# Why do accelerated charges emits a photon?

1. Sep 5, 2007

### Fisix

Is it only when the acceleration is negative? If yes, when it is positive it absorbs a photon?

2. Sep 5, 2007

### Demystifier

Perhaps you should first understand why classical (not quantum) charge emits electromagnetic radiation.

3. Sep 5, 2007

### ZapperZ

Staff Emeritus
This thread has been moved from QM sub-forum to the classical physics sub-forum. I agree that the OP needs to first understand this from the classical E&M perspective.

Zz.

4. Sep 10, 2007

### Andrew Mason

This is an important question and, remarkably, one that has no definitive answer as far as I can tell.

A charged particle that is accelerated by gravity does not emit an em wave/photon. The only other way to accelerate a charged particle is to apply a mechanical or electro-magnetic force to it over some distance. And a mechanical force is nothing but an em force. Thus, the observational evidence amounts to:
a charged particle does not emit an electro-magnetic wave or photon except when it receives electro-magnetic energy. ​

So this leads to a reformulation of the question: does the charged particle emit a photon because it accelerates? Or does it emit a photon in response to receiving a photon, which incidentally causes it to accelerate?

The question seems to be still unresolved: See this site, for example.

AM

5. Sep 10, 2007

### olgranpappy

This very cute problem is addressed to some extent by Peierls in his little book "Surprises in Theoretical Physics".

6. Sep 10, 2007

### lugita15

What is your basis for that assertion?

7. Sep 10, 2007

### olgranpappy

Apparently, the principle of equivalence...

Consider a man standing on a scale on earth. The scale reads, say, 170 pounds.

Consider a man in completely empty space in a rocketship accelerating at 9.8m/s^2. The man is standing on a scale. The scale reads 170 pounds.

Now, consider the exact same situation but now with a point charge stapled to the man's nose.

If the principle of equivalence holds then... well, shouldn't the stationary point charge in a gravitational field be radiating? Regardless of whether or not it "should", it does not.

8. Sep 11, 2007

### Andrew Mason

Further to what olegranpappy has said, according to the principle of equivalence, a charged particle accelerating in freefall in a gravitational field is locally equivalent to a charged particle at rest in an inertial frame of reference. Since a charged particle at rest in an inertial frame does not radiate, the same must hold true for the charged particle in freefall. This is consistent with all observation.

A charged particle that is stationary in a gravitional field is locally equivalent to a charged particle accelerating in a gravity-free space. So it should radiate. This is a bit of a problem, because as olegranpappy points out, it is not observed. Here is a paper that explains why this occurs - it suggests that the radiation is a relative phenomenon that depends on the relative acceleration between the observer and the charge:

AM

Last edited: Sep 11, 2007
9. Sep 11, 2007

### lugita15

But I thought that principle of equivalence contradicts special relativity, and by consequence, classical electromagnetic theory as well.

10. Sep 11, 2007

### olgranpappy

Hopefully not too much, since special relativity and classical electromagnetic theory are correct.

11. Sep 11, 2007

### Andrew Mason

The General Theory and the Special Theory of Relativity are perfectly consistent. The principle of equivalence is the cornerstone of GR. Maxwell's equations are consistent with both - certainly with SR. In fact, the belief that Maxwell's equations are valid in all inertial frames of reference was one of the things that drove Einstein to develop SR.

AM

12. Sep 11, 2007

### Andrew Mason

I should have added after: "So it should radiate"... "IF accelerating charges radiate because they are accelerating.

AM

13. Sep 12, 2007

### olgranpappy

Oh, I finally found that book I was talking about earlier. Here's some highlights:

Although the formula for the rate at which a small object loses energy via radiation is
$$\frac{dW}{dt}=\frac{2e^2}{3c^3}a^2$$
where a is acceleration. And this gives the right answer for total loss. It can not be the right expression for instantaneous rate of loss of energy.

blah blah blah. For periodic or other types of motion we can use instead the expression
$$\frac{dW}{dt}=\frac{-2e^2}{3c^2}v\dot a$$
where v is velocity and $$\dot a$$ is "jerk".

Unfortunately, further thought shows that this rewriting does not resolve the paradox... actually, it gets a bit complicated. A careful discussion was given by Rohrlich and Fulton in Ann. Phys. 9, 499 (1960). But the paradox was not truly resolved until arguments by D. Boulware came along (unpublished as of 1979, but given by Peierls in his book).

14. Sep 12, 2007

### idea2000

Hi,

I was reading this thread and I have an additional question to ask. If an electron and a proton were accelerating at the same rate, would they emit the same photon?

Last edited: Sep 12, 2007
15. Sep 13, 2007

### Andrew Mason

The force on, or rate of momentum change of, the proton would be 1800 times greater than that for the electron. This means that the electromagnetic interaction needed to cause this (ie. the momentum of the photon) would have to be 1800 times greater.

AM

16. Sep 13, 2007

### idea2000

So, it would take really high frequencies of light to accelerate a proton? And would the proton re-emit this frequency of light as well?

A related question:
If we were to accelerate a proton using a very weak magnetic field, would it emit radiation?

17. Sep 13, 2007

### cesiumfrog

You're saying that if an apparatus stands stationary to the Earth, it can not detect radiation from co-supported charges (this much seems sensible, else where would the energy come from: no other part of the system has given up change). But just to clarify:

If a charge is released to fall freely past the apparatus (and vice-versa), is radiation then detected? If the apparatus is placed on (say) an accelerating train, will it continue not to detect radiation from co-accelerated charges?

Actually, my understanding was that the field lines of a constantly accelerated charge merely "droop". This isn't radiation if you can stay stationary with respect to charge and the direction of acceleration. I think the equivalence principle here is unscathed (until you add boundary conditions, but that feels like cheating).

Last edited: Sep 13, 2007
18. Sep 13, 2007

### olgranpappy

classically you would use the same formula to determine the power radiated, there's no reference to mass in that formula.

19. Sep 14, 2007

### Andrew Mason

There is a difference between a charge in free fall in a gravitational field and a charge accelerating due to a non-gravitational force (ie. an electromagnetic force). In the first case it does not radiate. In the second, it does.

I am not sure I understand what you mean here.

AM

20. Sep 14, 2007

### cesiumfrog

Do you have evidence of that? How do you define "radiate"?
I say that to a classical detector-apparatus, "radiation" means any time-varying field at the apparatus.

Now, if an apparatus inside an elevator is stationary with respect to some charge, regardless of whether the elevator is "standing in a planet's gravitational field" or "accelerating constantly in space" the apparatus will detect no radiation (because in both cases the field at the apparatus is time-independent). Hence I disagree with your statement that there is some problem with the equivalence principle.

(As for the rough shape of the field inside the elevator, it happens to droop -- in both cases as per the equivalence principle -- as illustrated here. Actually if you're interested in the exact shape, you should be able to distinguish the two cases due to the differing global geometry.)