photons and magnetic attraction

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How does photons cause magnetic attraction?

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They interact with charged particles.

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Photons do not cause magnetic attraction. First, the classical EM field consists of very rapidly oscillating electric field, and an associated --also very rapidly oscilating-- magnetic field perpendicular to the E field, and both perpendicular to the direction of propagation of such EM field. Particles are predicted to oscillate in response by the theory, and so it is confirmed experimentally. That's how antennae work. It's oscillation, rather than overall attraction.

Photons, OTOH, are quanta of such EM field. They can be absorbed, emitted, scatter... They never result in overall attraction either. They just change the state of electrically charged particles by making them change their energy and momentum.

I hope that helps.

Edited by joigus
minor correction
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14 minutes ago, joigus said:

Photons do not cause magnetic attraction. First, the classical EM field consists of very rapidly oscillating electric field, and an associated --also very rapidly oscilating-- magnetic field perpendicular to the E field, and both perpendicular to the direction of propagation of such EM field. Particles are predicted to oscillate in response by the theory, and so it is confirmed experimentally. That's how antennae work. It's oscillation, rather than overall attraction.

Photons, OTOH, are quanta of such EM field. They can be absorbed, emitted, scatter... They never result in overall attraction either. They just change the state of electrically charged particles by making them change their energy and momentum.

I hope that helps.

Thanks for a more detailed explanation.  +1

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10 minutes ago, studiot said:

Thanks for a more detailed explanation.  +1

Thank you. Feel free to add anything I may have missed.

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5 hours ago, joigus said:

Photons do not cause magnetic attraction. First, the classical EM field consists of very rapidly oscillating electric field, and an associated --also very rapidly oscilating-- magnetic field perpendicular to the E field, and both perpendicular to the direction of propagation of such EM field. Particles are predicted to oscillate in response by the theory, and so it is confirmed experimentally. That's how antennae work. It's oscillation, rather than overall attraction.

Photons, OTOH, are quanta of such EM field. They can be absorbed, emitted, scatter... They never result in overall attraction either. They just change the state of electrically charged particles by making them change their energy and momentum.

I hope that helps.

I wonder if this may be about virtual photons as "force carriers" in magnetism. They do that, don't they, in QED? (I never learnt this stuff).

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5 minutes ago, exchemist said:

I wonder if this may be about virtual photons as "force carriers" in magnetism. They do that, don't they, in QED? (I never learnt this stuff).

That's a very interesting question I don't have a totally rigorous answer to. You are totally correct in making that distinction. So-called real photons are indeed those that can be detected with, eg, a photodetector. In some sense they're "somewhere there."

Virtual photons, OTOH, are only those that appear in QED calculations as taking part in intermediate states that only are relevant to the effect of calculating the initial and final states of real particles, including other photons, or the vacuum.

If you're familiar with Feynman diagrams, think of the external legs of the diagram as real particles. In between legs you have internal lines connecting vertices. These vertices represent local interactions via virtual particles that mediate the interaction, like in this image taken from Wikipedia representing electron-electron scattering:

Now, this is all very interesting, but what happens when a big chunck of magnet attracts/repels another magnet? And when two charged pieces of matter do a similar thing? Can we extrapolate that picture somehow?

I'm not aware that anybody has taken QED to do the detailed calculations on, say, a big piece of ferromagnetic material. But here's one reasoning that I think is very convincing to see why it has to be the case that it's virtual particles that are doing the job. One of the characteristics of virtual particles that we learn from QED is that they violate Einstein's energy-momentum constraint. They can do so because they're only allowed to exist for a very ephemeral time lapse consistent with HUP.

The way to see that is that a particle at rest cannot emit a photon and start moving from the recoil. This is a very surprising consequence of special relativity. And here's the proof.

