Andeh Posted January 25, 2012 Posted January 25, 2012 So virtual particles can be thought of as having an "energy debt". The greater the energy of the virtual particles, the shorter ammount of time it exist for. I've never discreetly read this...but I'm assuming that if a virtual particle has an energy debt of planck energy, than it will exist for planck time. Therefor planck energy is the maximum energy that a virtual particle can "owe", since having more would mean existing for shorter than planck time, which is impossible. Am I correct to assume this?
swansont Posted January 26, 2012 Posted January 26, 2012 The energy debt concept is a bit of classical thinking creeping into quantum mechanics, but it's based on the Heisenberg Uncertainty Principle. [math]\Delta{E}\Delta{t}>\hbar/2[/math] Both approaches get you to Planck's constant, but that's because the Planck mass (and thus energy) is relatively large. Conceptually, the Planck units use G, which doesn't really come into play in most QM situations.
timo Posted January 26, 2012 Posted January 26, 2012 The virtual particles that I know conserve energy and momentum on every vertex in the diagram. I don't quite see how you can properly speak of an "energy dept" in this context. 1
derek w Posted January 27, 2012 Posted January 27, 2012 question?Is total amount of energy in universe at present equal to total amount of energy released in big bang. Virtual particles must be sharing the energy from their source.Or is the universe gaining energy.
yaik7a Posted January 27, 2012 Posted January 27, 2012 I don't think the universe can simply gain energy, correct me if I'm wrong though.
Widdekind Posted January 29, 2012 Posted January 29, 2012 The virtual particles that I know conserve energy and momentum on every vertex in the diagram. I don't quite see how you can properly speak of an "energy debt" in this context. I understand, that when quantum 'particles' have, locally, negative energy, i.e. [math]E(\vec{x}) < V(\vec{x})[/math]; then their wave-functions decay away exponentially in time [math]\Psi \propto e^{-\frac{V-E}{\hbar}t}[/math]. So, I understand, that "virtual" particles, being "borrowed" into existence, from the zero-energy vacuum, on a "Heisenberg energy loan", propagate whilst decaying away exponentially quickly. Qualitatively, therefore, I understand that "virtual" particles can only propagate a distance [math]d \approx c \Delta t \approx \hbar / E[/math], before that exponential decay, of their wave-functions, "drives" the particles out of existence [math]\Psi \rightarrow 0[/math]. You said, "virtual particles...conserve energy-and-momentum, at every vertex, in a Feynmann diagram" ? I understand, that, "in the ultimate end", every interaction winds up, after-the-fact, having conserved energy-and-momentum. However, I understand, that, during interactions -- i.e. as they are "in progress", and the virtual particle, mediating that interaction, is propagating, across the FD, from the emitter ("emission vertex"), to the absorber ("absorption vertex") -- force-carrying quanta can be, and are, "borrowed" into existence. That "energy debt", on the force-carrier, explains the short range, of the residual Strong force, between nucleons, mediated by massive pions [math]\left( E \approx 140 MeV\right)[/math]. By analogy, if the force-carrier is likened to a USPS mail delivery truck, which carriers an "letter", i.e. interaction information, from one particle (emitter) to another (absorber); then a massive pion, "borrowed" into existence, is like a USPS mail truck, "borrowed" from the "garage", which has to be "brought back before noon". Accordingly, the "letter", representing transferred momentum-and-energy, can only be transmitted over short ranges, before the carrier is "due back". But, momentum-and-energy, is strictly conserved, amongst the emitter, the actual amount of force transmitted, and the absorber: [math]E_e \longrightarrow E_e' + \delta E \longrightarrow E_e' + \left( E_a + \delta E \right)[/math] When you say "energy-and-momentum is conserved at every vertex", I understand, that you are referring to the quantities [math]E_e \longrightarrow E_e' + \delta E[/math] (emission vertex); and [math]\delta E + E_a \longrightarrow E_a'[/math] (absorption vertex). But, that interaction energy [math]\delta E[/math] may be "carried", "conveyed", i.e. mediated, by a massive boson, "borrowed on a Heisenberg energy loan", of mediator mass-energy [math]\Delta E \ne \delta E[/math] having nothing, directly, to do with the amount of interaction energy [math]\delta E[/math]. The latter is strictly conserved, as you said; the former is "borrowed", as the OP said (as I understand): [math]mediator \; mass-energy \; \ne \; interaction \; energy[/math] [math]\Delta E \ne \delta E[/math]
swansont Posted January 29, 2012 Posted January 29, 2012 You said, "virtual particles...conserve energy-and-momentum, at every vertex, in a Feynmann diagram" ? I understand, that, "in the ultimate end", every interaction winds up, after-the-fact, having conserved energy-and-momentum. However, I understand, that, during interactions -- i.e. as they are "in progress", and the virtual particle, mediating that interaction, is propagating, across the FD, from the emitter ("emission vertex"), to the absorber ("absorption vertex") -- force-carrying quanta can be, and are, "borrowed" into existence. That "energy debt", on the force-carrier, explains the short range, of the residual Strong force, between nucleons, mediated by massive pions [math]\left( E \approx 140 MeV\right)[/math]. Energy and momentum are conserved at the vertex. Not "in progress". The "energy debt" concept assumes classical notions rather than QM ones, namely the HUP.
