How do Photons exert Electrostatic forces ?

[Widdekind]

If photons must be laterally polarized, how can photons, exchanged along the line between two particles, explain the longitudinal "pushing" & "pulling", along that line, caused by Electrostatic forces ??

[swansont]

The Poynting vector, which tells you the direction of energy transport (and thus momentum) is E x B.

 

http://en.wikipedia.org/wiki/Poynting_vector

http://en.wikipedia.org/wiki/Poynting%27s_theorem

[Bob_for_short]

If photons must be laterally polarized, how can photons, exchanged along the line between two particles, explain the longitudinal "pushing" & "pulling", along that line, caused by Electrostatic forces ??

Nohow. It is not the transverse photons that act along the line. It is the electrostatic Coulomb force = the property of charges, not photons.

The Poynting vector, which tells you the direction of energy transport (and thus momentum) is E x B.

The pointing vector for electrostatic field equals zero (B=0). It makes sense only for propagating waves, not for static fileds.

Edited June 11, 2010 by Bob_for_short

[Widdekind]

The Poynting vector, which tells you the direction of energy transport (and thus momentum) is E x B.

 

http://en.wikipedia.org/wiki/Poynting_vector

http://en.wikipedia.org/wiki/Poynting%27s_theorem

 

Wikipedia says:

 

"In a propagating sinusoidal electromagnetic plane wave of a fixed frequency, the Poynting vector oscillates, always pointing in the direction of propagation."

 

So, such could account for particle-particle "pushing" (repulsion). Yet, what about "pulling" (attraction) ? (How could a photon, propagating one way, have "backwards pointing" Poynting momentum ?)

 

And, if photons (S=1) mediate EM interactions, does that mean that EM interactions always involve the exchange of Angular Momentum ?

[swansont]

The pointing vector for electrostatic field equals zero (B=0). It makes sense only for propagating waves, not for static fileds.

 

The OP asks about photons. Not static fields.

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Consecutive posts merged

So, such could account for particle-particle "pushing" (repulsion). Yet, what about "pulling" (attraction) ? (How could a photon, propagating one way, have "backwards pointing" Poynting momentum ?)

 

It won't. But that's limited to real photons. Are you asking about virtual photons?

[Widdekind]

The OP asks about photons. Not static fields.

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Merged post follows:

Consecutive posts merged

 

It won't. But that's limited to real photons. Are you asking about virtual photons?

 

Please tell me -- would "virtual photons" explain the electrostatic attraction, between oppositely charged particles ? Could "virtual photons" have "backwards momentum", so that the emission of the same would pull the emitter towards the target (a little like a "reverse rocket motor", which pulls the particle in the same direction as the (photon) mass-energy propelled away), and the absorption of the same would pull the target towards the emitter ?

[darkenlighten]

I would like some clarification from Widdekind....What electrostatic force are you referring to? Because the way you are phrasing the question imo doesn't make much sense. What forces acting on what?

 

Re-reading your above post. Are you just talking about the force between two charged particles (Coulomb force) in your original question....if so what does the photon have to do with anything?

[Widdekind]

(Virtual) photons mediate all Electromagnetic phenomena, including the "Electrostatic Coulomb force". That's what confuses me -- how can two, static, point charges, each carrying opposite "signs" of electrical charge, attract each other, by exchanging (virtual) photons ??

 

Explaining how two point charges, each carrying the same "sign" of electrical charge, could repel each other, by exchanging (virtual) photons, is easy -- as Swansont pointed out, photons carry momentum, so that the two positive point charges, could "shoot" photons between themselves, and each time they "fired" a photon, they'd be "kicked backwards" (a little like a rocket), and each time they absorbed the other charge's photons, they'd be "kicked backwards" again. In such a way, two same-signed charges would repel each other.

 

But, how can you use photons, to explain Electrostatic attraction ?

[swansont]

http://www.xs4all.nl/~johanw/PhysFAQ/Quantum/virtual_particles.html

[Widdekind]

Thanks for the link !