A massive particle has 4-momentum that we can write as,

$\left(E,c\boldsymbol{p}\right)$

satisfying Einstein's constraint,

$\left(mc^{2},\boldsymbol{0}\right)$

Thin, of a charged massive piece of matter that's sitting somewhere, and consider the rest frame. We have a 4-momentum,

$\left(mc^{2},\boldsymbol{0}\right)$

And now it emits a photon. We have --by virtue of energy-momentum conservation,

$\left(mc^{2},\boldsymbol{0}\right)=\left(E,c\boldsymbol{p}\right)+\left(\hbar\omega,c\hbar\boldsymbol{k}\right)$

But the total 4-momentum is the very same 4-vector it was before, in particular it must satisfy the same Einstein relation, so that,

$\left(E+\hbar\omega\right)^{2}-c^{2}\left(\boldsymbol{p}+\hbar\boldsymbol{k}\right)^{2}=m^{2}c^{4}$

The corresponding pieces must satisfy their respective Einstein relations, because someone seeing a piece of matter out there doesn't know it has just spat out a photon. Let's suppose the same goes for the photon. So,

$\left(E+\hbar\omega\right)^{2}-c^{2}\left(\boldsymbol{p}+\hbar\boldsymbol{k}\right)^{2}=m^{2}c^{4}$

$\overset{m^{2}c^{4}}{\overbrace{E^{2}-c^{2}\boldsymbol{p}^{2}}}+\overset{0}{\overbrace{\hbar^{2}\omega^{2}-c^{2}\hbar^{2}\boldsymbol{k}^{2}}}+2\hbar E\omega-2c^{2}\hbar\boldsymbol{p}\cdot\boldsymbol{k}=m^{2}c^{4}$

so that,

$2\hbar E\omega-2c^{2}\hbar\boldsymbol{p}\cdot\boldsymbol{k}=0$

and inevitably,

$E\omega=-\hbar c^{2}\left|\boldsymbol{k}\right|^{2}$

which is impossible if both energies are positive. So photons emitted by a particle at rest seem to be an impossibility unless we admit the possibility that the electromagnetic field carries with it allowance for these ephemeral modes of propagation.

As soon as the charged particle starts accelerating, then --I assume-- both real and virtual photons are being exchanged. The detailed picture being far more complicated than this.

I forgot to say. The last step is on account that,

$\boldsymbol{p}+\hbar\boldsymbol{k}=\boldsymbol{0}$

due to 3-momentum conservation.

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If I may simplify further ...

The classical model of light treats it as a wave or a 'corposcule', but not necessarily a photon.

The Quantum Mechanical model introduces the concept of a well defined photon, while QED, being a quantized field theory has, by necessity, mediator particles, virtual photons, for the Electromagnetic and other fields.
( and of course magnetism is an electromagnetic phenomenon )
This does not preclude real photons from the QED model, of course.

Edited by MigL
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40 minutes ago, MigL said:

simplify further ... mediator particles, virtual photons, for the Electromagnetic and other fields.
( and of course magnetism is an electromagnetic phenomenon )

Right, and that was my simple comment above:

8 hours ago, Genady said:

They interact with charged particles.

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57 minutes ago, MigL said:

If I may simplify further ...

The classical model of light treats it as a wave or a 'corposcule', but not necessarily a photon.

The Quantum Mechanical model introduces the concept of a well defined photon, while QED, being a quantized field theory has, by necessity, mediator particles, virtual photons, for the Electromagnetic and other fields.
( and of course magnetism is an electromagnetic phenomenon )
This does not preclude real photons from the QED model, of course.

10 minutes ago, Genady said:

Right, and that was my simple comment above:

8 hours ago, Genady said:

They interact with charged particles.

And I have absolutely no objection to any of your explanations, of course. It's just that, within the context of the OP question, @exchemist posed a very interesting --very much related-- meta-question about virtual particles perhaps being responsible for what we perceive as the force between charged particles and their currents. It's because of its interest, and there not being any direct derivation --that I know of-- from QED that gives you the law of force between macroscopic charges or currents, that I sketched that kind of argument. It's not completely general, of course.

