Does an electron moving along a geodesic radiate?

1. Feb 22, 2010

jnorman

layperson here, so please correct any misconceptions i have on this.

an electron will emit photons if it is accelerated (including changes to either velocity and/or direction of travel).
acceleration occurs if the electron experiences a force.

since GR indicates that gravity is not a force, but rather the shape of spacetime, the electron does not experience a force, per se, in its own frame as it travels through a region of spacetime warped by the presence of mass, since it only traveling along a spacetime geodesic (which it interprets as a "straight" line), even though it appears to an outside observer that the electron has experienced a force because its direction of travel appears to be changing.

so, does an electron moving through a gravitational field radiate in the same way it would moving through a magnetic field? if so, does that have any implications on whetehr gravity is a "force" or not?

2. Feb 22, 2010

PhilDSP

Your first statement may be more or less true macroscopically when a charge is present but a lot goes on on the atomic or smaller level that in effect brings Quantum Mechanics into the picture to consider details involving the near field and far field radiation and the de Broglie wave associated with a particle. Even if a particle accelerates its near field radiation can be recaptured resulting in no release of far field radiation.

There are at least several configurations of electron accelerated motion that are known not to radiate based on analysis using the Maxwell Equations. Drs. Herman Haus, Geodecke, Abbott and at least one other have studied this and presented papers with their findings.

3. Feb 22, 2010

bapowell

This is a great question! I don't have an answer, but I'd like to add a couple thoughts. First, my feeling is that the gravitationally accelerated electron would radiate. Of course, that would seem to contradict the equivalence principle, because it would imply that an electron at rest in the absence of gravity should be radiating. However, I think the catch is that the equivalence principle is not applicable to the free falling electron. The equivalence principle only holds locally -- the free falling electron lives in an inertial frame only as the size of this frame approaches zero. But it seems to me that a frame containing an electron with its electrostatic field cannot be made arbitrarily small, and so even a freely falling electron "samples" the surrounding space with its electric field. Due to the non-uniformity of earth's gravitational field, points in the surrounding space will live in different inertial frames, and therefore the equivalence principle will not be applicable here.

If that reasoning is true, than an electron in free fall in a uniform gravitational field should not radiate, because in this case the inertial frame can be made arbitrarily large.

4. Feb 22, 2010

Naty1

5. Feb 23, 2010

vwilmot

Question : Does an electron moving through a gravitational field radiate ?

Answers to date are all bits of theories - what has happened to experiment in physics ?!

6. Feb 23, 2010

PhilDSP

One of the problems with experimental verification is that, since the universe it not devoid of matter, all massive particles everywhere are continuously being accelerated by other massive particles as they move in relation to each other. So if it were true that charged particles always radiated due to acceleration then by now, most charged particles will have radiated away their own existence.

7. Feb 23, 2010

vwilmot

But surely charged particles can absorb photons as well as radiate photons ?
However granted it may be difficult to show if some photon radiation is due to gravity rather than other causes ?

8. Feb 23, 2010

Staff: Mentor

Accelerated observers can detect radiation that unaccelerated observers do not. So presence or absence of radiation is not frame invariant.

9. Feb 23, 2010

jnorman

whoa - is this really true? either photons are emitted or not, right? how can one observer see photons that another observer cannot? confusing...

10. Feb 23, 2010

Jonathan Scott

As a rough analogy, consider how you would detect that electromagnetic radiation. You would detect it by the way in which it shakes a charge. If there is relative acceleration between the source and the test charge, then you can either interpret the shaking as being due to radiation from the motion of the source or due to the accelerated motion of the charge within a static field.

The "photon" model can be confusing and misleading. We often get an impression that all electromagnetic interactions involve charges firing photons at each other. There are obviously cases where real photons clearly get fired off, for example during transitions between atomic energy levels. However, the sort of interactions in this case would be better described as local interactions between charges and fields, which may be described in terms of virtual photons, and this description works from either point of view.