 

Can virtual photon emission, by electrons in atoms, be regarded as the (quantum equivalent of) the Larmor radiation, of the electrons, accelerating around in the attractive potential well of their nucleus ? Can you view the Hydrogenic wave functions, and their ensuing "electron clouds", as a "Bohr-model atom plus virtual photons" ??

 

Subatomic particles do not just sit around being subatomic particles. They are beehives of activity. An electron, for example, is constantly emitting & absorbing photons. These photons are not full-fledged photons, however. They are a now-you-see-it-now-you-don't variety. They are exactly like real photons, except that they don't fly off on their own. They are re-absorbed by the electron, almost as soon as they are emitted. Therefore, they are called "virtual photons"... They are virtually photons. The only thing that keeps them from being full-fledged photons is their abrupt re-absorption by the electron that emits them.

 

In other words, first there is an electron, then there is an electron and a photon, and then there is an electron again. This situation is, of course, a violation of the conservation law of mass-energy... The reason this can happen, according to the theory, is the famous Heisenberg Uncertainty Principle... There is a reciprocal uncertainty of time and energy. The less uncertainty there is about the time involved in a subatomic event, the more uncertainty there is about the energy involved in the event (and the other way around)... B/c of this uncertainty, the balance books kept in the conservation law of mass-energy are not upset. Said another way, the event happens, and is over so quickly, that the electron can get away with it...

 

Virtual photons differ from real ones, in that their rest mass is not zero; only zero rest mass photons can escape. There are two ways of looking at virtual photons mathematically. In the first (old-fashioned perturbation theory), the mass of a virtual particle is the same as the mass of a real particle, but energy is not conserved. In the second (Feynman perturbation theory), energy-momentum is exactly conserved but the virtual particles do not have physical mass.

 

Gary Zukav. The Dancing Wu Li Masters, pp. 222-224.

If the electron needs to reabsorb its own virtual photons, their emission time must be comparable to the electron's crossing time, for its particular orbit. Thus, solely to order-of-magnitude, and using the H.U.P., the time & energy uncertainty, in these virtual photon self-absorption events is:

 

[math]\Delta t \approx \frac{\Delta x}{\Delta p / m_e} \approx \frac{m_e}{\hbar} \Delta x^2[/math]

 

[math]\Delta E \approx \frac{\hbar^2}{m_e \, \Delta x^2}[/math]

Thus, the energy uncertainty, "dumped" into the virtual photon field, is comparable to the K.E. of the emitting electron, and, hence (Virial Theorem), to the total bound-state energy of the emitting electron, [math]\Delta E \propto E[/math]. Could such an energy-uncertainty causes higher energy, excited electronic states, to "bleed over", into other orbitals of similar energy, and help account (Overlap Integrals, Transition Matrix Elements) for the spontaneous transitions, of excited electrons, into lower energy orbitals ?

 

 

 

Emission / Absorption as "Promotion / Demotion" of (Virtual / Real) photons

 

If we provide the necessary energy, for a virtual photon to become a real photon w/o violating the conservation law of mass-energy, it does just that. That is why an excited electron emits a real photon... The jettisoned photon is one of the electron's virtual photons, that suddenly finds itself with enough energy to keep going, w/o violating the conservation law of mass-energy, and it does. In other words, one of the electron's virtual photons suddenly is "promoted" to real photon.

 

ibid., pg. 224.

This strongly suggests, that photon absorption represents the "demotion", of an incident real photon, into a "bound" or "hang-around" virtual photon. Perhaps an electron's cloud of virtual photons is the repository housing all the incident photons it has ever absorbed ?? What happens, to the virtual photon cloud, when the electron's wave function discontinuously "re-adjusts", during the wave function collapse caused by a Measurement ??

Edited September 4, 2010 by Widdekind

[Widdekind]

http://www.xs4all.nl/~johanw/PhysFAQ/Quantum/virtual_particles.html

 

Thanks again again for the link, it is very informative. Gary Zukav says essentially the same thing, albeit qualitatively, w/o the mathematical rigor:

 

Electrons are always surrounded by a swarm of virtual photons. (There are other virtual particles in the cloud of virtual particles surrounding an electron, but photons are the most common among them.)

