``for another observer in a frame of reference that is moving, at the same pace as the charge,the charge appears stationary and apparently should not emit photons.''. There is a slight ambiguity in this statement. Let me first assume you mean an observer that is moving at a uniform speed that happens to coincide, at a given time, with the speed of the charge. The charge then indeed appears stationary, but this is not a reason why it should not emit radiation. This depends on *acceleration* not on velocity. If your observer is at fixed velocity, then the acceleration of the charge is the same in this reference frame as in any other, and thus the charge will radiate.

If, on the other hand, you consider an accelerated observer, then in deed, the apparent acceleration of the charge will be zero. But then, there is no reason to compare the two reference frames and expect them to be equivalent. A charge observed in an accelerated frame does not satisfy Maxwell's equations, but rather Maxwell's equations transformed to this accelerated frame. No contradictions then arise.

``for another observer in a frame of reference that is moving, at the same pace as the charge,the charge appears stationary and apparently should not emit photons.''. There is a slight ambiguity in this statement. Let me first assume you mean an observer that is moving at a uniform speed that happens to coincide, at a given time, with the speed of the charge. The charge then indeed appears stationary, but this is not a reason why it should not emit radiation. This depends on *acceleration* not on velocity. If your observer is at fixed velocity, then the acceleration of the charge is the same in this reference frame as in any other, and thus the charge will radiate.

If, on the other hand, you consider an accelerated observer, then in deed, the apparent acceleration of the charge will be zero. But then, there is no reason to compare the two reference frames and expect them to be equivalent. A charge observed in an accelerated frame does not satisfy Maxwell's equations, but rather Maxwell's equations transformed to this accelerated frame. No contradictions then arise.

Derek, you have been spending too long in engineering departments if you are going to bandy words like 'deceleration' around ;)

@F Leyvraz's reply is very good, but I fear will not entirely satisfy you.  However the ambiguity he has identified in what you say is where I believe the issue arises from.

Deceleration is a word that most physicists (quite rightly) try to expunge from the vocabulary of our students, for the simple reason that it is ambiguous and leads to all kinds of confusion.  The physics of the situation you describe is that all accelerating charges emit light (c.f. changes of direction such as cyclotron emission).  Here again, we need to be slightly careful, because what we are really concerned about is acceleration that is observed in the frame of reference, i.e. we are explicitly NOT considering free-fall.

When thought of using the above conceptual framework, the solution becomes far more clear, and as @F Leyvraz points out, there is no reason to expect the frames to be equivalent when there are abrupt forces applied to the charge (irrespective of whether these forces are also applied to the observer).  The application of the force (which I think is what you are referring to with the deceleration that you mention) breaks the symmetry of the system.

The only other issue that may be pertinent here is the Doppler shift between the frames of reference as they are brought into alignment, and the effect that this has on the photon wavelengths.

Emission of photos is conversion of energy during transition  from high energy state to low energy state.Where as accelerated frame of reference could not influence the energy state of other frame .Hence in my humble view it will not occur.

Dear Derek,

it is a delicate matter....

the only thing to be sure about is that non uniform accelerated charges emit photons no matter which frame of reference we are considering. This is also how a dipole radiation works. When radiation is emitted, is emitted, there is no way to say from any reference frame that it is not emitted. The existance of a free photon is not a matter of coordinate systems or reference frames, it does exist if somehow I'm able to detect it..then its frequency will depend on reference frames etc..

That uniform accelration produces radiation is not so clear...according to electrodynamics and Schwinger, it should not...

According to the Einstein Equivalence principle it should not, otherwise it should radiate also while still on the surface of Earth... (I totally disagree with the Einstein equivalence principle)

Regarding free falling in a gravitational field, it should not radiate since a free falling body is considered in an inertial reference frame...and this has the same properties according to many as a GALILEAN inertial system.

I don't share this point of view since according to my opinion a free falling system doesn't have the properties of a GALILEAN inertial system.

For sure in a GALILEAN inertial system in deep space (not experiencing accelerations), it would be very difficult to experience a charge radiating...

There are interpretations that affirm that a charge in free fall in a gravitational field radiate and this would constitue a counterexample against the Einstein Equivalence Principle. The radiation formula would be the Larmor's.

The other sure thing is that since the gravitational field of Earth is not strong, an experiment certifying a radiation would not be so easy due to the weakness of the field...

