Find final velocity of partical traveling at relativistic speeds under constant force over set distance without time

Question

What equation do I need to find the final velocity of a particle traveling at relativistic speeds given some initial velocity (say 0.3C) and exposed to a constant force. The relationship between momentum and velocity are not linear as the mass is effectively increasing as energy is added to the system. I would prefer an equation that uses an acceleration over some distance as opposed to some time which is the typical norm.

Answer 1

Well, you can simply use the definitions involved. Let the particle have mass $m$, postion $\mathbf r$ and velocity $\mathbf v$, and let the force acting on it be $\mathbf F$.

The particle's momentum is $\mathbf p=\gamma m\mathbf v$, where $\gamma=\frac{1}{\sqrt{1-\mathbf v^2/c^2}}$ is the usual Lorentz factor. The force, by definition, is $\mathbf F=\frac{d\mathbf p}{dt}$. If we then take the scalar product of that with the velocity, it gives $$\mathbf F\cdot\mathbf v=\frac{d\mathbf p}{dt}\cdot\mathbf v,$$

and if we remember that $\mathbf v=\frac{d\mathbf r}{dt}$, then we can write that as

$$\mathbf F\cdot\frac{d\mathbf r}{dt}=\frac{d\mathbf p}{dt}\cdot\mathbf v.$$

Integrating over some time $t_i$ to $t_f$, remembering that the force is constant, we get

$$\int\limits_{t_i}^{t_f}\mathbf F\cdot\frac{d\mathbf r}{dt}\,dt=\int\limits_{t_i}^{t_f}\frac{d\mathbf p}{dt}\cdot\mathbf v\,dt,$$

$$\int\limits_{\mathbf r_i}^{\mathbf r_f}\mathbf F\cdot d\mathbf r=\int\limits_{\mathbf p_i}^{\mathbf p_f}\mathbf v\cdot d\mathbf p,$$

$$\mathbf F\cdot (\mathbf r_f-\mathbf r_i)=\int\limits_{\mathbf p_i}^{\mathbf p_f}\mathbf v\cdot d\mathbf p.$$

On the left hand side, we have force times displacement, and on the right hand side, a function of velocity only. This integral is a little tricky to calculate, but with some work, we get that

$$\mathbf F\cdot (\mathbf r_f-\mathbf r_i)=(\gamma_f-\gamma_i)mc^2,$$

with the respective Lorentz factors. This is enough to answer your question, solving for the final velocity. But it's interesting that, if we remember that the energy of a relativistic particle is given by $E=\gamma mc^2$, then we can write

$$\mathbf F\cdot (\mathbf r_f-\mathbf r_i)=E_f-E_i,$$

Which is just equating the work to the change in energy of the particle. Not bad!

Also, you commented that standard "Newtonian" kinematics doesn't work in relativity, yes it does. You, again, just have to remember the definitions.

The constant acceleration is $\mathbf a=\frac{d\mathbf v}{dt}$. Taking the scalar product with the velocity and integrating on time we get

$$\int\limits_{t_i}^{t_f}\mathbf a\cdot\frac{d\mathbf r}{dt}\,dt=\int\limits_{t_i}^{t_f}\frac{d\mathbf v}{dt}\cdot\mathbf v\,dt,$$

$$\int\limits_{\mathbf r_i}^{\mathbf r_f}\mathbf a\cdot d\mathbf r=\int\limits_{\mathbf v_i}^{\mathbf v_f}\mathbf v\cdot d\mathbf v,$$

$$\mathbf a\cdot(\mathbf r_f-\mathbf r_i)=\frac{1}{2}(\mathbf v^2_f-\mathbf v^2_i),$$

which is the standard result, independent of any dynamics (relativistic or not) that may be going on.