A Special Relativity question

Question

My problem is simple. imagine you have a proton at rest at the origin and an electron traveling along the x-axis with the speed of $0.9c$. (this is our lab frame) now I want to calculate the force on the electron when the electron is at the $x=+d$. I faced a contradiction:

one way of calculating the force is simply $F=Eq$ in which: $$E=\frac{e}{4\pi\varepsilon d^2 }$$

the other way is going to the electron's frame: in the new frame, we have an electrical field, which is smaller than the previous field by a factor of $1/\gamma^2$ (because of length contraction); and for moving to lab view we don't need anything because the longitudinal component of force won't change.

so now we have two different values for force, why? what's my mistake?

Answer 1

In order to abide by the conventions used in Resnick's book, I had to assume that the primed frame is attached to the proton so that we have: $$x'=d'\space , \space E'_x=\frac{e}{4\pi\epsilon_0 d'^2}$$ According to Resnick, the electric field of the proton (in the place of the electron) is calculated as follows from the viewpoint of the electron:

$$E_x=\frac{e\gamma (x-ut)}{4\pi\epsilon_0\left [\gamma^2(x-ut)^2+y^2+z^2 \right]^{\frac{3}{2}}}\space.$$

Since $y=z=0$, we have: $$E_x=\frac{e}{4\pi\epsilon_0\gamma^2(x-ut)^2}\space.\tag{*}$$ On the other hand, the Lorentz transformation asserts that: $$x=\gamma(x'+ut')\space,\space t=\gamma(t'+ux'/c^2)\space.$$ Using $x'=d'$ and $t'=0$, we get: $$x=\gamma d'\space,\space t=\gamma ud'/c^2\space.$$ Substituting the above terms in (*) gives: $$E_x=\frac{e}{4\pi\epsilon_0\gamma^2(\gamma d'-\gamma d'u^2/c^2)^2}=\frac{e}{4\pi\epsilon_0 d'^2}=E'_x\space.$$ Therefore, the forces are measured the same.