It is doubful too that a charged orbiting satellite would radiate....at very low frequency though...according to the period of its orbit....

Did you try searching the RG threads before posting the question?  It is essentially a duplicate. 

A "uniformly" accelerated charge, without start or stop to the acceleration, does not radiate.  You can only talk about induction in that case, which is very observer dependent.  A 3rd derivative is required for radiation.  There is an article on this by Feynman. 

The idea that accelerated charges radiate is a sloppy interpretation of periodic waves that can be expressed as a summation of sin/cosine functions (a la Fourier).  These functions have infinite derivatives which all have the same form and just change phase.  So if there is a 2nd derivative at all, there must be a 3rd, 4th, etc.

Robert,

here the question is also a bit different...it is regarding that any observer would experience photon emission. That is for sure...if there is photon for one observer there is for everybody.

No that's not different, it was discussed in many earlier posts.  Only in the case of induction is the energy transfer observer dependent.  If it radiates, it radiates in all frames.  I've pointed out the flaw in this thread, and now I am not going to stay and argue.  It is silly to keep asking the same question.

@Robert: this thread talks about observer dependence and not necessarily uniform acceleration. In that case, the answer seems straightforward.

As to your statement: `` A 3rd derivative is required for radiation.'' Where does that come from? It depends on what you mean by radiation. Let me define it as the flow to infinity of energy. Then the existence of radiation is linked to the presence of a term in the fields that goes as 1/r. Such a term in the field of a moving point charge appears whenever the charge is accelerated. No 3rd derivative needed. See the formula for the Liénard-Wiechert potentials.

If, on the other hand, you wish to work out what force is exerted on the point charge as it emits the radiation, then indeed third derivatives are required. On the other hand, this treatment involves coupling the EM field to a point charge, which involves singularities, since we are essentially dealing with the charge's self-field. So I would not trust such a calculation. The Liénard-Wiechert result is much simpler, as it only says which fields are produced by the (linear) Maxwell equations when the source is a current generated by a charge with a given acceleration.

As to experimental verification: if you look at the results of computing the radiation for linear acceleration, in terms of the Liénard-Wiechert potentials, you find that the radiatioin is utterly negligible even in the case of the most powerful linear accelerators. See Jackson, Section 14.2. So verification does not seem feasible.

As to the Equivalence Principle, I have argued in earlier poosts that the argument is unconvincing, since the equivalence principle is a local statement: the equivalence only holds ``in a sufficiently small neighbourhood''. But in order to work out whether something radiates, you need to have a neighbourhood at least as large as the wavelength. Since this is not infinitesimal, but in fact of the order of the distance of the accelerated charge to its horizon, I do not see that he EP applies.

Finally, the reference to the article by Feynman would be nice.

Dear Gyan,

Experimentally, an accelerated electron certainly radiates: a common apparatus to produce X-rays involves emittiing fast electrons and letting them hit a solid surface. The acceleration due to the shock is considerable and radiation is emitted. Theoretically, the emission of radiation by accelerated point charges also follows from Maxwell's equations in classical electrodynamics. The ``electron in a curved orbit, which does not radiate'' I assume is a reference to electrons in the ground state of an atom: these cannot be said to be in a ``curved orbit'' or, indeed, to be moving in any way. Rather, they are in a quantum state and the theory describing their radiation is quite different.

Vahid: emphatically not. If you read the answers to the questions, you will see that we all agree that emission will either happen or not happen in all reference systems. There may be some disagreement as to whether radiation will or not be emitted, but all agree that, if emitted, it is observed by all observers, and if not, it is not observed by any observer.

This question can be confusing when referring to photons, because a photon is commonly depicted as a discrete packet (particle) or quantum of electromagnetic radiation. For example, when an atom changes energy levels, it emits a photon that can later be reabsorbed by another atom.

With something like Bremsstrahlung or radiation emitted from decelerating a charged particle, a discrete particle is not produced. Instead, the electromagnetic radiation is emitted over a range of angles and wavelengths, i.e. it is dispersive. Note however that this is relative to a stationary observer’s frame of reference. For a more direct approach with basic electrodynamics, I would look into Liénard–Wiechert potentials using advanced and retarded fields.

In short, the problem isn’t whether the particle emits anything in different frames. It is instead how the field evolves and how different observers/frames see this.

Correct me if I am wrong, electron might not move as per observer but field is moving respective to observer, can this account to emmision of photon even when observer sees electron at rest in changing field?

