# Decay rate as a function of velocity in special relativity

The mean lifetime of a muon at rest is $$2.2 \mu s$$. Assuming a muon to have this precise lifetime at rest, travelling at, say, $$0.9c$$, will live longer by the factor $$\gamma=\frac{1}{\sqrt{1-0.9^2}}=2.29$$ So, the velocity affects the decay rate. Now, if I accelerate a group of $$N$$ muons from rest [$$t=0$$] to a target velocity of $$0.9c$$ [$$t=T$$ in lab frame] by applying a constant electric field to them, how can I determine the number of muons that would decay before the target velocity is achieved, i.e. in the interval $$0?

Using units where $$c=1$$ we can write the Lagrangian of a single particle in an electromagnetic field as $$L = \frac{1}{2} m u^\mu u_\mu + q u^\mu A_\mu$$ where $$u^\mu$$ is the particle's four-velocity and $$A^\mu$$ is the four-potential.
Assuming flat spacetime with a particle moving only in the x direction through a uniform E field in the x direction and using a standard Minkowski inertial frame we have $$x^\mu = (t(\tau),x(\tau),0,0)$$ where $$\tau$$ is the proper time and will be suppressed, but partial derivatives with respect to $$\tau$$ will be indicated with dots.
So the four-velocity is $$u^\mu = \left( \frac{\dot t}{\sqrt{-\dot t^2 + \dot x^2}}, \frac{\dot x}{\sqrt{-\dot t^2+ \dot x^2}} ,0,0 \right)$$ and the four potential is $$A^\mu = (E \ x,0,0,0)$$ so the Lagrangian simplifies to $$L = \frac{1}{2}m - \frac{q E \ x \ \dot t}{\sqrt{-\dot t^2 + \dot x^2}}$$
If I made no mistakes in the math then the resulting Euler-Lagrange equations are $$\ddot t = -\frac{2 \dot t^3}{x \ \dot x}$$ $$\ddot x = \frac{-3 \dot t^2+ \dot x^2}{x}$$
You can solve these for $$x(\tau)$$ and $$t(\tau)$$ where $$\tau$$ is directly proportional to the number of half-lives.