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.
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 $$E_i = -F_0^{\; i}.$$
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 $\gamma$.
This is best addressed by considering the transforms in infinitesimal form where the algebra becomes linear and simpler. The transforms for the coordinates, coordinate differentials and components of momentum and energy are given by: $$Δ𝐫 = 𝞈×𝐫 - 𝞄t + 𝝴, \hspace 1em Δt = -α𝞄·𝐫 + τ, \tag{1}$$ $$Δ(d𝐫) = 𝞈×d𝐫 - 𝞄 dt, \hspace 1em Δ(dt) = -α𝞄·d𝐫, \tag{2}$$ $$Δ𝐩 = 𝞈×𝐩 - α𝞄E, \hspace 1em ΔE = -𝞄·𝐩, \tag{3}$$ where $α = (1/c)^2$ and where $𝞈$, $𝞄$, $𝝴$ and $τ$ are the infinitesimal forms, respectively, of rotation, boost, spatial translation and time translation. It's the Lorentz boosts you want, but we can answer the question more comprehensively.
The transform induced on the differential operators is obtained by requiring that $$d𝐫·∇ + dt\frac{∂}{∂t}$$ be invariant. Applying transforms: $$\begin{align} 0 &= Δ\left(d𝐫·∇ + dt\frac{∂}{∂t}\right) \\ &= Δ(d𝐫)·∇ + d𝐫·Δ(∇) + Δ(dt)\frac{∂}{∂t} + dtΔ\left(\frac{∂}{∂t}\right) \\ &= (𝞈×d𝐫 - 𝞄 dt)·∇ + d𝐫·Δ(∇) + (-α𝞄·d𝐫)\frac{∂}{∂t} + dtΔ\left(\frac{∂}{∂t}\right) \\ &= d𝐫·(-𝞈×∇) - dt𝞄·∇ + d𝐫·Δ(∇) + d𝐫·\left(-α𝞄\frac{∂}{∂t}\right) + dtΔ\left(\frac{∂}{∂t}\right) \\ &= d𝐫·\left(Δ(∇) - 𝞈×∇ - α𝞄\frac{∂}{∂t}\right) + dt\left(Δ\left(\frac{∂}{∂t}\right) - 𝞄·∇\right), \end{align}$$ and separately equating components to zero, we get: $$ Δ(∇) = 𝞈×∇ + α𝞄\frac{∂}{∂t}, \hspace 1em Δ\left(\frac{∂}{∂t}\right) = 𝞄·∇. \tag{4} $$
The velocity $𝐯$ satisfies the differential equation $d𝐫 = 𝐯dt$. Applying transforms, we find: $$ Δ(d𝐫) = (Δ𝐯)dt + 𝐯Δ(dt) \\ 𝞈×d𝐫 - 𝞄 dt = (Δ𝐯)dt + 𝐯(-α𝞄·d𝐫) \\ 𝞈×𝐯dt - 𝞄 dt = (Δ𝐯)dt + 𝐯(-α𝞄·𝐯dt) \\ 𝞈×𝐯 - 𝞄 = Δ𝐯 - α𝞄·𝐯𝐯. $$ Thus: $$Δ𝐯 = 𝞈×𝐯 - 𝞄 + α𝞄·𝐯𝐯. \tag{5}$$
The moving time derivative, that evaluates position-dependent and time-dependent quantities along a body's trajectory, is given through the chain rule by: $$\frac{d}{dt} = \frac{∂}{∂t} + 𝐯·∇.$$ Applying transforms, we get: $$\begin{align} Δ\left(\frac{d}{dt}\right) &= Δ\left(\frac{∂}{∂t}\right) + (Δ𝐯)·∇ + 𝐯·(Δ∇) \\ &= 𝞄·∇ + (𝞈×𝐯 - 𝞄 + α𝞄·𝐯𝐯)·∇ + 𝐯·\left(𝞈×∇ + α𝞄\frac{∂}{∂t}\right) \\ &= 𝞄·∇ - 𝞄·∇ + 𝞈×𝐯·∇ + 𝐯·𝞈×∇ + α𝞄·𝐯\left(\frac{∂}{∂t} + 𝐯·∇\right). \end{align}$$ After noting the cancellation $$𝞈×𝐯·∇ + 𝐯·𝞈×∇ = 𝞈×𝐯·∇ - 𝞈×𝐯·∇ = 0$$ this reduces to: $$Δ\left(\frac{d}{dt}\right) = α𝞄·𝐯\frac{d}{dt}. \label{6}\tag{6}$$