 

If two electrons come close enough to each other, close enough so that their virtual-photon clouds overlap, it is possible that a virtual photon that is emitted from one electron, will be absorbed by the other electron... The close the electrons come to each other, the more this phenomenon occurs. Of course, the process is two-way, with both electrons absorbing virtual photons that were emitted by the other.

 

This is how electrons repel each other. The closer two electrons come, the more virtual photons they exchange. The more virtual photons they exchange, the more sharply their paths are deflected. The "repulsive force" between them is simply the cumulative effect of these exchanges of virtual photons, the number of which increases at close range, and decreases at a distance. According to this theory, there is no such thing as action-at-a-distance -- only more & fewer exchanges of virtual photons. These interactions (absorptions and emissions) happen on location, so to speak, right there, where the particles involved are located.

 

The mutual repulsion of two particles, of the same charge, like two electrons, is an example of an electromagnetic force. In fact, according to QFT, an electromagnetic force is the mutual exchange of virtual photons. (Physicists like to say, that the electromagnetic force is "mediated" by photons). Every electrically charged particle continually emits & re-absorbs virtual photons, and/or exchanges them with other charged particles. When two electrons (two negatively charges) exchange virtual photons, they repulse each other. The same thing happens when two protons (two positive charges) exchange virtual photons. When a proton and an electron (a positive charge & a negative charge) exchange virtual photons, they attract each other.

 

Therefore, since the development of QFT, physicists generally have substituted the word "interaction" for the word "force". (An interaction is when anything influences anything else.)

 

Gary Zukav. The Dancing Wu Li Masters, pp. 224-226.

The author goes on to describe the Strong Force, as an exchange of virtual pions between nucleons, in essentially similar terms.

[Widdekind]

Can virtual photons be described, as having positive momentum (by which "(EM) forces" are exerted), but imaginary rest-mass-energy (being "off-mass-shell"), so that their total energy (& energy density, of the associated virtual photon "cloud" shadowing quantum particles) is zero: E² = E₀² + (pc)² = 0 ?

 

Imaginary rest-mass-energy, would translate into imaginary frequency (E = hf), which would make virtual photons fizzle away, exponentially, in time (e^ift --> e^-ft), explaining why they never venture far from the particles producing them. (This might mean, that the lowest frequency virtual photons, would last the longest, and have the "longest reach" (??).)

[J.C.MacSwell]

Can virtual photons be described, as having positive momentum (by which "(EM) forces" are exerted), but imaginary rest-mass-energy (being "off-mass-shell"), so that their total energy (& energy density, of the associated virtual photon "cloud" shadowing quantum particles) is zero: E² = E₀² + (pc)² = 0 ?

 

Imaginary rest-mass-energy, would translate into imaginary frequency (E = hf), which would make virtual photons fizzle away, exponentially, in time (e^ift --> e^-ft), explaining why they never venture far from the particles producing them. (This might mean, that the lowest frequency virtual photons, would last the longest, and have the "longest reach" (??).)

 

from above:

"explaining why they never venture far from the particles producing them"

 

Doesn't that imply a distance limit to the Coulomb forces?

 

And if they fizzle away exponentially in time, why is it an inverse square law?

[Bob_for_short]

...Electrons are always surrounded by a swarm of virtual photons...

 

...If we provide the necessary energy, for a virtual photon to become a real photon w/o violating the conservation law of mass-energy, it does just that. That is why an excited electron emits a real photon... The jettisoned photon is one of the electron's virtual photons, that suddenly finds itself with enough energy to keep going, w/o violating the conservation law of mass-energy, and it does. In other words, one of the electron's virtual photons suddenly is "promoted" to real photon...

 

This strongly suggests, that photon absorption represents the "demotion", of an incident real photon, into a "bound" or "hang-around" virtual photon. Perhaps an electron's cloud of virtual photons is the repository housing all the incident photons it has ever absorbed ??...

 

In my pet theory this bla-bla about virtual photons surrounding an electron is represented in a concrete formula (solution): the electron is a part of quantum oscillators. The whole system is called an "electronium". When you push the electron, the state of oscillators change - they get excited that corresponds to appearing real photons.

