Lorentz transform of force 
If a particle of mass $m$ and velocity $v$ is moving due to a constant electric force what would the force be in the the frame where the particles velocity is 0?

To try and solve this I used the four force and did a Lorentz transform of the four momentum. However I got different answers in each component of the force and if this scenario was taken as one dimensional I got no change in the force. So I was wondering how to find a equation relating the new force to the old force.
 A: You cannot just transform $d \bf p \rm/dt=q(\bf E  + v \wedge B \rm )$, as it is not a tensorial equation. The tensorial form of this equation is
$$\frac {d p^\mu }{d\tau } = -\frac q mp^\lambda F_\lambda^{\; \mu} $$
The tensorial nature of this equation guarantees it is valid in any coordinate system. Turning back now to your question, we can use this equation to calculate the force in the coordinate system that is momentarily comoving with the particle. In this coordinate system,  the momentum fourvector $p^\mu$ reduces to $(m,0,0,0)$ and consequently the equation reduces to $$ \frac {d p^\mu }{d\tau } = -q F_0^{\; \mu}. $$ Replacing the components of the EM field tensor $F_\lambda ^{\; \mu}$ by the corresponding electrical and magnetic field components (in the momentarily comoving frame!), we get $$ \frac {d p^0 }{d\tau} =0 \\ \frac {d p^i }{d\tau} = q E_i\ \ ,i=1,2,3   $$ with $E_i$ being the three components of the electrical field. This means that the particle will move according to the classical laws in the momentarily comoving frame, but you need of course first to calculate the components of the electrical field in this frame. In order to do this, you plug in your $\bf E$ and $\bf B$ components in your EM field tensor $F_\lambda^{\; \mu}$. You transform the field tensor using the Lorentz transformation, what will allow you to recuperate the searched $$E_i = -F_0^{\; i}.$$
A: The Lorentz force must be transformed in the same way as other forces in special relativity.
Avoiding a tensor treatment, you can say that
$${\bf F'} =  {\bf F_{\parallel}} + \frac{1}{\gamma}{\bf F_{\perp}},    $$where $\gamma$ is the usual Lorentz factor and the subscripts refer to the components of the Lorentz force in the rest frame that are parallel and perpendicular to the relative velocity between the rest frame and moving frame and the "unprimed" frame is the rest-frame of the particle.
However, I don't understand your question. A particle which is subject to a constant force will not be moving with a constant velocity except at some instantaneous time. Are we meant to assume that the velocity arises only from the acceleration due to the electric field so that we can assume that the electric field and velocity are parallel? If so, then you can see from my equation above that the Lorentz force on the particle is unchanged. The reasoning is that the magnetic field, that must be present in the rest frame of the particle, exerts no force since ${\bf v} \times {\bf B}=0$ and ${\bf E_{\parallel}'}={\bf E_{\parallel}}$. Any component of the electric Lorentz force that is in fact perpendicular to ${\bf v}$ in the primed frame will be increased (in the absence of a magnetic field in the primed frame) by a factor of $\gamma$.