Also, charges in currents are already moving. But if they're moving at constant speed, there would be no reason --from a classical POV-- why they should radiate. Only accelerating particles should radiate. I'm sure the relativistic argument I presented here can be extended to particles moving at constant relative speed. You need some kind of mechanism that's equivalent to virtual particles at some point.

The gist of it is: If you assume charged particles to start moving because of the exchange of a particle, it's impossible to assign an energy and momentum to this carrier of the interaction that's consistent with special relativity. Therefore, there must be so-called off-shell particles. That is, particles that violate Einstein's relation. Those are virtual particles.

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7 minutes ago, joigus said:

Therefore, there must be so-called off-shell particles. That is, particles that violate Einstein's relation. Those are virtual particles.

It is perfectly correct, but I have a caveat to add. I think that all photons are virtual. We never observe photons directly, only their interactions with some detectors. The so-called real photons are just virtual photons that last so long that their deviation from the shell is immeasurably small.

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3 minutes ago, Genady said:

It is perfectly correct, but I have a caveat to add. I think that all photons are virtual. We never observe photons directly, only their interactions with some detectors. The so-called real photons are just virtual photons that last so long that their deviation from the shell is immeasurably small.

I'm not sure that's tenable, or necessarily true, or at least plausible. It could be. At present in QFT "virtual particles" is nothing but a fancy name for quantum mechanical amplitudes that cannot be made consistent with the on-shell condition, but must be included in the calculations. Nobody has seen a virtual particle, and I'm sure nobody will. In that sense, they might perhaps be comparable to the interior of a BH. Is it really there? I don't know, and I can't even think of a way to measure what's in there and report to the experimental physicists outside. You could say "I think ordinary space is made up of swarms of black holes that appear and disappear so rapidly that the deviation of their effect is immeasurably small."

Similarly, I can't even conceive of a way to have a virtual particle do it's "virtual particle job" --participating in the amplitude off-shell-- and be possible to measure.

You can perhaps always modulate a discontinuous pattern of behaviour by using an ad hoc sigmoid curve that does the trick.

But remember virtual particles come in all kinds of flavours, masses, and other quantum numbers. So the concept is not to do with gauge bosons per se.

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49 minutes ago, joigus said:

I'm not sure that's tenable, or necessarily true, or at least plausible. It could be. At present in QFT "virtual particles" is nothing but a fancy name for quantum mechanical amplitudes that cannot be made consistent with the on-shell condition, but must be included in the calculations. Nobody has seen a virtual particle, and I'm sure nobody will. In that sense, they might perhaps be comparable to the interior of a BH. Is it really there? I don't know, and I can't even think of a way to measure what's in there and report to the experimental physicists outside. You could say "I think ordinary space is made up of swarms of black holes that appear and disappear so rapidly that the deviation of their effect is immeasurably small."

Similarly, I can't even conceive of a way to have a virtual particle do it's "virtual particle job" --participating in the amplitude off-shell-- and be possible to measure.

You can perhaps always modulate a discontinuous pattern of behaviour by using an ad hoc sigmoid curve that does the trick.

But remember virtual particles come in all kinds of flavours, masses, and other quantum numbers. So the concept is not to do with gauge bosons per se.

It is not my original idea. It took me some time to find where I got it from, but here it is:

Quote

All photons that we detect actually interact with electrons in detectors such as the eye. They must all then, in some sense, be virtual! How can this be? We know that particles that are offshell have the range over which they can propagate limited by the extent to which they’re off-shell. If we see photons that have travelled from distant stars they have to be pretty close to being on-shell. We’ve seen before that when a particle is on-shell we hit the pole of the particle’s propagator. Therefore photons from Andromeda, visible on a moonless night, must be so close to the pole that there can’t be any observable effects from being off-shell.

