electromagnetism - Quantitative relation between two charges moving in parallel in two frames of reference

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 $d$.

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  In the frame where the charges are stationary. We have:

  $$m_0 a_0=\frac{q^2}{4\pi\epsilon_0d}$$
  
  


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  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.

Answer

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.