For non-relativistic limit, we should expect that our usual intuition regarding the accelerating charges would hold, and a falling charge should radiate with an acceleration. On the other hand, in general relativity a falling charge simply follows a geodesic of the earth curved space-time; it is not accelerating relative to freely falling frames, and so is not accelerating in a meaningful sense.
@ Bharat
I suggest you to first read the Equivalent principle in general relativity. Let me put it simply for you in the words of Einstein.
"we assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system. "— Einstein, 1907
For a charge to radiate due to its motion, it must accelerate or decelerate.
Case 1: Eg in Large Hadron colliders where charged particles are accelerated using strong electrical fields. - An example of acceleration in the local reference system
Case 2: If a charged particle's trajectory falls in the vicinity of a massive body, say a star, or our sun. Then the resulting geodesic of the trajectory won't be flat. A non-flat geodesic implies the presence of gravitational potential. Also, the resulting particle will accelerate or decelerate depending on where it is in the gravitational potential. - An example of acceleration in a gravitational field.
Now, the equivalence principle says the two cases are equivalent! There is no physical experiment which can be done to distinguish the two cases. It means, all laws of physics will remain the same in two cases.
Force manifests itself from the acceleration. It does not matter whether you accelerate your motorbike with a=9.8 ms^-2, or, you fall in earth's gravitational field with g = 9.8 ms^-2. The two cases are physically equivalent!
In the statement of your question: "simply follows a geodesic of the earth curved space-time" - "acceleration" is embedded in it.
To respond to your question: Thus, either both cases are equally meaningful, or equally meaningless. Depends on your nomenclature!
Hope that clarifies.
Dear Bharat Kumar,
I am not from field, but I have suggestion… As you expect only in ideal condition falling charge in the Earth’s gravitational behave like charge which are under acceleration… But in real case perhaps do not exist freely falling of charges because the Earth has a radiation (From thermal gradient + sunlight reflections) which acts again freely falling..
Best Regards,
Attila
This is perhaps one of the most interesting questions for a long time, in that it touches on classical relativity, quantum principles, and observer theory, which means it is a deep question about how physics should be conducted theoretically and empirically. This question alone could occupy a full career in real research.
As with many such questions, there are several ambiguities that need to be resolved before any physically consistent answer can be given. In my experience, these ambiguities arise because of undeclared conditioned imagery in the questioner's mind. The same goes for most respondents. The key in such questions is to establish proper contextuality wherever it is needed.
Here goes.
A: Do you want to discuss the question according to classical mechanics only? if so, then good luck, I have nothing to say. Classical mechanics is not a complete description of the real world of experience and therefore has limited scope here.
B: To answer this question properly, several contexts have to be established. First, WHO is looking at the experiment? HOW has it been established that a "free" charge is there in the first place? The question says nothing about that at all. You should tell us what sort of equipment allows what sort of observer to assert that there is a "free charge". Second, you need to tell us something about the observer. Are they freely falling or are they held in a gravitational field, as would be the case on the surface of the earth. These are VERY different observational contexts.
Third, what sort of detectors are being used? To monitor a charge so that you could say "it is falling", you presumably need something that monitors position, but that is entirely not simple, either theoretically or empirically. Then you need to tell us how you would monitor radiation. Certainly not in terms of waves but in terms of yes/no (= photon) detectors. Signal detection in curved spacetime is a big, relatively unexplored topic. In my view, that is what "quantum gravity" should mean. The fascinating development of relativistic quantum information theory is an indicator of how deep this topic is. The theory of localized detection in quantum field theory (for that is really what this question points to) is relatively underdeveloped, the Unruh effect being just one manifestation of the issues involved.
I would stress that such discussions are not best phrased using classically conditioned concepts, as these can mislead badly. For instance, it is pure conjecture to talk about "a particle moving from A to B", when there are no detectors in between A and B. All that the experimentalist will do and see in such experiments is press buttons (effectively) to trigger a beam, and subsequently look at signal detectors to see if something has happened in them. Everything else is mental imagery.
