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Sure, it's also a question of time scales. On a time scale of a typical nuclear reaction a neutron or a pion is a very stable particle and thus can be treated almost always as such.
I think the intended reference was p.722, http://journals.aps.org/prd/abstract/10.1103/PhysRevD.22.722mfb said:"Rizzo Phys. Rev. D22 (1980) 389"
mfb said:"Hey what does the star mean?" "It is off-shell".
Haelfix said:If you like, you can draw one of the possible reactions as H --> Z l l, where the real Z is treated as an unstable particle that subsequently decays (in another diagram that you draw) and we reconstruct the usual peak (which we can interpret in the usual way as a pole of our S matrix process). The internal Z is a virtual particle as per your own definition!
I think I understand now in my own words what happens in the ##H\to WW^*## decay. I added the following two sections to the collection of definitions in the Insight article The Physics of Virtual Particles.A. Neumaier said:So for me there is something more to be understood clearly enough that I can explain it in my own words.
A difference between resonances and virtual particles is that resonances are actual states in the Hilbert Space, i.e. they are present even non-perturbatively.vanhees71 said:The same applies to unstable particles. Strictly speaking they are never particles but resonances and as such appear in the description of scattering matrix elements as internal lines.
I agree that this debate is in a sense "about nothing", but many people, even professionals do not think as you seem to believe they do. They think virtual particles are really physical states, not just perturbative labels and that the proton is really a "sea" of quarks and gluons, rather than this being an approximation model.Also the debate about the "inner structure of protons" is funny. I think, the whole debate is much ado about nothing or say about sloppy language in the QFT community. Any practitioner of QFT,...
We can get answers because the series is an asymptotic series, i.e. a specific kind of divergent series where the early terms do give a good approximation. In such a series the series approaches the full result up until the point where the error is of order ##e^{-\frac{1}{\lambda}}##, with ##\lambda## the expansion parameter. At this point it begins to diverge.bhobba said:I think the much more interesting issue is the series is asymptotically divergent. How can we get answers from a divergent series?
DarMM said:A difference between resonances and virtual particles is that resonances are actual states in the Hilbert Space, i.e. they are present even non-perturbatively.
IMO this whole discussion is pointless. All that is ever measured are interactions(that's why the meaningful quantities come from the vertices in the diagrams), never "real particles" in the idealized infinite time of the free field theory nor "virtual particles". The detections, whether one refers to them as clicks, dots or tracks or scattering cross sections all refer to interactions and it is nonsensical to debate whether the particles exist as it will only depend on the kind of book-keeping one previously decides to use based on their particular prejudices.atyy said:So this Physics FAQ is wrong?
http://www.desy.de/user/projects/Physics/Quantum/virtual_particles.html
"Then, the use of virtual particles as a communication channel is completely consistent with quantum mechanics and relativity. That's fortunate: since all particle interactions occur over a finite time interval, in a sense all particles are virtual to some extent."
Presumably the answer should be that virtual particles do not exist, whereas resonances are short lived states in Hilbert space which can be seen by a suitable measurement?
It tells (just as wikipedia) the ''popular science'' story based on the silent (but meaningless) identifications mentioned in post #58, not the ''research level'' one which has precise conventions. In addition, the Physics FAQ makes some other meaningless statements such as ''A virtual particle with momentum p corresponds to a plane wave filling all of space''. In fact it treats virtual particles as if they had wave functions (which would make them on-shell).atyy said:So this Physics FAQ is wrong?
As discussed before, there is no clear line between a W* that is clearly off-shell and a muon that is treated as real particle, despite its finite lifetime and corresponding uncertainty in the invariant mass of its decay products. Yes, the muon is a much better approximation to a real particle with a proper mass, but it is still just an approximation.A. Neumaier said:It tells (just as wikipedia) the ''popular science'' story based on the silent (but meaningless) identifications mentioned in post #58, not the ''research level'' one which has precise conventions. In addition, the Physics FAQ makes some other meaningless statements such as ''A virtual particle with momentum p corresponds to a plane wave filling all of space''. In fact it treats virtual particles as if they had wave functions (which would make them on-shell).
