As I understand it, what we think of as the electrostatic attraction between a positive and negative charge is caused by the exchange of virtual photons. Now consider what happens when a positive charge emits a virtual photon that is absorbed by the negative charge. Since the positive charge and negative charge move towards each other )or their trajectories are altered in that fashion), I can only understand this if the virtual photon carries momentum whose vector points in the opposite direction of its travel. Conservation of momentum would then make the positive charge move towards the negative charge when the virtual photon is emitted and would make the electric charge move towards the positive charge when it is absorbed.
On the other hand when two negative charges or positive charges repel each other, the virtual photon would carry positive momentum.
Or is this not a good way to think about it at all?
Honestly I don’t remember my quantum electrodynamics too well from school, but I know you have to sum over a range of momenta for the virtual particles, including values that seem physically impossible (i.e. that violate the normal relationship between momentum and energy). Also, thinking of one photon being exchanged is only the lowest-level approximation to the interaction, you can also have multiple photons or photons splitting into other particles which annihilate back into photons and so forth.
I know I always found it more comfortable to think of all that virtual stuff as just a useful formalism for working out the mathematics of the interaction. I’m not sure it’s wise to ascribe too much reality to the virtual particles.
Someone who knows this stuff way better than me will no doubt be along shortly.
Well, assuming that you are ascribing reality to the virtual photons and not just considering them a mathematical trick, but if the trick works, I’ve never seen the argument for saying it’s not “real”. I mean, by the same token, at the time of the Copernican debate many folks said that the heliocentric model was a neat mathematical trick, but obviously not “really” true.
I don’t know quantum field theory all that well, but from what I do know it sounds very plausible that virtual particles could have negative momentum and that’s because they are mathematical fictions.
Section 9 of this article explains why it’s sketchy to talk about virtual particles as if they were a real physical phenomenon.
I can’t help but feel like there should be more to it, though. If Old Guy’s take is essentially correct, then one should be able to write down the tree level calculation for the scattering amplitude, which will include some integral over momentum, and then one should be able to show that the dominant contribution comes from where the photon has a negative momentum. (Or positive for the case of like charges.)
But I’m not quite sure what “negative” momentum would mean here (we could say “opposite the direction of propagation”, but how do you define the direction of propagation independently from the value of the momentum?), and I don’t remember enough to know how to write down the equations without consulting my old quantum field theory textbook, which I don’t have handy.
I guess I’m saying I’d love to see a more mathematically detailed explanation, even if it’s beyond the scope of what the OP really wanted. I imagine it shouldn’t be that hard (for someone much more in-practice at QED than me) if we’re only talking about tree level calculations, although God knows how you post the equations on this forum. Link to images?
Well, virtual photons aren’t “real” in the sense that we can in any way observe or measure them directly, else they wouldn’t be virtual. In contrast to the heliocentric model in which direct measurements of the relative positions of the Sun and planets match the predictions of the model with good precision (much better and cleaner once Kepler’s laws of planetary motion give the model of elliptical orbits), QED offers unseen and indeed unobservable mechanics that offers a precise and consistent result. The mathematics works, and clearly represents some “real” phenomenon, but our conception of photons as discrete particles imparting momentum via some analog of physical impulse is probably not representative of what is happening. Then again, we may just not be able to actually conceive of what happens in quantum mechanics in terms of our normal everyday experience.
You can’t just do the calculation over one of the charged particles. You have to include both. And the end result is that momentum is traded between the two so that it’s properly conserved. The virtual photon is a bookkeeping entity–it has no measurable effect and it used only to simplify the computation.
For those who are not physicists, virtual particles are like the carry digits when adding a column of numbers. The carry digits aren’t needed for anything other than making it easier to complete the calculation.
It’s even worse than that with virtual photons, they’renot even part of quantum field theory per se, they’re just representative of certain terms in a certain method of performing calculations.
In quantum mechanics something like quantum superpostion is something that cannot be directly observed (though it has clear physical consequences), but it’s pretty trivial to show that a quantum system can be in a state that is not an eigenstate of an observable that you are seeking to measure. Therfefore it’s perfectly sensible in the context of quantum mechanics to talk about superpostion.
With virtual particles though even within the context of quantum field theory it’s debatable whether they should be considered anything more than mathematical fictions as they’re not something that can be derived directly from the postulates of the theory.
Isn’t Hawking radiation from black holes the conversion of one virtual particle of an anti-particle pair into a real particle? In that case virtual particles would have to have some reality to them. I don’t suppose we could somehow get to keep a negative momentum one though.