To answer your question on an intuitive level, I would guess a classical field theory answer is that classical outgoing radiation waves would be generated as a charged particle fell into any potential well. That is why classical mechanics cannot account for the stability of Bohr-model atoms. On a quantum theory level, I would say that an electron bound to a proton would NOT radiate electromagnetic radiation (the hydrogen atom is stable in relatively weak gravitational fields), unless it was in a very strong curvature regime. Then more detailed modelling would probably predict that there would be disassociation of the particles with the emission of electromagnetic radiation. But that would be a great topic for several years' of research.
Finally, a quantum optics experiment (where quantum degrees of freedom are relatively minimal) such as the double slit experiment or Stern-Gerlach experiment conducted in a gravitational field environment would undoubtedly show that outcome probabilities WERE affected by that field. The extent would depend on how closely the laboratory was localized sufficiently (if it were free falling) so that the equivalence principle would hold. In the idealized limit of an extremely localized laboratory, the classical prediction of general relativity is that there would be no changes in outcome probabilites. But contextuality is everything here, even in classical mechanics: the non-localization of objects (such as moons) in Newtonian gravitational theory leads to the prediction of the Roche limit, at which point extended infalling objects begin to break up due to gravitational tidal stresses.
A good, deep question, posed incorrectly (but understandably so).
“…Second, you need to tell us something about the observer. Are they freely falling or are they held in a gravitational field, as would be the case on the surface of the earth. These are VERY different observational contexts …”
- for the thread’s question seems all the same where observers are.
“…Third, what sort of detectors is being used? To monitor a charge so that you could say "it is falling", you presumably need something that monitors position, but that is entirely not simple…”
- here is no problem, the “freely falling charge”, for example, can be a charge on a satellite, which (and so the charge also) freely falls rather long time.
“…Then you need to tell us how you would monitor radiation…”
- that should be, of course, corresponding photon detector.
“…On a quantum theory level, I would say that an electron bound to a proton would NOT radiate electromagnetic radiation…”
- the thread’s question doesn’t relate to the Bohr-model atoms, here accelerated free [besides the gravity] charge radiates, and in the case above there is the situation that is very similar to synchrotron radiation.
Thus with a great probability charges radiate photons when fall freely in a gravity field, gravitational accelerattion doesn’t differ in this case from accelerations that are caused by any other force. Though, of course, an experimental confirmation of this inference (and the opposite also, though) is rather hard problem, first of all – because of real gravitational fields [gradients] are comparatively small, when a weak signal is necessary to detect on a large background/noise.; for example at experiment on an orbit – that are natural radioactivity and cosmic radiation, Earth magnetic field, etc., etc., etc.
Cheers
@ George,
You are right but also I think you have dragged the question unnecessarily. The question simply implies how does the general notion of acceleration corresponds to a motion in curved space-time? I am confident that equivalent principle is more than enough to answer it.
Bharat asked a simple yes-or-no question and he deserves a simple yes-or-no answer: No.
According to the principle of equivalence, there is no difference between a charge in free fall in a gravitational field and an unaccelerated charge inside a rocket ship that is accelerating. The unaccelerated charge has no preferred direction in which to radiate momentum and energy, so it doesn't radiate.
Emission of electromagnetic energy by an accelerated particle is a quantum phenomenon that happens independently of the cause (electric, magnetic or gravitational) that produces acceleration. The Theory of Reference Frames proves quantum emission happens at highest speeds. From the theoretical viewpoint it is therefore possible that a particle accelerated by a gravitational field is able to emit quantum e.m. energy but because of the fact that the gravitational field is weakest, it is practically hardest that a particle accelerated by a gravitational field is able to reach highest speeds that are necessary for the emission of quanta of e.m. energy. It follows that emission happens generally when particle is accelerated by electric or magnetic fields.
