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How does Cherenkov radiation work?

  1. Jun 5, 2006 #1
    Hi.
    The way I understand it, if a charged particle passes through a medium, it disrupts the EM field in that medium. And that causes the medium's electrons to emit photons in order to restore equilibrium in the positions of the disrupted electrons. Right?
    However, I'm not sure where the fact that electrons travel faster than the speed of light in that particular medium comes in. Also, how come higher frequencies of the emitted photons are more intense than lower frequencies? (basically, all of the colours should be represented, but we can only see brilliant blue)
     
    Last edited: Jun 5, 2006
  2. jcsd
  3. Jun 5, 2006 #2

    Morbius

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    raul_I,

    It's the same principle that creates the "shock front" which causes a sonic boom
    as a plane flies faster than the speed of sound.

    Think of the track of the particle [ or the plane, if you're more confortable with that];
    for all the electrons that the particle disturbed that are behind the current position
    of the particle, have relaxed to the ground state and emitted photons.

    Those photons have been traveling out at the speed of light in the medium. If the
    particle is traveling at less than the speed of light in the medium, any photons
    produced now can't ever hope to catch up with the ones that were emitted earlier.

    However, if the particle is travelling faster than light in the medium - then the photons
    emitted now can "catch up" to the ones that were emitted previously.

    Imagine we have a bunch of marathon runners in the back of a pickup truck, and we
    let them out one by one along the way. It also takes a finite amount of time for the
    runners to climb out of the truck. If the truck is traveling slower than the runners run,
    then the later runners can never catch up to the first ones out, since the first ones
    got a head start.

    However, if the truck speeds out faster than the runners, the truck can get way out
    front of the runners, and the latest runner to get out and run, starts up just as the
    previous runners are running by where he was let off. The truck speeds off again.

    If you keep doing that, then all your runners will be running side-by-side. But in order
    to do that - you have to have the truck race ahead of the runners to prepare for the
    next runner to get out of the truck.

    The only way the truck can race out ahead of the runners is for it to go faster than
    the runners.

    Dr. Gregory Greenman
    Physicist
     
  4. Jun 6, 2006 #3
    Ok, the "shock front" thing makes sense. But I still don't understand why higher frequencies are more intense than lower ones. (Sorry, if it is mentioned in your explanation, I just don't see it)
     
  5. Jun 6, 2006 #4
    ... and come to think of it, I can't entirely cope with the idea of a speeding truck either :) . Why would the photons need to catch up with each other in the first place?
     
  6. Jun 6, 2006 #5

    eep

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    are the runners travelling in the same direction that the truck is moving?
     
  7. Jun 6, 2006 #6

    Morbius

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    Yes.
    Dr. Gregory Greenman
    Physicist
     
  8. Jun 6, 2006 #7

    Morbius

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    raul,

    In order to form a wave front that you can see.

    If the charged particle causing the excitation is traveling subluminal [ slower than light ],
    then all you will have is a stream of single photons of random phase with respect to
    each other. They will destructively interfere with each other due to the random phase.

    You aren't going to see individual photons - you need to see a bunch of them. Even
    then, the bunch of photons that you see can't have random phases - they have to
    be phase-aligned so that they form a wave front that you can see.

    The faster the particle, the greater the distance it travels in a given time, hence the
    greater the number of atoms that the particle encounters and excites into giving off
    photons per unit time. The greater the number of photons given off per unit time -
    the greater the intensity.

    Dr. Gregory Greenman
    Physicist
     
  9. Jun 6, 2006 #8
    So the faster the particle goes, the higher the frequency of the emitted photon? How's that? I understand that the speed light should remain the same in the medium. I don't see how the frequency of light depends on the rate of disturbance in the EM field (the speed of the passing charge).
     
  10. Jun 6, 2006 #9

    Morbius

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    raul,

    From CERN - the European nuclear laboratory:

    http://rd11.web.cern.ch/RD11/rkb/PH14pp/node26.html

    The integrand in the integral for [tex]dN/dI[/tex] is inversely proportional the the square of the wavelength.

    Dr. Gregory Greenman
    Physicist
     
    Last edited: Jun 6, 2006
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