Once more: the fact that an accelerated electron emits radiation (whether or not you want to call it photons: in the classical case, you might think of a large number of coherent photons) is a consequence of Maxwell's equations. Maxwell's eqautions, as they stand, are valid for inertial observers. For an accelerated observer, Maxwell's equations simply are not valid, and so there is no reason to think that an electron ``at rest'' with respect to the accelerated observer would not emit radiation. There is no equivalence between accelerated frames and inertial ones. A car accident is less harmful to the inertial observer who sees it than to the non-inertial driver. That asymmetry is obviously real and any attempt to act as if the points of view of the accelerated driver and the non accelerated bystander are equivalent, will end in disagreement with reality.

It means that some photons have been emitted.

Some observer is able to detect them some others can't, non because they don't have the right devices to detect them but because for them such photons don't exist due to the characteristic of the detector's motion.

Let alone Maxwell equations or tensors. You cannot make radiant energy emitted from a source disappear for a observer, otherwise you violate the conservation priciples.

If I understand, dr Low's remark nailed the problem. However, I don't think that, if his question is put in the form: "What does it mean to say that one sees light?" it can be easily answered.

In fact answers ought to differ according to the Poynting theorem, or to restricted or general relativity theory. In this context probably also dr Greentree's issue on the Doppler shift (http://www.loreto.unican.es/Carpeta2012/AJP%28Hecht%29Einstein2011.pdf ?) becomes pertinent.

Robert,

“By this I mean what property of the Maxwell tensor, or, if you prefer, the electric and magnetic fields, characterises radiation?”

The radiation emitted from Bremsstrahlung can be defined by the Poynting vector, which is equal to the rate of energy transferred per unit area.

S = (1/μ0)E x B

Since the electromagnetic tensor is defined by E and B components, you simply apply a Lorentz transformation: F’ = ΛFA, where A is the transpose of the Lorentz transformation (Λ). For a charged particle emitting radiation, the Poynting vector becomes

S = nsin2(θ) (q2a2)/(4πr2c3)

In certain frames, it is possible to make non-zero components relative to the lab frame disappear. However, this requires for other components that are zero in the lab frame to become non-zero, i.e. the electromagnetic energy is conserved overall.

Just found this fascinating discussion here that shows if we dig deeper into this question there are some open questions:

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

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

Any thoughts?

They are reconciled by the fact that the vacua of observers in nonuniform relative motion are not equivalent-therefore they will not detect the same number of particles. This is the basis of Hawking's calculation pertaining to emission of radiation by black holes and of Unruh's subsequent work. Cf. for a review: http://arxiv.org/abs/0710.5373

Robert,

Yes the Poynting vector only provides part of the electromagnetic stress-energy tensor. However, in the case of a decelerating charge, the emitted radiation at a single point in space at a single point in time will move outwards in a radial direction. With an electromagnetic wave, a time-dependent change in the E field at such point will in return change the curl of the B field. So in a sense, if you know the Poynting vector and the relation between E and B for electromagnetic waves, then you can describe the entire electromagnetic stress-energy tensor.

However, the equation I provided for the case of a decelerating charge is only 1st order (i.e. the dipole). You can derive higher orders through the lorentz transformation or other methods.

@Derek: the link you quote is interesting, but does contain some equivocations. An important one, as far as I can see, concerns the issue of what it means to say tha a system radiates. Trying to evaluate the power which the point charge exerts against radiation reaction is perhaps highly interesting from a theoretical viewpoint, but since it all happens in the intimate regions of the electron's self field, it is not really unambiguous, or for that matter, observable. On the other hand, the fields generated at a large distance from the accelerated charge ae observable. And they undoubtedly carry energy at a finite rate to infinity when the charge is accelerated. There is no issue of 3rd derivatives there. This is unrelated to radiation reaction, and only refers to the fields generated by a given motion of a point charge, which can be evaluated by so-called retarded potentials, also known as Lienard-Wiechert potentials.

Another questionable statement, I felt, was the one stating that ``there is no absolute acceleration, jst as in special relativity there is no absolute velocity''. This may make sense for purely gravitational systems, where everything is in free fall, but is emphatically not the case when non gravitational interactions enter the picture, as they do, for example, in the instance of the car accident I mentioned earlier. What hurts the driver are the electromagnetic interactions which correspond to shocks, and these are not just an effect of geometry.

