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Proton-electron interactions interms of photons

  1. Sep 19, 2008 #1
    When 2 electrons come close the "throw" photons at each other causing them to move to opposite way.
    All I want to know is if you had a proton and electron coming close, do they "throw" photons outwards to come closer?
    Sorry this is probably an easy question!
  2. jcsd
  3. Sep 19, 2008 #2


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    Nah, it's not an easy question. Electrons don't "throw" photons at each other. Protons and electrons don't throw photons away from each other--or anything like that really.

    The whole "throwing" analogy is just a way to get the point across that the force between two charged particles is not instantaneous--even though in freshman physics we say it is instantaneous (Coulomb's Law).
  4. Sep 19, 2008 #3
    More specifically, I think it's trying to convey the idea that the force is mediated by some particle- the photon. You can come up with some pretty sad pictures to make analogies of attractive forces like this (boomerang, anyone? :biggrin:) but they're really just to give you some idea of how exchanging particles leads to that kind of effect- once you understand that the photons mediate the interaction, that's as much as you can gain from them, and you shouldn't try and push them too far.
    (Anyone who actually understands QFT in any depth should feel free to shoot me down if necessary btw!)
  5. Jan 1, 2009 #4
    I know what you really want answered, so I will answer it.

    Quantum Field Theory shows how the mediations of Photons can make particles attract or repel electromagnetically. The key part in all of this is that the photon is 'virtual' meaning that it has defined momentum, but no defined position in space. Here's how it's done.

    A virtual particle, like a virtual photon, is one that does not precisely obey the
    m2c4 = E2 − p2c2 relationship for a short time. In other words, their kinetic energy may not have the usual relationship to velocity — indeed, it can be negative.

    The energy carried by the particle in the propagator can be negative. This can be interpreted simply as the case in which, instead of a particle going one way, its antiparticle is going the other way, and therefore carrying an opposing flow of positive energy.

    Since the virtual photon is a particle with fully defined momentum, it has the potential to be found anywhere, because of the uncertainty principle. 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.

    __0_ --->____________0__ <---o -photon
    __p+________________ e-__

    Since the change in energy of the proton is positive and virtual photons conserve energy, the photon’s energy must be negative. It is going backwards in time. This means it is going the other way as if emitted by the electron:

    __0_ ---> ___________0__ o---> -photon
    __p+________________ e-__

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

    __0_ ---> ______<---_0__ o---> -photon
    __p+_______________ e-__

    with a virtual photon traveling backwards in time moving away from the electron.
    Last edited: Jan 1, 2009
  6. Feb 8, 2009 #5
    So the photon is being emitted by both the electron and the proton at the same time?

    Also, why does this ALWAYS happen? I mean, if the reason it does happen is because the photon can be anywhere, why is it always behind the photon when a proton and electron interact? (as far as I know they never repel each other)
  7. Feb 8, 2009 #6
    You're questions are the ones I thought of at first too. It turns out that in a sense, the emission of the photon behind the proton is emitted after the electon emits the photon because the energy of the emission of ther photon behind the proton is drawn from the energy of the 1st emission. The concept can be reversed. The proton could emit the 1st photon, and so on. As far as why this happens, and what is the reason for all of this, that is the real question. Quantum Field Theory is what I have explained so far, but it doesn't provide an answer to the question. It might interest you to know that string theory doesn't answer it either. The only answer I have been able to find is the one I propose in my research. The answer is fairly complicated, and involves the idea of space-time instead of particles. Get rid of virtual particles, and put in space-time, you have these interesting higher dimensional spheres that shrink and grow.
  8. Feb 9, 2009 #7
    So there are 2 different photons? Or just one photon which is emitted by both the electron AND the proton?
  9. Feb 9, 2009 #8
    hmmm - and just what exactly stimulates the emission of the photon(s) in the first place? do the particles somehow "know" they are near each other? or are all particles continously emitting a cloud of virtual photons, and if so, where is all that energy coming from? in the end, dont the particles have to exchange a "real" photon to convey momentum, and how do the virtual photons excite the emission of an actual photon?
  10. Feb 9, 2009 #9
    I'll kill two birds with one stone here.

