Proton + Photon = ???
May 3, 2010 11:25 AM
We've all heard about what happens when an electron absorbs a photon (it jumps orbits) but what happens when a proton absorbs a photon? I googled it to no avail.
I'm no physicist, I am just trying to get through Griffiths' "Introduction to Elementary Particles". What I understand so far leads me to believe you may produce a Delta particle in some cases. I don't know what else you may end up with, though.
posted by Fruny at 11:48 AM on May 3, 2010
posted by Fruny at 11:48 AM on May 3, 2010
You are thinking about an electron bound to a nucleus in an atom. The "jumping orbits" refers to exciting the system as a whole up to a higher energy state. People like to think of the electron "orbiting" the proton because electrons are very light compared to protons, so the situation is somewhat analogous to the Earth orbiting the Sun (well, a bit closer to Jupiter or Saturn orbiting the Sun, but whatever). So you might think of external forces primarily affecting the electron, but it's important to keep in mind that you're really dealing with a two body problem, and treating the electron as "orbiting" a fixed point is only an approximation.
It may help to consider positronium or muonic atoms instead. The energy levels will be different because the masses involved are different (in fact, you can reduce the problem to an equivalent one-body problem by considering the reduced mass of the system), but otherwise everything works basically the same. It just so happens that in these systems the mass difference between the involved particles is smaller.
posted by dsword at 11:56 AM on May 3, 2010
It may help to consider positronium or muonic atoms instead. The energy levels will be different because the masses involved are different (in fact, you can reduce the problem to an equivalent one-body problem by considering the reduced mass of the system), but otherwise everything works basically the same. It just so happens that in these systems the mass difference between the involved particles is smaller.
posted by dsword at 11:56 AM on May 3, 2010
Protons have energy states within the nucleus, as well. Of course, it takes a much higher energy photon to excite them.
posted by stevis23 at 12:15 PM on May 3, 2010
posted by stevis23 at 12:15 PM on May 3, 2010
The usual description takes the point of view that the nucleus is massive, stable and unchanging, while all the energy levels that take place occur because of jumps of electrons from one orbit to the other. This description is incomplete. When a photon is absorbed by an electron/atom, the momentum carried by the photon is transferred to the recipient (producing a recoil). Since the atom is a bound system, it is the entire atom that recoils ... and not the individual electron that is said to have absorbed the photon and changed orbit.
As dsword mentioned, it might be useful to consider positronium or muonic atoms instead. In these cases, there is not one electron (or muon) that jump orbit while the other component stays unchanged. Here, it is the entire "atom" that absorbs the energy.
So, while the standard explanation/description given is that the electron absorbs a photon and jumps to a new orbit, a better explanation/description might be that it is the entire atom that absorbs a photon, resulting in a change in its energy levels, with one electron finding itself into a higher orbit. Energy levels available are quantized (i.e. discrete) and we do associate a particular energy with a particular combination of orbits - hence, when there is a change in energy, we say that a given electron changed orbit ... and say that the electron "absorbed" the photon ... but this is a simplification.
It may actually help to consider more complicated objects, like molecules. They have energy levels associated with 1) electronic orbits, 2) vibration modes, 3) rotational modes, and 4) nuclear energy levels, as well as some translational energy. When a molecule absorbs a photon (of the appropriate energy), one of the energy levels changes (with also a change in translational energy). To simplify, we sometimes associate the absorption of the photon as having been done by one given particle (with a change in orbit of that particle, be it an electronic or nuclear energy level change) ... but really, it is the entire molecule that is affected.
posted by aroberge at 1:33 PM on May 3, 2010
As dsword mentioned, it might be useful to consider positronium or muonic atoms instead. In these cases, there is not one electron (or muon) that jump orbit while the other component stays unchanged. Here, it is the entire "atom" that absorbs the energy.