11. Feb 23, 2010

elect_eng

This is a great question, and personally I've never heard it posed before. Ordinarily you'd expect the radiation of energy (photons) for an accellerating charge. The case of a parabolic freefall, or accelleration in a linear direction is less clear, but it's clear that a circular motion should radiate photons with a frequency matched to the orbital frequency. At least, this is clear if gravity is not the cause of motion. However, if gravity is the cause of motion, the answer is less clear. Based on intuition, it seems that radiation should still occur as seen by a distant observer. However, if this is true, the orbiting object would then start spiraling in to the source of the gravity. This would mean that the object is no longer in a free-fall type of orbit (i.e. no longer on a geodesic), but has an external drag force on it. Very confusing, indeed.

The question can be restated by asking whether an electrically charged moon, orbiting a planet will radiate, and then spiral in to the planet. I guess one would need to solve Einstein's equations to find out for sure. I hope a GR expert here can answer this with a clear logical proof based on principles.

Last edited: Feb 23, 2010
12. Feb 23, 2010

Fredrik

Staff Emeritus
It's called the Unruh effect. And yes, it is confusing. I haven't seen the math, but I've been told that one way to think of this radiation (at least if the observer is doing constant proper acceleration forever) is as Hawking radiation from the Rindler horizon (the region of spacetime behind the accelerating observer from which light can't reach him).

I think it also has to do with the definition of particles in relativistic QM. (Chapter 2 of Weinberg's QFT book). It's very mathematical, so I won't repeat it here. I'll just say that the definition relies heavily on Lorentz transformations, and my point is that maybe this means that "particle" isn't the appropriate concept to use when we try to describe things using non-inertial coordinate systems.

It could also be another indication that it's better to think of QM not as a description of the real world, but as just a set of rules that tells us how to calculate the probabilities of possible results of experiments.

13. Feb 23, 2010

Staff: Mentor

Yes, this is my understanding also.

14. Feb 23, 2010

jfy4

***i still havent figured out how to do latex for these posts? can someone tell me how to type math in this forum please.

acceleration four vector is the covariant derivative of the proper four velocity, which equals zero for the geodesic equation. or, a geodesic is a curve whos tangent vector u obeys that thing i wrote before. so i guess there is no acceleration, and no radiation then, right?

15. Feb 23, 2010

bapowell

I'd like to add a bit to Fredrik's explanation. In order for two observers to agree on the vacuum (the no particle state), they must both be able to decompose the field into positive and negative frequency modes (the usual Fourier decomposition of the field in flat space). This decomposition is possible because in flat space (ie inertial observers), the eigenfrequencies, are, well, eigenvalues of the time translation operator -- time translation is uniquely defined for all observers (up to Lorentz transformations). Then, the vacuum (the no particle state) is defined as "no positive frequency excitations". However, once we go to curved spacetime, or accelerated observers, the mode frequencies become time dependent. There is no longer a unique time variable by which to define eigenfrequencies, and the decomposition into strictly positive/negative frequency modes is impossible. Instead each mode is really a mixture of positive/negative frequencies. Therefore, the vacuum of the inertial observer (which corresponded to no positive frequencies excitations) becomes a state with a mixture of positive/negative frequencies -- a state that the observer sees filled with particles.

It is a bit prohibitive to work out the math here, but it really is an elegant result. You ultimately show that the accelerated observer sees a rotated version of the inertial observer's state space, with a correspondingly different vacuum. I have references if anyone is interested...

16. Feb 23, 2010

bapowell

Just use the tag tex, enclosed with [].

17. Feb 23, 2010

jfy4

[TEX]\nabla_{u} u=0[/TEX]

if this works that should say it.

18. Feb 23, 2010

Fredrik

Staff Emeritus
You can use "noparse" tags when you want to say that [noparse]$$\nabla_u u=0$$[/noparse] produces the output $$\nabla_u u=0$$. Hm, it looks like writing TEX in uppercase letters doesn't work. (A good tip is to always preview).

To see what I did here, click "quote".

19. Feb 23, 2010

jfy4

thank you thank you thank you!

20. Feb 23, 2010

AEM

I'm interested in the references. Please post a couple of them.