Proton-electron interactions interms of photons

[mmang]

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!

 

[olgranpappy]

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!

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).

 

[muppet]

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? ) 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!)

 

[benk99nenm312]

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:

__0___________________0__
__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.

 

[Unredeemed]

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:

__0___________________0__
__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.

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)

 

[benk99nenm312]

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.

 

[Unredeemed]

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.

So there are 2 different photons? Or just one photon which is emitted by both the electron AND the proton?

 

[jnorman]

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, don't the particles have to exchange a "real" photon to convey momentum, and how do the virtual photons excite the emission of an actual photon?

 

[benk99nenm312]

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.

 

[QuantumBend]

Photons seem to know a lot: where to travel, when to travel, what path to take.

How they knowing this? - no good.

 

[Unredeemed]

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.

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?

 

[benk99nenm312]

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 separate particles, but the net energy will always be zero.

 

[QuantumBend]

Why electron probables receive photon is 1? Photon travels 3 space all ways not ONLY to electron - no good.

 

[Unredeemed]

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 separate particles, but the net energy will always be zero.

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?

 

[malawi_glenn]

Why electron probables receive photon is 1? Photon travels 3 space all ways not ONLY to electron - no good.

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

 

[malawi_glenn]

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?

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.

 

[tiny-tim]

Welcome to PF!

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?

Hi mmang! Welcome to PF!

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! )

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 says, don't take virtual photons too seriously …

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? ) 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.

"mediate" and "exchange" are just words that physicists use to make the maths look more familiar.

 

[Vanadium 50]

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

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.

 

[Fredrik]

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.

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.

 

[Unredeemed]

Right, now I'm confused. So what IS the answer?

 

[tiny-tim]

hmm … what was the question? …

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)

So there are 2 different photons? Or just one photon which is emitted by both the electron AND the proton?

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?

Ah, I see now. Thanks very much.

So no photon is actually created in the interaction?

That's right! As Fredrik says …

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.

Virtual photons are maths, not physics … representation, not reality …

no photon is actually created (or involved in any way) in the interaction.

 

[Unredeemed]

hmm … what was the question? …





That's right! As Fredrik says …


Virtual photons are maths, not physics … representation, not reality …

no photon is actually created (or involved in any way) in the interaction.

Then why were virtual particles theorized in the first place? :s

 

[tiny-tim]

virtual particles

Then why were virtual particles theorized in the first place? :s

They're …

… a graphic representation of the individual terms in a series expansion of a mathematical expression for a probability amplitude.

Different fields use different virtual particles in that mathematical expression …

the electromagnetic field uses photons …

they appear as annihilation and creation operators in the Dyson expansion.

(The Dyson expansion has coefficients, like binomial coefficients but much more complicated, and Feynman diagrams are a graphic way of adding up the contributions to those coefficients.)

 

[benk99nenm312]

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.

i know exactly where your coming from. I don't always explain myself well, but everything I have said has been theorized, and some of which has been proven. I will explain how I make sense.

First of all are my diagrams. Yes, my diagrams. You will not find these diagrams in any book you have read, but that does not make them false. You talk about the mathematical expansion, and all of the mathematical representations, but there is a bottom line to perturbation theory. There is an energy loss. This energy loss is given out in virtual particles, the very name of the solution of the mathematical expansion you are talking about. This is not just me talking. I have had many confirm that there is an energy loss in perturbation theory. This energy loss is the key to my diagrams. I have developed these diagrams to correctly represent a more pictorial version of a mathematical concept. The diagrams do follow the rules of physics. they are not incorrect or invalid in anyway.

You say there is a point when one can take virtual particles to far. I take it that you think i have more than reached this point. I will retort by saying that there is a point where virtual particles are taken to insignificantly. If they exist in math, then they exist in nature. The fact that they are a mathematical solution to an expansion does not mean they can't be adressed pictorially.

Second is my virtual photon energy problem. You state correctly that the photons do not come in pairs. This is a tough subject, because we are not the only ones to argue on the interperitation that fact. My reasoning is that the photon can go both forwards and backwards in time, which by CPT conjugation leads us to a positive or negative energy. So you see, as I stated earlier, there is only one photon in my diagrams. OPne of my posts I actually mispoke in. It happened to be the one you made sure to copy on your post. Since then I have corrected myself. I treat it as two, but it is the one and the same photon that is produced.

You are right about the paths. We can't say which path the photon took. The momentum is the definite part. I said in one post that the photon knows which path to take. What I should have said is that in the end, if we were to observe, say, a real photon, the photon would choose a specific path, and we wouldn't know why. For a virtual particle, it is my belief that you are right. It would take all paths to get from point A to B. My diagrams have no way of showing that.

I know many are still stuck on the fact that a virtual particle doesn't really exist.
If you are completely convinced of this, there is nothing I can say to change your mind. The point is that I think we have forgotten something. When we find that an electron and a proton attract each other, there has to be a reason. You can give me the field concept of things, but a field is just another word for ether. What is this field made of. Photons. Why can't we see them. Is it because they don't exist, is it because they are a solution to an expansion equation? I don't think so. There has to be something there that is responsible for the attraction and repulsion of matter. We have found that we can represent them with virtual particles, but in the end, there has to be something doing the work that we can't see. Virtual particles can't just exist to complete an equation. They have to be there because they have to cover the energy loss that is found in nature. This is why I treat them the way I do.

 

[malawi_glenn]

Well we don't talk about our own theories and developments here (unless they are published in peer-reviewed journals), so your post(s) have been reported.

You are constantly referring to your thoughts and developments, and not answering from the paradigm of contemporary physics.

 

[benk99nenm312]

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.

I get what you're saying. This is one of those things where I think that the virtual photon still has a well defined momenta because if it were not a well defined momentum, the particle would have a relative position. This is not what we observe when we look at vacuum fluctations, where virtual particles seem to appear out of the blue.

I do see your point though. That's interesting.

 

[Unredeemed]

If someone disagrees with benk99nenm312 could they please give an explanation as to how the exchange of a virtual photon leads to the attraction of an electron and a proton?

My physics background is not at all strong and when someone says that virtuals particles are "a graphic representation of the individual terms in a series expansion of a mathematical expression for a probability amplitude" it literally means nothing.

Thanks in advance,
Jamie

 

[tiny-tim]

virtual particles

when someone says that virtuals particles are "a graphic representation of the individual terms in a series expansion of a mathematical expression for a probability amplitude" it literally means nothing.

It means that to calculate one individual term in the Dyson expansion of a field, we need to add lots of numbers obtained in different ways, and we keep track of those numbers by using a diagram for each possible different way …

and each diagram looks like a lot of particles colliding, with conservation of energy momentum and spin.

 

[benk99nenm312]

If someone disagrees with benk99nenm312 could they please give an explanation as to how the exchange of a virtual photon leads to the attraction of an electron and a proton?

My physics background is not at all strong and when someone says that virtuals particles are "a graphic representation of the individual terms in a series expansion of a mathematical expression for a probability amplitude" it literally means nothing.

Thanks in advance,
Jamie

I had a thought on how to answer your question and I wrote it down. What I said makes sense, but it is not proven. I'm sorry if I mislead you or anyone else.

 

[Unredeemed]

It means that to calculate one individual term in the Dyson expansion of a field, we need to add lots of numbers obtained in different ways, and we keep track of those numbers by using a diagram for each possible different way …

and each diagram looks like a lot of particles colliding, with conservation of energy momentum and spin.

So then how does a proton attract an electron?