# Velocity of particle due to acceleration. Special relativity

I understand how to do transformations of four velocities, acceleration and so on under Lorentz boosts. However after all I have learned in special relativity I still don't know how just an ordinary particle's velocity changes due to an external force. That is if I'm given a particle with some velocity and we apply a force in the same direction of the velocity or in any direction for a zero initial velocity. How would I find the velocity at time t later in my frame.
For Newtonian mechanics it is obviously $$v = v_0 + at$$. Can someone point me in the right direction or to some literature on the topic.

You just need to apply $$\vec{F}=\frac{d}{dt}(\gamma m\vec{v})$$. For example suppose the second case you say, that I apply a force in some direction, let's say the x-direction, to a particle with zero initial velocity. You would get $$F_x=m\frac{d}{dt}(\gamma v_x)$$ $$0=\frac{d}{dt}(\gamma mv_y)$$ $$0=\frac{d}{dt}(\gamma mv_z)$$
From the second equation you get $$m\gamma v_y=constant=0$$, because at $$t=0$$, $$\gamma=1$$ and $$v_y=0$$. From this, since $$\gamma\ge1$$ always, you get $$v_y=0$$ for all $$t$$. The same can be done for the $$z$$ component, and you get $$v_z=0$$ for all $$t$$.
For the x component, just as in Newtonian mechanics, things can get as complicated as you want if the force has some dependence on time or position, you will be faced to solve a very ugly integral. Let's say that the force is constant, then $$\frac{F_x}{m}=\frac{d}{dt}(\gamma v_x)$$ and $$\gamma v_x=\frac{F_x}{m}t+[\gamma v_x]|_{t=0}=\frac{F_x}{m}t$$ You can now substitute $$\gamma$$ $$\frac{v_x}{\displaystyle\sqrt{1-\frac{v_x^2}{c^2}}}=\frac{F_x}{m}t$$ and solve for $$v_x$$.