# How does a particle's acceleration relate to the emitted photon?

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jactor
I understand that any accelerating charged particle will emit a photon. But I do not understand how the rate of acceleration relates to the photon that is emitted.

For example:
If a proton is falling to Earth in a vacuum at 9.8 m/s^2, then what are the characteristics of the photon that will be emitted? And how do the characteristics differ from another proton 1) accelerated at a different rate and 2) stopped sooner in its fall than the first?

Further, if this weren't in a vacuum and the proton reached terminal velocity, then why does a photon stop getting emitted after terminal velocity is reached?

Edit: Additionally, what effect would a change in rate of acceleration have on the photon? (e.g. accelerating faster as it gets closer to the Earth)

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Mentor
I understand that any accelerating charged particle will emit a photon. But I do not understand how the rate of acceleration relates to the photon that is emitted.
An accelerated charged particle will emit electromagnetic radiation, but that’s not the same thing as emitting a photon. The electromagnetic radiation emitted by accelerated charges does not in general correspond to a single-photon state.

We’re better off treating radiation from accelerated charges using classical electrodynamics (no photons, just waves, particle motion described by classical trajectories), and even then there’s much more to the problem than the simple statement “accelerated charges radiate” would suggest. Google for “Larmor formula” and “Lienard-Wiechert potential” to get started.

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• jactor
An accelerated charged particle will emit electromagnetic radiation, but that’s not the same thing as emitting a photon.
Would you be able to cite a source for this? My understanding is that a photon is a common representation of light. And when representing light with photons, then any EM wave would consist of photons.

The electromagnetic radiation emitted by accelerated charges does not in general correspond to a single-photon state.

We’re better off treating radiation from accelerated charges using classical electrodynamics (no photons, just waves, particle motion described by classical trajectories), and even then there’s much more to the problem than the simple statement “accelerated charges radiate” would suggest. Google for “Larmor formula” and “Lienard-Wiechert potential” to get started.

I'm strictly interested in what the differences will be in light emitted between the cases in my example. I'm not too concerned with whether we represent the light as waves or photons.

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Would you be able to cite a source for this? My understanding is that a photon is a common representation of light. And when representing light with photons, then any EM wave would consist of photons.
Any text on classical EM would do. E.g. Introduction to Electrodynamics by Griffiths, chapter 11.

There is simply no such thing as a photon in classical EM, where light is modeled as an EM wave obeying Maxwell's equations. In particular, an EM wave does not consist of photons.

Photons arise in QED (Quantum Electrodynamics), which is a different theory of light, where they do not represent classical EM radiation.

• vanhees71
jactor
Right, but as long as we are using photons as the representation of light, then any charged particle being accelerated would produce a photon? Edit: or multiple photons

• weirdoguy
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Right, but as long as we are using photons as the representation of light, then any charged particle being accelerated would produce a photon? Edit: or multiple photons
I'd say not really. First, once you step away from classical EM into quantum mechanics, it's not so clear what you mean by "accelerated". There are no classical particle trajectories in QM. Remember that QM originated to explain why the electron orbiting the hydrogen nucleus did not radiate away its energy.

Second, in QED charged particles produce virtual photons all the time, so to speak: that's how the EM force/interaction is mediated. What you need, therefore, is a QED model for the scenario of an accelerated classical particle. There are a number of hits if you search for this, such as:

https://arxiv.org/abs/quant-ph/0407162

And there is a discussion here:

https://physics.stackexchange.com/questions/79932/accelerated-charge-in-qed

• PeterDonis and jactor
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Photons are not "the representation of light" but a very special state of the electromagnetic field, which is not so easy to prepare as it seems. Of course today we have the possibility to use, e.g., parametric down conversion to prepare with great accuracy and high intensity heralded single photons.

If you have accelerated charged particles you rather prepare something like a coherent state (like a radio wave emitted from an antenna with a more or less harmonically oscillating charge-current distribution or visible light emitted by a laser pointer). Other common light sources as the Sun are emitting thermal radiation, i.e., an electromagnetic-field state in or close to thermal equilibrium with the material making up the source.

It's much better to think about light in terms of electroamagnetic waves than in some oversimplified photon picture of too many popular-science books.

