Force between 2 Point Charges Across Frames

[silverrahul]

TL;DR

The force between 2 moving point charges in the lab frame should be 1/ϒ times the force in their rest frame. But my calculations are showing 1/ϒ^2 times the force in rest frame. Where am i going wrong ?

So, i was trying to calculate, the net force between 2 point charges in their rest frame, and in a frame where they are moving.

So, assume, there are 2 point charges each of charge +q.
They are r distance apart from each other and moving parallel to each other with a speed v relative to a lab observer.So , in the unprimed frame, i.e. the rest frame of the point charges.
Net force of repulsion between them F = q²/4πε₀r²Whereas in the primed frame , i.e the lab frame,
Net force ( including the magnetic force ) = F' = q²/4πε₀r² - μ₀q² v²/4πr²

Upon rearranging the terms of F', it results in F' = F / ϒ²

But, it was my understanding that F' should be equal to F / ϒ
Can someone please point out where i am going wrong ?

 

[jartsa]

What kind of electric field is measured when charge q flies past a measuring device at speed v?

The maximum value is: $E=\gamma(v)*q/4πε_0r$

 

[A.T.]

No , i think it should be E=q/4πε₀r²

That's for the rest frame of the charge. If you transform the E field into frame where the charge moves, it gets length contracted, and thus stronger perpendicular to the motion, but weaker in line with the motion.

From: https://www.researchgate.net/figure/Electric-Field-of-a-Charged-Particle_fig2_2177605

 

[silverrahul]

That's for the rest frame of the charge. If you transform the E field into frame where the charge moves, it gets length contracted, and thus stronger perpendicular to the motion, but weaker in line with the motion.From: https://www.researchgate.net/figure/Electric-Field-of-a-Charged-Particle_fig2_2177605

Thanks a lot for this . I had no idea about this, i will look it up to see if it helps

 

[Dale]

Summary:: The force between 2 moving point charges in the lab frame should be 1/ϒ times the force in their rest frame. But my calculations are showing 1/ϒ^2 times the force in rest frame. Where am i going wrong ?

Whereas in the primed frame , i.e the lab frame,
Net force ( including the magnetic force ) = F' = q2/4πε0r2 - μ0q2 v2/4πr2

Your transformations for the E field are incorrect. See here for details: https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#The_E_and_B_fields

In this case since in the rest frame $\vec B=0$ and since $r$ is perpendicular to $v$ (meaning also that $\vec E$ is perpendicular to $v$) we have: $$\vec E'=\gamma(\vec E + \vec v \times \vec B)-(\gamma - 1)(\vec E \cdot \hat v) \hat v= \gamma \vec E$$ $$ \vec B' = \gamma \left(\vec B - \frac{\vec v \times \vec E}{c^2} \right) - (\gamma - 1) (\vec B \cdot \hat v)\hat v = - \frac{\gamma}{c^2} \vec v \times \vec E$$ so then $\vec F' = q \vec E' + q \vec v \times \vec B'$ which gives $$F'= q \gamma \vec E + q \vec v \times (-\frac{\gamma}{c^2} \vec v \times \vec E) = q \gamma \vec E - q \gamma \frac{ v^2}{c^2} \vec E = \frac{q}{\gamma}\vec E$$