# Kinetic energy change in virtual photon exchange

1. Dec 13, 2005

### El Hombre Invisible

I understand (vaguely) how momentum is exchanged via virtual photons in electromagnetic interactions. But how do the changes in kinetic energy occur?

For instance, a proton and electron are at rest separated by a distance x. The proton emits a virtual photon and recoils in the direction of the electron with momentum p. The electron absorbs the virtual photon and recoils in the direction of the proton with momentum -p. But the kinetic energy of the both the proton and the electron have increased. From whence this naughty energy appear?

If my example doesn't work, please replace with any other where a force acting on a charge changes its energy.

2. Dec 13, 2005

### lalbatros

From the potential energy.

3. Dec 13, 2005

### El Hombre Invisible

BUT... this is precisely my problem. Let me illustrate.

Proton and electron at rest, separated by distance x. The virtual photon of momentum p is emitted from the photon causing a recoil of -p. Potential energy is determined by distance, so to change potential energy the proton has to move. To move the proton needs kinetic energy, and a change in momentum necessitates this. But if the kinetic energy comes from the change in potential energy we have a first-mover type problem. Potential energy doesn't change until the proton moves, so how can it supply the kinetic energy to cause the motion to change the potential energy?

It gets worse for me. If the proton recoils at the instant the virtual photon is emitted, it will take T = x/c for that virtual photon to be absorbed, during which time the proton is moving and so the distance from the electron is moving, meaning the potential of the electron is changing. But the electron doesn't move until the photon is absorbed. So for x/c the change in the potential energy function of the electron does not cause a change in kinetic energy - this doesn't happen until momentum is transfered.

It's doing my head in. I could wait several years and learn it, but... I'm not going to.

Lastly, so far as I can tell, electric potential energy arises from the exchange of virtual photons. That is, as the electron nears the proton, the virtual photons it absorbs carry a higher momentum since the time between emission and absorption is less. In a sense, the potential of the electron increases with the energies of the virtual photons it absorbs (or could absorb). So I'm not even sure 'from the potential energy' is much different from saying 'from the virtual photons', which just brings me back to my original question.

I should have said, I know next to nothing about QED or QFT, and not much more about QM. We've studied old quantum theory, simple solutions to 1D time-independant SEs, quantum theories of matter (gases, conduction, etc). Almost no particle physics though.

4. Dec 13, 2005

### Staff: Mentor

How about this: the (classical) electrical potential energy is the total energy of the cloud of virtual photons that surrounds (say) a pair of charges? This is a half-assed guess on my part, because I have only a very slight acquaintance with QED, so I hope someone who actually knows QED can correct or expand on this if necessary.

5. Dec 13, 2005

### lalbatros

Hello El Hombre,

It looks like the quantum and the photons are only there for the decor.
Your question about time taken by the photon can be translated to classical terms.
Have a look at the classical field theory and retarded potentials, for example.

When one of the particles sets in motion, the available potential energy is converted in kinetic energy and -additionally- into instationary field energy. In a sense this make your question even worse. But this how things works: not only is the particle accelerated but it also radiates and dissipates energy. This why classical atoms would collapse. This why Rydbergs states collapse too. Nevertheless, energy is conserved: (kinetic energy + potential energy + field energy) remains constant. Therefore, the potential energy produces kinetics energy and field energy in another spectral domain (unstationary).

6. Dec 13, 2005

### El Hombre Invisible

Similar to my picture, so it's good to see someone more experienced than I am thinking along similar lines. I'm thinking there might be some problems considering it all of the virtual photons surrounding the pair of charges. Your classical potential energy is determined by your distance from the charge in question, so would seem to exclude those at other distances (which makes sense - if the electron is at distance r from the proton, virtual photons appearing at distance r + $$\Delta$$r aren't going to effect its mechanical energy). It seems more likely to me that your electrical potential energy is the supply of virtual photons coincidental with the charge's position (i.e. those with precise energy $$E = \frac{h}{4\pi c\Delta t}$$), from which the position dependancy of potential energy arises.

But I still don't see how this energy is transferred if all of the photon's energy arises from the HUP. The answer will, I'm sure, be blindingly simple, but those are the things I usually can't see.

7. Dec 13, 2005

### El Hombre Invisible

Hi

You are right insofar as I can ask the same question in classical physics. However, I think the answer in classical physics is: we don't know. The laws of electrostatic forces were known well enough, but the actual mechanism behind them is, AFAIK, the realm of QED and none other. The mechanism in QED by which momentum is transfered I (kind of) understand and am not aware of an equivilent in EM. But what is converted to kinetic energy and how, I don't know.