 

On the other hand I need a Coulomb interaction term too. It is another, independent property of charges (apart from emitting photons) and is not reduced to exchange of "virtual" photons. I my model each electronium has its own oscillators, like each atom has its own electrons and energetic levels. So the charge attraction/repulsion is another feature of charges.

 

In the standard QED this question is quite vague due to initially decoupling the "mechanical" and electromagnetic properties of a change and considering the coupling by the perturbation theory. Factually nobody has written the solution for a real electron surrounded by the virtual photons. Such a solution has to include the "mechanical" and the "wave" variables. I advances such a solution from physical and mathematical reasoning as an ansatz in http://arxiv.org/abs/0811.4416 and http://arxiv.org/abs/0806.2635

Edited September 7, 2010 by Bob_for_short

[Widdekind]

(thanks for the replies)

 

If w.h.t. (1) F = dp/dt; (2) F = -dV/dx; then could you consider, that the "electro-static potential (V)", which an electron "emits", and whose gradient defines the electro-static force-field emanating from said electron, as a "cloud" of virtual photons, of varying spatial density, so that their "electro-magnetic pressure" (P ~ dp/dt) generates the "push" associated with the electro-static potential (V) ?? In symbols, can you equate -dV/dx = dp/dt (of the virtual photon "cloud") ??

[Bob_for_short]

(thanks for the replies)

 

If w.h.t. (1) F = dp/dt; (2) F = -dV/dx; then could you consider, that the "electro-static potential (V)", which an electron "emits", and whose gradient defines the electro-static force-field emanating from said electron, as a "cloud" of virtual photons, of varying spatial density, so that their "electro-magnetic pressure" (P ~ dp/dt) generates the "push" associated with the electro-static potential (V) ?? In symbols, can you equate -dV/dx = dp/dt (of the virtual photon "cloud") ??

 

No! A charge does not emit any potential V( r )! V( r ) is a potential energy of interaction of two charges and r is their relative distance. There is no other meaning of the potential energy.

Edited September 7, 2010 by Bob_for_short

[Widdekind]

http://www.xs4all.nl/~johanw/PhysFAQ/Quantum/virtual_particles.html

 

Thanks again for the link. I understand, that charged particles are always 'emanating' or 'spinning out' virtual photon fields, 'out to infinity'. So, charge particles in interaction are immersed fully within their partner's 'far flung virtual photon field'. And so, even if the partner particle is 'in front', that partner particle's virtual photons can impinge upon the other particle 'from the back'.

 

Could you consider point particles, to 'emanate' spherical waves -- [math]e^{i k r} / r[/math] -- of both 'incoming' & 'outgoing' ([math]\pm k[/math]) varieties ?? (With a finite aged universe, would, technically, virtual photon fields have 'had time to establish' themselves, beyond their local Hubble Volume ??)

[Widdekind]

According to Zukov (PP), when particles interact, electromagnetically, each particle is absorbing virtual photons, from the other particle's entrained cloud, of virtual photons. And, when oppositely charged particles bind together, virtual photons, from the interaction, are 'promoted', into real, actual, actualized photons -- which promptly carry away the binding energy / momentum. And more, the afore-cited link seems to say, that a pair of equal-but-oppositely directed virtual photons, are involved in electrostatic interactions (of attraction). Does that mean, that one pair of virtual photons, one initially associated to each interacting particle, 'merge' & 'meld', 'forging', from two virtual photons, one real photon ?? (In a crude, qualitative-only, mathematical analogy, the sum of two imaginary numbers/functions, can yield a single real number/function.) And if so, by reversibility, an impinging, ionizing real photon, upon liberating the once-bound electron, 'splits', into a pair of 'equal-but-opposite' virtual photons, one associated to each free charged particle ??

 

[Widdekind]

from above:

"explaining why they never venture far from the particles producing them"

 

Doesn't that imply a distance limit to the Coulomb forces?

 

And if they fizzle away exponentially in time, why is it an inverse square law?