So, while the standard explanation/description given is that the electron absorbs a photon and jumps to a new orbit, a better explanation/description might be that it is the entire atom that absorbs a photon, resulting in a change in its energy levels, with one electron finding itself into a higher orbit. Energy levels available are quantized (i.e. discrete) and we do associate a particular energy with a particular combination of orbits - hence, when there is a change in energy, we say that a given electron changed orbit ... and say that the electron "absorbed" the photon ... but this is a simplification.
It may actually help to consider more complicated objects, like molecules. They have energy levels associated with 1) electronic orbits, 2) vibration modes, 3) rotational modes, and 4) nuclear energy levels, as well as some translational energy. When a molecule absorbs a photon (of the appropriate energy), one of the energy levels changes (with also a change in translational energy). To simplify, we sometimes associate the absorption of the photon as having been done by one given particle (with a change in orbit of that particle, be it an electronic or nuclear energy level change) ... but really, it is the entire molecule that is affected.
posted by aroberge at 1:33 PM on May 3, 2010
When an atomic nucleus (in a magnetic field) absorbs a photon, you can get nuclear magnetic resonance. These photons are in the radiofrequency range of the EM spectrum.
posted by sararah at 2:05 PM on May 3, 2010
posted by sararah at 2:05 PM on May 3, 2010
The original question was about photon-photon interactions, not photons with atoms (or electrons).
Photon-photon interactions are called two-photon physics or gamma-gamma physics.
Photons do not couple directly with each other because they have no charge. Light waves pass through each other without interference. Photons do not collide.
But they can interact in higher level processes or create mass.
posted by JackFlash at 2:23 PM on May 3, 2010
Photon-photon interactions are called two-photon physics or gamma-gamma physics.
Photons do not couple directly with each other because they have no charge. Light waves pass through each other without interference. Photons do not collide.
But they can interact in higher level processes or create mass.
posted by JackFlash at 2:23 PM on May 3, 2010
There's a lot of separate issues that have been alluded to (and somewhat mixed up) in the above comments.
The example you've given above, of a photon hitting an electron and knocking it to a higher energy level, is what physicist call an excitation of a bound state. Essentially, it's energetically favourable for an electron and a nucleus to be nice and close to each other (forming an atom) than it is for them to be an arbitrary distance apart. The difference in energy between these two states (particles close together vs. particles far apart) is what's known as the binding energy.
When a photon with less energy than the binding energy comes along, it can happen that (as you say) the electron gets bumped up to a higher orbital. These orbitals are examples of excited states — "intermediate" states of the system in which the system has somewhat more energy than the bare minimum energy possible, but not enough energy to fly apart. Most bound systems will have at least one of these intermediate states, though it's possible to concoct systems which don't have any.
So what happens when a photon strikes a proton? Well, if the proton is bound to other protons and neutrons in a nucleus, the nucleus (being itself a bound state of several particles) can be bumped up to an excited state. It's not as easy to visualize this excitation as acting on just one particle; very roughly speaking, the protons and neutrons are so tightly bound together that any energy added to one quickly gets shared among the rest. It's more like hitting a drumhead than hitting a billiard ball; the energy gets spread out to the rest of the system rather than just going into one particle.
If the proton itself is alone, and not bound to any other nucleons, then what happens? Well, the proton itself happens to be made up of smaller particles (quarks) that are bound together, and if you give these sub-particles enough energy then they'll get excited up to a higher energy level. This is the Delta particle that was referred to above; like the nuclear excitations I mentioned above, it's best viewed as a collective excitation of all the constituent particles in the proton rather than an excitation of any particular particle inside. This requires a fairly high-energy photon, though — a gamma ray with a wavelength on the order of the diameter of the proton itself.
Finally, what happens if you have a photon hitting an electron that's just sitting in empty space, unbound to anything else? Generally, it'll just bounce off like a billiard ball, albeit a highly relativistic billiard ball. This process is called Compton scattering; generally, the electron ends up taking a fraction of the photon's energy (smaller for less energetic photons, greater for more energetic photons), and the photon heads off in a different direction. Compton scattering also occurs when photons without enough energy to cause the proton to transition to a Delta particle (i.e. most photons) impact a free proton.