The question is widlely discussed in the web and seems most definitively have been answered by a work:
Bryce S DeWitt, Robert W Brehme : Radiation damping in a gravitational field. Annals of Physics, Volume 9.2 1960, pp. 220-259.
To Franklin: The answer is yes.
I copy a sentence from the abstract of this work:
"The particle tries its best to satisfy the equivalence principle in spite of its charge. It is only prevented from doing so (i.e., from following a geodetic path) because of the fact that, contrary to the case of flat space-time, the electromagnetic Green's function in a curved space-time does not generally vanish inside the light cone, but gives rise to a “tail” on any initially sharp pulse of radiation."
So the authors share Bharat's opinion that the equivalence principle prevents a point charge from radiating if it would follow a geodetic line exactly.
Ulrich, you say DeWitt and Brehme "share Bahrat's opinion that the equivalence principle prevents a point charge from radiating ...". That was precisely my point as well. And in that case, the answer to Bharat's question, "does it radiate?" is no. So why do you say the answer is yes?
By the way, would you please attach the DeWitt paper? Thanks.
Franklin,
the full statement was that the charge would not radiate if it would follow a geodetic line exactly. As DeWitt and Brehme point out in detail the charge caries with it a Coulomb field which produces reaction forces and prevents the charge from following a geodetic line precisely and by this mechanism lets the charge radiate and changes the answer to yes. Of course, the quantitative aspect of the matter is not clear from this and it may well be that 'yes' may mean 'negligible in all conceivable practical circumstances'.
Hope this answers your question to your satisfaction.
Ulrich, thank you for clarifying.
But doesn't it seem strange to you that the radiation might even be 'negligible in all conceivable practical circumstances', instead of identically zero? Because then an experiment can be done, in principle, that would distinguish between an accelerated reference frame and a gravitational field, in violation of the equivalence principle. Unless there is some reason why such an experiment could not be physically realized. (That's why I was hoping you could attach the paper.)
Glad that the discussion converged to my initial opinion. But now it is making me wonder, Why an accelerating particle actually radiates be it in an accelerator or in a curved space-time? I never thought about it earlier. As already mentioned radiation is a quantum phenomenon; the electrons must transit to excited states in either accelerating scenario.
I see acceleration is always connected with a curved spacetime. It needs to specify that that connection is in GR the result of a mathematical model and just other mathematical models are possible. Besides particle accelerators can be circular and curved, for technical reasons, but they can be also linear. There is no general physical reason that connects necessarily an acceleration with the curvature of spacetime. In GR gravitational acceleration, that has central symmetry, is connected with the curvature of spacetime as per a tensor mathematical model. Changing the mathematical model also the curvature of spacetime becomes non necessary. I agree, quanta of electromagnetic energy are generated by the transit of electrons between two different energy levels and it happens generally for electrons that are constrained inside atom. If electron is free (i.e. non constrained inside atom) and accelerated (in a curved or linear accelerator) in that case the Theory of Reference Frames proves accelerated free electron emits two quanta of e.m. energy at highest speeds, and precisely a first quantum at the physical speed of light and a second quantum, exactly equal to the first, at the critical speed. I think in order to have a clear and complete idea of physics it needs to know, and in case to criticize, also other physico-mathematical models that are proposed by contemporary physics and not to stop at usual models that are proposed by postmodern physics. Sites like RG have just this function.
@ All, I have created a different post in order to focus at the key aspect of this discussion. As I understand there is more than general relativity in this problem. The link can be found here: https://www.researchgate.net/post/Radiation_mechanism_in_accelerated_bodies_and_isolated_charges
Actually accelerating charges don't radiate; even though Lamor's formula is correct.
http://www.mathpages.com/home/kmath528/kmath528.htm
as Feynman pointed out, one needs a change in acceleration.
Here is another twist to the question. Whether a moving charge radiates or does not is not observer independent.