    Unredeemed asked if there are two photons. The answer that physicists will give you is that there was never any photon to begin with, that the virtual particle is just a representation of the energy loss. This is unclearly described in perturbation theory. The answer that is more clear and possibly better is that there was only one photon throughout the whole interaction, but it has the ability to take on the roll of what might seem to be two photons because it is virtual. This is why I said what I said earlier, I just forgot to mention the fact that it was really the same virtual photon. So, in essence, the photon that was emitted by the electron was the same photon that was emitted by the proton. This obviously unusual incident is why virtual particles were even theorized at all.

    jnorman asked what stimulates the photons to emit at these times and in this manner. How do the photons know? This is one of the key questions in modern physics. Photons seem to know a lot: where to travel, when to travel, what path to take. The list goes on, but the answers come short. There is no modern explanation for this.

    As far as the continuous emission of virtual photons, that is interesting, and possibly true. No one today would know.

    As for the last idea, the one that states you would need a real photon to create momentum, this on is known and easy to answer. The electromagnetic force is governed by virtual particles. They have fully defined momentum, and when they are exchanged between other particles, they change the momentum of the particles too. They can cause the particles to recoil or come together based on the amount of energy they have. It's like throwing a baseball in a vacuum. You would throw the baseball forward, but because you threw the positive-energy baseball, you would be sent backwards. If you threw a negative energy baseball, you would be sent forwards. Technically, the baseball analogy is a little confusing because it has mass, but I assure you, the meaning of the metaphor is still solid. You don't need a real particle to create momentum. All you need to create momentum is an energy balance. That is what perturbation theory is about, and my previous post illustrates this too.

    If your still confused, feel free to ask me, I'll try to explain whatever I can.
  11. Feb 9, 2009 #10
    How they knowing this? - no good.
  12. Feb 9, 2009 #11
    If this is true then the proton and electron gain the energy of two photons when, in fact, only one is emitted. How does this conserve energy? Where did the whole other photon's energy go?
  13. Feb 9, 2009 #12
    QuantumBend- There is no answer is to this question. We have no idea how they know which path to take, we just know that they know. That is part of producing a theory of everything.

    Unredeemed- You ask a good question. The fact of the matter is that virtual photons, like every other virtual particle, come in pairs; one with positive energy, one with negative. When we look at the situation, we do find the energy to be conserved.

    Consider the following. An electron and a proton are at rest. This will never occur in reality, but for general purposes, it will suffice.

    __e- ________________ p+__

    The electron emits a virtual photon, and it starts to gain momentum towards the proton. If we consider this, we find that the photon's energy must be negative. Why? If the electron had emmitted a positive energy photon, it would have recoiled, just like in the baseball analogy. The fact that it moved forward means that the photon's energy must be negative in order to conserve momentum.

    __0_ --->____________0__
    __e-________________ p+__

    Now, the photon that was just emmitted appears behind the proton as if the proton had emmitted it. We can treat the situation as if the proton emmitted the photon in the opposite direction of the electron because the photon was found behind the proton. If it was found behind the proton, and we treat the proton to have emitted it, then we must reason that the proton emmitted it in the direction it was found.

    __0_ ---> ___________0__ o---> -photon
    __e-________________ p+__

    Now, the proton starts to head towards the electron. This means that the proton has recoiled, because it has emitted the photon in the other way. From this final step, we can conclude that the photon that the proton emmitted was positively energetic.

    __0_ ---> ______<---_0__ o---> -photon
    __e-_______________ p+__

    Now we look back. The photon's energy that the electron emmitted was negative, and the photon that the proton emmitted was positive. This means that energy is conserved.

