Q of radiating electron according to classical theory

[mbigras]

Homework Statement

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 anyone 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.

Homework Equations

Q = \frac{\omega_{0}}{\gamma}

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?

 

[TSny]

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

That looks good to me.

I feel confused about part b...
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?

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.

 

[mbigras]

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

 

[TSny]

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

 

[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}?

 

[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?

 

[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.

 

[TSny]

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

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.

 

[yucheng]

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.

Would you elaborate why there should be an extra factor of 2 for equation of Q given in post #5?

 

[yucheng]

Either ways I got an explanation at http://hafner.web.rice.edu/pdfs/201_solutions_6-11.pdf

 

[TSny]

Would you elaborate why there should be an extra factor of 2 for equation of Q given in post #5?

Look at the first equation on page 6 of your link in post #10. Note that $\omega = 2 \pi \nu$ where $\nu$ is the frequency (in Hz). This gives an expression for the energy $E$ that is twice the expression for $E$ given in post #5. This factor of 2 contributes to $Q$, so the correct expression for $Q$ should be twice that given in post #5.