(On preview: good, the original question is in fact asking about photon-proton interactions. I got scared there for a moment that I'd typed all that enough for nothing.)
posted by Johnny Assay at 2:48 PM on May 3, 2010
The example you've given above, of a photon hitting an electron and knocking it to a higher energy level, is what physicist call an excitation of a bound state. Essentially, it's energetically favourable for an electron and a nucleus to be nice and close to each other (forming an atom) than it is for them to be an arbitrary distance apart. The difference in energy between these two states (particles close together vs. particles far apart) is what's known as the binding energy.
When a photon with less energy than the binding energy comes along, it can happen that (as you say) the electron gets bumped up to a higher orbital. These orbitals are examples of excited states — "intermediate" states of the system in which the system has somewhat more energy than the bare minimum energy possible, but not enough energy to fly apart. Most bound systems will have at least one of these intermediate states, though it's possible to concoct systems which don't have any.
So what happens when a photon strikes a proton? Well, if the proton is bound to other protons and neutrons in a nucleus, the nucleus (being itself a bound state of several particles) can be bumped up to an excited state. It's not as easy to visualize this excitation as acting on just one particle; very roughly speaking, the protons and neutrons are so tightly bound together that any energy added to one quickly gets shared among the rest. It's more like hitting a drumhead than hitting a billiard ball; the energy gets spread out to the rest of the system rather than just going into one particle.
If the proton itself is alone, and not bound to any other nucleons, then what happens? Well, the proton itself happens to be made up of smaller particles (quarks) that are bound together, and if you give these sub-particles enough energy then they'll get excited up to a higher energy level. This is the Delta particle that was referred to above; like the nuclear excitations I mentioned above, it's best viewed as a collective excitation of all the constituent particles in the proton rather than an excitation of any particular particle inside. This requires a fairly high-energy photon, though — a gamma ray with a wavelength on the order of the diameter of the proton itself.
Finally, what happens if you have a photon hitting an electron that's just sitting in empty space, unbound to anything else? Generally, it'll just bounce off like a billiard ball, albeit a highly relativistic billiard ball. This process is called Compton scattering; generally, the electron ends up taking a fraction of the photon's energy (smaller for less energetic photons, greater for more energetic photons), and the photon heads off in a different direction. Compton scattering also occurs when photons without enough energy to cause the proton to transition to a Delta particle (i.e. most photons) impact a free proton.
(On preview: good, the original question is in fact asking about photon-proton interactions. I got scared there for a moment that I'd typed all that enough for nothing.)
posted by Johnny Assay at 2:48 PM on May 3, 2010
Sorry, my mistake above. I misread photon-photon, but you asked about proton-photon. I'm blaming it on fonts.
posted by JackFlash at 4:10 PM on May 3, 2010
posted by JackFlash at 4:10 PM on May 3, 2010
Mostly concurring with the above answers: what you describe is what happens when an atom absorbs a photon. In fact it's only approximately what happens. Changing the orbit of an electron about a nucleus changes the position/velocity/wavefunction of both the electron and the nucleus, and a transition in one electron changes all the electron-electron interactions so that the wavefunctions of the "spectator" electrons all change too. But those changes are all small, since the electron whose orbit changes has a mass thousands of times smaller than the mass of the nucleus, and since the electron whose orbit changes exerts only a small part of the force on any of the spectator electrons. For most transitions the approximation that only one electron orbit changes is good enough to be useful.
You have a suggestion above that nuclear magnetic resonance is that happens when a nucleus absorbs a photon. I would disagree: nmr is also a process where photons are absorbed and emitted by the entire atom, as a collective object. There the transition in the atom is, in the same sort of approximation, a change in the angles between the spin of the nucleus, the spins of the outermost electrons, and the planes of the electrons' orbits. NMR transitions use low energy (audio- or radio-frequency) photons because the changes involve the magnetic fields of the subatomic particles, whose interactions carry less energy than those involving the electric fields. But because of the way that different sorts of spin mix in quantum mechanics, you don't get to say "aha! I've flipped the spin of a nucleus!" unless you add some extra experimental cleverness.
