# Accelerated charged particles

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1. Jun 3, 2012

### LitleBang

Accelerated charged particles radiate photons. If I am accelerating along with the particle will I see it radiate?

2. Jun 3, 2012

### Jano L.

Great question! I do not know how to treat this question in quantum theory, but it is good to think of this first in classical theory.

In classical theory, accelerated charged particle does not radiate photons. It radiates electromagnetic waves of E, B, which will manifest themselves by altering the motion of the surrounding particles.

If we were moving along the accelerated particle, we would be in a non-inertial frame. Here the concept of E, B fields is not so clear - if we define them by Lorentz formula, they do not obey Maxwell's equations but some modification due to acceleration. However, some influence on the surrounding particles would be observable as well. So in the accelerated frame, I think we can say there is some wave propagation too, it just is not the same as in inertial frame.

If we imagined that the charged accelerated particle instantaneously released another particles, it seems that the particle would have to have variable mass that jumps as the photon is released.

3. Jun 3, 2012

### LitleBang

In find it difficult to believe I would see radiation because the charged particle would have an associated mass gain that I would not detect. For it to radiate I would have to see a mass loss.

Another interesting question, how does Bremsstrahlung radiation produce both a positive and negative wave? What interaction does an accelerated charged particle have with space that produces them?

Last edited: Jun 3, 2012
4. Jun 3, 2012

### Bob S

Great question. In looking at the classical radiation of a low velocity (non-relativistic) accelerating charged particle in a vacuum, the power radiated is given by
$$-\frac{dW}{dt'}=\frac{e^2a^2}{6\pi \epsilon_o c^3 }$$ where a is the acceleration and c is the velocity of light. The radiated power is at the time of emission (t'), and not retarded time. See Panofsky and Phillips Classical Electricity and Magnetism eqn (19-19). Since the velocity is not a parameter, it is independent of velocity. So you would see radiation in any inertial frame. Jackson Classical Electrodynamics gives the same result in Gaussian units. Could you be referring to the Unruh radiation? See http://en.wikipedia.org/wiki/Unruh_effect

5. Jun 3, 2012

### DrZoidberg

The equivalence principle says that there is no way to distinguish between gravity and acceleration without an external reference.
I don't think that gravity will cause a charge to emit photons since that would mean you could produce an infinte amount of energy.
So I think a charge won't emit a photon in the accelerated frame of reference.

6. Jun 3, 2012

### LitleBang

Bob I don't think your response really applies here. I'll restate in a different way. If I am a stationary observer with respect to an accelerated electron, that electron will gain mass and radiate. If I am being accelerated at exactly the same rate as the electron it will not gain mass nor will it radiate with respect to me.

7. Jun 3, 2012

### LitleBang

Could someone address the question of Bremsstrauhlung radiation as posed in post #3? It was the source of the OP.

8. Jun 4, 2012

### Jano L.

It is good to look at the things first from the point of view of inertial frame.

The word mass usually refers to either invariant or relativistic mass in this inertial frame.

There if the particle is steadily accelerating, relativity implies increase of energy and hence of relativistic mass. No loss. The loss can happen only if the particle releases another particle. If this happens in inertial frame, it happens in the accelerated frame as well.

If we do not believe in photons and have just EM field, loss of rest mass is implausible. The electrons were never observed with lesser mass than standard m = 9.1E-31 kg.

9. Jun 4, 2012

### the_emi_guy

Regarding radiation from *linearly* accelerated charge, DrZoidberg's input is significant.
As far as I know, this is still an open question and gets a fair amount of attention.
See:
Radiation from a Uniformly Accelerated Charge
Stephen Parrott
University of Massachusetts at Boston

He argues that the strong equivalence principal is violated.
Others argue the opposite (see his references).

10. Jun 4, 2012

### LitleBang

I hope there is no disagreement that any particle or mass accelerated WRT an observer gains mass according to relativity or that an accelerated charged particle or any mass emits radiation no matter how small the acceleration. That said there is no mechanical explanation for Bremsstruhlung radiation. In the case of two balls rolling toward each other we can predict both direction and velocity of the collision of the two because of the mechanical aspect of the event. In the case of Bremstrauhlung we can't explain the nature of any interaction with space that causes it. If someone is aware of any such please point me to it. Is it a thorn in the side of physics?

