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You'll have to forgive me if this question is too wrapped up in "classical" thinking.

I've read that electrons and protons interact by trading photons, but this only raises more questions. What determines the energy and direction of the emitted photon? How often can a particle emit a photon? How often can a particle absorb a photon? Can one particle emit and absorb multiple photons at once? Where does the energy to emit a photon come from? Is the destination of the photon somehow pre-determined or is the photon simply emitted in the hopes of being absorbed?

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  • $\begingroup$ Electrons, protons and photons are merely states of the same quantum mechanical object, which we call a quantum field. That quantum field has the properties of a) having (almost?) pure electron states, b) having pure photon states, c) almost pure proton states and d) allowing for an interaction between these states which is, MOSTLY mediated by the exchange of virtual photons and can lead to the release of a real photon state under the correct initial conditions for one electron and one proton state. But whatever happens, we are always talking about a single object interacting with itself. $\endgroup$ – CuriousOne Aug 28 '14 at 4:00
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Earlier answers seem to be off-topic, since their authours talk about real photons, while the question is asked about the picture of virtual photons which serve as interaction mediators for the electromagnetic interaction, even in the electrostatic case.

The most important thing to settle beforehand is that the picture of interaction by exchanging virtual particles assumes essentially quantum arrangement. That means some different way of thinking about reality and processes. I advice the popular book
Feynman. QED, The Strange Theory of Light and Matter
that explains it the best way. Here I tell very roughly the most important points.

In classical physics you can think that processes proceed as they are described as the time goes by. In quantum physics, you think by the following pattern 1:

  1. You imagine the process as a whole, from its initial state to its final state.
  2. For this process as a whole, you calculate a complex number called the probability amplitude (it is just a word, don't think of its sense). In the Feynman's book it is called 'arrow' for simplicity.
  3. You imagine all other possible processes that give exactly the same final state. For them you repeat steps 1 and 2. Sometimes you can skip very complicated processes because they give very small numbers.
  4. You add all probability amplitudes, and only after that you decide, whether the process has taken place at all.

For the interaction by the exchange of virtual particles, this means that the absorber has the same importance as the emitter. It is the presence of the absorber "in the right place and time" that makes the whole process possible at all.

Added later: Also for the words emitting and absorbing, they are used in some figurative sense, since the virtual photons are emitted and absorbed within the bounds of a single quantum process, and cannot reach some detector, for example. Also, the temporal order of interactions can switch depending on the viewpoint, so the roles of the emitter and the ovserver can switch as well. More about that in Feynman's.

Now we are ready to go through your questions.

What determines the energy and direction of the emitted photon?

The positions and velocities of both the emitter and the absorber. After some perplexed 4-dimensional algebra, that comes down to the usual Coulomb and Biot-Savart laws.

Notice that for two static charges the energy of the photon would be 0! Such photons would transfer only momentum, until at least one of charges would gain some speed. That corresponds to the fact that the Coulomb force does not produce power if the charge does not move.

How often can a particle emit a photon?

As often as it needed to make the interaction of needed strength, for given emitter and absorber.

How often can a particle absorb a photon?

As often as it needed to make the interaction of needed strength, for given emitter and absorber.

These two questions lead to the question "how often two particles actually exchange photons?" That is calculated by the value of action of the whole process (which you have considered on the step 1). Very roughly, you can take the energy of interaction $E$, the interval of time $\Delta t$, and then the action per that time would be $A=E\,\Delta t$. This action can be attibuted (very roughly) to the interchange of $n$ photons where $A=2\pi\hbar n=hn$ and $2\pi\hbar=h$ is the Plank's constant.

You see that the closer charges are, the more photons they exchange, and as the time goes by, more and more photons run between them. For macroscopic charges and distances, the number of photons would be very large, so the interaction feels smooth as the classical physics tells. For elementary particles flying by, it is not unusual to exchange only one photon (or none at all), which is one of the most interesting processes for the particle physics.

Can one particle emit and absorb multiple photons at once?

In the Quantum Electrodynamics (QED), no. In some other interactions, it is sometimes possible, for example, one gluon (which is charged with color charge) can emit two other gluons at once.

Where does the energy to emit a photon come from?

From the energy of the charged particle. But remember, the energy of a photon can be 0 (see above). So it is not needed to have some spare energy to take part in interactions. And sometimes the charged particle can get energy, if the other charged particle gives it.

Is the destination of the photon somehow pre-determined or is the photon simply emitted in the hopes of being absorbed?

The destination is determined: it is the absorber. But it is not pre-determined in some temporal sense, because the quantum process happens as a whole, and not by some consequtive stages. Just when the absorber happens to be there to catch the photon, it is emitted.

If the absorber does not happen to be there, actually the photons are emitted anyway. But that is a very special case: all these photons go back to emitter. They do not take any energy or momentum, and their very existence would be unobservable, but they show themselves in some subtler phenomena, being known as radiative corrections.


