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Clarification regarding Cherenkov radiation

  1. Jan 22, 2012 #1
    I only recently began reading about Cerenkov radiations and I've got so far as reading how the photons are emitted (which I think isn't very far!) Now I have a few questions from that point on:

    a. I take it the charged particle that zips through the dielectric polarises the atoms and when electrons return to the ground state after the particle has passed, they release photons.
    Now the way I see it that means 'polarisation' not only refers to a shift in the denser regions of the electron cloud but also implies electrons jumping to higher energy levels. How exactly does that happen?

    b. Once the photons are emitted, there seems to be something to the effect that these 'photons do not interfere destructively, rather constructively when the particle is travelling at a velocity > c/n
    I don't quite understand that part either! (Perhaps it has something to do with the shock front?)

    c. Lastly, why does the radiation confine itself to the 'blue' region?

    While simple verbal explanations will suffice, I really appreciate any maths you can inject in explaining this!

    Thanks a lot.
     
  2. jcsd
  3. Jan 24, 2012 #2

    Andy Resnick

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    Cerenkov radiation can be produced by the movement of bare charges through a dielectric, but also the movement of dipoles as well- the essential consideration is the speed of the particle through the medium. It is the moving particle that emits Cerenkov radiation, not the induced polarization of the medium (AFAIK).

    The radiation is emitted in a cone, similar to that of a shock wave. The cone angle θ is given by cos(θ) = 1/βn, where β= v/c. It's important that the medium is dispersive n = n(ω). The duration of the flash is given by Δt = r/v (tanθ_2 - tanθ_1), where the two angles represent the two cone angles corresponding to the detector limits (e.g. 400nm to 700nm), and 'r' is the distance from the axis of the moving particle.

    I could not find any derivation of a spectrum, but the radiation output is given in the classical theory as dW/dl = e^2/c^2 ∫(1-1/β^2n^2) ωdω. The spectrum depends on the material refractive index- n(ω)>1/β for emission.

    I can write more later... does this help?
     
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