What are space-like and time-like virtual photons?

[Hans de Vries]

?? what is?

Do you mean QED is a real physical process?
If so, what does that have to do with whether virtual particles are real?

By "it" I mean the first (tree-level) term as in the preceding post:

Since the first (tree-level) term is something like 99% accurate in QED one
might assume that it corresponds "for 99%" with a real physical process.

Since the physics is mostly determined by this term you can't say that it is
something artificially mathematics which only makes sense if all terms are
added together. Further, simply checking the involved math gives a "real"
physical process.

If so, what does that have to do with whether virtual particles are real?

Virtual particles are not real particles. As I said in post https://www.physicsforums.com/showpost.php?p=2550632&postcount=21". You can't equate
them with particles with a deBroglie wavelength with the usual momentum/mass
relationship. The mass would be imaginary, the speed would be FTL. One gets
these unphysical properties only if one insist in calling the electromagnetic field
involved in the scattering process a "particle". The electromagnetic field itself is
nothing special.

Regards, Hans

 

[Hans de Vries]

No, but Hans is right. Looking at the sci.* archives since 1992, there is a strong correlation between people asking for virtual particles and people looking for FTL or other exotic effects. After all, the "invariant mass" of these particles is imaginary, blah blah. One needs to show the full equation, with the propagator of, say, the W boson or the photon, to convince people that the mass we use is not the one in the "virtual four-momentum" and that the "virtual four-momentum" is just an integration parameter.

Indeed.

Here you have Tom van Flandern's response to me defending his
"The speed of gravity is much higher as the speed of light" theory
using virtual particles:

http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/msg02932.html

If you substitute "virtual photon" for "classical graviton" and "charge"
for "mass", the same reasoning applies to electrodynamics. Virtual photons,
as you might have heard, are alleged to have infinite speed to "explain"
this unexpected behavior.

That what one gets by using a too "frivolous" jargon , ... Regards, Hans

 

[Frame Dragger]

Indeed.

Here you have Tom van Flandern's response to me defending his
"The speed of gravity is much higher as the speed of light" theory
using virtual particles:

http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/msg02932.html




That what one gets by using a too "frivolous" jargon , ...


Regards, Hans

That seems like a personal bias looking for an excuse on Fandern's part, not a mistaking of the jargon. Anyway, I'm not arguing devil's advocate on this when I agree that the language of QM is messy. On this forum, some other members made several good counterpoints. The first was, "Too late, nothing to be done", with, "Physics takes in a span of history that must equal some crossover and confusion of terms". I find both convincing and probably it's better to avoid arguing with people who base a theory of FTL action on something that MEANS, "Not Real". Hence, "Virtual" reality, vs. "ACTUAL" reality. It should be instinctive for people...

 

[tiny-tim]

By "it" I mean the first (tree-level) term as in the preceding post …

I still don't get it …

are you saying that the first (tree-level) term of QED is real, but QED isn't, or that they're both real?

Virtual particles are not real particles.

Are you saying only that they're not particles, or are you saying that they're not particles and they're not real?

Since the physics is mostly determined by this term you can't say that it is something artificially mathematics which only makes sense if all terms are added together.

Yes I can. And I do.

A theory which is 99% correct is WRONG!

A theory which says that a projectile follows a polygon (rather than a curve) is WRONG, even if it does give 99% accuracy.

And, more importantly, if two theories each say that the other is only 99% accurate, then one of them is wrong!

(And if you accept that one of those two theories is better, and that it doesn't contain a feature that the other contains, then that feature is not only not real, it isn't even a part of the better theory!)

Approximating QED by a first-level approximation is exactly as bad as approximating an integral by a step-thingy, or an orbit by epicycles … QED does not contain its own "approximations", any more than Newtonian mechanics contains epicycles.

Further, simply checking the involved math gives a "real" physical process.

What does that sentence mean?

 

[arivero]

? And, what is happening here? Veteran PF inhabitants willing to engage in flame war?

Approximating QED by a first-level approximation is exactly as bad as approximating an integral by a step-thingy, or an orbit by epicycles … QED does not contain its own "approximations", any more than Newtonian mechanics contains epicycles.

I note the careful wording "exactly as bad as". Actually, I can not see what it is implied by "exactly". Does it mean that it all cases we can put a control of the error due to the approximation? That we can stablish conditions for convergence to the real value (real as in "real line", no real as in "reality")? That we can see all the three cases from the point of view of truncation of a series expansion?