    But remember what I said earlier. The electron and the proton emmitted the same virtual photon. This leads to something you may have read about. A photon is its own antiparticle. You can treat it as two seperate particles, but the net energy will always be zero.
  14. Feb 10, 2009 #13
    Why electron probables receive photon is 1? Photon travels 3 space all ways not ONLY to electron - no good.
  15. Feb 10, 2009 #14
    Ah, I see now. Thanks very much.

    So no photon is actually created in the interaction?

    Also, if the photon is its own antiparticle, why does it not annihilate with itself?
  16. Feb 10, 2009 #15


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    You are missing the point here, first of all, these photons doesn't behave as photons from a flash-light. They are emitted by a charged particle and HAVE TO be reabsorbed within a very very small fraction of time. Either the virtual photon will be absorbed by another charged particle, or it will be absorbed by the particle which emitted it (see this diagram)

    http://upload.wikimedia.org/wikipedia/en/thumb/e/e4/Electron_self_energy.svg/151px-Electron_self_energy.svg.png [Broken]
    Last edited by a moderator: May 4, 2017
  17. Feb 10, 2009 #16


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    Because it has no coupling to itself, it only couples to charges. Photons are uncharged electrically, but couples to charges.

    Gluons, the force carrier of the strong interaction, can couple to themselves, since they also carries the charge with they are coupling to. Gluons carry color and have color, and they couple also to quarks which has color.
  18. Feb 10, 2009 #17


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    Welcome to PF!

    Hi mmang! Welcome to PF! :smile:

    The photons are called "virtual particles" … that means they don't exist, except in the maths.

    They are a useful mathematical concept which helps to explain perturbation equations in quantum field theory … the annihilation and creation operators of photons are used in the algebraic expansion that approximates the field, but that's all that's meant by "mediated by photons".

    The field moves the electrons, and "virtual photons" are just a mathematical "currant-bun" approximation of the physical field.

    Things that exist in reality are called "real" (the clue's in the name! :wink:)

    If you work out the energy-momentum equations, you find that the photon would always have to travel faster than light … that would mean that causality would break down, and you couldn't say whether it was leaving one photon to go to the other, or vice versa.

    As muppet :smile: says, don't take virtual photons too seriously …
    "mediate" and "exchange" are just words that physicists use to make the maths look more familiar. :smile:
  19. Feb 10, 2009 #18

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    I was with you until this point. One reason I disagree is that for a virtual particle one integrates over all possible momenta, so it doesn't have a well-defined momentum. (And before you say it's determined by the ingoing and outgoing kinematics, remember that you can have multiple photon exchange plus you can - indeed must - have additional soft radiation).

    Also, like muppet points out, it's easy to take the concept of virtual particles too far. They are not real, and they do not have the properties that real particles have. They are a way of describing excitations in a field - in this case, the electromagnetic field.
  20. Feb 10, 2009 #19


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    Benk99nenm312, you've been saying some rather strange things in this thread. It would take too long to comment on all of them, so I'll just focus on some of the things you said in the post I'm quoting.

    The question of how a photon "knows" what path to take doesn't really make sense. I consider virtual particles to be nothing more than a graphic representation of the individual terms in a series expansion of a mathematical expression for a probability amplitude. Even if we (with absolutely no justification) decide to interpret the little pictures (Feynman diagrams) that we can associate with each term in the series as a description of "what really happens", we still can't say that the virtual particles take one specific path. Since we integrate over all paths in the mathematical expression, the only reasonable intepretation of this kind is that the virtual particles take all paths.

    Virtual photons do not come in pairs. Where did you get that idea?

    I also can't make sense of your pictures, and most of the conclusions you have come to based on your pictures are wrong. It seems to me that you're taking the description of an interaction in terms of virtual particles much too seriously. If you'd like to see how a QFT book explains why two electrons repel each other, I suggest you take a look at this book. You can use the "search inside this book" feature to search for the words "like charges repel". The explanation is on pages 30-31.
  21. Feb 10, 2009 #20
    Right, now I'm confused. So what IS the answer?
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