Does an electron moving along a geodesic radiate?

[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?

 

[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.

 

[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.

 

[Naty1]

try here
https://www.physicsforums.com/showthread.php?t=65767&highlight=electrion+radiate

and here
https://www.physicsforums.com/showthread.php?t=370495&highlight=electron+radiate

 

[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 ?!

 

[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.

 

[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 ?

 

[Dale]

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

 

[jnorman]

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

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

 

[Jonathan Scott]

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

The basic answer is that even radiation is relative (loosely speaking).

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.

 

[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 into 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 into 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.

 

[Fredrik]

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

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.

 

[Dale]

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.

Yes, this is my understanding also.

 

[jfy4]

***i still haven't 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?

 

[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...

 

[bapowell]

***i still haven't figured out how to do latex for these posts? can someone tell me how to type math in this forum please.

Just use the tag tex, enclosed with [].

 

[jfy4]

***i still haven't 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?

Just use the tag tex, enclosed with [].

$$\nabla_{u} u=0$$

if this works that should say it.

 

[Fredrik]

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".

 

[jfy4]

thank you thank you thank you!

 

[AEM]

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...

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

 

[bapowell]

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

A good review that includes the full derivation is http://arxiv.org/abs/0710.5373" . The early paper by Davies (P.C.W. Davies, J. Phys. A8, 609 (1975)) is short and easy, and lays out the basic idea as well. Unruh's paper (for whom the effect is named), W.G. Unruh, Phys. Rev. D14, 870 (1976) is a bit longer and more general, and goes into substantial detail.

Also, the book "Quantum Fields in Curved Space" by Birrell and Davies discusses the effect in great clarity, and is a great reference in general.

 

[George Jones]

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

Some additions to the excellent references given by bapowell.

Chapter 9, Quantum Field Theory in Curved Spacetime, in Sean Carroll's GR text Spacetime and Geometry: An Introduction to General Relativity has a good treatment of the Unruh effect.

Another good treatment is Chapter 8, Unruh effect, from Mukhanov and Winitzki's text Introduction to Quantum effects in Gravity. I think this is the only text devoted to semi-classical quantum gravity pitched at the level of senior undergrads/beginning grads.

 

[actorres]

The solution to this problem can be found in the following paper:

C. Almeida and A. Saa, Am. J. Phys. 74, 154 (2006)
"The Radiation of a uniformly accelerated charge is beyond the horizon: A Simple derivation"
Free access: http://arxiv.org/PS_cache/physics/pdf/0506/0506049v5.pdf

In short, the answer is that inertial observers will surely detect radiation emitted by free-falling charges, while comoving observers will see only a static electric field. The "paradox" is solved by noting that comoving observers have no access to the radiation field of a uniformly accelerated charge (a horizon emerges).

The paper is pedagogical, easy to read and very instructive.

 

[closet mathemetician]

Here's something I just thought of: the photon is the boson that mediates the electromagnetic force. When electrons radiate, they are gaining and losing photons. The graviton (theoretically) is the boson which mediates the gravitational force. So wouldn't gravitational acceleration be the action of gravitons upon an electron? Well, electrons don't absorb and emit gravitons, hence, no radiation from gravitational acceleration.

 

[bapowell]

Well, electrons don't absorb and emit gravitons, hence, no radiation from gravitational acceleration.

Well, the discussion wasn't about gravitational radiation (gravity waves) but whether or not an electron radiates electromagnetically.

 

[closet mathemetician]

The question was whether an electron radiates electromagnetically due to acceleration by gravity. How does an electron radiate electromagnetically? By absorbing and emitting photons. But acceleration due to gravity would involve an interaction with gravitons, not photons. So if a graviton was interacting with an electron, one wouldn't expect electromagnetic radiation.

 

[bapowell]

An electron doesn't need to be 'seeded' by electromagnetic radiation to radiate. Classically, an accelerated electron from a non-constant force radiates electromagnetic. The specific source of the acceleration is unimportant. For example, if I could tether an electron to string and swing it around in a circle, it would radiate electromagnetically.

 

[bcrowell]

Here's something I just thought of: the photon is the boson that mediates the electromagnetic force. When electrons radiate, they are gaining and losing photons. The graviton (theoretically) is the boson which mediates the gravitational force. So wouldn't gravitational acceleration be the action of gravitons upon an electron? Well, electrons don't absorb and emit gravitons, hence, no radiation from gravitational acceleration.

