Does the equivalence principle hold for charged particles?

[pianoplayer]

I'm new here so this may be an old question. The equivalence principle states (roughly) that one can't distinguish between an accelerating frame and a uniform gravitational field. But an accelerated charged particle radiates. Thus the EP seems to imply that a stationary charged particle in a gravitational field will radiate, but this doesn't happen. The other possibility is that an observer in a frame accelerating along with a charged particle sees no radiation. This seems reasonable, but this question still seems to be the subject of active debate. The question appears to boil down to whether radiation is frame independent (i.e., seen by all observers) or frame dependent (not seen by an accelerated observer). The latter view is advanced in a paper by Shariati in Found. Phys. Lett. 2 (1999) 427–439, entitled "Equivalence principle and radiation by a uniformly accelerated charge" and the former view is held by Parrott in his paper "Radiation from a uniformly accelerated charge and the equivalence principle" in Found. Phys. 32 (2002) 407-440. This guy claims that the EP fails for charged particles. Any ideas on who is right?

 

[Meir Achuz]

I believe that the EP does fail for charged particles. It is just intended for gravitiation in the absence of any EM effects. Einstein tried for 30 years to include EM and GR in a unified theory and failed. Poor fellow did not believe in QM.
He stopped his quest 50 years ago today.

 

[Ich]

Funny, exactly the same question puzzled me for a while.
I still don´t know the answer, but I think that radiation is frame independent. There must be some other rules.

 

[JesseM]

It seems that answering this question is actually quite tricky, and that our current theories of electromagnetism may not be up to the job. See the lengthy discussion here:

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

Here are the first three paragraphs:

One of the most familiar propositions of elementary classical electrodynamics is that "an accelerating charge radiates". In fact, the power (energy per time) of electromagnetic radiation emitted by a charged particle is often said to be strictly a function of the acceleration of that particle. However, if we accept the strong Equivalence Principle (i.e., the equivalence between gravity and acceleration), the simple idea that radiation is a function of acceleration becomes problematic, because in this context an object can be both stationary and accelerating. 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. Since the upward force holding the object in place at the Earth's surface does not act through any distance, the work done by this force is zero. Therefore, no energy is being put into the object, so if the object is radiating electromagnetic energy (and assuming the internal energy of the object remains constant) we have a violation of energy conservation.

Of course, we could question the claim that no work is being done by the force holding the object in place. Indeed if we imagine a capsule in freefall, and within that capsule an object being accelerated in such a way that it maintains a constant altitude relative to the outside gravitating source, we would say, inside the capsule, we had done work on the object as we increased its velocity relative to the capsule, even though from the outside standpoint of the gravitating source the object is stationary and no work has been done on it. This is not too surprising, since work and kinetic energy are understood to be relative concepts, but it seems to lead to the puzzling conclusion that electromagnetic radiation must also be a relative concept. The familiar relativity of kinetic energy corresponds to the symmetry between different frames of reference, which is to say, we can always find a system of inertial coordinates with respect to which any given object (at a given instant) has zero kinetic energy. Our consideration of charged particles in a gravitational field seems to suggest similarly that we can always find a system of coordinates (at least locally) with respect to which a charged particle (at a given instant) does not radiate - even though the particle may be radiating at that instant with respect to some other system of coordinates.

It's also possible to question whether the equations of electrodynamics really do imply that an accelerating charge necessarily radiates. Surprisingly, this is still an open question for the classical theory. The difficulty is in knowing how to correctly account for the influence of a charged particle on itself. Remember that two electrons repel each other with a force (statically) proportional to the reciprocal of the square of the distance between them. This is traditionally understood in terms of each particle interacting with the field of the other particle. The intensity of each electron's field increases to infinity as the distance goes to zero (assuming point-like particles), so the force with which an electron is repelled increases to infinity as it approach the location of an electron - but therein lies a conceptual difficulty. According to this description, each electron is located in a place where there is an infinite force of repulsion against electrons!

 

[Ich]

WOW
So we don´t know?

 

[pervect]

Funny, exactly the same question puzzled me for a while.
I still don´t know the answer, but I think that radiation is frame independent. There must be some other rules.

I think that radiation (in the sense of "photon number") is "frame independent" only for inertial frames. There's no good short name for an arbitrary coordinate system, but photon number is not conserved by general coordinate transformations. Look at unruh radiation for an example.

So it doesn't surprise me that one observer claims the photon number is zero and the other observer claims the photon number is not zero if one of the observsers is in an inertial frame and the other is in an accelerating coordiante system - because "photon number" isn't conserved by that sort of coordinate transformation.