If you're comfortable with the idea of an electron jumping orbits, there is a useful model for thinking about what happens when a nucleus does absorb a photon. You might remember that the reason for talking about "inner" and "outer" electrons is that electrons obey the exclusion principle. A given electron orbit can only hold two electrons, one with each spin; once the orbit is full, any extra electrons have to go into a higher, less tightly bound orbit. Well, protons and neutrons in the nucleus also obey the exclusion principle, and so their orbits within the nucleus also pile on top of each other in the same way as the orbits of the electrons pile onto the atom. The approximation that protons and neutrons (together, nucleons) are independent particles which can jump between well-defined paths is not so good as the approximation that atomic electrons are independent particles jumping between their well-defined orbitals, but it's good enough to expand into a quantitatively useful statement of what happens when a nucleus absorbs a gamma ray.
Finally there's the possibility of a photon with so much energy that it excites a nucleon directly, instead of just rearranging the nucleons inside of a nucleus. Some other folks have already alluded to this above: the first excited state of the nucleon is a particle called the delta, Δ. You might expect based on the explanations above that the Δ could be described as a proton with one of its three quarks in some sort of an excited orbit. It turns out this is not a good approximation: there's not any "shell model" approximation for the quarks in a nucleon that gives quantitatively useful results. It turns out that it is quantitatively useful to describe a Δ as an unstable "atom" made of a nucleon orbited by a pi meson. So a reasonable hand-waving model of the Δ resonance is that a photon creates a quark-antiquark pair in the vicinity of a proton, over the (brief!) lifetime of the Δ particle the four quarks and one antiquark arrange themselves into a nucleon and a pion, and then the pion falls off.
posted by fantabulous timewaster at 7:14 PM on May 10, 2010
You have a suggestion above that nuclear magnetic resonance is that happens when a nucleus absorbs a photon. I would disagree: nmr is also a process where photons are absorbed and emitted by the entire atom, as a collective object. There the transition in the atom is, in the same sort of approximation, a change in the angles between the spin of the nucleus, the spins of the outermost electrons, and the planes of the electrons' orbits. NMR transitions use low energy (audio- or radio-frequency) photons because the changes involve the magnetic fields of the subatomic particles, whose interactions carry less energy than those involving the electric fields. But because of the way that different sorts of spin mix in quantum mechanics, you don't get to say "aha! I've flipped the spin of a nucleus!" unless you add some extra experimental cleverness.
If you're comfortable with the idea of an electron jumping orbits, there is a useful model for thinking about what happens when a nucleus does absorb a photon. You might remember that the reason for talking about "inner" and "outer" electrons is that electrons obey the exclusion principle. A given electron orbit can only hold two electrons, one with each spin; once the orbit is full, any extra electrons have to go into a higher, less tightly bound orbit. Well, protons and neutrons in the nucleus also obey the exclusion principle, and so their orbits within the nucleus also pile on top of each other in the same way as the orbits of the electrons pile onto the atom. The approximation that protons and neutrons (together, nucleons) are independent particles which can jump between well-defined paths is not so good as the approximation that atomic electrons are independent particles jumping between their well-defined orbitals, but it's good enough to expand into a quantitatively useful statement of what happens when a nucleus absorbs a gamma ray.
Finally there's the possibility of a photon with so much energy that it excites a nucleon directly, instead of just rearranging the nucleons inside of a nucleus. Some other folks have already alluded to this above: the first excited state of the nucleon is a particle called the delta, Δ. You might expect based on the explanations above that the Δ could be described as a proton with one of its three quarks in some sort of an excited orbit. It turns out this is not a good approximation: there's not any "shell model" approximation for the quarks in a nucleon that gives quantitatively useful results. It turns out that it is quantitatively useful to describe a Δ as an unstable "atom" made of a nucleon orbited by a pi meson. So a reasonable hand-waving model of the Δ resonance is that a photon creates a quark-antiquark pair in the vicinity of a proton, over the (brief!) lifetime of the Δ particle the four quarks and one antiquark arrange themselves into a nucleon and a pion, and then the pion falls off.