What is regarded as existent (i.e., modeled by a state rather than symbolically by an internal line) in a particular experimental situation is in borderline cases a matter of modeling. But since it changes the predictions (factorized or unfactorized computations in the simulation!), one can in principle distinguish the two situation - though (due to limited data and/or simulation accuracy) perhaps not in practice.mfb said:there is no clear line between a W* that is clearly off-shell and a muon that is treated as real particle
Your model never influences physics. It would be like claiming 2 meters is a fundamental concept of nature if you use one model above 2 meters and a different one below that just because they are better approximations.A. Neumaier said:What is regarded as existent (i.e., modeled by a state rather than symbolically by an internal line) in a particular experimental situation is in borderline cases a matter of modeling. But since it changes the predictions
(factorized or unfactorized computations in the simulation!), one can in principle distinguish the two situation - though (due to limited data and/or simulation accuracy) perhaps not in practice.
The decay width is an observable - it does not care about the model you use. Every unstable particle has a decay width, if you measure the invariant mass of its decay products precise enough you will always get different values for different decays.A. Neumaier said:In any case, it makes no sense (except in sloppy talk) to model a particle as off-shell and then claim it has a lifetime or another spatio-temporal property since the latter are properties of on-shell particles only. If one wants something to have a lifetime one must model it as on-shell (even when the sloppy talk says otherwise).
An observed decay width always needs for its interpretation a model involving a process observed in time, hence a state, hence a real particle (resonance). A virtual particle never has a decay width since it does not make formal sense to talk about its temporal properites. Thus one cannot model a decay width by a virtual particle.mfb said:The decay width is an observable - it does not care about the model you use.
I'm not talking about virtual particles now, I'm discussing the "on-shell"/"off-shell" difference which does not have a clear dividing line. The W* in a Higgs decay is clearly off-shell, while muons in decays are considered as on-shell. And there is no sharp line dividing those two cases. You can treat one as off-shell and one as on-shell, fine, but that is just your model used, it is not the physics behind it. You can also treat the muon as off-shell particle (with way too much computing effort) or the W* as "not so far away from on-shell particles" (with a huge error).A. Neumaier said:An observed decay width always needs for its interpretation a model involving a process observed in time, hence a state, hence a real particle (resonance). A virtual particle never has a decay width since it does not make formal sense to talk about its temporal properites. Thus one cannot model a decay width by a virtual particle.
So what is the same physics behind the two different treatments?mfb said:You can treat one as off-shell and one as on-shell, fine, but that is just your model used, it is not the physics behind it.
Well, and the good thing is that it doesn't matter, how you call the diagrammatic elements of a Feynman diagram symbolizing a mathematical expression to calculate an S-matrix element. You just calculate it, square it, multiply it with the appropriate kinematical factors, and you get (within a model, of course, because without a model, you'd have no Feynman diagram to begin with) a lifetime/decay width or cross section.mfb said:I'm not talking about virtual particles now, I'm discussing the "on-shell"/"off-shell" difference which does not have a clear dividing line. The W* in a Higgs decay is clearly off-shell, while muons in decays are considered as on-shell. And there is no sharp line dividing those two cases. You can treat one as off-shell and one as on-shell, fine, but that is just your model used, it is not the physics behind it. You can also treat the muon as off-shell particle (with way too much computing effort) or the W* as "not so far away from on-shell particles" (with a huge error).
What is different? After a long time, you get stable particles flying away. W* or muon are "just" our description how those particles got created.A. Neumaier said:So what is the same physics behind the two different treatments?
Someone in this thread argued that there was a difference between those categories I think.vanhees71 said:How call them, on shell or off shell is completely irrelevant.
After a long time, we are all dead, and physics no longer matters.mfb said:After a long time, you get stable particles flying away. W* or muon are "just" our description how those particles got created.
No. A ##W^*## is never responsible for the creation of these particles, while a ##W## can be. For getting created is a process in time, and this requires for its description a state - honce (on the level of the particle language) an on-shell particle.mfb said:W* or muon are "just" our description how those particles got created.
But we get leptons from both in a Higgs decay.A. Neumaier said:No. A ##W^*## is never responsible for the creation of these particles, while a ##W## can be.
And there is the "on-shell" again. No unstable particle is "exactly on-shell". Not the W, and not even the muon.For getting created is a process in time, and this requires for its description a state - honce (on the level of the particle language) an on-shell particle.
No. We get leptons from Higgs, and we calculate it either in partially resolved time as ##H\to Wll## and ##W\to ll## or unresolved in time as ##H\to llll##. The ##W^*## is only a formal book-keeping device, not something existing in time, hence nothing from which we can gat anything - except figuratively.mfb said:But we get leptons from both in a Higgs decay.