 

Gary Zukav says charged particles are 'always emitting & absorbing' virtual particles, during interactions. Now, first, if virtual photons had energy, then, since [math]\Delta E \Delta t \approx \hbar / 2[/math], the energetic virtual photon would have a finite lifetime, and, hence, a finite range. So, since the EM force is potentially infinite range, such cannot be the case. And, second, real particles constantly emitting & absorbing virtual photons would constantly lose/gain energy. Such seemingly suggests, that virtual photons carry momentum, but no energy -- as 'force carriers', they're all force-momentum, no energy. If attractive interactions involve pairs of virtual photons, then, perhaps the deconstructive interference, of two opposite virtual photons, amounting to a DC null photon (E=p=0), which is 'on mass shell', is the point of entry, of that pair, from virtuality, into birthed reality?? (A 'promoter' virtual photon detector, would need to 'match' the incident virtual photon, with an appropriately opposite pair.)

 

[Widdekind]

I seem to have been confused. In post 10 & post 11, I cite Gary Zukav, who wrote:

 

Subatomic particles do not just sit around being subatomic particles. They are beehives of activity. An electron, for example, is constantly emitting & absorbing photons. These photons are not full-fledged photons, however. They are a now-you-see-it-now-you-don't variety. They are exactly like real photons, except that they don't fly off on their own. They are re-absorbed by the electron, almost as soon as they are emitted. Therefore, they are called "virtual photons"... They are virtually photons. The only thing that keeps them from being full-fledged photons is their abrupt re-absorption by the electron that emits them.

 

In other words, first there is an electron, then there is an electron and a photon, and then there is an electron again. This situation is, of course, a violation of the conservation law of mass-energy... The reason this can happen, according to the theory, is the famous Heisenberg Uncertainty Principle... There is a reciprocal uncertainty of time and energy. The less uncertainty there is about the time involved in a subatomic event, the more uncertainty there is about the energy involved in the event (and the other way around)... B/c of this uncertainty, the balance books kept in the conservation law of mass-energy are not upset. Said another way, the event happens, and is over so quickly, that the electron can get away with it...

 

Virtual photons differ from real ones, in that their rest mass is not zero; only zero rest mass photons can escape. There are two ways of looking at virtual photons mathematically. In the first (old-fashioned perturbation theory), the mass of a virtual particle is the same as the mass of a real particle, but energy is not conserved. In the second (Feynman perturbation theory), energy-momentum is exactly conserved but the virtual particles do not have physical mass.

 

...

 

If we provide the necessary energy, for a virtual photon to become a real photon w/o violating the conservation law of mass-energy, it does just that. That is why an excited electron emits a real photon... The jettisoned photon is one of the electron's virtual photons, that suddenly finds itself with enough energy to keep going, w/o violating the conservation law of mass-energy, and it does. In other words, one of the electron's virtual photons suddenly is "promoted" to real photon.

 

...

 

Electrons are always surrounded by a swarm of virtual photons. (There are other virtual particles in the cloud of virtual particles surrounding an electron, but photons are the most common among them.)

 

If two electrons come close enough to each other, close enough so that their virtual-photon clouds overlap, it is possible that a virtual photon that is emitted from one electron, will be absorbed by the other electron... The close the electrons come to each other, the more this phenomenon occurs. Of course, the process is two-way, with both electrons absorbing virtual photons that were emitted by the other.

 

This is how electrons repel each other. The closer two electrons come, the more virtual photons they exchange. The more virtual photons they exchange, the more sharply their paths are deflected. The "repulsive force" between them is simply the cumulative effect of these exchanges of virtual photons, the number of which increases at close range, and decreases at a distance. According to this theory, there is no such thing as action-at-a-distance -- only more & fewer exchanges of virtual photons. These interactions (absorptions and emissions) happen on location, so to speak, right there, where the particles involved are located.