posted by fantabulous timewaster at 7:14 PM on May 10, 2010
Several things can happen. First, if the energy of the photon is too low, the photon could just scatter off of the proton - this is especially true if it is a "free" proton, which is just an ionized hydrogen atom; however, there isn't a huge amount of H sitting around - it is the most metallic element, and as such, quickly acquires an electron and bonds with something else (often another hydrogen atom) to make a molecule.
Given this, in a theoretical case where you have a free proton, the will just scatter off of the much larger (and more energetic) proton in most cases (if the energy of the electron isn't sufficiently high). Keep in mind, however, that a proton is a hadron, and as such, is made up of 3 quarks - it is not a fundamental particle, like the electron. Given enough energy, you could produce (a very improbable, due to the energy required and the number of vertices in the Feynman diagram) an interaction such that the photon scatters off an electron (or a virtual one), and the electron is energetic enough to produce deep inelastic scattering, which was first used to prove the existence of quarks. You *could* - although it is very unlikely - have the scattered electron hit the proton and produce a neutron and an anti-electron neutrino. A similar process, known as inverse beta decay, occurs *within* some radioactive atoms, but here the proton isn't free, and it "swallows" one of it's own electrons.
For photons interacting with protons in a nucleus, it is possible to change the nucleus's total spin state (which has to do with adding up all the spins in the nucleus using nasty Clebsch-Gordan coefficients and Lie Algebra and is in no way as simple as adding two ups to a down and getting an up spin). But in this case, the photon isn't interacting with a single proton, but rather a collection of protons and neutrons in the nucleus to change its overall energy state. In this respect, photon-nucleus scattering is a little similar to photon/atomic electron scattering in that there is a discrete number of energy levels in the nucleus (none in between), and a photon, if it has the EXACT energy required, can cause a transition from one energy level to another. In most cases, there is no definitive answer to your question, because as with all quantum mechanics, it is probabilistic in nature and you can't know what will happen until the interaction occurs. For the most part, protons do not often interact with protons individually.
posted by physicsphil at 12:36 AM on May 20, 2010
Given this, in a theoretical case where you have a free proton, the will just scatter off of the much larger (and more energetic) proton in most cases (if the energy of the electron isn't sufficiently high). Keep in mind, however, that a proton is a hadron, and as such, is made up of 3 quarks - it is not a fundamental particle, like the electron. Given enough energy, you could produce (a very improbable, due to the energy required and the number of vertices in the Feynman diagram) an interaction such that the photon scatters off an electron (or a virtual one), and the electron is energetic enough to produce deep inelastic scattering, which was first used to prove the existence of quarks. You *could* - although it is very unlikely - have the scattered electron hit the proton and produce a neutron and an anti-electron neutrino. A similar process, known as inverse beta decay, occurs *within* some radioactive atoms, but here the proton isn't free, and it "swallows" one of it's own electrons.
For photons interacting with protons in a nucleus, it is possible to change the nucleus's total spin state (which has to do with adding up all the spins in the nucleus using nasty Clebsch-Gordan coefficients and Lie Algebra and is in no way as simple as adding two ups to a down and getting an up spin). But in this case, the photon isn't interacting with a single proton, but rather a collection of protons and neutrons in the nucleus to change its overall energy state. In this respect, photon-nucleus scattering is a little similar to photon/atomic electron scattering in that there is a discrete number of energy levels in the nucleus (none in between), and a photon, if it has the EXACT energy required, can cause a transition from one energy level to another. In most cases, there is no definitive answer to your question, because as with all quantum mechanics, it is probabilistic in nature and you can't know what will happen until the interaction occurs. For the most part, protons do not often interact with protons individually.
posted by physicsphil at 12:36 AM on May 20, 2010
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posted by jardinier at 11:30 AM on May 3, 2010