Equivalence principle implies uniformly accelerated charge doesn't radiate?

the_emi_guy

Could supply a scholarly reference to this radiation that connects it to our classic problem?


If it is easy then it surely has been done. Can you supply a reference?

the_emi_guy

You stated in post #10 that accelerating charges do radiate (and supplied no references).
Here you are referring to Schwinger's book to say they do not radiate??

andrien

I am saying they do radiate but one has to be careful.it is shown there that one must take into account the beginning and ending of acceleration.
page 391-395

the_emi_guy

I have attached a paper on Hawking-Unruh radiation which includes a discussion on its possible use in resolving our paradox.

To me it shows how non-trivial this problem is that we are dipping into quantum mechanics in order to explain it.

Attached Files

unruhrad.pdf (101.8 KB, 7 views)

atyy

I haven't seen Schwinger's paper, but Eriksen and Gron in the second of their series of articles say that Schwinger considered constant eternal coordinate acceleration, which is not possible in special relativity because it leads to superluminal motion. Eriksen and Gron do consider constant eternal proper acceleration, and find that the charge does radiate.

Later in their series they reach the same technical conclusion regarding the observability of the radiation as the Almeida and Saa article that the_emi_guy (#6) and I (#5) posted.

My view is that there is no paradox in the first place, since it arises from an abuse of the equivalence principle (which applies locally in curved spacetime). Nonetheless, the question is interesting as a classical special relativistic electrodynamics problem.

andrien

see post# 20.

the_emi_guy

This is not the problem that we are discussing. We are not discussing a particle that is at rest and then begins accelerating, this is a change in acceleration (from 0 to some positive value).

andrien

it is always there,and one should carefully include it for not getting non-sense results.

Andrew Mason

You need a time dependent electric field or time dependent magnetic field and that seems to depend on what reference frame you are in. If the equivalence principle means anything, there is neither in the reference frame of the charge.

AM

atyy

It radiates and loses energy. Its path is not geodesic. It is not in free fall because it is acted on not only by the gravitational field, but also its own electromagnetic field. The equivalence principle does not apply because radiation is non-local. In GR, curved spacetime is only locally flat. If you look at second derivatives of the metric, then you will detect curvature which cannot be transformed away, and the equivalence principle does not apply.

In GR, when we write the equivalence principle, roughly speaking we have to use the form of the laws of physics in which only first derivatives or lower appear. This is because higher derivatives are more "non-local" in the sense that each time you take a derivative, you take a difference of values at different points. Technically, the derivatives all exist at a point after taking the limit, but this is the idea behind higher derivatives being considered "non-local".

Jano L.

I agree with Andrew, the situations are not the same. However, the circular motion has the additional complication - the radial field cannot be transformed away.

I think better example is the fall of a charged sphere due to gravity in the vicinity of the Earth surface, where the gravitational field is almost uniform. Consider two frames:

1) In the frame of the sphere, there is no gravitational force. The principle of equivalence and experience leads us to believe that both the motion of the sphere is uniform and its fields are given by the standard retarded solution of the Maxwell equations. That is, the sphere is at rest and the EM field in its vicinity is static.


2) In the frame of the Earth, there is gravitational force. The sphere accelerates, but this motion is referred to non-inertial frame of reference. Therefore there is no easy way to find the EM fields of the sphere. The only clue we have is to go back to the situation 1), find the field there and get back to 2) by coordinate transformation.

Now, since the field in 1) is frozen into the sphere, it seems very probable that the "free fall transformation" will make it into EM field which is comoving with the sphere.

Of course, the field will be time-dependent, as the charge moves. But I think we should not call it radiation. This is because in ordinary usage of that word, one has in mind some sort of propagation of phase or signal from the source to infinity, and in this case, there is such propagation. If the fall happens in the z- direction, in the planes xy no propagation happens. In the direction z, there is propagation of the sphere, but this is motion with slow velocity, not an EM wave.

Andrew Mason

So you appear to be saying that there is a real difference between a charge in circular gravitational orbit and a charge at rest in an inertial frame of reference. If that is the case, the same would apply to an electric dipole would it not? How would the principle of equivalence ever apply?

Why is radiation non-local? Is there something preventing it from being absorbed locally?

AM

atyy

Yes, there is a real difference. I'm not sure about the electric dipole specifically, but as a rule of thumb one cannot naively apply the equivalence principle to charged particle trajectories. For example, one might be inclined to think about a "free falling" charge - but generally a charge cannot be free falling because it is acted on by the field it produces. Also, the equivalence principle fails in non-local situations. How much non-locality can the EP tolerate? A rule of thumb is the second derivatives are too non-local for the EP to apply. One can heuristically see this by noting that at a point in curved spacetime, it is possible to make the metric and its first derivatives look the same as in flat spacetime, but if spacetime is curved the second derivatives cannot be transformed away. Thus the EP holds at all points in curved spacetime (if one looks at the metric and first derivatives), and it also fails at all points in curved spacetime (if one looks at second derivatives and higher). Roughly, higher derivatives are considered more non-local than lower derivatives, because they involve more differences of quantities at different spacetime points.

Radiation is non-local because it involves second derivatives. Also, the boundary conditions are important in solving such problems - roughly, one cannot solve a problem involving radiation by confining the charge to a small cabin and not looking outside.

So how does one apply the EP to charged particles? In the Lagrangian for all known charged particles, there are at most first derivatives coupled to the metric. So one can apply the EP to the Lagrangian by asserting that the form of the Lagrangian in curved spacetime is the "generally covariant" form of the Lagrangian in flat spacetime. This is known as the "minimal coupling" prescription.

There's a discussion about similar issues in section 24.7 of Blandford and Thorne http://www.pma.caltech.edu/Courses/p...2006/text.html.

mfb

This is thermal radiation. Gravity just helps to compress and heat stuff, it does not generate the radiation itself.

Andrew Mason

Does that mean that a charge in free-fall will always have a non-symmetric electric field - right down to the smallest distances? That seems to be something that is testable. It seems also to violate the principle of equivalence because it shrinks the meaning of 'local' to an arbitrarily small region.

I think the dipole is important because one could always create an electric dipole by separating + and - charges in a neutral body. Many molecules (eg. water) have an electric dipole moment. If a dipole in free-fall radiates, then a great deal of matter in freefall must radiate. Mind you, there is likely a quantum mechanical threshold for radiation for an electric dipole ie. quantum harmonic oscillator.

AM

atyy

Yes, in principle the EP applies only to an arbitrarily small region - a point, and even then only as long as one does not look at spacetime curvature.

I don't know the quantitative answer, but here's my guess. Let's first ignore gravity and the equivalence principle, and do classical electrodynamics in flat spacetime. If we put water in a car and accelerate it, we generally do not detect radiation, so water can be treated as effectively neutral and classical at our level of experimental accuracy, ie. we don't worry about water violating classical electrodynamics even though we don't detect radiation when we accelerate it. For this reason, at the same level of accuracy for which water in a car is considered neutral, we also don't expect to detect any violation of the equivalence principle when water is accelerated by a gravitational field.

But at some level, one might think the non-neutrality of water at small scales does come into play. I don't know the answer for what one might expect then. For example, is it a limiting factor in our ability to do extremely precise equivalence principle tests?

Andrew Mason

I think you would have to reduce the water temperature to close to absolute 0 as well. Otherwise the thermal radiation would predominate. So it would be a pretty difficult thing to test.

AM