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Physical explanation for power broadening

  1. Jul 10, 2014 #1
    I have been looking into broadening mechanisms and I'm stuck at trying to provide a physical explanation for power broadening. I get how the math shows that at high intenseties the decay rate goes through the roof due to saturation, but how does this increased decay rate manifest in a spread of generated frequencies? Are the electrons reexcited or decaying while between ground and excited states?
     
  2. jcsd
  3. Jul 10, 2014 #2

    mfb

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    The excitation and deexcitation has "less time". Imagine a wavepacket with a shorter length: it has to have a broader frequency distribution (the mathematical "uncertainty principle" for fourier transformations). The same happens here.
     
  4. Jul 12, 2014 #3
    The precision with which you can define the frequency of a wave depends on the number of cycles. If you have 10 cycles, you can define the wave length or frequency to ~10%, 100 cycles to ~1%, 1000 cycles to ~0.1% and so on.

    A strongly damped wave or a short pulse has a small number of cycles. A fast decay means strong damping.

    Mathematically, in order to produce a short wave pulse you have to overlay waves with many frequencies. The spread of frequencies increases the shorter the pulse. A single frequency wave would have to be infinitely long in space and in time.

    (this is the same thing mfb said, in more words)
     
  5. Jul 13, 2014 #4
    Thank you for both answers, they helped a lot :-)
     
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