We don't have a theory of quantum gravity, so there's not really much point in bringing gravitons into the discussion. The question is purely classical.

However, if yo want to view it in terms of a hypothetical quantum theory that includes both gravity and electromagnetism, then electrons certainly would emit and absorb gravitons; every object with mass would do so, because gravitational interactions would be described in such a theory as an exchange of gravitons.

 

[256bits]

The solution to this problem can be found in the following paper:

C. Almeida and A. Saa, Am. J. Phys. 74, 154 (2006)
"The Radiation of a uniformly accelerated charge is beyond the horizon: A Simple derivation"
Free access: http://arxiv.org/PS_cache/physics/pdf/0506/0506049v5.pdf

In short, the answer is that inertial observers will surely detect radiation emitted by free-falling charges, while comoving observers will see only a static electric field. The "paradox" is solved by noting that comoving observers have no access to the radiation field of a uniformly accelerated charge (a horizon emerges).

The paper is pedagogical, easy to read and very instructive.

Isn't the falling electron and comoving observer both in an inertial field according to Einstein's falling elevator thought experiment?
And the inertial observer is now standing on say Earth in a gravitational field, or accelerating according to the electron's viewpoint?
Is it all frames of reference, and semantics as to who is at rest and who is accelerating?

Would then not a free falling observer who is now inertial ( elevator again) see a stationary electron in a gravitational field emit radiation and the observer stationary with respect to the electron see the static field?

 

[PAllen]

Discussions of this are complicated among other things, by the following:

1) Who is measuring for radiation? e.g. what is their relative motion relative to the charge.

2) What mix of classical vs. QED physics to we use? An electron is not classical, yet often this question is implictly asking about classical effects. Classically, a 'ball of charge' can continuously lose energy due to radiation, not only 'no problem' but required for some configuration. However, quantum theory says no way, ground states and quantization get in the way.

3) How is radiation measured? I found a note from Professor Steven Carlip to the effect that antenna design is crucial: any particular antenna will detect some effects which you probably don't want to consider radiation, while failing to detect some that you do. In effect, just as total energy is a tricky global phenomenon in GR, radiation is a tricky global phenomenon classically (it is actually greatly simplified in some ways in QED).

A specific observation is that an electron moving along a geodesic measured by detector moving along a (nearby) geodesic is completely different from the case of measurement by a 'static' detector on a planet surface.

 

[cosmik debris]

This is indeed an interesting question, even the premiss that accelerating charges radiate seems debatable. Some authors say that linear acceleration is not enough and there has to be some sort of repetitive motion. I found this article one of the better discussions on the various ideas.

http://mathpages.com/home/kmath528/kmath528.htm

 

[PAllen]

This is indeed an interesting question, even the premiss that accelerating charges radiate seems debatable. Some authors say that linear acceleration is not enough and there has to be some sort of repetitive motion. I found this article one of the better discussions on the various ideas.

http://mathpages.com/home/kmath528/kmath528.htm

I am generally a fan of mathpages. However, the following statements involve several of the confusions I got at in my prior post:

"For example, a charged object at rest on the Earth's surface is stationary, and yet it's also subject to a (gravitational) acceleration of about 9.8 m/sec2. It seems safe to say (and it is evidently a matter of fact) that such an object does not radiate electromagnetic energy, at least from the point of view of co-stationary observers. If it did, we would have a perpetual source of free energy. "

(I should also note that a question about this very area was my introductions to physics forums, where I found two peer reviewed, puplished papers from 2010 coming to opposite conclusions about issues related to this. Even more interestingly, the one most disputed here was the more professionally reviewed one, published in Annalen der Physik. The other one was published in a journal for science teachers.).

The problem with the mathpages quote above is that classically, there is no problem with conversion of mass/energy to radiation. Classically, the process of an electron sitting on a planet radiating could amount to continuous conversion of the mass of the planet to radiation, the process assymptotically stopping when the mass is exhausted (and then there is no more proper acceleration producing radiation). Quantum mechanically, one has completely different expectations - just as continuous instability of atoms predicted classically is wrong, such continuous radiation of for a stationary (but properly accelerating) charge is subject to limitation. Yet further complication is that there is no *exact* formulation of QED+gravity; and the classically predicted effects are *many* orders of magnitude too small to detect - so experimental observations are irrelevant.