the S frame the charges are both moving in the $x$ direction at speed $v$, and we can write $$F_y=\frac{dp_y}{dt}$$ Transverse momentum is Lorentz-invariant, so $p'_y=p_y$. But $dt=\gamma(v) dt'$. This because in the S' frame the charges have just been released from rest, but the time interval during which they pick up $dp_y$ is dilated in the S frame, where the charges are moving at speed $v$. So we have $$F_y=\frac{dp_y}{dt}=\frac{1}{\gamma}\frac{dp'_y}{dt'}=\frac{1}{\gamma}F'_y=\sqrt{1-v^2/c^2}\ F'_y$$ So we see that in the S frame, in which the charges are moving, the force is less than in the S' frame, in which they are almost stationary. Clearly the force in the S frame approaches zero as $v$ approaches $c$. This can be attributed to time dilation.

We've reached this result by looking only at the left hand side of the Lorentz force equation. Now look at the right hand side.

In the S' frame the only external field in the vicinity of X is an electric field, $E'_y$ in the $y$ direction, due to Y. So $$F'_y=q E'_y.$$ To find the force in the S frame, we need to use the transforms of field components, of which the relevant ones are $$E_y=\gamma\left(E'_y+vB'_z \right)= \gamma\ E'_y$$ $$B_z=\gamma\left(B'_z+\tfrac{v}{c^2}E'_y \right)= \gamma\tfrac{v}{c^2}E'_y$$ So in the S frame, the Lorentz force on X is in the $y$ direction and given by $$F_y = q(E_y-v B_z)=q \gamma \left(1-\tfrac{v^2}{c^2}\right)E'_y =\gamma \left(1-\tfrac{v^2}{c^2}\right)F'_y=\sqrt{1-\tfrac{v^2}{c^2}}F'_y$$So we've reached the same result as before, but with the insight that the change in repulsive force is not simply due to an attractive magnetic force, but an attractive magnetic force, $-\gamma\tfrac{v^2}{c^2}F'_y$, that is greater than the increase, $(\gamma -1)F'_y$, in repulsive electric force due to $E_y$ being greater than $E'_y$! To first order in $v^2/c^2$ these two quantities are $-\tfrac{v^2}{c^2}F'_y$ and $\tfrac 12 \tfrac{v^2}{c^2}F'_y$, so at any speed $v$ of the charges for which the magnetic attraction is significant, so too is the increase in electric repulsive force.

You might argue that, far from providing insight, the electric and magnetic fields approach is an obfuscation. There is simply an electromagnetic field; transforming components from frame to frame is sometimes a necessary calculating evil, but in the present case it can be avoided if all we're interested in is what happens to the force between the charges!

Answer 2

Indeed the charges will experience a magnetic force. It can be considered a result of the Lorentz transformation of their Coulomb repulsion. Your conclusion that electric and magnetic force balance out in the limit of the speed of light is also correct.

Answer 3

According to "Darwinian Lagrangian - Interacting particles" the force between two charged particles (written in SI units) is $$\begin{align} \mathbf{F} &=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\hat{\mathbf{r}} \\ &+\frac{\mu_0}{4\pi}\frac{q_1q_2}{r^2}\frac{1}{2} \{\mathbf{v}_1(\hat{\mathbf{r}}\cdot\mathbf{v}_2) +\mathbf{v}_2(\hat{\mathbf{r}}\cdot\mathbf{v}_1) -\hat{\mathbf{r}}[\mathbf{v}_1\cdot\mathbf{v}_2+3(\mathbf{v}_1\cdot\hat{\mathbf{r}})(\hat{\mathbf{r}}\cdot\mathbf{v}_2)] \} \end{align}$$ where $q_1$ and $q_2$ are the charges, $\mathbf{v}_1$ and $\mathbf{v}_2$ are the velocities of the two prticles, and $\hat{\mathbf{r}}$ is the unit vector between them.

The first term is the well-known electric Coulomb force between the two particles.

The second (very long) term is a magnetic force between the two particles. Roughly speaking, this magnetic force is attractive when $q_1\mathbf{v}_1$ and $q_2\mathbf{v}_2$ are parallel. And it is repulsive when they are antiparallel.

But remember that this Darwinian formula is still an approximation valid only for slow velocities ($v_1,v_2 \ll c$). So don't expect correct results when $v_1\to c$ or $v_2\to c$.

Having said that, you see when both $v_1$ and $v_2$ approach $c$, then the magnetic force becomes similar in size to the Coulomb force. (Remember $\mu_0=\frac{1}{\epsilon_0c^2}$ to verify this.)

Answer 4

The real question that needs to be asked is this: "according to whom?" The special theory of relativity states that all inertial observers' points of view of the world are equivalent.

According to an observer in the laboratory frame, charges moving at a constant velocity possess both an electric and a magnetic field. The magnetic field of such a moving charge can be calculated without too much difficulty (see my answer to a similar question here: Conceptual question about special relativity in electrodynamics), and it can be shown that if we know the moving charge's electric field $\mathbf{E}$, then $$\mathbf{B} = \frac{\mathbf{v}}{c^2}\times \mathbf{E}.$$

Thus, an observer in the lab will say that each of the moving charges is affected by force due to both the electric and magnetic fields of the other charge.

However, if we looked at this from the point of view of one of the charges (or perhaps an observer moving with the same speed as the charges), both these charges would appear to be at rest with respect to each other, and therefore there would be no magnetic field there. The charges would interact with a purely electric field, $\mathbf{E}'$, which is different from $\mathbf{E}$!

As I've shown in the linked answer, you can compute the electric and magnetic fields experienced by the second charge due to the first as show that:

$$F_E = \gamma \, F'_E \quad \text{ and} \quad F_B = -\gamma\frac{v^2}{c^2}F'_E \quad \implies\quad F_\text{lab}=\gamma \left(1 - \frac{v^2}{c^2}\right) F'_E= \frac{F'_E}{\gamma},$$

where $\gamma = 1/\sqrt{1-v^2/c^2}$, and $F'_E$ is the Coulomb force in the particle's rest frame.

Using the above relation, you should also be able to prove why imposing the magnetic and electric forces being the same on each of the charges leads to $v = c$. (See the end of the answer I linked.) I doubt it's a coincidence, but I think it needs to be interpreted as saying that the force due to the magnetic field of a moving charge can never be equal to the force due to its electric field or greater than it (since $v$ can never be equal to $c$ for a massive object).

A simple reason for this I can think of is the following: as I've shown above, the force due to the magnetic field is opposite in direction to the force due to the electric field. (Like charges repel, but like "currents" attract.) Now imagine that there could be some constant speed $v$ such that their magnitudes could be equal. This would mean that both these forces would cancel each other out; i.e. there would be some "privileged" speed $v$ where the particle would not experience a net force. But this violates the very idea of an inertial frame! Therefore it must not be possible for any sensible $v$. And indeed, using the formula for $F_\text{lab}$ above, you can see that the measured force seems to reduce, but it never goes to zero for, since $v<c$. So I suppose -- for what it's worth -- you could say that this is "due" to special relativity.