Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: DoctorBeaver on 19/02/2008 09:47:34
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I've read in various sources that, according to QFT, virtual particles affect the strength of forces due to their interaction with real particles.
Over longer distances I can see that their influence would average out. But, surely, as they can have any energy except that of a real particle, over shorter distances their influence must be variable to quite a large extent. So how do short-range forces keep an exact strength? Is there some limiting factor?
I started wondering this when I was reading about Higgs particles giving mass to the Weak Gauge bosons. I understand the concept of spontaneous symmetry breaking, so is that the limiting factor? Is ≈250GeV the cut-off point for mass acquisition of W & Z bosons?
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While I am not an expert in all facets of this subject, I think that we must be careful not to take the picture of "particle" too far into the quantum realm. We envision 2 persons on rafts at sea, tossing a ball back and forth. By reason of the impact of the ball each time, an impulse is imparted to the catcher-thrower, with the result that there is a net repulsion between the rafts. Of course, using this classical example, we concluded that that repulsion is discontinuous in time. Using that kind of an idea at the quantum level, however, is probably not valid. It is doubtful that we can properly speak of a particle having a definite position apart from an operation to measure it.
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OK, so let's say the probability of 2 particles being further apart than 2 others.
Let's take the Weak Force as an example. A Weak Gauge Boson's path is long enough for it to be influenced by n virtual particles. Those virtual particles can have any energy so long as it isn't the same as a real particle of the same type. The greater the value of n, the greater the probability of the interactions averaging out. If, on the other hand, the WGB's path is only long enough for it to encounter 1 virtual particle, then the strength of that interaction can take any value.
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Doesn't anyone know? [:(]
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Beaver I hope this helps although I do not fully understand the question in your last paragraph maybe you could rephrase it.
Short range forces do not keep an "exact" strength, the uncertainty principle always applies. For example particle masses are variable depending on the lifetime of the particle. The shorter the life of the particle the wider is the peak that reperesents the mass in an accelerator experiment in fact the width of the mass peak is the measure of the particle lifetime.
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The last sentence - OK, I'll try to re-phrase it.
It is the W and Z bosons' interactions with Higgs particles that gives them their mass. The further a boson travels, the greater its possibility of encountering, and interacting with, a Higgs. It follows, therefore, that the further it travels, the more Higgs particles it will interact with.
The more interactions there are, the greater the probability that the strength of the interactions will average out. It's akin to rolling 2 dice. The average total for rolling 2 dice is 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). After a lot of rolls, your average score will be around 7. But by just rolling the dice once you could score 2 or 12. Similarly, just 1 interaction with a Higgs could add mass that is far from average unless all Higgs add the same amount of mass; that being the Weak Scale mass. If that is the case then something limits the total amount of mass that can be acquired to that of the Weak Scale mass. What is that something? Is it spontaneous symmetry-breaking?
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The answer hinges on the difference between quantum mechanics and quantum field theory, the latter of which is used to describe the physics of elementary particles. In quantum mechanics you sum over all the paths a particle might take to calculate the probability that a particle might be in a given place; there is no concept of "the path the particle took". In quantum field theory, you sum over all the paths a particle might take *and* all possible interactions it can have; there is no concept of "only interacted twice".
For the case of virtual particle momentum you talked about it gets a little more
complicated. The formalism requires you to sum (actually integrate) over all possible momentum, however high. Some such integrals will be ill defined, however (give infinite answers). In such cases, the sum/integral effectively restricted to a finite range of momenta, leaving out momenta over a certain value. The particular maximum
is arbitrary, as long as it is much higher than any momentum you are dealing with, with the strength of the interactions (ans values of the masses, etc..) being adjusted, as this maximum momentum is adjusted to keep all the answers the same.
A second point is that the idea of "interacting with the Higgs giving mass" doesn't really tally with your picture of interacting particles. The picture you present is
what I would call "perturbative": particles move along and every so often they interact, with some fixed probability per interaction such that having "many" interactions is unlikely. From the quantum field theory point of view you can try and apply perturbation theory for excitations about the vacuum (lowest energy particle/field configuration). It doesn't always converge (i.e. the case of interacting 10 times may be way more important than the case of 1; 100 more than 10 etc...), but it might be a good picture. For the standard model, it is a good picture for electromagnetism and the weak force, but not for the strong force (QCD) at low momenta. Adding the Yukawa couplings to the higgs (the way the W and Z are given mass), however, is a dramatic enough change to the physics that it cannot be described by perturbation theory, and instead should be thought of as a change in the vacuum; if you want to have a mental picture of a W moving along and rarely interacting with a higgs, it should be a W that already has mass.
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Thanks for that, Chris. I'll have to read it a few times to try to absorb it.
1 point I have picked up on, though is "...with some fixed probability per interaction such that having "many" interactions is unlikely."
If the probability is fixed, doesn't that mean that 100 interactions are as likely as 1?
Surely, if having many interactions is unlikely then every time a particle interacts the probability of it having further interactions must decrease.
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You have to remember that in the quantum domain nothing happens until you "see" it happen ie some interaction actually measures the state of the particle. in between the particle effectively experiences a tiny bit of every interaction that might have happened over every route possible that the particle might have happened to produce the result when you know it. I know it sounds completely loopy you just can't say what actually happens in between it is a pointless question.
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You have to remember that in the quantum domain nothing happens until you "see" it happen ie some interaction actually measures the state of the particle. in between the particle effectively experiences a tiny bit of every interaction that might have happened over every route possible that the particle might have happened to produce the result when you know it. I know it sounds completely loopy you just can't say what actually happens in between it is a pointless question.
That's 1 interpretation.