I don't quite understand why people (Ich apparently isn't alone) think that "radiation" is coordiante-independent. I'm not sure if it's an actual difference in belief, or whether their defintions don't match mine.

This is one of those questions I've been meaning to dig into more "someday". Meanwhile I will happily continue to believe that radiation is not coordinate independent unless someone can show me some reason why it should be.

 

[pmb_phy]

I'm new here so this may be an old question. The equivalence principle states (roughly) that one can't distinguish between an accelerating frame and a uniform gravitational field. But an accelerated charged particle radiates. Thus the EP seems to imply that a stationary charged particle in a gravitational field will radiate, but this doesn't happen.

The detection of radiation is frame dependant. This is true for both uniform and non-uniform g-fields. This is a topic which comes up often here so I placed some references online. See http://www.geocities.com/physics_world/misc/falling_charge.htm

The question appears to boil down to whether radiation is frame independent (i.e., seen by all observers) or frame dependent (not seen by an accelerated observer). The latter view is advanced in a paper by Shariati in Found. Phys. Lett. 2 (1999) 427–439, entitled "Equivalence principle and radiation by a uniformly accelerated charge" and the former view is held by Parrott in his paper "Radiation from a uniformly accelerated charge and the equivalence principle" in Found. Phys. 32 (2002) 407-440. This guy claims that the EP fails for charged particles. Any ideas on who is right?

The former is correct. Parrot is incorrect.

Pete

 

[Chronos]

Pete is correct. It is frame dependent. A charge will appear to radiate any time it is accelerating with respect to the observer. It does not matter which one is accelerating, the charge or the observer. An observer coaccelerating with a charge will not observe it to radiate. If acceleration were a manifestation of absolute motion, all charges would appear to radiate and the intensity would fluctuate seasonally due to variations in Earth's orbital acceleration.

 

[selfAdjoint]

Pete is correct. It is frame dependent. A charge will appear to radiate any time it is accelerating with respect to the observer. It does not matter which one is accelerating, the charge or the observer. An observer coaccelerating with a charge will not observe it to radiate. If acceleration were a manifestation of absolute motion, all charges would appear to radiate and the intensity would fluctuate seasonally due to variations in Earth's orbital acceleration.

Seems like this would be an easy effect to confirm experimentally. Has it been? If not, why not?

 

[Stingray]

It seems that answering this question is actually quite tricky, and that our current theories of electromagnetism may not be up to the job.

It is tricky, but Maxwell's theory is fine as long as we don't try to force it to accommodate the fantasy of (classical) point particles.

Anyway, the textbook derivations (which are incomplete) say that a charge with nonzero 3-acceleration with respect to an inertial frame in flat spacetime will appear to radiate in any such frame. This result does not apply to a charge sitting on the surface of the earth. As others have mentioned, even the definition of radiation becomes nontrivial in accelerated frames (or ones in curved spacetime).

It is also not true that all accelerated charges radiate. The statement that they do is a simplification. The actual conditions are extremely complicated (they're not even completely known except in implicit form), and highly dependent on non-standardized definitions.

Although this is personal preference, I actually prefer not to use the word "radiation" very much. It is a concept which only makes sense far away from the source(s), so it often makes things very difficult to discuss.

 

[Ich]

I don't quite understand why people (Ich apparently isn't alone) think that "radiation" is coordiante-independent. I'm not sure if it's an actual difference in belief, or whether their defintions don't match mine.

It´s just ignorance. I don´t know much about the topic, so I stick to the notion that photons either exist or don´t exist, just like massive particles (I hope they do at least). Guess I should read more about it.

 

[pmb_phy]

It´s just ignorance. I don´t know much about the topic, so I stick to the notion that photons either exist or don´t exist, just like massive particles (I hope they do at least). Guess I should read more about it.

If you've noticed, I've try to restrict myself to saying that its the "detection" of radiation that is coordinate dependant and stayed away from assertions such as a photon exists etc. The Am. J. Phys. articles covers this part in one of the sections. I'll quote it later. All in all its good to restrict yourself to what you can measure rather than what "exists" or not.

Pete

 

[pianoplayer]

A closely related question is whether a charged particle radiates when in freefall in a gravitational field. The consensus seems to be that it wouldn't. But would an observer not in freefall see it radiate? The EP would seem to imply that no observer would see radiation. Does this make sense?