8. Dec 13, 2005

### Physics Monkey

El Hombre,

The key point when dealing with virtual particles is that they are not "on the mass shell". What this bit of jargon means is that the energy of a virtual particle isn't related to its momentum in the same way that you are familiar with for real particles. In particular, the energy of that virtual photon your proton emits is not given by $$E = p c$$ like for real photons. Instead, the energy and momentum of the virtual photon are determined by conservation of energy and momentum applied to the emission. I would encourage you to give it a try and see what you come up with for the photon energy.

With this in mind, you can take each of these virtual processes and associate an amplitude with it. To calculate physical quantities you have to sum the effects of each virtual process. In doing so, you find that the less "real" the virtual particle is, the less it contributes to the final result. The end result is that by summing contributions from these virtual processes one can obtain the Coulomb energy between two classical charges. I can show you the details of the calculation if you're interested.

Last edited: Dec 13, 2005
9. Dec 14, 2005

### Wishbone

I LOL'ed
________________

10. Dec 14, 2005

### lalbatros

El Hombre,

It is a good start point if you agree that in classical theory your question is answered simply by recognizing that fields can carry energy and exchange energy with particles. Then we are simply faced to a translation problem.

Now, from your discussion, it strikes me that nowhere the electrostatic field receives any quantum description. Would it not be the way to solve your question? In QFT, how would you described something like an electrostatic potential, or similar. Since, classically, the electrostatic potential is the source of energy you are looking for, I would expect that its QFT description would translate that somehow. I would guess that mathematically this reduces to something like a Fourier description of an electrostatic potential.

But I see another small difficulty with your question, regarding the uncertainty principle. Indeed you assumed your particles are at rest in the initial conditions. Probably this description has to be refined a little bit. And this rises another question: in QFT, are stationary fields possible?

Finally, my naïve picture: the static field might be grossly represented by a sea of photons with a broad spectrum and a space density increasing near the particles you considered. Might that not help to picture your question? Would that not indicate that real photons are involved instead of virtual photons?

If you can go forward with these naive comments, I would appreciate if you could point me to an introductory text on QFT where such ideas are developped in a more exact/mathematical fashion. I feel very unconfortable with mathematical physics text: I most often feel completely disconnected from reality (maybe an epr effect too).

thanks,

michel

11. Dec 14, 2005

### lalbatros

I was expelled from a lecture about banana orbitals ...
I could not stop laughing ...

I guess "retarded potentials" made you jump a state ?

12. Dec 14, 2005

### El Hombre Invisible

I'm not sure I can because I'm not sure I follow you. I understand that the momentum of the photon is determined by the conservation of momentum in the proton-photon system, so if a proton feels a recoil of $$\Delta p$$, the photon will have a momentum of $$-\Delta p$$ and all is well.

This recoil will correspond to a change in kinetic energy of the proton of:

$$E_{kin} = \frac{p^2}{2m_p}$$

How can this determine the energy of the photon? Let's say this recoil increases the kinetic energy of the proton. I don't see how I can then derive the energy of the photon on the basis of conservation of energy alone.

Can I have one more hint?

This sounds promising. Your statement about how "real" the virtual photon is suggests that the photon does have a real contribution as well as a virtual one, which at least gives some mechanism for transferring energy (though doesn't help me understand how it is conserved). On your point about how one can obtain the Coulomb energy from the virtual contributions, do you mean that the electric potential is essentially the sum of the virtual-only contributions to the process? That would mean that, classically, the potential energy converted to kinetic energy is the sum of the virtual contributions to the photon energies exchanged? That would also mean that these virtual photons permanently become real kinetic energy, at which point I conclude I have, again, misunderstood.

Thanks for all your help, not least for pointing out that I was using the wrong equations.

Last edited: Dec 14, 2005
13. Dec 14, 2005

### El Hombre Invisible

Hi Michel. This is what jtbell and I were discussing earlier and part of my last reply to Physics Monkey. In trying to translate the classical idea of electrostatic potential to QED, the virtual photons themselves would seem the prime candidate. The problem I have with this is that I was of the impression that all of the 'borrowed' energy of a virtual photon must be paid back - it cannot be converted to real kinetic energy. This may be my problem.

