# Q of radiating electron according to classical theory

1. Aug 18, 2013

### mbigras

1. The problem statement, all variables and given/known data
According to classical electromagnetic theory an accelerated electron radiates energy at the rate $Ke^{2}a^{2}/c^{3}$, where $K = 6*10^{9} Nm^{2}/C^{2}$, $e =$ electronic charge, $a =$ instananeous acceleration, and $c =$ speed of light.

a) If an electron were oscillating along a straight line with frequency $v$ (Hz) and amplitude $A$, how much energy would it radiate away during 1 cycle? (Assume that the motion is described adequately by $x = A\sin{2 \pi v t}$ during any one cycle.)

b) What is the $Q$ of this oscillator?

c) How many periods of oscillation would elapse before the energy of the motion was down to half the initial value?

d) Putting for $v$ a typical optical frequency(i.e., for visible light) estimate numerically the approximate Q and "half-life" of the radiating system.

2. Relevant equations
$$Q = \frac{\omega_{0}}{\gamma}$$

3. The attempt at a solution
For part a, I took the integral of the rate that the energy radiates from 0 to $\frac{1}{2v}$. So the energy radiated during 1 cycle is $\frac{8 \pi^{4} v^{3} A^{2} K e^{2}}{c^{3}}$ J

I feel confused about part b. I'm given the rate the energy radiates and from that I think I should find $\omega_{0}$ and $\gamma$ which will tell me about $Q$. By knowing how much energy is being lost I can imagine how that tells you about the damping but right now I don't see how they're related. Something I was thinking was to integrate the given rate:
$$\int \frac{dE}{dt} dt = \int \frac{K e^{2}}{c^{3}} \frac{d^{2} x}{d t^{2}} dt$$
$$E = \frac{1}{2} \frac{K e^{2}}{c^{3}} \left( \frac{dx}{dt} \right)^{2} + constant$$
Now it's starting to look like a familiar differential equation...but really, I'm not sure what going on here. I think my main question is: How is the quality of an oscillatory system related to the rate that it losses energy due to damping?

2. Aug 18, 2013

### TSny

That looks good to me.

See http://en.wikipedia.org/wiki/Q_factor#Definition_of_the_quality_factor for how Q relates to the energy stored in the oscillator and the energy loss per cycle that you found in part a.

3. Aug 18, 2013

### mbigras

Thank you for that formula. What is the energy stored in an oscilating electron?

4. Aug 18, 2013

### TSny

It's just the energy of a simple harmonic oscillator.

5. Aug 18, 2013

### mbigras

Thats what I feel also. There is an electron that's oscillating around so it has some total energy and can be thought of as a harmoic oscilator. But now there's this concept where the energy is dissapating and from looking in the back of the book:
$$Q = 2 \pi \frac{E}{8 \pi^{4} v^{3} A^{2} K e^{2}} = \frac{mc^{3}}{4 \pi v K e^{2}}$$
so
$$E = m\left(Av\pi \right)^{2}$$
and that feels like it should look familiar but it doesn't. How is that like $\frac{1}{2} k x^{2}$?

6. Aug 18, 2013

### TSny

E = KE + PE. You can calculate E at any point of the cycle, say at x = 0. How would you express KE and PE at the instant x = 0?

7. Aug 18, 2013

### mbigras

KE = $\frac{1}{2}m \dot{x}^{2}$, when t = 0 the PE = 0 and the KE = m(Av*pi)^2. Right on TSny thank you.

8. Aug 18, 2013

### TSny

I differ by a factor of 2 in the expression for KE at x = 0.

I also differ by a factor of 2 in the expression for Q that you gave in post #5. But it's easy to drop such factors, so I could be off.