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I am sure there is a simple explanation for my confusion, but I am a little puzzled:

We are dealing with two parallel electron cannons that each produces a straight beam of electrons. They are placed at a distance d next to each other, which means there will be a repulsive electrostatic force between the electrons of the two beams. Also, according to the Maxwell equations, every moving electron will produce a magnetic field which then exerts another force on the other beam's moving electrons (similar to the effect that two currents in wires will cause an attractive force between the wires).

However, if we change the reference frame such that we are moving at the same speed alongside the electrons, there will be no magnetic field and no Lorentz Force. We are only left with the electrostatic force.

Do you guys know what might be the fallacy?

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There is no fallacy, you're just not being particularly careful. You need to include both the electric and magnetic forces of the right magnitude and a covariant result drops out. (Of course historically it went the other way around: people noticed that frame changes were messed up unless the transformation laws were different and this led to the development of special relativity.)

For simplicity let both beams be very thin and have equal uniform charge density $\rho$ in the rest frame and suppose they run exactly parallel separated by a distance $l$. Let the velocity of the beams be $v$ in the lab frame.

Rest frame:

Taking the usual gaussian pillbox gives the electric field of one beam at the location of the other as

$$ \vec{E}_\text{rest} = \frac{\rho}{2\pi\epsilon_0 l} \hat{r}, $$

where $\hat{r}$ is the unit vector directed away from the source beam. Thus the force on a single particle in the second beam (charge $q$) is:

$$ \vec{F}_\text{rest} = \frac{\mathrm{d}p}{\mathrm{d}t_\text{rest}} = \frac{\rho q}{2\pi\epsilon_0 l} \hat{r}. $$

Lab frame:

The charge density of the beam is enhanced by the relativistic $\gamma=1/\sqrt{1-v^2/c^2}$ factor. Thus the electric field is:

$$ \vec{E}_\text{lab} = \frac{\gamma \rho}{2\pi\epsilon_0 l} \hat{r}. $$

There is also a magnetic field of magnitude

$$ B = \frac{\mu_0 \gamma\rho v}{2\pi l} $$

and directed so as to produce an attractive force. Plugging these in the Lorentz force formula

$$ \vec{F}_\text{lab} = q (\vec{E}_\text{lab} + \vec{v}\times\vec{B}) = q\left( \frac{\gamma \rho}{2\pi\epsilon_0 l} - \frac{\mu_0 \gamma\rho v^2}{2\pi l}\right) \hat{r} = \frac{\rho q}{2\pi\epsilon_0 l}\gamma\left(1 - \epsilon_0\mu_0 v^2\right) \hat{r}. $$

Using $\epsilon_0 \mu_0 = c^{-2}$ this reduces to $\vec{F}_\text{lab} = \gamma^{-1} \vec{F}_\text{rest}$ which, on noting the relativistic time dilation $\mathrm{d}t_\text{lab} = \gamma \mathrm{d}t_\text{rest}$, is exactly right! Note that I've used the fact that the force is orthogonal to the velocity implicity when writing the Lorentz transformation law for the force. You can prove the covariance for general motions using the covariant formulation of EM.

Lesson:

Relativity and electromagnetism go together like hand and glove!

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and that, kids, is the story of the annum mirabilis. –  Jim Jul 17 '13 at 16:04
    
@Jim ...only with all of Einstein's brilliance replaced by 108 years of hindsight and assuming that which he had to prove. :) –  Michael Brown Jul 17 '13 at 16:07
    
Doesn't electrons accelerate under these forces? –  Mostafa Jul 17 '13 at 21:49
1  
@Mostafa Yes they will. I've done the analysis neglecting the acceleration for simplicity. You can either imagine this is only looking at a snapshot in time or that, instead of particle beams, you have long thin charged rods which are mechanically constrained to stay in position. But this is just for simplicity. You can certainly talk about accelerated objects in special relativity. –  Michael Brown Jul 18 '13 at 0:34

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