 

The mutual repulsion of two particles, of the same charge, like two electrons, is an example of an electromagnetic force. In fact, according to QFT, an electromagnetic force is the mutual exchange of virtual photons. (Physicists like to say, that the electromagnetic force is "mediated" by photons). Every electrically charged particle continually emits & re-absorbs virtual photons, and/or exchanges them with other charged particles. When two electrons (two negatively charges) exchange virtual photons, they repulse each other. The same thing happens when two protons (two positive charges) exchange virtual photons. When a proton and an electron (a positive charge & a negative charge) exchange virtual photons, they attract each other.

 

Therefore, since the development of QFT, physicists generally have substituted the word "interaction" for the word "force". (An interaction is when anything influences anything else.)

 

Gary Zukav. The Dancing Wu Li Masters, pp. 224-226.

 

According to the HUP, [math]\Delta E \, \Delta t \geq \hbar / 2[/math]. So, the maximum distance that an 'energy borrowed (i.e., virtual)' photon can propagate, away from the emitter, is [math]c \times \Delta t \approx \hbar / \Delta E[/math]. Thus, at greater distances, only weaker 'virtual' photons can be borrowed -- small [math]\Delta E[/math] -- long enough, to reach that remote location. The whole point of 'promotion' appears, then, to merely mean 'repaying energy debt'. If two charged particles can 'repay the debt' on the 'Heisenberg loan', on some 'virtual' photon, then the photon can be 'promoted', 'liberated' from 'debt slavery' [sic] to 'independence'.

 

The fundamental principle, which I did not comprehend, is the Quantum, non-Classical, 'detailed' violation, of strict Energy Conservation, per the HUP.

Edited April 16, 2011 by Widdekind

[Widdekind]

According to the HUP, [math]\Delta E \, \Delta t \geq \hbar / 2[/math]. So, the maximum distance that an 'energy borrowed (i.e., virtual)' photon can propagate, away from the emitter, is [math]c \times \Delta t \approx \hbar / \Delta E[/math]. Thus, at greater distances, only weaker 'virtual' photons can be borrowed -- small [math]\Delta E[/math] -- long enough, to reach that remote location. The whole point of 'promotion' appears, then, to merely mean 'repaying energy debt'. If two charged particles can 'repay the debt' on the 'Heisenberg loan', on some 'virtual' photon, then the photon can be 'promoted', 'liberated' from 'debt slavery' [sic] to 'independence'.

 

The fundamental principle, which I did not comprehend, is the Quantum, non-Classical, 'detailed' violation, of strict Energy Conservation, per the HUP.

 

Were we to please permit an over-extended analogy...

 

Charged fundamental particles, are a little like 'submarines'... which constantly fire 'virtual photon torpedos'... in various different directions... and which 'time the torpedos', based upon how much energy ([math]\Delta E \implies \Delta t[/math]) they 'borrow from the energy bank'.

 

And now, if those 'virtual photon torpedos', fired in some direction, actually 'hit something' -- interacting, with another charged fundamental particle -- then they 'explode' ('promoted', into 'actualized reality' -- albeit, immediately [and destructively] absorbed). For, that other particle, by adjusting its momentum-and-energy, can pay back the 'debt to the energy bank'.

 

And, so, in some sense, those 'virtual photon torpedos', are like Cash-on-Delivery (CoD) 'debt instruments', bearing the words 'bearer owes the energy bank 1 eV'. Thus, they are 'fired' from one charged fundamental particle; propagate through space; and 'hit' another particle, who 'picks up the tab, and pays the bill' (the ultimate 'passing the buck'). Again, at that point, the photon is immediately 'actualized', 'promoted' from virtuality to reality -- and promptly absorbed, deflecting the other particle, and so accounting for the EM charge interaction.

 

If this picture is appropriate, then charged fundamental particles are constantly 'emitting messengers', with which they are constantly 'probing their environment', constantly querying "is their a charge over there? or, how about over that-a-way" ?? And, if any of their 'CoD virtual photon torpedos' actually 'hit' other particles, those particles 'pick up the tab, & pay back the bank', thereby 'actualizing' the photon, from virtual-to-real, and then absorbing the ensuing momentum/energy, accounting for the EM interaction. (Again, such virtual photons, being destroyed upon actualization, amount to a private point-to-point communication, whose former virtuality, and ensuing instantaneous reality, are utterly unobservable, to all other particles, who are "not involved", which is why the photons, underlying the EM interactions, are never observed, except by their effects upon charged particles, whose paths are visibly deflected by those photons.)