Quantum Field Theory for the Gifted Amateur, Tom Lancaster and Stephen J. Blundell, 2014, Oxford University Press

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I understand what you were attempting Joigus, but those 'on-shell'/'off-shell' descriptors don't really cut it as definitions for me.
Neither does Genady's idea that all photons are virtual because they all interact with electrons.

To me, a real particle is measurable; a virtual particle is not.
By my definition, then, a virtual particle is below the measurable threshold of one quantum of action, while a real particle is above that same threshold.

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4 hours ago, Genady said:

It is not my original idea. It took me some time to find where I got it from, but here it is:

Quantum Field Theory for the Gifted Amateur, Tom Lancaster and Stephen J. Blundell, 2014, Oxford University Press

Ok. Yes, I have no problem in accepting that actual photons hitting my eyes are not precisely sitting on a mathematical line that's an idealized mathematical object. All experiments have error bars.

14 minutes ago, MigL said:

I understand what you were attempting Joigus, but those 'on-shell'/'off-shell' descriptors don't really cut it as definitions for me.
Neither does Genady's idea that all photons are virtual because they all interact with electrons.

To me, a real particle is measurable; a virtual particle is not.
By my definition, then, a virtual particle is below the measurable threshold of one quantum of action, while a real particle is above that same threshold.

There are many ways that we can think of to try to come to terms with this --perhaps uncomfortable-- concept. I remember some words by Sidney Coleman to that effect, from his Harvard lectures on QFT. We know what the mathematics of the theory says. From the mathematics of QFT we know more or less qualitatively --or in some reasonable cases-- that classical trajectories are the most likely, because they are at saddle points of the action. We also know that any event with a continuous distribution of probabilities has a chance zero of happening. That doesn't mean it's impossible, but it does mean that --under reasonable assumptions of continuity and differentiability-- right next to it are infinitely many events that are almost 'as much zero chance of happening.'

Edited by joigus
minor modification
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4 hours ago, joigus said:

Ok. Yes, I have no problem in accepting that actual photons hitting my eyes are not precisely sitting on a mathematical line that's an idealized mathematical object. All experiments have error bars.

There are many ways that we can think of to try to come to terms with this --perhaps uncomfortable-- concept. I remember some words by Sidney Coleman to that effect, from his Harvard lectures on QFT. We know what the mathematics of the theory says. From the mathematics of QFT we know more or less qualitatively --or in some reasonable cases-- that classical trajectories are the most likely, because they are at saddle points of the action. We also know that any event with a continuous distribution of probabilities has a chance zero of happening. That doesn't mean it's impossible, but it does mean that --under reasonable assumptions of continuity and differentiability-- right next to it are infinitely many events that are almost 'as much zero chance of happening.'

I'm following this discussion as far as I can. There is an article by Strassler here that seems to me to say virtual particles are qualitatively different from real ones: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/. and perhaps should never have been given a label containing the word "particle".

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1 hour ago, exchemist said:

I'm following this discussion as far as I can. There is an article by Strassler here that seems to me to say virtual particles are qualitatively different from real ones: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/. and perhaps should never have been given a label containing the word "particle".

I've skimmed through the article, and as far as I can see, this is more or less what I meant by,

12 hours ago, joigus said:

At present in QFT "virtual particles" is nothing but a fancy name for quantum mechanical amplitudes that cannot be made consistent with the on-shell condition, but must be included in the calculations.

They don't appear in the measurements; they appear in the calculations.

It's been claimed that when we hit a particle, and we produce jets of other particles, what we're doing is giving these amplitudes enough energy so that the virtual mode gets on-shell, so to speak, and can then be measured. That's, again, a way of speaking. But it's not too far off the mark, I would say.

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27 minutes ago, joigus said:

They don't appear in the measurements; they appear in the calculations.

True. But not distinct enough for me.

