# Radiated energy by decelerating (relativistic) point charge

1. May 7, 2013

### taha.hojati

1. The problem statement, all variables and given/known data
A point charge q is moving relativistically with constant speed $\beta$ along the x-axis.
At t = 0 a constant decelerating force F is applied in opposite direction of
its velocity. If the charge stops after traveling a distance d, find the total radiated
energy.

2. Relevant equations
I found these equations in Griffith's book:

Total radiated power by a relativistically moving point charge (Lienard's eq.):
$$P = \frac{\mu_0 q^2 \gamma^6}{6 \pi c} (a^2 - |\frac{\beta \times a }{c} | )$$ Here again $\beta$ is the the velocity of the charge (in SI units), and a is its acceleration.

Newton's law of motion for charged particle (with radiation reaction):
$$a= \tau \dot{a} + \frac{F}{m}$$ here $\tau = \frac{\mu_0 q^2}{6 \pi c }$, ($m \tau \dot{a}$ is the abraham-lorentz formula for the radiation reaction).
This equation is from Griffith's book but I am a little confused by it because in relativity we have $F_{net} = \frac{d}{dt} \gamma m v \neq m a$.

3. The attempt at a solution

First I tried to find acceleration and velocity based on time by solving the differential equation ==>
$$a = -\frac{F}{m} e^{t/\tau} + \frac{F}{m}$$
$$\Rightarrow v= \beta + \frac{F}{m} ( t - e^{t/\tau})$$
From here I wanted to find $t_{final}$ based on d and then calculate the total energy radiated by integrating the Lienard's equation, but I am unsure since it seems like I am forgetting all I learned about relativity in my mechanics class (also the integral is hard).

So I thought I can approach from conservation of energy aspect. So total kinetic energy is $KE_{initial} = (\gamma-1) m c^2$. Now part of this energy is radiated and the rest is spent on resisting the force F. So can I just write $E_{rad} = KE_{initial} - F d$ ? That would make everything easier. Thanks in advance!

2. May 7, 2013

### hexidecimel10

You know that before the force is applied there is no radiation. You also know that the acceleration is in the opposite direction of the velocity. The cross product of any two col-linear vectors is 0 so you can drop the cross product term.