Angular momentum coupling with graviton exchange?

[bcrowell]

I know there are many deep issues that come up in attempts to do quantum gravity, but I suspect that the following is not deep but easily explained by someone who knows more about this than I do.

Suppose two neutrons are interacting gravitationally. The Feynman diagram that I would naively draw for this process would show two neutron world-lines with a graviton being exchanged. I.e., it would be exactly like the H-shaped diagram I'd use to show two particles interacting electrically, except that the photon is replaced by a graviton. But the graviton version obviously doesn't work, because at a given vertex, it's impossible to couple the angular momenta correctly; spins 1/2, 1/2, and 2 don't satisfy the triangle inequality. This would seem to prove that no particle with spin less than 1 can interact gravitationally, which is obviously wrong.

So what does the correct Feynman diagram look like?

 

[Kraflyn]

Hi.

Yes, I understand it now. Fairly simple and still quite interesting question.

My imagination devised a trivial solution while I was making coffee... Well, please allow me to present it. It might prove correct in the end maybe, and we will definitely have something to chatter about. Here it is.

Suppose a test particle is hit by an incoming graviton. Suppose also that at the very moment the scattering occurs, test particle also emits a graviton. So gravitation is in this picture a double simultaneous scattering. If both incoming and outgoing gravitons have exactly opposite spins, the gravitational scattering depicted here would not actually change test particle's spin.

So, to advance this trivial idea further, one might introduce a new kind of line into Feynman diagrams: a double line. It consists of two particles moving in opposite directions with generally different 4-impulses. Rule for that kind of line: spins sum to 0.

This way gravitons would be allowed to interact with all kinds of particles. And all particles would be forced to have mass: one cannot affect another without the instantaneous gravitational reaction. So all particles couple with gravitons.

Furthermore, simultaneous scattering happens at the very same location. Therefore the process is simultaneous for all Lorenz frames. Actually, it is simultaneous for all frames, obviously, not just Lorenz ones.

This trivial idea is so trivial that it might actually be good.

I am also interested to see if there already exists a proper standard explanation of gravitational scattering off fermions.

Cheers.

 

[Kraflyn]

Hi.

Uh, there is another possible explanation, but too exotic one... Here it is. Whenever we arrive at situation when some conservation law is violated, we invoke virtual particles. It does wonders for Feynman diagrams... So, gravitons are definitely virtual as mediators within Feynman diagrams. Mediators are generally off-shell, not obeying $m^2 = E^2 - p^2$ relation. Virtual mediators can also move with super-luminal velocities and perform other miracles, already now within our current understanding of particle processes... So... What if virtual particles were not required to conserve spin either? This would create a host of exotic processes theoretically, not observed experimentally ever... Well, it's just a little thought, aimed at amusing us for a while. Plus, who knows, it might spark a new idea in someone's mind. I do not advise clinging to this exotic proposition of mine too much. It is obviously wrong... But what if it was allowed for gravitons? What would that mean then... Let's brain-storm over it!

Cheers.

 

[member 11137]

I know there are many deep issues that come up in attempts to do quantum gravity, but I suspect that the following is not deep but easily explained by someone who knows more about this than I do.

Suppose two neutrons are interacting gravitationally. The Feynman diagram that I would naively draw for this process would show two neutron world-lines with a graviton being exchanged. I.e., it would be exactly like the H-shaped diagram I'd use to show two particles interacting electrically, except that the photon is replaced by a graviton. But the graviton version obviously doesn't work, because at a given vertex, it's impossible to couple the angular momenta correctly; spins 1/2, 1/2, and 2 don't satisfy the triangle inequality. This would seem to prove that no particle with spin less than 1 can interact gravitationally, which is obviously wrong.

So what does the correct Feynman diagram look like?

Well, I don’t know the answer to the question and I have also no personal theory concerning that question. Because of that my intervention is just a free speculative remark but I do it with the hope and the intention to bring some constructive element for the answer you are looking for.

Gravitation has the reputation to be a universal interaction = it acts everywhere. But in stopping its description to that sentence might induce a misleading classical representation of gravitation.
In fact, gravitation propagates and it does it at the same speed than light. Translating some concepts of the Quantum Theory approach into the theory of relativity brings at least one result: the “graviton” as a spin 2 particle carrying the gravitational interaction (indeed!).
Now let us consider the whole situation and try to summarize it. “Is not any region fulfilled with a gravitational field, far from any source, an analog of a spin 2 particles bath?”
I see another problem with the Feynman’s diagram: gravitons are always considered as “extern” particles interacting with neutrons, coming from the outside of the region occupied by the neutron. This is the transcription of a very classical Newtonian vision at the atomic scale. Does it really work like that at this scale? Why did nobody suggest that gravitons are automatically a part of any other particle like quarks and gluons are automatically a part of any neutron?

So: no solution but one more question!

 

[Haelfix]

Why do you say that its impossible to couple the angular momentum.. The total angular momentum doesn't only depend on the spin components...

What is probably true (but I would have to work out) is that for certain directions/polarizations of the incoming neutrons, you will have vanishing amplitudes. But then I don't see why it should vanish in general.

 

[bcrowell]

Why do you say that its impossible to couple the angular momentum.. The total angular momentum doesn't only depend on the spin components...