 

[pianoplayer]

OK, here's something for pmb_phy. After a little Googling I came up with this recent title and abstract, which states that (classical) radiation is observer independent. The author also states that "A freely falling charge in curved spacetime does not move along a geodesic and therefore radiates." The paper can be found at
http://arxiv.org/abs/gr-qc/9909035. This seems to contradict what a bunch of you guys are saying. Any reaction?


Notes on covariant quantities in noninertial frames and invariance of radiation in classical and quantum field theory
Hrvoje Nikoli´c
Theoretical Physics Division, Rudjer Boˇskovi´c Institute, Croatia
August 25, 2004

Abstract
A local observer can measure only the values of fields at the point of his own
position. By exploring the coordinate transformation between two Fermi frames, it is shown that two observers, having the same instantaneous position and velocity, will observe the same values of covariant fields at their common instantaneous position, even if they have different instantaneous accelerations. In particular, this implies that in classical physics the notion of radiation is observer independent, contrary to the conclusion of some existing papers. A “freely” falling charge in curved spacetime does not move along a geodesic and therefore radiates. The essential feature of the Unruh effect is the fact that it is based on a noninstantaneous measurement, which may also be viewed as a source of effective noncovariance of measured quantities. The particle concept in Minkowski spacetime is clarified. It is argued that the particle concept in general spacetime does not depend on the observer and that there exists a preferred coordinate frame with respect to which the particle number should be defined.

 

[pmb_phy]

OK, here's something for pmb_phy. After a little Googling I came up with this recent title and abstract, which states ...

Thanks for the reference. However at this time I'm in a power down more where I've stopped reading everything to do with science/physics. I kinda need a break after doing this for 7 years straight.

..that (classical) radiation is observer independent. The author also states that "A freely falling charge in curved spacetime does not move along a geodesic and therefore radiates." The paper can be found at
http://arxiv.org/abs/gr-qc/9909035. This seems to contradict what a bunch of you guys are saying. Any reaction?

They do seem to contradict each other. At this point it comes to a few reasons such as each uses a different set of postulates or each uses a different interpretation of results etc.

A “freely” falling charge in curved spacetime does not move along a geodesic and therefore radiates.

This comment makes no sense to me. An uncharged object with no external forces on it (i.e. a free object) need not follow a geodesic in a curved spacetime so I don't see why a charged particle would be expected to. However in a flat spacetime all objects must follow geodesics and therefore the path taken by a charged object must as well. Note that you can't think of a point charge as a pointlike object since the field itself is treated as part of the object itself and the field is not a local object.

Pete

 

[Chronos]

The are differing opinions about accelerating observers:

http://arxiv.org/abs/gr-qc/9903052
Classical roots of the Unruh and Hawking effects
Authors: Massimo Pauri, Michele Vallisneri

This paper has a pretty decent citation history.

 

[Qoo]

Teasers about Equivalence Principle: Charged Particles and Radiation

Often people asked if charged particles will radiate or not in the 2 diff
reference frames of "thought experiment" suggested by Einstein and
seemingly argued that Equivalence Principle did not apply due to a
"dilemma" (I'm going to show you that this dilemma is fictitious) that
the man in the accelerating lift will see nothing radiating from the charge
while the outsider will see it radiate.

I'm here to tell you how this Equivalence Principle should be carefully
applied as afterwards you can see not just it applies for charged
particles and radiation but also any other physical phenomena so far
discovered by human kind.

To detect radiation from a charged particle, the man in the lift MUST
rely on a sensor of certain kind (maybe in his brain) that can interact
with the EM wave emitted from the charge, before he can tell that
the charge radiates. Now this sensor must also carry charges so as to
detect the radiation. During the "thought" experiment, you the outsider
"see" that the charged particle in the lift, even though it's fixed in certain
solid state in the lift, it will still radiate because it's accelerating away
from you the outsider ...

However, do you, being now the outsider, also see that the charges in
the sensor MUST also radiate? Now, you get the answer. The man
in the lift, by using that sensor alone, can never detect any radiation
from the charged particle; because for the outsider, even though he
observes radiations from the charge and the sensor, these two will
counter each other and the sensor will still read nothing. To the man
in the lift, he still cannot detect any radiation from the charge no
matter what experiment he carries out, given that the experiment
must be carried out within the world of the lift!

The second "thought" experiment will be like this: the charge in the lift
is not fixed to certain solid, but freely in the air. When the lift accelerates,
to the outsider, no radiation is observed from the charge. But the sensor
which is fixed w.r.t. the lift is now radiating. Hence the sensor will still
read and the man can detect the charge radiation and "think" that the
charge really radiates by dropping towards the floor due to "gravity."