Yes, I am quite certain that my example is not sensible. It certainly doesn't seem very sensible from a classical point of view. Classically speaking, the electron has a negative potential energy and no kinetic energy - a zero mechanical energy. How, then, did it get there? Presumably by radiating its kinetic energy away. The electron cannot move away from it since it can only gain kinetic energy by lowering its potential further. In which case, in QM terms, it is confined and so MUST have kinetic energy. So, yes, my attempt at simplification may have over-complicated the question.

Absolutely - that is AFAIK how it is. However the mechanics involved don't follow easily from that for me (i.e. how the energy is supplied to the particles).

I don't see how. A real photon, as Physics Monkey pointed out, has energy-momentum relationship E = pc. Where would this energy come from? It cannot come from the change in kinetic energy of the proton, since this is far less than pc.

Who, me? If I could I assume I would not be asking the question. Someone else?

14. Dec 14, 2005

### erickalle

On a simpler note.
Something else which is doing my head in.
When an electron is moved away from a proton the convention says that the electron is raised to a certain potential. But has anybody around here had a thought for the poor proton? Is its potential not raised?
If we take a metal surface to the proton the result will be that the proton generates a mirror image and will be flying off to this image.

15. Dec 14, 2005

### El Hombre Invisible

Yes, it is.

????

This isn't very related to my question, is it...

16. Dec 14, 2005

### Physics Monkey

El Hombre,

Thanks!

Now, on to your question. The energy of the system before the virtual photon was emitted is $$m_p c^2$$, and the energy of the system after the virtual photon is emitted is $$m_p c^2 + \frac{p^2}{2 m_p} + E_\gamma$$, where I assumed the proton was initially at rest and the recoil was small so that non-relativistic formulas apply. Energy is conserved by the emission process so all you have to do to find $$E_\gamma$$ is equate the two and solve. It's completely analagous to the way you found the momentum of the photon. Don't be shocked if the answer looks weird, remember that the photon is virtual.

Last edited: Dec 14, 2005
17. Dec 14, 2005

### lalbatros

I cannot understand this statement:

In QFT just like in classical field theory, there should be a term for the potential energy. Isn't it?

Again, my naïve guess is that the -hypothetical- static field would be made up of a wide spectrum of photons. The accelation of the electron would produce a change is the field energy as well as the particle energy. The fields energy is not zero in the initial conditions, but it is stationary, as far as this makes sence in QFT (like for the question). After a short time, the total field energy should have decreased, and additonally the spectrum should have changed and become partly unstationary.

Now concerning my request for some reading:

I assumed that if you had such questions you might also have some interresting reading. Maybe also a small chance to find a readable one: one that puts the maths behind the scene instead of on the stage.

18. Dec 14, 2005

### El Hombre Invisible

Really? Well this is just the negative of the kinetic energy I mentioned earlier. So your answer is that I don't have to do anything to derive the photon energy from that - it just is it ('xcept times -1).

So if the energy-momentum relationship is simply:

$$E = -\frac{p^2}{2m_p}$$

then two virtual photons of the same momentum but emitted from charges of different masses will have different energies. Is there any idea why virtual photon energy depends on the mass of the emitter, other than conservation of energy?

And, of course, there are the two more obvious and no doubt related questions: how can the energy be negative, and where does $$\Delta E\Delta t = h/4\pi$$ come into it?

Thanks for all your help. I'm getting there.

19. Dec 15, 2005

### El Hombre Invisible

Hi

I read this in Wiki and it seemed to answer my question:

To consolidate this with my current (mis)understanding, since the virtual photon is a particle with fully defined momentum, it has the potential to be found anywhere. In the case of a proton and electron at rest:

p_____________________e

the proton may emit a virtual photon and recoil towards the electron. To conserve momentum, the photon has a momentum in the direction from the electron to the proton. Spatially it can be found anywhere, even behind the electron:

__p->_________________e-<-~~~~

Since the change in energy of the proton is positive and virtual photons conserve energy, the photon enery is negative. This means it is going the other way as if emitted by the electron:

__p->_________________e~~~~->+

This causes a recoil on the electron, so we have:

__p->____________<--e__

and energy and momentum are conserved.

Is this a vaguely decent layman's overview?

It would seem to me then that the only HUP inequality that matters is

$$\Delta p\Delta x >= \frac{h}{4\pi}$$

A change of momentum p will arrive at distance x from the emitter.

Last edited: Dec 15, 2005
20. Dec 15, 2005

### erickalle

Just a thought which popped up. It will be popped elsewhere.