The relevant question is here. The accepted answer may have explained my question in a descriptive manner. However, I want to see how things are related quantitatively.

Imagine we have two charges $q$ moving parallel to each other. The distance between them is $d$.

  • In the frame where the charges are stationary. We have: $$m_0 a_0=\frac{q^2}{4\pi\epsilon_0d}$$

  • In the laboratory frame, the charges also experience a force caused by the magnetic field which is generated by the other charge: $$B=\mu_0\frac{qv}{2\pi d}$$ The total force is: $$F=\frac{q^2}{4\pi\epsilon_0d}-\mu_0\frac{q^2v^2}{2\pi d}=\frac{m_0}{\sqrt{1-v^2/c^2}}a$$

There is also the relation of $a_0$ and $a$ that relate these two equations of motion. However, it seems I cannot get the right result.

Any help in figuring out how to relate these two situations would be appreciated.


1 Answer 1


Actually it's not that difficult (but a neat problem), there's only one crucial step in the development that I will show you, but let's start from a bit earlier.

Let's first write out the two forces on interest here (in terms of magnitudes, as we know already they're on parallel trajectories):

The coulomb repulsion between the charges: $$F_C = \frac{q^2}{4\pi \epsilon_0 r^2} $$ Now since we have moving charges, (hence a current), each charge will experience a Lorentz force in the magnetic field induced by the other charge, using Biot-Savart's law, we have for the magnetic field $B$ : $$B = \frac{\mu_0 I dL}{4\pi r^2}$$ Which should be interpreted as (by analogy to a electrical circuit) "the magnetic field felt at a distance $r$, induced by a wire length $dL$ carrying current $I$".

Note $\mu_0$ is the permeability of vacuum, $\epsilon_0$ is the corresponding permittivity.

The Lorentz force: $$F_L = qBv = \frac{\mu_0 q^2 v^2}{4\pi r^2}$$ where the $I dL$ term in B field is replaced using: $I=q/dt$ and $dL/dt = v$, which gives finally $$I dL=qv.$$

The two forces together (with $F_C$ the repulsive force here): $$\sum F =\frac{q^2}{4\pi \epsilon_0 r^2} -\frac{\mu_0 q^2 v^2}{4\pi r^2}$$ It is clear that the resulting effect of the two forces depends on the strength of the $B$ field which in turn then depends on the velocity of the two charges, if they are slow, the repulsion dominates. So in order to be able to compare their relative strengths, we need a slight rearrangement for the $\sum F$:

$$F_{tot} =\frac{q^2}{4\pi \epsilon_0 r^2} \left(1-v^2 \epsilon_0 \mu_0\right) $$ where $$v^2 \epsilon_0 \mu_0=\frac{F_L}{F_C}$$ But we know that $\epsilon_0 \mu_0 = c^{-2}$ the inverse squared of speed of light in vacuum, substituted in the total force expression: $$F_{tot}= F_C \left(1-\frac{v^2}{c^2}\right) = \frac{F_C}{\gamma^2} $$ Where $\gamma$ is the relativistic factor $(1-v^2/c^2)^{-1/2}$ or often called Lorentz factor, which relates the measurements performed in different inertial frames moving at $v$ with respect to one another.

Last note: if the charges are not in vacuum, the relative permittivity $\epsilon$ and permeability $\mu$ are included in the expressions.

  • $\begingroup$ Thanks very much, my first equation is totally a stupid mistake. I still cannot figure out where my second equation goes wrong. I try to use Ampere's circuital law that is $2\pi r B=\mu_0 I=\mu_0 qv$ The unit is not right though. Can't I use Ampere law in this situation? $\endgroup$ Commented Aug 7, 2014 at 13:43
  • $\begingroup$ According to your derivation. I would claim that the acceleration observed by the observer "ride on one charge" is $\gamma^3$ of the observer in the laboratory frame. Is this a general case in special relativity(I am not familiar with this)? $\endgroup$ Commented Aug 7, 2014 at 13:48
  • 1
    $\begingroup$ @luming to your first question, no, as we're not dealing with a closed curve here (through which the current would run), if we did then you could apply ampere's law $\oint_C \mathbf{B} \cdot d\mathbf{l}=\mu_0 I_{\rm enclosed}$. Instead here we want to define the contribution to B field of each charge moving at v, so Biot-Savart's law has to be used. By the way $I=dq/dt$ and not $qv$, be careful. $\endgroup$
    – Ellie
    Commented Aug 7, 2014 at 14:31
  • $\begingroup$ @luming regarding your 2nd comment, yes that is correct but it is only for rectilinear motions that the observed acceleration boils down to $a=\gamma^3 A$ which is indeed the case here with our electrons, for more general velocities, it is slightly more complicated. Just read a bit on Lorentz transformations in special relativity and you will get familiarized to it rather quickly. $\endgroup$
    – Ellie
    Commented Aug 7, 2014 at 14:48
  • $\begingroup$ Surely the total force on a charge should be reduced by $\gamma$, not by $\gamma^2$. Why do you assume that the Coulomb fores between the two charges is unchanged when viewed from the moving frame? Aren't the electric fields transformed such that $E^2 - c^2B^2$ is invariant? $\endgroup$
    – ProfRob
    Commented Jan 21, 2017 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.