11. Jun 4, 2012

### Bob S

Bremsstrahlung results from the collision and deflection of an incident charged particle (electron, muon) with a heavy charged nucleus. The "acceleration" (actually deflection) results in the emission of quanta (e.g., continuous x-ray spectrum). The recoil of the heavy nucleus takes away mainly momentum and not energy from the incident particle. The energy loss of the incident particle is nearly the same as the sum energy of the emitted quanta (bremsstrahlung). Bremsstrahlung cannot occur without the collision of the incident charged particle with another charged particle. This is different than synchrotron radiation, which is due to the deflection of a fast charged particle in a strong magnetic field.

[added]The theory of bremsstrahlung is well known, and is presented in many books on nuclear physics and high energy physics. The best thorough review is in Heitler's 1954 book "The Quantum Theory of Radiation".

Last edited: Jun 4, 2012
12. Jun 4, 2012

### LitleBang

How the acceleration occurs is totally irrelevant to the argument at hand. Do you fell you have answered my question? I hope not. Quoting unrelated information from a book is not an answer.

13. Jun 4, 2012

### jartsa

Let's say a charged particle flies through a pipe, bouncing on the walls, like in the first picture here: http://en.wikipedia.org/wiki/Triangle_wave

The emitted wave should have the same triangular shape as the particle path.

Here are two alternative descriptions what is happening:

1: Bremsstrahlung is emitted during each collision.

2: Many continuous sine waves are radiated, which when added together form a triangle wave, like in the animation at the same wikipedia page.

Surprisingly 1 and 2 are just altarnative ways to say the same thing.

Conclusion: the particle was vibrating all the time, and emitting a wave that mirrored the vibration of the particle. So we have here perfectly normal generation of EM-wave.

Last edited: Jun 4, 2012
14. Jun 4, 2012

### LitleBang

I do not disagree with anything you say but it still does explain the why of Bremsstrauhlung radiation. An electron accelerated toward an anode of an electron gun produces radiation exactly like Bremsstrauhlung radiation. Why does accelerating a charged particle through a total vacuum produce that radiation.

Last edited: Jun 4, 2012
15. Jun 4, 2012

### jfy4

I was under the impression that the accelerating charge radiating formulae were derived under the assumption of an inertial frame. I think we should be careful when we are non-inertial.

16. Jun 4, 2012

### jartsa

It's possible, but unlikely, that I'm wrong.

I derived this accelerating charge "formula": The wave waves like the charge vibrates.

Is it wrong? Is Bremsstrahlung really different from radio waves?

17. Jun 4, 2012

### jartsa

I would say that the electron vibrates, and therefore radiates! It's a longitudinal vibration, and Fourier transform gives the frequencies.

In this case it seems to be not so clear what should be Fourier transformed. But that's because constant acceleration is somewhat unphysical. So we Fourier transform electrons path in a circuit that contains a vacuum tube.

By the way I added a final conclusion to the post number 13.

Last edited: Jun 4, 2012
18. Jun 4, 2012

### LitleBang

Jartsa, all radiation is the same and all radiation comes from the same thing. The only time a particle does not radiate is at absolute zero. The only reason I bring up Bremsstrauhlung radiation is because it is a certan method of producing photons. Larmor Radiation, ( http://en.wikipedia.org/wiki/Larmor_radiation ) basically explains the production of all radiation. The only reason for this thread is to search for an understanding of not how the photon is produced but the interaction of space with a charged accelerated particle that creates them. If we don't figure that out we'll never understand the Universe.

19. Jun 4, 2012

### jfy4

I was just imagining that if you were a detector, and you were accelerated towards an electron, that if you looked at your references for formulae to use in that case, the commons one's weren't built for that situation. I expect the formula for accelerating charges in non-inertial frames must be modified. I bet you could go through $\dot{u}^\alpha=\frac{q}{m}F^{\alpha\beta}u_{\beta}$ and put in a non-inertial metric and see what comes out. But it's just my intuition... :shy:

20. Jun 4, 2012

### cosmik debris

I think that both these statements are debatable. These days mass is an invariant and relativistic mass is not used as it is misleading. Some will disagree of course. There is also some debate as to whether a uniformly accelerated charge radiates, and someone has provided a reference for this. The operative word here is "uniformly".

A similar situation is a charged particle sitting on the kitchen table. It is accelerating because it is not following a geodesic in spacetime. Does it radiate? In the frame where the charged particle is at rest it does not, or does it?