1 It is important to note that quantum physics can be represented in several ways mathematically equivalent. Here I tell only the Feynman Path Integral picture, which is the most natural for the story about virtual particles. But some explanations would sound wrong and would turn on the different side, if one would start with Schrodinger picture, for example.

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  • $\begingroup$ One should probably mention that talking of emitting and absorbing photons is a just-so story designed to put the Feynman diagrams into non-technical language. Virtual particles (as "exchanged" here) are not empirically detectable, so there is no reason known to me to believe that they exist as real particle do, and one should not take the intuitive picture of charges emitting and recieving photons too seriously. Otherwise, good answer! $\endgroup$ – ACuriousMind Aug 28 '14 at 14:51
  • $\begingroup$ @ACuriousMind Added a few words about that in the middle of the answer. $\endgroup$ – firtree Aug 28 '14 at 15:03
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I think you are confusing two different subjects. Let me explain, and I'll give you the simplified version. If you ask I'll update with a more advanced explanation.

There are four fundamental forces in the universe. Strong Nuclear, Weak Nuclear, Electromagentic, and Gravity. Electrons and protons mainly interact by the electromagnetic force, since this is much stronger at that range (the other forces are there, just much smaller in comparison to this power). The proton and electron will pull on one another actually (opposites attract). Now if we have electrons interacting then they will push (likes repel).

As for photon emission, this is a little different. It is more easily understood from a stance of energy conservation. As you probably know, we have to conserve all energy in a system. This means that $$E_{before}=E_{after}$$ Always.

You may or may not know, but electrons are arranged in orbits around their nuclei. The higher orbits have higher energy states. So let's be simple and consider the hydrogen atom. This is an electron circling a proton (simplest case). If this electron is in an excited state then it can be in the $n=2$ or $n=3$ orbit (can be caused by heating, passing an electrical current through, or other forms of adding energy). But in physics we know that things like to be in the lowest energy state. So this electron will want to return to it's $n=1$ state. So let's look at the situation where the electron goes from an $n=2$ state to an $n=1$ state. We also know that the energy in orbit is $\frac{E_R}{n^2}$ where $E_R = -13.6eV$. So let's look at the two energies. $$E_2 = E_1$$ $$-\frac{13.6eV}{4} = -13.6eV$$ Now this can't be correct, we are missing energy. But we do know that in this effect that a photon is released to make up for it (more advanced topic). So really we have $$E_2 = E_1 + E_{\gamma}$$ $$E_2 = E_1 + \frac{hc}{\lambda}$$ Now we can see that we are able to account for the energy of the transition. And we are actually able to find the wavelength of the light emitted. (I'll leave that to you. Hint: $hc=1240eV nm$)

So yes, this atom is emitting a photon (particles are a different). This is actually how we do spectral analysis. Since these photon emissions end up being quite unique, we can identify which atoms they came from. Another way to excite this atom is to actually inject a photon into it (again, this is simplified). This is the beginning of quantum mechanics actually, and this is a form of radiation.

Getting the destination of the photon is MUCH more difficult to do (and is statistical in nature). I am afraid to answer some of your questions fully, because they are difficult subjects. And some involve statistics. Some things are "predetermined" like the wavelength (colour) of the emitted photon, but others are difficult to deal with, like direction.

I hope this answers your question. Feel free to comment if I am unclear and I will try my best to fix those areas.

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    $\begingroup$ The EM force is the result of charged particles exchanging photons. This is related, though slightly different, to spontaneous emission, which you have described. $\endgroup$ – innisfree Aug 28 '14 at 13:15
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By trading photons? I'm not sure if I missed something everybody else knows, but an electron(-) stays near a proton(+) because opposites attract.

Unless you meant, the photon emitted by an electron moving into a lower energy shell, interacting with the electrons of another atom.

To answer some of your questions:

Where does the energy to emit a photon come from?

As far as I know, an electron is emitted by an electron as it moves to a lower energy shell. Electron emission

Is the destination of the photon somehow pre-determined or is the photon simply emitted in the hopes of being absorbed?

The photon has no "hopes" of being absorbed, as it is inanimate. To answer the question though, I believe the directionality of a photon in completely random.

How often can a particle emit a photon?

There should be no set "time lag" between photon emission events, it just happens whenever an electron changed to a lower energy state. It can occur multiple times at once. It all depends on how many electrons are in a state where they can go to a lower energy state.

How often can a particle absorb a photon?

Again, no set "cooldown" time, it can absorb a photon as many times as the electrons in its electron cloud can go to a higher energy state.

Can one particle emit and absorb multiple photons at once?

Yup. The processes I described can occur simultaneously, many times per second.

I should also mention that atoms emit photons because they are shedding energy. You can add energy to an atom by heating it (think of a stove, and why it glows when hot) and by absorbing photons.

You can read more on the subject here, here, here, and something about reflection here, Also, here's a Google search that might help.

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    $\begingroup$ The EM force is the result of charged particles exchanging photons. $\endgroup$ – innisfree Aug 28 '14 at 13:13

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