For example, atom nuclei don't appear in measurements either; they appear in the calculation of a spread of recoiled alpha particles. The latter also don't appear in the measurements; they appear in calculations of trajectories from the gold screen to the detectors. Etc.

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30 minutes ago, Genady said:

True. But not distinct enough for me.

For example, atom nuclei don't appear in measurements either; they appear in the calculation of a spread of recoiled alpha particles. The latter also don't appear in the measurements; they appear in calculations of trajectories from the gold screen to the detectors. Etc.

Well, I don't know. A measurement is not just a particle hitting a screen. You can separate isotopes by their mass, for example. There's your nucleus. You certainly cannot centrifuge virtual particles. They do contribute to the particle's mass, but you can't separate them and analyse them as independent things. That's why people sometimes talk about a 'cloud of virtual particles' around the real, detectable particle. Again, a pictorial handle for the concept, not quite what it 'is' in the theory.

Edited by joigus
minor correction
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2 minutes ago, joigus said:

Well, I don't know. A measurement is not just a particle hitting a screen. You can separate isotopes by their mass, for example. There's your nucleus. You certainly cannot centrifugue virtual particles. They do contribute to the particle's mass, but you can't separate them and analyse them as independent things. That's why people sometimes talk about a 'cloud of virtual particles' around the real, detectable particle. Again, a pictorial handle for the concept, not quite what it 'is' in the theory.

Yes, they appear in fewer cases than nuclei, but still the same virtual particles appear in all Feynman diagrams. Regarding separating and analyzing them individually, one word: quarks.

I understand what you mean and do not disagree. It is just that the distinction is not good enough for me to decide that they are profoundly different. Like we say here often, it's a model, just like everything else.

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12 minutes ago, Genady said:

It is just that the distinction is not good enough for me to decide that they are profoundly different.

And I totally understand your dissatisfaction, believe me. It's part of the conceptual issues of QM. Is it there when I'm not measuring it? What does it mean that now it is? How does the formalism incorporate that fact?

It's not that one virtual particle contributes to the mass of the measured particle, it is rather that all the wildest things you can imagine the particle as possibly doing within the strictures of HUP are somehow contributing to the measurable properties of the particle. And all of these ghostly presences interfere in every which way to produce the observed behaviour. That't what's difficult to swallow. But that's what the quantum formalism tells us. How to wrap your head around it in a pictorial way that helps you understand it better is another matter.

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2 hours ago, Genady said:

True. But not distinct enough for me.

For example, atom nuclei don't appear in measurements either; they appear in the calculation of a spread of recoiled alpha particles. The latter also don't appear in the measurements; they appear in calculations of trajectories from the gold screen to the detectors. Etc.

Evidence of them does appear in measurements, though. Almost all non-trivial observations in science are to some extent indirect. Even elusive things like neutrinos can be shown to be real by patient enough observation.  But there is no known observation that can show the presence of a virtual particle, whether directly or indirectly. What experiment could be devised that could show whether virtual particles did or did not exist? I don't think anybody has ever tried to do that, and that's because physicists know it would be a futile exercise.

Edited by exchemist
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7 minutes ago, exchemist said:

But there is no known observation that can show the presence of a virtual particle, whether directly or indirectly. What experiment could be devised that could show whether virtual particles did or did not exist? I don't think anybody has ever tried to do that, and that's because physicists know it would be a futile exercise.

Isn't it simply a consequence of their definition as particles that do not show in input or output? If they do show, they are not virtual, by definition.

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Just now, Genady said:

Isn't it simply a consequence of their definition as particles that do not show in input or output? If they do show, they are not virtual, by definition.

I'm not sure what you man by no input or output. If you mean they have, even in principle, no measurable properties, then that seems to make my point. No measurable properties, even in principle  => unreal, surely?

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Just now, exchemist said:

I'm not sure what you man by no input or output.

I mean input and output in particle reactions. External legs on Feynman diagrams. Virtual particles are, by definition, internal lines in the latter.

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