Of course they are on-shell, on a mass shell with a complex rest mass, as explained in the companion Insight article. That's what defines an unstable particle or resonance. They have an associated eigenstate for the analytically continued Hamiltonian in the second sheet of the Riemannian surface defined by the resolvent. In terms of the original Hilbert space they have a state formed by a superposition of states in the continuous spectrum of ##H## (the system Hamiltonain in the rest frame of the unstable particle), with energies corresponding to a continuous range of masses - which may be visible as a resonance peak. Thus their real effective mass (defined by ##m=H/c^2##) is unsharp, in the same sense as the position of a particle in a beam is unsharp. But this doesn't make unstable particles off-shell.mfb said:No unstable particle is "exactly on-shell". Not the W, and not even the muon.
In this process as written, time is not resolved at all, and the S-matrix elements (that only describe the in-out behavior, not the behavior at finite time) are computed in terms of Feynman integrals described by virtual particles in which no muon is present. Thus the muon is gone from the description - there is no occasion to talk about it. If you want to have a more detailed picture in time you either need to factor the S-matrix in terms of on-shell but unstable intermediate particle states such as a muon, or you need to work in terms of a hydrodynamic or kinetic description where states are abstracted from the CTP (Schwinger-Keldysh) formalism. In the latter case, the muon appears as an on-shell resonance state.mfb said:Fine, then we use different definitions for "on-shell". Forget about the word. You can include the muon in the Higgs decay for ##H \to e^-e^+ (e^- \nu_\mu \bar\nu_e) (e^+ \bar\nu_\mu \nu_e) ##, brackets for clarity. Is the muon gone?
Hi @A. Neumaier:A. Neumaier said:Nothing virtual happens. The dry facts are that two real particles are created from gravitational energy (from two gravitons or from an external gravitational field), not from the vacuum. One particle escapes, the other is absorbed. A valid description is given on p.645 of the book
B.W. Carroll and D.A. Ostlie, An Introduction to Modern Astrophysics, 2nd. ed., Addison Wesley 2007.
No. A static field is not made up of photons. It is made up of the fields of the particles in the sphere, which in turn is due to the fact that every charged physical particle carries with it an electromagnetic field - in a good approximation a Coulomb field.Buzz Bloom said:the particles that make up the contents of the static electric field about this sphere? Are they virtual photons?
Buzz Bloom said:What are the particles that make up the energy of the static gravitational field? Are they virtual gravitons?
Hi @A. Neumaier:A. Neumaier said:No. In a quantum field theory of gravity and matter, all massive physical particles also carry a gravitational field; everything is completely analogous. A static gravitational field behaves quantum mechanically in the same way, except that it can also produce two photons, because photons are their own antiparticles. Or two neutrinos, etc.. There is no energy barrier for photon production since photons are massless, but to achieve a noticeable effect, the field has to be extremely strong.
No. The loss of energy is encoded in the distortion of the metric of the 3-dimensional space when moving the time of slicing spacetime into 3D spaces at fixed times. Note that in a space-time view, the latter corresponds to dynamics!Buzz Bloom said:does this mean that when a particle pair is created, and the particles do not annihilate each other, the corresponding loss of energy in the field is manifested by a corresponding change in the distortion of space-time?
Hi @A. Neumaier:A. Neumaier said:No. The loss of energy is encoded in the distortion of the metric of the 3-dimensional space when moving the time of slicing spacetime into 3D spaces at fixed times. Note that in a space-time view, the latter corresponds to dynamics!
Of course, if the mass of a black hole is reduced, the gravitational field everywhere also weakens.Buzz Bloom said:the energy in the entire field outside the EH has been reduced by the energy of the created particle pair?
Hi @A. Neumaier:A. Neumaier said:Of course, if the mass of a black hole is reduced
The pair is initially created (spontaneously, according to Born's rule of quantum mechanics) near the horizon from the field of the black hole, which subtracts energy corresponding to two masses from the black hole. One of the particles returns into the black hole, while the other leaves. The balance is one lost particle mass. Maybe this link helps.Buzz Bloom said:he only thing that crosses the EH is one of the created pair particles going from the outside to the inside. How can this reduce the mass inside?