 

Otherwise, the 'virtual photon torpedo' presumably 'fizzles out like a dud'... and the firing particle 'goes back to the way it was', 'undoing the recoil of firing the torpedo in the first place'. (Perhaps all those 'recoils', from constantly 'borrowing from the bank', and 'firing photon torpedos every which way', accounts for the Schrodinger-equation spreading of Wave Functions ???)

[lemur]

I could understand the analogy of potential energy with savings and kinetic energy as spending/consumption; but how can dumb objects hold and keep track of debts? Wouldn't that require memory? I suppose you could say that a compressed spring has memory, which is related to its ability to store energy as potential, but how could energy be put into a spring or other storage medium with constraints on how that energy gets "repaid?" Doesn't a physical system necessarily have to look for the path of least resistance where energy gets expressed?

[farmboy]

Were we to please permit an over-extended analogy...

 

Charged fundamental particles, are a little like 'submarines'... which constantly fire 'virtual photon torpedos'... in various different directions... and which 'time the torpedos', based upon how much energy ([math]\Delta E \implies \Delta t[/math]) they 'borrow from the energy bank'.

 

And now, if those 'virtual photon torpedos', fired in some direction, actually 'hit something' -- interacting, with another charged fundamental particle -- then they 'explode' ('promoted', into 'actualized reality' -- albeit, immediately [and destructively] absorbed). For, that other particle, by adjusting its momentum-and-energy, can pay back the 'debt to the energy bank'.

 

And, so, in some sense, those 'virtual photon torpedos', are like Cash-on-Delivery (CoD) 'debt instruments', bearing the words 'bearer owes the energy bank 1 eV'. Thus, they are 'fired' from one charged fundamental particle; propagate through space; and 'hit' another particle, who 'picks up the tab, and pays the bill' (the ultimate 'passing the buck'). Again, at that point, the photon is immediately 'actualized', 'promoted' from virtuality to reality -- and promptly absorbed, deflecting the other particle, and so accounting for the EM charge interaction.

 

If this picture is appropriate, then charged fundamental particles are constantly 'emitting messengers', with which they are constantly 'probing their environment', constantly querying "is their a charge over there? or, how about over that-a-way" ?? And, if any of their 'CoD virtual photon torpedos' actually 'hit' other particles, those particles 'pick up the tab, & pay back the bank', thereby 'actualizing' the photon, from virtual-to-real, and then absorbing the ensuing momentum/energy, accounting for the EM interaction. (Again, such virtual photons, being destroyed upon actualization, amount to a private point-to-point communication, whose former virtuality, and ensuing instantaneous reality, are utterly unobservable, to all other particles, who are "not involved", which is why the photons, underlying the EM interactions, are never observed, except by their effects upon charged particles, whose paths are visibly deflected by those photons.)

 

Otherwise, the 'virtual photon torpedo' presumably 'fizzles out like a dud'... and the firing particle 'goes back to the way it was', 'undoing the recoil of firing the torpedo in the first place'. (Perhaps all those 'recoils', from constantly 'borrowing from the bank', and 'firing photon torpedos every which way', accounts for the Schrodinger-equation spreading of Wave Functions ???)

 

Widdekind, is what you've said here (and in the last post) just an interpretation of the stuff you quoted by the Zukav dude, or was it based on your own personal knowledge also? I had never heard of this phenomena before this thread, but thought I was understanding it anyway, but then all this stuff about repaying debts between atoms didn't really seem to fit with what I had read before.

 

What I had understood up to that point was that the HUP allows the formation of these virtual photons because as well as a relationship between momentum and location there is also a relationship between energy and time. So essentially because the time is so short that allows us to have a greater uncertainty in the energy involved in the process (or maybe that is backwards) which allows the existence of the virtual photons without violating conservation of energy. This also explains why interactions decrease in strength the further the interacting particles are from one another. Is that incorrect (could easily be pure BS lol)?

