Clarification regarding Cherenkov radiation

In summary, Cerenkov radiation is produced by the movement of charged particles through a dielectric, and it has a characteristic cone-shaped emission spectrum.
  • #1
vhbelvadi
67
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 traveling 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.
 
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  • #2
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?
 

1. What is Cherenkov radiation?

Cherenkov radiation is a type of electromagnetic radiation that is emitted when a charged particle, such as an electron, passes through a medium at a speed greater than the speed of light in that medium.

2. How is Cherenkov radiation produced?

Cherenkov radiation is produced when a charged particle travels through a medium, such as water or air, at a speed greater than the speed of light in that medium. This causes the medium to become polarized, resulting in the emission of photons as the particle moves through the medium.

3. What is the significance of Cherenkov radiation?

Cherenkov radiation has been used in various scientific fields, such as particle physics and nuclear engineering, to detect and measure high-energy particles and radiation. It is also used in medical imaging techniques, such as positron emission tomography (PET) scans.

4. Is Cherenkov radiation harmful to humans?

Cherenkov radiation poses no direct harm to humans as it is a form of non-ionizing radiation. However, it can be an indication of the presence of high-energy particles or radiation, which can be harmful to human health.

5. Can Cherenkov radiation be seen by the naked eye?

Yes, Cherenkov radiation can be seen by the naked eye under certain conditions. When a high-energy particle passes through a medium, it emits a characteristic blue glow that can be seen in total darkness. However, this is only possible in specific environments, such as underwater or in nuclear reactors, and it may be difficult to observe without specialized equipment.

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