At the vertex, all three particles have the same r vector. So by conservation of momentum, isn't the incoming orbital angular momentum guaranteed to equal the outgoing orbital angular momentum? My understanding is that at every vertex, there is supposed to be conservation of charge, spin, and 4-momentum, even if there are particles that are off shell.

At this point, I can only envision three possible resolutions to this:
(a) Graviton vertices (always?) have more than 3 legs, so they don't need to satisfy a simple triangle inequality in terms of spin.
(b) Gravity does proceed through vertices with 3 legs, but these legs have 2 gravitons and one other particle, so that the simplest interaction between two neutrons might be something like a graviton box, rather than an H with a simple graviton exchange. But this would seem to run into problems with other conservation laws. E.g., at a neutron-graviton-graviton vertex, you'd be violating conservation of baryon number.
(c) I can cook up complicated diagrams that work. E.g., a neutron could emit a virtual Z or W, which could then participate in graviton exchange without violating a triangle inequality. But this kind of process would presumably be extremely weak, and it would seem to violate the equivalence principle.

So only (a) above seems viable...?

 

[Bill_K]

Similarly, how does electromagnetism (spin 1) couple to a spin 0 meson? That also looks impossible from the angular momentum point of view. Answer: through a derivative coupling, L = A_μj_μ where j_μ for a Klein-Gordon particle is like φ*∂_μφ, or in momentum space φ*k_μφ.

But wait, doesn't the gauge condition imply that A_μk_μ = 0? Answer: not off the mass shell. A spin 0 meson also experiences a Coulomb interaction, in which the photon acts effectively as spin zero.

The graviton couples to the stress-energy tensor -- for an electron this is like ψγ_(μ∂_ν)ψ. On the mass shell, the graviton obeys the gauge condition k_μh_μν = 0, which restricts it to spin-two behavior. The physical meaning of this is that gravitational waves couple in lowest order to a time-varying quadrupole moment, but the electron being spin-½ cannot have a quadrupole moment.

However off the mass shell, gravity also produces spin-1 and spin-0 effects, e.g. Newtonian attraction.

 

[Haelfix]

Right, so if this was say a decay vertex where an on shell graviton was produced, you would definitely have to worry about angular momentum conservation.

However, as the graviton is off shell (and a contact term not produced) the relevant information about conservation laws is encoded in the Ward identities, which you can compute and by direct inspection they do not necessarily vanish.

You would have to compute it to be sure, but I suspect that certain polarizations of the incoming neutrons do actually vanish, similar to the case of photon on photon scattering via graviton exchange (where beams that are parralel do not attract, and beams that are antiparralel do)

 

[Kraflyn]

Hi.

Well, as You pointed out earlier, virtual photon can have spin projection 0. So can graviton too. This solves the issue trivially too. This is due to the fact that real massless particles can only have helicity $\pm 1$, but virtual massles particles are off-shell and acquire mass through $q^2 \ne 0$. This way virtual mediators can have helicity 0. This is also true for virtual gravitons.

Cheers.

 

[Kraflyn]

... Although, there is something so very wrong with these virtual particles that can exist for only just a while... How on Earth and in Heaven can a short-living virtual particle send a signal from Andromeda?

Cheers.

 

[bcrowell]

Similarly, how does electromagnetism (spin 1) couple to a spin 0 meson? That also looks impossible from the angular momentum point of view. Answer: through a derivative coupling, L = A_μj_μ where j_μ for a Klein-Gordon particle is like φ*∂_μφ, or in momentum space φ*k_μφ.

But wait, doesn't the gauge condition imply that A_μk_μ = 0? Answer: not off the mass shell. A spin 0 meson also experiences a Coulomb interaction, in which the photon acts effectively as spin zero.

Interesting. Most of this goes over my head, but it's good to have an example involving known physics.

But can't I wriggle out of this by saying that the pion is composite, and the vertex is really a quark-quark-photon vertex?

 

[Kraflyn]

Hi.

Any progress on the subject?

Namely, on the subject of angular momentum conservation at vertex with 3 legs describing graviton production.

Cheers.

 

[bcrowell]

After some more thought and study, I think I understand this a little better, although still not 100%.

I argued in #6 that you have to have separate conservation of spin and orbital angular momentum at a vertex, since it's a point interaction, and all the particles have the same r. I think this is actually a valid classical argument (interactions at a point always conserve L if they conserve p), but not a valid quantum-mechanical argument, due to the uncertainty principle. The particles represented by a Feynman diagram are not necessarily localized at all, as you might think from the visual representation. I don't know if this is crucial to a full resolution of my OP, but it was certainly wrong.

Classically, static fields and near fields don't necessarily act like the classical analogs of spin-1 photons and spin-2 gravitons. For example, the static electric field of a point charge has zero momentum density everywhere, and therefore zero angular momentum density everywhere. I think this may be a partial and less highbrow way of suggesting what Bill_K was saying in #7.

Finally, I think there is a fundamental and simple reason why you can't require virtual particles to have the "right" spin if you want to produce static attraction or repulsion. These effects arise from interference between the no-exchange diagram and the one-particle-exchange diagram. (Baez has a nice explanation of this with crayons: http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html .) If the internal state of the emitting or absorbing particle changes, then you can't have this interference. Therefore the virtual particle has to have spin 0.