A freely falling charge in a true gravitational field may radiate for an infinitely far observer, because in that case a global description of both gravity and electromagnetism are needed, preventing applicability of the equivalence principle.
A freely falling charge will not radiate from the point of view of an observer that falls along with it on a close-by trajectory, as long as tidal forces remain unobservable. For in that case both are in the same freely falling local inertial system, where the charge is at rest. And charges at rest don't radiate.
Let me also point out that radiation is *not* an intrinsically quantum phenomenon. It is present in classical electrodynamics as well, a theory that preceded quantum mechanics. What quantum mechanics did was to add a mechanism to suppress radiation (electrons in atoms don't emit radiation unless they change energy level) and to give a reinterpretation of the origin of radiation at microscopic scales. Electromagnetic waves at the km wavelength scale and beyond are still best described outside of quantum mechanics.
C. Y. Lo is correct. The gravitational radiation of an object, and also the electromagnetic radiation of an object, has nothing to do with an observer. Once an object radiates gravitational or electromagnetic energy to the far zone (distance much greater than a wavelength), that energy is gone from the object, no matter who looks at it.
Re gravitating charge.
It is a curiosity and a greatly annoying fact that recent arguments on this site based on the assertion that , are usually defended on the grounds that to think otherwise is to introduce metaphysics into physics. Surely we can see objects all around us: they exist all the time and do things on their own.
The problem is, quantum process don't match that classical perspective. Without any reference to observers and how they do their observations, any suggestion that things happen on their own is by definition an unprovable assertion, and hence itself metaphysical in nature. How could you prove any assertion in physics without apparatus? To talk about `the gravitational radiation of an object' as if it exists regardless of observers is to do metaphysics. I could just as easily assert that objects fell because unobserved fairies pushed them.
The error stems from two facts: 1) a common unwillingness to recognise the respective roles of reductionism and emergence in physics, and 2) different observers may appear to get the same results. The (incorrect) logical deduction generally made on the basis of 1) and 2) is that physical properties `exist' in some real sense independently of observers. NO. All that can be said from empirical evidence is that in certain classes of experiment, there is consistency of observation. You cannot ignore the context in which those results were obtained. That is a central lesson in quantum mechanics. So if I have no means of detecting gravitational radiation of falling charges, I should not assert that such a thing exists. Of course, I may be motivated by theoretical speculation to look for it. That happened recently in the case of the LIGO experiments.
So does a falling charge radiate if it moves on a geodesic?
My answer: First, describe yourself. are you in a freely-falling laboratory or a rocket? Then tell me how you `know' there is a free charge there. Then tell me how you 'know' it is moving on a geodesic (define `geodesic' empirically). Then tell me how you are observing whatever radiation you are interested in.
It seems to me that many of the short glib answers I've read so far do none of those things.
I have the solution to this problem, and it is very simple! We missed it because we all deviated from the original questions by introducing too many arguments that trigger other discussions and misguided us.
First, let me say few things about our discussions in the blog. Some people made assertions based on opinions, without scientific deduction from current fundamental knowledge: this is not science! Instead, another group of physicists pointed out some fundamentals: K. Kassner made clear that experiments are not observer independent (True); Ulrich Mutze and Franklin Felber discussed that a freely falling charge cannot radiate under ideal conditions (True); Sergey Shevchenko discussed the setup of the experimental apparatus to make observations (True); George Jaroszkiewicz, more than anyone else pointed out that the true fundamentals of the theory and the experiments of the physical world (True)..
The true facts are:
1. Inertial frames (and we include GR) must provide the same physical observations, even for an accelerating charge. Any particle in free fall does not experience any acceleration (no tidal forces). Therefore a particle solely in a gravitational field does NOT have a gravitational acceleration.
2. The word “Acceleration” of a charge means that we need to have a reference frame to really say that the charge is accelerating. Without a reference frame we cannot say that something is accelerating.