I want to share some aspects not discussed in this thread.

1) The phenomenon of emission (continuous or photonic) is a causal process, so it is "absolute". There is no way that emitted radiation can be canceled if we accept the principle of causality. The phenomenon exists for all observers, inertial or non-inertial (including the comoving observer), although under certain conditions can't be actually observed. The comoving position and a low acceleration are the worst conditions to detect radiation.

On the other hand, radiation cannot be produced by the observer movement, whatever it is. A single charge only emits if it has acceleration with reference to an inertial frame.

2) Leyvraz's answers are very good indeed, except for something that is confusing.

Leyvraz wrote: "The charge then indeed appears stationary, but this is not a reason why it should not emit radiation. This depends on *acceleration* not on velocity. If your observer is at fixed velocity, then the acceleration of the charge is the same in this reference frame as in any other, and thus the charge will radiate".

I agree with the first part, the essential one, but not with the argument about acceleration, which is correct only for non-relativistic motion (Larmor). In the relativistic case (Liénard) the acceleration of the charge is different for inertial observers in relative motion, although it will never be null.

3) The radiation power (instantaneous) is Lorentz invariant, which means that all inertial observers will measure the same value. A comoving observer will measure a different not null value. So, Is wrong to believe that a comoving observer will see a perfect Coulomb field.

Hugo Fernández: thanks a lot for the clarification. I did express myself carelessly concerning the acceleration,

F. Leyvraz: "A car accident is less harmful to the inertial observer who sees it than to the non-inertial driver. That asymmetry is obviously real ... "

If the "inertial" observer is standing in front of the car and "observing" the car by being hit by it, then their subsequent injury exists in both the inertial and non-inertial frames.

One can argue that when a region contains a significant accelerating mass that interacts with a supposedly-inertial observer, that perhaps our initial definitions to do with "inertial" and "noninertial" frames no longer apply. We might want to argue that the accelerating observer sees spacetime to be curved and their nonaccelerating colleage sees it to be flat, but in reality, the fact that the accelerating observer has nonzero mass means that the existence of that noninertial mass should warp the spacetime geometry of the region, and the intrinsic curvature associated with this warpage (either the associated gravitoelectromagnetic field associated with the physical accelerated mass, or the backreaction on the metric associated with the forced change in velocity) should exist for both observers.

Since all observers then measure the region to be warped, a divergence between "flat" and "curved" descriptions is an indication that the "flat" description is diverging from reality.

@Eric Baird: even in the peculiar case you consider, there is no symmetry, in the sense that the damage to the observer and the damage to the car are unrelated. But what I had in  mind, obviously, was a car running, say, into a tree: someone looking at the car observes an acceleration of the car. On the other hand, if one puts oneself in the frame of the driver, the driver undergoes no acceleration, but the onlooker does, indeed he undergoes an acceleration altogether identical to that undergone by the driver from the onlooker's viewpoint. Nevertheless, in both reference frames, the onlooker does not suffer any major consequences, whereas the driver does.

Indeed, I did not consider the case in which the car's mass is such as to cause accelerations in the onlooker of the same order of magnitude as the accelerations caused by the accident. I was essentially concerned with arguing in a way as clear as possible that, under normal circumstances, in a space that is reasonably flat, there is no symmetry between accelerated and inertial frames, which seems to me to be the essential issue in answering the question.

Dear Dereck,

When you add two links it seems that this question is unsolved or with any difficulty to understand it. Let me to tell you that this is non true at all.

The main idea of electromagnetic radiation is very simple: if we make to depend the trajectory of the charged particle of the past time (curiously it could be also of the future time or non causal time as Feynman and Wheeler studied) and keeping in mind that from the retarded time to actual time the velocity of light is constant c ( Lienard and Wiechert make this work in a context out of relativity and previous to it). Then you obtain the Lienard-Wiechert potentials that you need more than to apply them tha usual Maxwell equations for obtaining the electromagnetic radiation.

As you can see here is not necessary at all to say what is the observer once the trajectory of the retarded time is assumed and this is the only part you need to take for the discussion. If you could chose one observer which could see the retarded time coinciding with the actual time (say to be the same and in one only instant) then no radiation could exist for it, but this is impossible in reasonable situations.

David J. Griffiths, Introduction to Electrodynamics, Prentice Hall, 1999

D.Baldomir and P.Hammond, Geometry of Electromagnetic Systems, Clarendon Press, 1996