Can you tell a charge radiates or not by NOT using a device with EM
interaction? To interact and know the existence of a target, you need
to play the same game of interaction.

Similarly, once trained up a bit for the thought experiment, you can
now easily distinguish among all the FALLACIES posted in the Internet
while they can hardly get into any Physical Review Letters. Examples
like the energy in any mass built-up after a long period of time in the
lift will smash things like crazy bombs while forgetting that the whole
lift and the person and any detector in the lift will also gain such crazy
energy in such crazy thought experiment and hence overlooking the
experiment in the lift in fact still cannot detect such crazy energy.
There can be any no. of examples of fallacy that you can draw from
if you apply the Principle wrongly.

We must be VERY CAREFUL when applying the Equivalence Principle.
All things happened in the lift must also be considered since they are
all accelerating away from the outsider. If the man in the lift really
carries out an experiment to read things, you have to "think" about
how the experiment will undergo and all the appliances will suffer the
same "accelerating" effect. You "cheat" if you suddenly let the man
in the lift obtains information or points of views from the outsider
(being a God-like being w.r.t. him alone).

As a last reminder, Einstein raised the thought experiment to demonstrate
the equivalency of respective observers and experimental results. Newton
knew that it applies to inertial frames because it's instinctive, but Einstein
boldly pushed it further into non-inertial frames. This move is a belief (as
Einstein didn't know any positive experimental result supporting his theory
during that time). And the result is that space-time is curved. Our feeling
of gravity alone is NOT sufficed to prove this reality. Readings from our advanced GPS system can prove this (ok, I won't tell you where I work).

Equivalence Principle is a very important principle in this universe for if it's
broken, what you see is not what I see and we can go into a monastery
and practise Zen and give up materialism. Well, this is another topic.

Qoo.

 

[yogi]

Does this make sense?

 

[pianoplayer]

Chronos: I thought this was a well-defined problem in classical physics. It's really surprising to me that there's such disagreement among experts.

Qoo: Where DO you work?

yogi: good question.

 

[selfAdjoint]

However, do you, being now the outsider, also see that the charges in
the sensor MUST also radiate? Now, you get the answer. The man
in the lift, by using that sensor alone, can never detect any radiation
from the charged particle; because for the outsider, even though he
observes radiations from the charge and the sensor, these two will
counter each other and the sensor will still read nothing. To the man
in the lift, he still cannot detect any radiation from the charge no
matter what experiment he carries out, given that the experiment
must be carried out within the world of the lift!

This paragraph is confusing to me. You change POV from inside to outside and apparently judge what the inside observer sees from what the outside observer sees him doing. But it seems to me that being very careful as you justly caution, we should not mix up the experience of these two observers but keep them separate. Compare the analogous case of an astronaut falling into a black hole and a distant observer watching the fall. They have two completely different reports of what happens.

 

[Garth]

Qoo welcome to these Forums with such a thought provoking post!

The question of how the measurement is made always has to be taken into account in these gedankens and thank you for pointing out how it affects the 'electron in a lift' situation.

The question I ask is: "If a freely falling electron radiates as observed by a supported, 'stationary', observer, where does the energy of that radiation come from?" Similarly: "A supported electron on a laboratory bench is also accelerating according to the equivalence principle. Should that electron radiate? If not why not? If so, where does the energy of that radiation come from?"

Garth

 

[pervect]

I probably should think about this more, but right off the top of my head---

An accelerating obsserver is not in an asymptotically flat space-time, so he cannot calculate a total energy via standard methodology.

An observer in asymptotically flat space-time looking at a charge on the surface of the Earth can calculate the total energy - of course, he also notes that the charge does not radiate.

[add]
Think, for instance, about calculating the energy of a photon in flat space-time. From the non-accelerated observer's perppective, this is easy (assuming that space-time is flat and non-expanding, it yields a constant answer). From the accelerated observer's perspective, one has to deal with the fact that the photon might well disappear from his universe by passing through the Rindler horizon!

 

[Chronos]

Does this make sense?

Yes. It was a very nice explanation. In a universe where the laws of physics [all of them] are the same in every reference frame, every observations is necessarily frame dependent. An event that appears to occur to an observer in one reference frame [like photon emission] may not appear to occur to an observer in a different reference frame. This is not a paradox and both observers are correct - even though they disagree on what, if anything, happened.