 

So yeah basically my problem was that I'm not sure where you are getting all this stuff about atoms borrowing money from one another and paying it back, has that information come from somewhere else (something you have studied but which hasn't been stated so far in this thread) or is that perhaps what was meant by zukarov in the sections quoted, and I have just failed to understand it.

 

Very interesting topic by the way, particle physics is something I know nothing about really, something I should probably try to rectify.

Edited April 17, 2011 by farmboy

[Widdekind]

Energy is not conserved, in a detailed sense -- moment by moment, virtual particles can occur, "borrowing" energy from the background (the Zero-Point Energy ??), and existing for some small amount of time, consistent with the HUP.

 

Most of the time, those virtual particles, apparently, "fizzle out" at the end of their HUP-allowed lifetime. However, if they encounter a real particle, which can supply the energy that they've "borrowed", then the virtual particles can be "promoted" into real particles (per Zukav).

[Widdekind]

Energy is not conserved, in a detailed sense -- moment by moment, virtual particles can occur, "borrowing" energy from the background (the Zero-Point Energy ??), and existing for some small amount of time, consistent with the HUP.

 

Most of the time, those virtual particles, apparently, "fizzle out" at the end of their HUP-allowed lifetime. However, if they encounter a real particle, which can supply the energy that they've "borrowed", then the virtual particles can be "promoted" into real particles (per Zukav).

 

Whenever two particles interact, even when one is an "energy-borrowed" virtual particle, the pair are entangled. And, according to the more detailed account, in the link, provided by Swansont (thanks again!), virtual photon emission occurs in pairs of opposite-but-equal momentum ([math]\pm k[/math]); and, charged particles interact, with the 'appropriate' virtual photon, of that pair, in order to effect the appropriate EM interaction (e.g., electrostatic repulsion (same charge), or attraction (opposite charge)).

 

Thus, what such seemingly suggests, is the following:

 

1. a charged particle "borrows [math]\Delta E[/math]", and emits a virtual photon, with which the emitting particle is entangled
   
2. that virtual photon has "outward" (positive) & "inward" (negative) momentum components
   
3. if that virtual photon "fizzles away", without interacting with any other charges, then, due to the entanglement, the emitting particle "pops back" to the way it was
   
4. otherwise, if that virtual photon interacts with another charge, then that charge "pays the [math]\Delta E[/math] bill", promotes the 'appropriate' aspect of the virtual photon into reality, and absorbs the 'appropriate' amount of momentum (moving towards, or away, from the emitting particle, per opposite, or same, charge)
   
5. and, that fact is 'communicated', to the emitting particle, through the entanglement; therefore, and thereby, the emitting particle "pops" into the appropriate position (towards, or away, from the absorbing particle, per opposite, or same, charge)
   

This writer is still unclear, on how photons can have "backwards" momentums. For, according to Zukav, virtual photons are exactly the same as real photons, except that "they're in energy debt mode" (my words), having been "borrowed" into existence, "on [math]\Delta E[/math] loan". And, if so, then their 'ought' to be such a thing, as a "backwards" real photon, too (shine your flashlight, in the direction, that you want to go -- a 'puller', not a 'pusher', thruster).

 

Note, too, that, according to this picture, the actual 'detailed' motion, of charges, 'accelerated' by interactions, via virtual photons, with other charges, is not smooth and continuous. Instead, it occurs, via successive quantum jumps, as the charges simultaneously emit paired [math]\pm k[/math] virtual photons, with which they are, having interacted, entangled; and, then, 'ghost out' into corresponding 'pushed' & 'pulled' states. Those then wave-function collapse, if/when the emitted virtual photons actually interact, with other charges (which promote the virtual photons, into reality, and then absorb the 'appropriate' energy/momentum). Thus, the Classical differential equations, represent only the 'macroscopic Classical continuous limit', of the actual discrete, 'Quantum-hop-skip-and-jumping', which is rather more of a series of staccato 'snaps'.

Edited April 21, 2011 by Widdekind