Electric field due to a moving charge

Question

I am having some problems involving the force that a source moving with speed $v$ along the $x$-axis would exert on a test charge at the $x$-axis.

Moving to the frame of the source charge, we got that the electric field it exerts is $$E' = kq/x'^{2}.$$

Now, moving back to the lab frame, and considering that $x'$ above is contracted, since it was at the frame of a moving charge, we got $\vec{E}'_{\parallel} = \vec{E}_{\parallel}$, we got that:

$$ E = kq/x'^{2} = kq \frac{\gamma^{2} }{x^{2} } \implies F = kq Q\frac{\gamma^{2} }{x^{2} }$$

This is wrong, but I don't know why. :/

Where is my error? How to solve this question?

Answer 1

If you are going to do electromagnetism in relativity then you should use the appropriate four-vector and tensor framework. The EM field tensor for a stationary point charge at the origin of the primed frame is given by: $$F^{\mu' \nu' }=\left( \begin{array}{cccc} 0 & -\frac{k q x'}{\left(x'^2+y'^2+z'^2\right)^{3/2}} & -\frac{k q y'}{\left(x'^2+y'^2+z'^2\right)^{3/2}} & -\frac{k q z'}{\left(x'^2+y'^2+z'^2\right)^{3/2}} \\ \frac{k q x'}{\left(x'^2+y'^2+z'^2\right)^{3/2}} & 0 & 0 & 0 \\ \frac{k q y'}{\left(x'^2+y'^2+z'^2\right)^{3/2}} & 0 & 0 & 0 \\ \frac{k q z'}{\left(x'^2+y'^2+z'^2\right)^{3/2}} & 0 & 0 & 0 \\ \end{array} \right)$$

When you transform this to the unprimed frame you get: $$ F^{\mu \nu }=\left( \begin{array}{cccc} 0 & -\frac{k q \left(v^2-1\right) (t v-x)}{\left(t^2 v^2-2 t x v+x^2-\left(v^2-1\right) \left(y^2+z^2\right)\right)^{3/2}} & -\frac{k q y}{\sqrt{1-v^2} \left(\frac{(x-t v)^2}{1-v^2}+y^2+z^2\right)^{3/2}} & -\frac{k q z}{\sqrt{1-v^2} \left(\frac{(x-t v)^2}{1-v^2}+y^2+z^2\right)^{3/2}} \\ \frac{k q \left(v^2-1\right) (t v-x)}{\left(t^2 v^2-2 t x v+x^2-\left(v^2-1\right) \left(y^2+z^2\right)\right)^{3/2}} & 0 & -\frac{k q v y}{\sqrt{1-v^2} \left(\frac{(x-t v)^2}{1-v^2}+y^2+z^2\right)^{3/2}} & -\frac{k q v z}{\sqrt{1-v^2} \left(\frac{(x-t v)^2}{1-v^2}+y^2+z^2\right)^{3/2}} \\ \frac{k q y}{\sqrt{1-v^2} \left(\frac{(x-t v)^2}{1-v^2}+y^2+z^2\right)^{3/2}} & \frac{k q v y}{\sqrt{1-v^2} \left(\frac{(x-t v)^2}{1-v^2}+y^2+z^2\right)^{3/2}} & 0 & 0 \\ \frac{k q z}{\sqrt{1-v^2} \left(\frac{(x-t v)^2}{1-v^2}+y^2+z^2\right)^{3/2}} & \frac{k q v z}{\sqrt{1-v^2} \left(\frac{(x-t v)^2}{1-v^2}+y^2+z^2\right)^{3/2}} & 0 & 0 \\ \end{array} \right)$$

Here we notice two things. First, in the x-component of the E field we have a factor of $(v^2-1)$ in the numerator. So instead of multiplying by $\gamma^2$ you should be dividing by $\gamma^2$. The other thing is that the field is now a function of time. This should be expected because in this frame the charge is moving.

So, if we restrict our interest to the x axis by setting $y$ and $z$ to $0$ we get: $$ F^{\mu \nu }=\left( \begin{array}{cccc} 0 & \frac{k q \sqrt{(x-t v)^2}}{(t v-x)^3 \gamma ^2} & 0 & 0 \\ -\frac{k q \sqrt{(x-t v)^2}}{(t v-x)^3 \gamma ^2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$ This shows both the correct placement of the $\gamma$ factors as well as the time dependence.