3. An accelerating charge radiates, and this radiation does not need to be treated at the quantum level.
The above three facts are true but one of them is incomplete.
Let me show that we get into an experimental contradiction using solely the above three facts. Let’s design an ideal experiment: A freely falling charge in a gravitational well. Two observers in two different inertial frames (fact 1): one co-moving with charge along the geodesic, and another along the same geodesics, but far away, say at infinity. The first observer, in free fall with the charge, will measure no acceleration (GR), since the charge appears stationary to him (fact 2). Therefore the observer records no radiation from the charge (fact 3). The second observer at infinity measures that the charge is accelerating with respect to him (fact 2), since the charge and the observer at infinity have different values of the gravitational potential (their relative motion is accelerated, nevertheless they are inertial according to GR). Therefore, in principle, the second observer “should” notice that the charge radiates (fact 3). But both observers are in inertial frames and therefore must observe the same physics (fact 1), i.e., either there is radiation (it must be observable in both frames) or there is not. Apparently, there is a contradiction for the two inertial observers, since the one in free fall observes no radiation from the charge, while the other one at infinity observes the charge radiates. The truth that we missed here is in the incompleteness of “fact 3”: We were misguided in all discussions about a charge, that accelerates, radiates. This is not the whole story! What are Liénard–Wiechert potential, Larmor formula, and Cyclotron radiation telling us? Exactly, a charge, accelerated by an external electromagnetic field, radiates. So, it is the presence of the electromagnetic field the very source of the charge’s acceleration (gravity is negligible), and it is this type of interaction that truly triggers the emission of electromagnetic radiation. Therefore, an accelerating charge under gravity, without an external electromagnetic field, will not be able to radiate. The radiation mechanism is entailed to the presence of an electromagnetic field that accelerates the charge: i.e., The interaction of a charge coupled with an electromagnetic field, that accelerates the charge, generates the electromagnetic radiation we observe in any inertial frame.
This is analogous to the spontaneous decay of some particles in the vacuum: Why is this happening? Well, if the vacuum was without fluctuations, then no particle will spontaneous decay. But the vacuum has fluctuations that trigger the stability of a particle and make it to decay (ref., Casimir’s effect, QFT).
I hope this clarifies and answers the original question!
P.S. For a more detail discussion see: http://www.mathpages.com/home/kmath528/kmath528.htm
Thank you Feynman!
Sante
What a coincidence! I found an article published on ArXiv today about the same topic. http://arxiv.org/pdf/1608.07177.pdf
to Daniel Sasso:
Daniel, your comment is more than clear - it is true!
Regards!
Speaking of arXiv, here's an interesting and well-written article from over a decade ago:
https://arxiv.org/ftp/physics/papers/0510/0510115.pdf
that shows that the Larmor power formula can be derived within a factor of order unity by simply estimating the rate of work done by a charge in causing the effective mass of its own electrostatic field to accelerate. That is, energy is lost from an accelerating particle at the Larmor rate because it is doing work against its own electric field.
The original question then becomes: Does a charged particle falling in a gravitational field have to do any work to cause its own electrostatic field to accelerate, or does its electrostatic field fall by itself with the particle. The principle of equivalence says, I believe unambiguously, that the electrostatic field falls with the particle, the particle does no work on its own field, and therefore, no, there is no radiation.
Dear Anatolij,
I reply with delay to your comment. Thanks for your positive appraisal. Anyway don't worry, the downvote that you have received in actuality is for me. I don't worry, I think to be the most downvoted researcher in RG simply because I say and write things that others don't know and don't want to understand. I hope to increase my primacy with this comment.
Best regards.
If this phenomena exists, then should exists a poynting theoreme relating gravity and electromagnetism. Such a thing is not know. The acceleration of gravity is an especial one: it is a relative acceleration that can be localy canceled by an inertial observer. Constant acceleration is exclusive of gravity, no electric field can do it on a single charge.
Article RADIATION OF AN ACCELERATED CHARGE
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