• jactor and PeroK
jactor
@PeroK and @vanhees71 Appreciate your responses, I will definitely dig into that topic further.

If we drop back to wave theory of light, is there a simple answer to my original example? Is it just the speed of light compared to rate of acceleration to determine what the wavelength/frequency of EM wave will be?

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What do you mean by "drop back"? I'd say today we rather are at a point to realize that on a fundamental level the most comprehensive description of all matter is indeed in terms of "waves" or "quantum fields.

The wavelength and frequency of the em. wave of course depends on, how this wave is made. E.g., if you have an antenna with a harmonically time-dependent current you'll get an em. wave with a frequency given by the current, and the wavelength is determined by the dispersion relation of this wave in the medium it travels (in the vacuum it's just ##\lambda=c/f##, where ##f## is the frequency of the wave, i.e., the frequency of the harmonically time-dependent current.

Further the Maxwell equations, describing how the waves are emitted due to the charge-current distributions are linear, and that's why you can describe the waves from sources of arbitrary time dependence by the superposition of waves of all frequencies, i.e., a Fourier integral.

Another representation is in terms of a socalled "retarded Green's function", which directly gives a relation between the electromagnetic charge and current distributions and the emitted electrodmagnetic field in terms of an integral. In this picture you can think of the radiation field as the superposition of spherical waves radiated out from any point, where accelerated charges are. "Retarded" means that if you observe the light at some place it is the superposition of waves emitted from each point of the source at a time needed by the light signal to reach your point of observation, i.e., if the point in question is at a distance ##r## from you, you see the wave emitted at time ##r/c##. That's why what we see as the Sun is in fact the Sun how it looked about 8 min or so before. For some star the light may be emitted millions of years ago.

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If we drop back to wave theory of light, is there a simple answer to my original example? Is it just the speed of light compared to rate of acceleration to determine what the wavelength/frequency of EM wave will be?
In general, classical EM was unable to explain the frequency distribution of EM radiation from, for example, an ideal black body. This was known as the ultraviolet catastrophe. This again relates to the orgin of QM, which was developed to explain this.

In general, therefore, classical EM can't answer this question.

• vanhees71
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Would you be able to cite a source for this? My understanding is that a photon is a common representation of light. And when representing light with photons, then any EM wave would consist of photons.
It's tempting to think that an EM wave would consist of photons the way that a river consists of water molecules moving by, but it's not right. The relationship between photons and EM waves is much more complicated than that and photons themselves do not behave anything like you'd expect from the word "particle". To understand photons properly requires quantum electrodynamics, a subject that won't be encountered outside of a graduate-level PhD course. For an I-level thread, the best source I can offer is http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf
If we drop back to wave theory of light, is there a simple answer to my original example? Is it just the speed of light compared to rate of acceleration to determine what the wavelength/frequency of EM wave will be?
The only simple case I know of is a particle undergoing simple harmonic motion, oscillating back and forth. In this case the frequency of the emitted radiation is the frequency of the oscillation of the particle; there are many good visualizations online and once you've looked at one it will be intuitively obvious what's going on.

Some other posters here may be aware of other similarly straigthforward examples, but in general for more complicated motion you'll need to start with the Larmor formula and Lienard-Wiechert potentials I mentioned above - and be aware that these problems generally are not trivial. Googling for "accelerated charge radiation" will bring up many good sources at various levels of complexity.

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• vanhees71, Doc Al and PeroK
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My understanding is that a photon is a common representation of light.

A photon is a common quantum mechanical representation of light. But it's not the only possible one. It's not common because any phenomenon involving light can be described using states that have a useful "photon" interpretation. It's only common because (a) there are phenomena of interest that do involve such states, but more than that, (b) it's become fashionable to say "photon" instead of "light" even though "photon" is often not a useful description of the phenomenon involving light that is being discussed.

• vanhees71
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For example:
If a proton is falling to Earth in a vacuum at 9.8 m/s^2, then what are the characteristics of the photon that will be emitted?
That's a bad example, because it's very controversial whether a charge in free gravitational fall radiates. That's because, by the equivalence principle, one expects it to be equivalent to an inertial charge without gravity.

• vanhees71
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