Widdekind Posted January 29, 2012 Posted January 29, 2012 Energy and momentum are conserved at the vertex. Not "in progress". The "energy debt" concept assumes classical notions rather than QM ones, namely the HUP. How do particles interact Weakly, via super-massive Weak bosons, of 80-90 GeV, without "borrowing" momentum-and-energy ?
swansont Posted January 29, 2012 Posted January 29, 2012 How do particles interact Weakly, via super-massive Weak bosons, of 80-90 GeV, without "borrowing" momentum-and-energy ? Why do they need to "borrow" energy?
Widdekind Posted January 30, 2012 Posted January 30, 2012 Why do they need to "borrow" energy? Where would a ~MeV solar neutrino get 80 GeV, to interact Weakly, with human neutrino detectors on earth, e.g. SNO, Super-K ? How can "the books balance", without "borrowing" ?
swansont Posted January 30, 2012 Posted January 30, 2012 Where would a ~MeV solar neutrino get 80 GeV, to interact Weakly, with human neutrino detectors on earth, e.g. SNO, Super-K ? How can "the books balance", without "borrowing" ? The books only have to "balance" to be in accord with [math]\Delta{E}\Delta{t}>\hbar[/math]
timo Posted January 30, 2012 Posted January 30, 2012 Virtual particles do not need to have an energy greater than the mass of the corresponding real particle. More precisely, they do not need to obey the relation that [math] E^2 = {\vec p}^2c^2 + m^2 c^4[/math], where m is the mass of the corresponding real particle and p is the momentum. They are virtual particles, not real particles. In fact, I believe it is better not to think of virtual particles as particles at all. But I am aware in an Internet forum there is no chance at all for this attitude to become acknowledged ("they are called particles, therefore they must be the kind of object I see in the pictures on Wikipedia!").
derek w Posted January 31, 2012 Posted January 31, 2012 -timo- is right,virtual particles are not particles,they are field disturbances that look like particles,they only appear when real particles are close to each other.
Widdekind Posted February 1, 2012 Posted February 1, 2012 In the CoM frame, for two colliding particles, having equal masses; if both particles impinge upon the point of impact, and if both are "deflected to the right" (say); then did not the virtual particle, effectively carry a momentum impulse, transverse to its direction of travel, from emitter to absorber ?
swansont Posted February 2, 2012 Posted February 2, 2012 In the CoM frame, for two colliding particles, having equal masses; if both particles impinge upon the point of impact, and if both are "deflected to the right" (say); then did not the virtual particle, effectively carry a momentum impulse, transverse to its direction of travel, from emitter to absorber ? In the CoM frame, how could both particles be deflected to the right?
derek w Posted February 2, 2012 Posted February 2, 2012 (edited) Please correct me if I am misunderstanding.That a virtual particle is an energy spike created by the interference of field waves,and that no actual particle is exchanged,but acts more like a spring pushing the real particles apart? If the particles were just drifting towards each other they would just pass through each other(no Doppler effect?). Edited February 2, 2012 by derek w
swansont Posted February 2, 2012 Posted February 2, 2012 Virtual particles are part of a model of interactions. The model works reasonably well.
derek w Posted February 2, 2012 Posted February 2, 2012 hmm.Think i will study up on waves a bit more.
Widdekind Posted February 2, 2012 Posted February 2, 2012 In the CoM frame, how could both particles be deflected to the right? Clarifying, in the CoM frame, each particle would veer to "its own right", yes ? But I may have mis-stated -- all virtual particle "pushes or pulls" could plausibly be modeled, with virtual particles, having momenta, either parallel to, or anti-parallel to, their direction of motion. All other angular scattering variabilities, could plausibly be accounted for, by effective "impact parameters", i.e. "off axis" collisions.
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