Frequency of EM waves produced by linearly accelerating charges

[cg0303]

I was wondering about EM waves produced by linearly accelerating charges, as opposed to oscillating charges.

With oscillating charges, the frequency of the wave depends on the frequency of the oscillation of the charge. But what determines the frequency of the wave produced by a linearly accelerating charge? How do they relate: i.e. does greater acceleration mean a higher frequency? And can anyone please direct me towards any materials that discuss this issue in detail?

 

[Vanadium 50]

But what determines the frequency of the wave produced by a linearly accelerating charge?

There is a range of frequencies.

 

[berkeman]

I was wondering about EM waves produced by linearly accelerating charges, as opposed to oscillating charges.

https://en.wikipedia.org/wiki/Cherenkov_radiation

 

[cg0303]

There is a range of frequencies.

And what determines these?

 

[Meir Achuz]

The frequency spread does not depend on the acceleration. It depends on the length, L, of the pulse of radiation caused by the acceleration. Fourier transforming the pulse leads to the distribution of frequencies
$$\left|F(\omega)\right|^2 =\left[\frac{\sin(\omega L/2c)}{\pi L\omega}\right]^2.$$

 

[Vanadium 50]

And what determines these?

The range covers the full spectrum (no pun intended) of allowed energies.

 

[sophiecentaur]

This made me think of Bremsstrahlung which a are X Rays generated as a fast electron comes to a halt. This figure is from the Hyperphysics link.

Interesting when you think of the quantum situation. I'm sure someone can help me out but those spectra are continuous and are generated by loads of electrons, slowing down. What about the photon that's generated by just one braking electron? Where does the left over Energy go? From the wide frequency range, perhaps there are multiple photons, as the electron slows down in steps from, say 50keV to zero.

 

[cg0303]

It depends on the length, L, of the pulse of radiation caused by the acceleration.

What exactly do you mean by that?

 

[Meir Achuz]

L appears in the equation. a does not.

 

[Meir Achuz]

"What about the photon that's generated by just one braking electron?"
In x-ray bremsstrahlung, many photons are emitted by a single electron, with the energy distribution you show.
This is a classical wave. For high energy bremsstrahlung like p+p->p+p+gamma at ~100MeV,
a single photon is produced in a QED process.

 

[cg0303]

L appears in the equation. a does not.

I meant more what you meant by the term 'length of the pulse'. Is that the length of time that the charge is accelerating?

 

[shjacks45]

https://en.wikipedia.org/wiki/Cherenkov_radiation

Accelerating charges.

 

[shjacks45]

Thinking of old style TV picture tubes where electrons are accelerated from 0 to up to 70KV. No evidence of radiation until it hit the front tube face, emitting x-rays when it hits the front face (~2 mA). No radiation from acceleration in other high voltage radar and TV tubes. Maybe synchrotron radiation from beam in magnetic field?

 

[vanhees71]

"What about the photon that's generated by just one braking electron?"
In x-ray bremsstrahlung, many photons are emitted by a single electron, with the energy distribution you show.
This is a classical wave. For high energy bremsstrahlung like p+p->p+p+gamma at ~100MeV,
a single photon is produced in a QED process.

An "X ray" as any other em. wave is not a stream of "many photons" but a coherent state, if you insist in using QED instead of classical electrodynamics to explain em. waves.

Though shalt not use (wrong) images! Particularly there should no appearance of the word "photon" in a classical-electromagnetism textbook or lecture.

 

[sophiecentaur]

a coherent state

Dodgy ground, I know but what is to say that the photons due to one collision will be in any way related to the photons from a next door collision? As with most other 'optics' we do our calculations based on coherence but it doesn't follow that all EM waves are coherent. Yes, for stable CW transmissions and from lasers but, even in those cases there is a bandwidth involved.
But I think it's a valid question as to what would be detected for a single electron impact with a target. I guess the answer is that a single photon detector would probably miss any more than one of the emerging photons.
But I heartily agree that bringing photons in at every opportunity is not the way forward and doesn't actually help in real understanding. But it's a culture, forced on us by our over trivialised Science education. We start with little bullets and most of us would rather have that than even the simplest wave explanation.

 

[Vanadium 50]

!

We have someone who is confused about classical EM. They posted in the classical EM section. And what do we do? Drag in photons. Is the problem any easier to solve with photons? Nope - but we drag them in anyway.

Because everything's better with photons.

 

[sophiecentaur]

Yes, I agree but . . . . without photons we need to explain, in a pictorial way, how the pulse period and the Energy get translated into a burst of frequencies. That involves some pretty sophisticated explanation which involves the Fourier transform of an isolated pulse and how that implies an infinite number of frequency components over a finite frequency range. I can see that as a big hill to climb before someone new to the topic can get closure on it.

The only alternative is (and it's tempting at times) to say that it's too hard so don't bother. I guess a half way house is to come back with that when you get unreasonable supplementary questions when the simple photon ideas fail.

Another problem is that a search about bremsstrahlung only seems to yield photon explanations. Perhaps it's better to talk in terms of Cyclotron radiation which can be easier to describe classically because the radiated spectrum has a main frequency component at the frequency of the orbit; it's a sort of antenna perhaps?

 

[Meir Achuz]

"Is that the length of time that the charge is accelerating?"
'Length' means length. If I meant 'length of time', I would have written 'length of time'.
Look at the units in the equation.

 

[Meir Achuz]

"is not a stream of "many photons" but a coherent state,"
Do you mean a coherent state many photons?
I didn't bring up photons, but they are coming out of your lamp right now.

 

[vanhees71]

Out of my lamp comes classical light. A single photon source is too expensive for me.

 

[berkeman]

Accelerating charges.

Deceleration is just acceleration with a minus sign...