I’m saying I’d like to saying I’d like to see someone do the calculation at tree-level with two charged particles (say an electron and a positron) and show that the momentum of the photon exchanged is in some sense negative (or that the negative momentum dominates the integral), whereas it’s positive for two electrons. It maybe wouldn’t even need to be the whole calculation to show this, just maybe the integral.
But I haven’t had to do real QFT calculations since taking QFT in grad school, so I don’t remember how to do this (and am a little confused about what “negative momentum” would mean here anyway).
Nevertheless, unless it’s possible to do that sort of calculation I don’t really feel like it’s fair to say “Yeah, the photon carries negative momentum.” If we’re just saying “Well, that’s a decent classical analogy, but it doesn’t really work like that” then that should be the answer.
I really wish I remembered this stuff better. Maybe someday I’ll get out my old textbooks and work through them again at a more leisurely pace then I was allowed in school.
Strictly speaking, if the particle ends up as a real particle, then it’s always real, and never “converted”. Then again, strictly speaking, a photon that travels from a distant star to my eye is also a virtual photon, since it interacts with other particles at both ends, and doesn’t propagate out to infinity. It’s just that, because the distance it propagated is so long, it’s constrained to behave very close to how a real photon would behave.
The trouble with this kind of statement is that you seem to be ascribing to Bohr’s concept that:
And even thogh this is probably correct it would mean that all popular science books would just be filled with math. Ultimately that could hurt funding, and possibly the Superconducting Super Collider would never have been built.
After this thread I did some investigating and found it can be shown that, within the constraints of the uncertainty principle, some virtual particles are close enough to their mass shells that they can be detected. Or as Or as Griffiths says in Elementary Particles:
ETA I should have read Chronos post b4 posting this
Yes, the hole expends enough energy via its tidal gravitation to convert the virtual particles to real particles. Therefore the hole created two particles but only got the energy of one back. Maybe someone else can correct me but I know of no reason why the particle wouldn’t keep the momentum direction that it started with.
I think though the issue is more complex than that, even if you were to interpret all the elements of quantum field theory physically then there’s still no reason that you’d assign a physical meaning to virtual particles though preumably you’d have to assign a physical menign to real particles as they are an empirical fact.
What happens when you don’t need to use petrubative solutions, should you still be considering virtual particles? Why is the system that is being peturbed more importnat than the system which we seek to find the solution for?
A real particle is, by definition, one that propagates to or from infinity, and a virtual particle is one that does not, but instead, begins and ends with interactions with other particles. It’s a global property, not a local one, and you can’t truly tell which is which unless you can see the entirety of time and space (though in practice, we make approximations based on distances that are far less than infinite). When people say that virtual particles are turned into real particles in Hawking radiation, what they mean is that the particle-pair originates in a process that’s normally associated with virtual particles.
It would, but if one of the particles escapes, it always has its momentum in the “correct” direction. The one that had its momentum in the “wrong” direction inevitably gets eaten.
The integral at tree level is trivial, owing to conservation at the vertices. That is, there is a delta function that picks out the only value that the photon four-momentum can take. What’s more, though, is that there isn’t a single virtual photon to speak of, as multiple diagrams contribute even at tree level.
Take an electron passing by a positron. In one tree-level diagram (the so-called t-channel), the electron and positron exchange a photon:
e- e-
\ /
\ /
\ /
|
| photon
|
/ \
/ \
/ \
e+ e+
In another (the s-channel), they annihilate:
e- e-
\ /
\ /
\____/
/ ph \
/ \
/ \
e+ e+
The momenta of the two photons above are both completely constrained and completely different. One cannot ask about the momentum of the virtual photon. Indeed, virtual particles are calculational aspects of perturbation theory, and the desire to picture one particle “tossing” a photon-like billiard ball over to another particle is a classical one, and asking classical questions about that billiard ball doesn’t get very far. (In the second diagram, there aren’t even any particles to toss or catch the virtual photon, as they annihilated to create the photon.)
How can we tell from QED that the force between an electron and a positron is attractive? The OP suggests that in classical physics one might look at the momentum exchanged between the particles, but here it seems that you can just plug in whatever momenta you want (that satisfy momentum conservation) and calculate the scattering amplitude.
The way I’ve always seen it done is to calculate a distance-dependent potential, and then take the gradient of it to find the force.
And to be clear, Pasta, time is horizontal on your diagrams? In any event, though, you could instead model an interaction between an electron and a mu+, to avoid the annihilation diagram.