 

[pervect]

Generally speaking, if one observer sees an event happening, all observers agree that the event did happen. It turns out, though, that the notion of a particle is somewhat fuzzy when viewed in the most general context (than of an accelerating observer or an arbitrary coordinate system). One example of this fuzziness is accelerating observers - one observer thinks a particle is a real particle, the other thinks it's virtual.

This "fuzzy nature" of particles is mentioned by Wald in his textbook, and one of the cited papers in this thread. It's really rather suprising, and definitely not intuitive. When you view particles as a consequence of certain symmetry groups, it's less surprising, I suppose.

 

[yogi]

So Chronos - if I surround a charged particle (e.g. an electron) with a Gaussian sphere collector and accelerate the sphere and the electron together, your saying that no matter how much radiation is emitted from the electron during the acceleration as measured (or calculated) by an observer at rest wrt space (an inertial observer), no photons will strike the collector because they are in the same acceleration frame?

 

[Chronos]

So Chronos - if I surround a charged particle (e.g. an electron) with a Gaussian sphere collector and accelerate the sphere and the electron together, your saying that no matter how much radiation is emitted from the electron during the acceleration as measured (or calculated) by an observer at rest wrt space (an inertial observer), no photons will strike the collector because they are in the same acceleration frame?

An observer comoving with the charged particle will not detect any radiation. An observer who is not comoving with the charged particle will detect radiation. The intensity will vary depending upon the relative difference in acceleration between their reference frames. So the long winded answer is yes. The photon, as pervect very astutely observed, is only virtual to an observer in a comoving reference frame.

 

[Qoo]

More teasers about E.P.

Again, here is one more teaser (not about charges):

Frame 1--Man in the Lift is the Observer O
(he doesn't know that he's in a lift but just finds that he's standing on a
scale not fixed to the ground with uniform gravity G acting on him)

Frame 2--Outsider is the Observer O'
(this guy sees that the lift is accelerating away from him with a magnitude
equals to G; he sees no gravity acting on the lift, the man, or the scale)

Now, to play the teaser, you have to imagine a remote control held by
the outsider O' who will then press a secret button that will make the lift
to decelerate all of a sudden with a magnitude of G continuously (even
after the lift will then accelerate towards the outsider). If you were the
man in the lift with mass M, what will you get from the readings of the
scale before, during, and after the pressing of the secret button?

Assume that the space in the lift is tall enough for all the dynamics
happening without hindrance.

Qoo.

 

[pianoplayer]

I'm inclined to go along with you guys (e.g., Qoo and Chronos) who argue that no radiation is seen by an observer accelerating with the charge. Here's a question: from a purely classical perspective, could one look at the problem as follows. The observer who's accelerating with the charge will see a static electric field -- no time dependence, no retarded potentials, and thus no radiation. Now a guy floating in space sees this accelerating charge move past. From his point of view, the electric field is not static. Could one sit down and do the calculation to show that he sees time-varying (transverse) components of an electric and magnetic field with a non-zero Poynting vector, i.e., EM radiation. If this can be demonstrated, it seems this would settle the argument without resorting to arguments about "fuzzy" particles, etc. Or perhaps it's not this simple.

 

[yogi]

So my Gaussian sphere collector heats up for everyone in the universe because they see it bombarded by photons - except for the guy riding on its surface who measures it ice cold!

 

[pianoplayer]

Apparently yes. Since the only way to measure the temperature of your Gaussian sphere collector is by the radiation it emits, we're back to where we were before.

 

[pervect]

I'm inclined to go along with you guys (e.g., Qoo and Chronos) who argue that no radiation is seen by an observer accelerating with the charge. Here's a question: from a purely classical perspective, could one look at the problem as follows. The observer who's accelerating with the charge will see a static electric field -- no time dependence, no retarded potentials, and thus no radiation. Now a guy floating in space sees this accelerating charge move past. From his point of view, the electric field is not static. Could one sit down and do the calculation to show that he sees time-varying (transverse) components of an electric and magnetic field with a non-zero Poynting vector, i.e., EM radiation. If this can be demonstrated, it seems this would settle the argument without resorting to arguments about "fuzzy" particles, etc. Or perhaps it's not this simple.

http://xxx.lanl.gov/abs/gr-qc/0006037

goes through this calculation, though I really have only glanced at it.

To write the Poynting vector at Rindler instant $$\mbox{\omega_0}$$ for the local observer who is
seated at (Xo, Yo,Zo), we can write everything in the instantaneous rest frame of the
source S at the retarded time and then use the Lorentz boost that transforms this frame
to the instantaneous rest frame of O (at the moment of observation).

They find that an accelerating observer sees only a pure electric field, and hence no Poynting vector.