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Q of radiating electron according to classical theory

  1. Aug 18, 2013 #1
    1. The problem statement, all variables and given/known data
    According to classical electromagnetic theory an accelerated electron radiates energy at the rate [itex]Ke^{2}a^{2}/c^{3}[/itex], where [itex]K = 6*10^{9} Nm^{2}/C^{2}[/itex], [itex]e = [/itex] electronic charge, [itex]a = [/itex] instananeous acceleration, and [itex]c = [/itex] speed of light.

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

    b) What is the [itex]Q[/itex] 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 [itex]v[/itex] a typical optical frequency(i.e., for visible light) estimate numerically the approximate Q and "half-life" of the radiating system.



    2. Relevant equations
    [tex]Q = \frac{\omega_{0}}{\gamma}[/tex]

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

    I feel confused about part b. I'm given the rate the energy radiates and from that I think I should find [itex]\omega_{0}[/itex] and [itex]\gamma[/itex] which will tell me about [itex]Q[/itex]. 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:
    [tex]\int \frac{dE}{dt} dt = \int \frac{K e^{2}}{c^{3}} \frac{d^{2} x}{d t^{2}} dt[/tex]
    [tex]E = \frac{1}{2} \frac{K e^{2}}{c^{3}} \left( \frac{dx}{dt} \right)^{2} + constant[/tex]
    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. jcsd
  3. Aug 18, 2013 #2

    TSny

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    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.
     
  4. Aug 18, 2013 #3
    Thank you for that formula. What is the energy stored in an oscilating electron?
     
  5. Aug 18, 2013 #4

    TSny

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    It's just the energy of a simple harmonic oscillator.
     
  6. Aug 18, 2013 #5
    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:
    [tex]Q = 2 \pi \frac{E}{8 \pi^{4} v^{3} A^{2} K e^{2}} = \frac{mc^{3}}{4 \pi v K e^{2}}[/tex]
    so
    [tex]E = m\left(Av\pi \right)^{2}[/tex]
    and that feels like it should look familiar but it doesn't. How is that like [itex]\frac{1}{2} k x^{2}[/itex]?
     
  7. Aug 18, 2013 #6

    TSny

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    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?
     
  8. Aug 18, 2013 #7
    KE = [itex]\frac{1}{2}m \dot{x}^{2}[/itex], when t = 0 the PE = 0 and the KE = m(Av*pi)^2. Right on TSny thank you.
     
  9. Aug 18, 2013 #8

    TSny

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    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.
     
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