Why Is the X-Component of the Electric Field Unchanged in a Moving Frame?

[sspitz]

I'm trying to understand why the x-component of electric field is the same in the rest frame and the frame moving in the x direction. I thought I should just be able to write the force four vector in the rest frame and transform it. Symbols with arrows are four vectors.

$$\vec{p}=\gamma(mv,mc)\\ \vec{f}=\frac{d\vec{p}}{d\tau}=\gamma(\frac{dp}{dt},\frac{dmc}{dt})$$
For the charge in its rest frame, just looking at the x-component
$$\gamma=1\\ \vec{f}_{x}=\frac{dp_{x}}{dt}=qE_{x}$$
Now I transform to the frame moving with velocity v in the positive x direction
$$\vec{f'}_{x}=\gamma'(qE_{x}-\frac{v}{c}\frac{dmc}{dt})$$
So, how does that prove that the x-component of x doesn't change? dmc/dt is not zero as far as I can tell.

 

[AvroArrow]

The key here is that the electric field does not depend on the frame of reference. This means that in both frames, the x-component of the electric field is the same. In the rest frame, you have the equation $\vec{f}_{x}=\frac{dp_{x}}{dt}=qE_{x}$ which tells us that the x-component of the force due to the electric field is equal to the x-component of the electric field times the charge. In the frame moving with velocity v in the positive x direction, the equation is $\vec{f'}_{x}=\gamma'(qE_{x}-\frac{v}{c}\frac{dmc}{dt})$, which is still equal to the x-component of the electric field times the charge, since $\frac{v}{c}\frac{dmc}{dt}$ is zero. Therefore, the x-component of the electric field is the same in both frames.