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 $0<t<T$?
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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.