Hi @A. Neumaier:A. Neumaier said:The pair is initially created (spontaneously, according to Born's rule of quantum mechanics) near the horizon from the field of the black hole, which subtracts energy corresponding to two masses from the black hole.
Note that the black hole is the whole nonlocal object including the field - not just the mass concentration in its center. Even classically, the Schwarzschild metric defining a class of isolated black holes is an extended object, not just the singularity, and also not just ending at the event horizon. (An observer crossing the event horizon doesn't notice anything special!)Buzz Bloom said:the physical mechanism that causes a subtraction of energy corresponding to two masses from inside the EH when the particle pair is created outside the EH. Do these two events, one inside and one outside the EH, occur simultaneously?
You could try the forum Special and General Relativity, where specialists on general relativity might answer. (If you do so, please place here a link to the new discussion.) But your questions are of a kind nobody can answer to your satisfaction since they are based on the assumption that quantum field processes can be dissected into elementary pieces. If you are trying to do this you only end up in subjectively animating virtual reality. Thus you are likely to generate only more confusion. If you'd study the matter yourself (by going to the formal literature on the subject) you'd find that as you get more insight you'll gradually change your view of which kind of questions one can reasonably expect to answer.Buzz Bloom said:Do you think it might be useful for me to start a new tread specifically on this question which might attract some additional PF participants into the discussion?
Hi @A. Neumaier:A. Neumaier said:If you'd study the matter yourself (by going to the formal literature on the subject) you'd find that as you get more insight you'll gradually change your view of which kind of questions one can reasonably expect to answer.
These are not classical events! Like everywhere in quantum field theory, one can interpret only the end result of complicated calculations. One cannot say more about the process than the analogy of flow mentioned in my previous post.Buzz Bloom said:a single event involves (1) changes at a distance from the center mass of the BH (particle creation outside the EH) and also (2) changes (the total vanishing of a matter particle and its mass) some distance away close to the center of the BH
A. Neumaier said:your questions are of a kind nobody can answer to your satisfaction
Hi @A. Neumaier:A. Neumaier said:Like everywhere in quantum field theory, one can interpret only the end result of complicated calculations.
A particle cannot self annihilate. Apart from that 1 and 2 are ok. But not 3. At the center of the black hole are no particles, only (semiclassically) a singularity corresponding to the gravitational field of a point mass. This mass is a parameter that will have decreased. What one has instead in full quantum gravity is unknown.Buzz Bloom said:Would it be OK to say that one "end result" of the Hawking Radiation phenomenon is the following?
1. One particle of a pair created by the HR phenomenon near the outside of the EH, which does not self annihilate, will have escaped from the vicinity of the BH.
2. The other particle of the pair will have crossed to the inside of the EH.
3. One or more particles which are constituents (quarks and/or gluons?) of the BH mass near the center of the BH will have disappeared without having had its/their mass converted to corresponding energy.
Hi @A. Neumaier:A. Neumaier said:A particle cannot self annihilate.
Well, I am feeling that I am finally making some progress. 2 out of 3 ain't so bad. ;-)A. Neumaier said:1 and 2 are ok
Is the above description based on QG, or might it also be a correct description based on a combination of particle physics together with GR, or perhaps based on just GR alone?A. Neumaier said:At the center of the black hole are no particles, only (semiclassically) a singularity corresponding to the gravitational field of a point mass.
It never will. The two particles created will have opposite momentum, hence fly in opposite directions and are very unlikely to meet again.Buzz Bloom said:the pair did not self annihilate.
The singularity corresponds to the classical GR description. Particles will fly into the singularity in a very short (external) time and disappear. The quantum version is unknown until we have a proper theory of quantum gravity plus matter.Buzz Bloom said:Is the above description based on QG, or might it also be a correct description based on a combination of particle physics together with GR, or perhaps based on just GR alone?
Thus here.Buzz Bloom said:The thread is:
Qs re Hawking Radiation – Part I
I copy the target link from the ''bookmark'' position at the bottom of each post into the clipboard, then mark the text for the link using the chain symbol in the edit toolbar, paste the text from the clipboard into the free space for the link address, and remove the trailing text ''/bookmark/''.Buzz Bloom said:How do you create a link like that?
Hi @A. Neumaier:A. Neumaier said:the ''bookmark'' position at the bottom of each post
So there are no particles as "force carriers", correct?mfb said:The fields.
https://en.wikipedia.org/wiki/Force_carriermfb said:Depends on your interpretation of "force carriers".