Negative momentum of virtual photons?

[Pasta]

@tim314

How can we tell from QED that the force between an electron and a positron is attractive? …it seems that you can just plug in whatever momenta you want (that satisfy momentum conservation) and calculate the scattering amplitude.

Good eye. You are right that in the above e-e scattering picture, you can’t tell attraction from repulsion. This is because Feynman diagrams are fundamentally momentum-space entities. The particles in the diagrams are completely de-localized plane waves. Thus, attractive and repulsive scattering is indistinguishable. Pictorially, all you see at “infinity” is this: (N.B.: the drawings below are not Feynman diagrams)

             /
        ____/__
---->--|       |
       | black |
       |  box  |
---->--|_______|
            \
             \

which could come from attraction:

             /
            /
---->----  /
         \/
         /\
---->----  \
            \
             \

or repulsion:

             /
            /
---->-------
         
         
---->-------
            \
             \

To talk about the direction of a force, one necessarily needs a spatial representation, which gets messy (and fundamentally non-perturbative). If you try to calculate something like this, it ends up looking more like non-relativistic QM, insofar as you must step backwards into the messy pre-Feynman-diagram portion of relativistic QFT. But the approach is like that hinted at by Chronos – calculate a potential from a fixed charge and take gradients. (Or, if you prefer an easier way out, simply note that the coupling constant changes sign. )

@Chronos

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.

Yes, time is horizontal. You are right that I could have chosen a muon as one of the particles, but (as you recognize) that would only have hidden the main point that the classical analogy is limited, at best.

[tim314]

Good answer again. Thanks

[Ring]

@Chronos

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.

So does this mean that the hole doesn’t have to expend energy to create the particle pair? If so then I don’t see what role the BH plays in this process other than to eat one of the particles.

But that would seem to mean that the absorbed particle would have to have negative energy (intrinsic negative energy not just negative with respect to the field.)

[Chronos]

But that would seem to mean that the absorbed particle would have to have negative energy (intrinsic negative energy not just negative with respect to the field.)

Yup.

[Asympotically_fat]

@Ring

So does this mean that the hole doesn’t have to expend energy to create the particle pair? If so then I don’t see what role the BH plays in this process other than to eat one of the particles.

But that would seem to mean that the absorbed particle would have to have negative energy (intrinsic negative energy not just negative with respect to the field.)

I believe Hawking radiation is essentially a frame effect caused by the way that the operator representing the particle number varies between more general types of coordinate transformatrions.

I guess from what I have read an infalling observer would NOT observe Hawking raditation just before they crossed the event horizon (though I admit the situation may be more complicated than that), instead the radiation is observed by a far away Schwarzchild observer.

I don’t know how exactly how it would break down if you wanted to describe it in terms of virtual particles, though I’ve certainly remember reading in Stephen Hawking’s book about how Hawking radiation is cause by virtual particle pairs and the infalling particle having negative mass. Whether that’s a great description of what’s going on even in terms of virtual particles is another thing.

So the black hole plays it part in providing the background that this can all happen against and what’s going on with the virtual particles probably isn’t worth bothering too much about unless you’re trying to perform calculations.

[Uncertain]

@Chronos

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.

So, are any photons real photons?

[Asympotically_fat]

@Uncertain

@Chronos

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.

So, are any photons real photons?

Real photons are an empirical fact.

My technical grasp of quantum field theory is not strong, but there’s a an important difference between the interactions being described on a Feynman diagram and a photon being absorbed by an eye. The former are reversible whereas the latter is irreversible (and the fact that it’s irreversible is important in quantum measurement theory). From what I understand also that the virtual photons usually represent the effects of static electromagnetic fields, whereas the photons emitted by a star which allow us to view that star are the result of a non-static electromagnetic field.

So I’d hazard a guess that viewing all photons that don’t start at infinity or propagate out to infinity is a serious case of model overstretch.

[Ring]

@Chronos

Yup.

Okay, I’ve read a ton of stuff on Hawking Radiation from textbooks, the internet, sci.physics etc., and the conclusion I’ve come up with is that there are as many interpretations of HR as there are physicists.

I’ve come across a number of guys that hold that the total energy of the infalling particle is negative, but that, nonetheless, its mass/energy is still positive. This I can live with. Negative mass/energy I couldn’t live with because there would be no need for either a black hole or the HUP to create nothing from nothing.