# How does Cherenkov radiation work?

1. Jun 5, 2006

### raul_l

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. Jun 5, 2006

### Morbius

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

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

3. Jun 6, 2006

### raul_l

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)

4. Jun 6, 2006

### raul_l

... 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?

5. Jun 6, 2006

### eep

are the runners travelling in the same direction that the truck is moving?

6. Jun 6, 2006

### Morbius

Yes.
Dr. Gregory Greenman
Physicist

7. Jun 6, 2006

### Morbius

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

8. Jun 6, 2006

### raul_l

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).

9. Jun 6, 2006

### Morbius

raul,

From CERN - the European nuclear laboratory:

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

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

Dr. Gregory Greenman
Physicist

Last edited: Jun 6, 2006