The force $𝐅$ and power $P$ satisfy the differential equations $$d𝐩 = 𝐅dt, \hspace 1em dE = Pdt.$$ Transform work transparently through differentials $$Δ(d𝐩) = 𝞈×(d𝐩) - α𝞄(dE), \hspace 1em Δ(dE) = -𝞄·(d𝐩). \tag{7}$$ Applying this, with the other transforms, we get: $$ Δ(d𝐩) = (Δ𝐅)dt + 𝐅Δ(dt) \\ 𝞈×(d𝐩) - α𝞄(dE) = (Δ𝐅)dt + 𝐅(-α𝞄·d𝐫) \\ 𝞈×𝐅dt - α𝞄Pdt = (Δ𝐅)dt + 𝐅(-α𝞄·𝐯dt) \\ 𝞈×𝐅 - α𝞄P = Δ𝐅 - α𝞄·𝐯𝐅, \\ $$ and $$ Δ(dE) = (ΔP)dt + PΔ(dt) \\ -𝞄·(d𝐩) = (ΔP)dt + P(-α𝞄·d𝐫) \\ -𝞄·𝐅dt = (ΔP)dt + P(-α𝞄·𝐯dt) \\ -𝞄·𝐅 = ΔP - α𝞄·𝐯P. $$ Thus, $$Δ𝐅 = 𝞈×𝐅 - α𝞄P + α𝞄·𝐯𝐅, \hspace 1em ΔP = -𝞄·𝐅 + α𝞄·𝐯P. \tag{8}$$
For all the quantities involving total time derivatives, the transforms can also be written as: $$ (Δ - α𝞄·𝐯)𝐯 = 𝞈×𝐯 - 𝞄, \hspace 1em (Δ - α𝞄·𝐯)\frac{d}{dt} = 0, \\ (Δ - α𝞄·𝐯)𝐅 = 𝞈×𝐅 - α𝞄P, \hspace 1em (Δ - α𝞄·𝐯)P = -𝞄·𝐅. \label{9}\tag{9} $$
The purely relativistic effects are marked by the parameter $α$. What's interesting is that power $P$ is seen as force $𝐅$ to a moving observer with a conversion by $α$. And, here, you thought the transform for force was a self-contained expression. Well, as you can see: it also involves some mixing up with the power.
The conversion of $Δ$ to $Δ - α𝞄·𝐯$ is directly connected to the $-α𝞄·𝐫$ "simultaneity shift" in the transform $Δt$ and to the time dilation effect of Relativity. This can be seen by applying the transform to $1 - α|𝐯|^2$: $$Δ\left(1 - α|𝐯|^2\right) = -2α𝐯·Δ𝐯 = -2α𝐯·(𝞈×𝐯 - 𝞄 + α𝞄·𝐯𝐯) = 2α𝞄·𝐯(1 - α|𝐯|^2).$$ Thus, $$Δ\sqrt{1 - α|𝐯|^2} = α𝞄·𝐯\sqrt{1 - α|𝐯|^2}, \tag{10}$$ which is the transform on the time dilation factor.
The dilation factor is more commonly written in reciprocal for as: $$γ = \frac{1}{\sqrt{1 - α|𝐯|^2}}, \tag{11a}$$ for which the transform is: $$Δγ = -α𝞄·𝐯γ. \label{10a}\tag{10a}$$
From this, it follows: $$γ(Δ - α𝞄·𝐯)(⋯) = Δ(γ(⋯)). \tag{11b}$$
Using this, ($\ref{6}$) can be rewritten as: $$Δ\left(γ\frac{d}{dt}\right) = 0,$$ and ($\ref{9}$) and ($\ref{10a}$) as: $$ Δ(γ𝐯) = 𝞈×(γ𝐯) - 𝞄(γ), \hspace 1em Δ(γ) = -α𝞄·(γ𝐯), \\ Δ(γ𝐅) = 𝞈×(γ𝐅) - α𝞄(γP), \hspace 1em Δ(γP) = -𝞄·(γ𝐅). $$ Therefore, $(γ𝐯,γ)$ have the same transform as does $(d𝐫,dt)$, while $(γ𝐅,γP)$ and $(𝐩,E)$ each has the same transform as the other does, and both have the same transform as does $(∇,-∂/∂t)$. That also means $𝐩·d𝐫 - E dt$ and $γ(𝐅·d𝐫 - P dt)$ are both invariants.
