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Imagine a medium in which an electron is moving at speeds faster than the speed of light in the medium. Within the same medium an observer moves at the same speed with the electron. Will the observer detect Cherenkov radiation from the electron in his frame? I think the answer is no because the electron is at rest with respect to the observer. However, the electron is still moving with respect to the medium (or rather the medium is rushing past the electron/observer in their frames) so there must be Cherenkov radiation even in the frame of the observer. I am unable to make up my mind regarding this. Can someone suggest a resolution?

Additional query: What if the electron was at rest and the observer moved faster than the speed of light in the medium? Would the observer detect Cherenkov radiation from the electron in his frame?

Qmechanic
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Dr. user44690
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3 Answers3

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There is only one reality. Different frames of reference are just different coordinate systems that measure that reality. That is, if an event (like an electron emitting light) happens in one inertial frame, it happens in all inertial frames. The frames may differ as to the location and time of the event (they give it different coordinates), or even as to the cause of the event (the electron moved through the medium, or the medium moved past the electron) but they all have to agree that the event occured.

So if an electron emits Cherenkov radiation in one inertial frame, it does so in all inertial frames. The frequency of the radiation, or its direction, may be a matter of dispute, but not the existence of the radiation.

Dale
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Eric Smith
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So you have an electron moving at $\beta =\frac v c > \frac 1 n$, which emits light into a cone with angle:

$$ \cos\theta = \frac 1 {n\beta} $$

I can describe the 4-velocity of a photon in the lab ($S$) as:

$$ q_{\mu} = (\gamma c, \gamma v \sin\theta, 0, \gamma v \cos\theta) $$

which cab be transformed into the electron frame ($S'$, with $c=1$):

$$q'_0 = \gamma^2(1-\frac {\beta} n)$$

with longitudinal component:

$$q'_3 = \gamma^2 (\frac 1 n - \beta) $$

and transverse component:

$$ q'_1 = \gamma\sqrt{\beta^2-\frac 1 {n^2}}$$

though you may want to check I did that right (edit: it was wrong, I fixed it). You can also convert it into a 3 velocity with:

$$ q'_{\mu} = (\gamma', \gamma' \vec \beta') $$

or

$$ \beta' = 1-\frac{n(1-\beta^2)}{n-\beta} $$

BTW: Phase,

$$ \phi = \vec k\cdot\vec x - \omega t = k^{\mu}x_{\mu}$$

is a Lorentz invariant, so you can always transform nodes in your wave, if needed.

JEB
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The electron may be at rest in your frame, but the medium with all of its charged constituents is not. They will deflect in the electrostatic field of the electron.

This is why you will see the charges around your electron radiating and not the electron itself.

fraxinus
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