 

[Frame Dragger]

? And, what is happening here? Veteran PF inhabitants willing to engage in flame war?




I note the careful wording "exactly as bad as". Actually, I can not see what it is implied by "exactly". Does it mean that it all cases we can put a control of the error due to the approximation? That we can stablish conditions for convergence to the real value (real as in "real line", no real as in "reality")? That we can see all the three cases from the point of view of truncation of a series expansion?

Oh come now, it's not a flame-war until the ad hominem attacks come to the fore. This is still a "heated disagreement" with undertones of, "Snarky". We're several orders down from Flame, unless we tunnel there by using personal epithets in the next minute or two. GO! (don't go. seriously, don't).

 

[Hans de Vries]

I still don't get it …

are you saying that the first (tree-level) term of QED is real, but QED isn't, or that they're both real? Are you saying only that they're not particles, or are you saying that they're not particles and they're not real? Yes I can. And I do.

A theory which is 99% correct is WRONG!

A theory which says that a projectile follows a polygon (rather than a curve) is WRONG, even if it does give 99% accuracy.

And, more importantly, if two theories each say that the other is only 99% accurate, then one of them is wrong!

(And if you accept that one of those two theories is better, and that it doesn't contain a feature that the other contains, then that feature is not only not real, it isn't even a part of the better theory!)

Approximating QED by a first-level approximation is exactly as bad as approximating an integral by a step-thingy, or an orbit by epicycles … QED does not contain its own "approximations", any more than Newtonian mechanics contains epicycles.What does that sentence mean?

Hi, Tiny Tim.

I think I did discuss the physics involved in "virtual photons". There is a lot
more to tell about that if you want so.

For the rest this is basically about the confusion caused by the jargon among
non-specialists.

If you call the EM field involved a "particle" then, according to the standard
rules for particles, it should have an imaginary mass and move faster then light.

That isn't really solved by calling it a "virtual particle" meaning to say that it
isn't really a particle. It causes more confusion with non-specialist for instance
thinking that "virtual particles are made out of virtual matter" because
"only virtual matter can move faster than light".

It is like calling a horse a car, but then, since it isn't a real car, calling it a
"virtual car" just to show that you realize that it isn't a real car. OK but it will
keep confusing people anyway and you'll get stories about how the Wild West
was conquered by people driving mystical virtual cars that could jump over
fences and that new virtual cars didn't come from the assembly plant but out
of the virtual car's own trunk...Regards, Hans

 

[Frame Dragger]

QUOTE=Hans de Vries;2553363]Hi, Tiny Tim.

I think I did discuss the physics involved in "virtual photons". There is a lot
more to tell about that if you want so.

For the rest this is basically about the confusion caused by the jargon among
non-specialists.

If you call the EM field involved a "particle" then, according to the standard
rules for particles, it should have an imaginary mass and move faster then light.

That isn't really solved by calling it a "virtual particle" meaning to say that it
isn't really a particle. It causes more confusion with non-specialist for instance
thinking that "virtual particles are made out of virtual matter" because
"only virtual matter can move faster than light".

It is like calling a horse a car, but then, since it isn't a real car, calling it a
"virtual car" just to show that you realize that it isn't a real car. OK but it will
keep confusing people anyway and you'll get stories about how the Wild West
was conquered by people driving mystical virtual cars that could jump over
fences and that new virtual cars didn't come from the assembly plant but out
of the virtual car's own trunk...


Regards, Hans[/QUOTE]

There is a reason for all of this "virtual particle" issue than just wanting to think of the universe in terms of oscillating Newtonian billiard balls. Perturbation Theory. It helps to get to "There" from "Here" if you get the metaphor, in a simpler model. Describing the field as being quantized at any given point is frankly easier than the alternatives for non-idealized problems.

This is more like calling a horse in the Wild West a "Mustang" (breed), which then came to be a car label. The general implication is there to carry the message of the original (speed, strength, etc), but it's clearly understood that at no point are the two Mustangs ACTUALLY related.

So it is with Virtual particles. They help with the math, and until a theory comes along that can do that AND more... virtual particles are going to be around (or not lol) for a while ($$\hbar$$) .

 

[tiny-tim]

what is the issue?

Hi Hans!

(btw, are you importing your posts from another application? they would be easier to read if you removed the imported line-breaks, which produce some very short alternate lines, on my screen anyway )

If you call the EM field involved a "particle" then, according to the standard rules for particles, it should have an imaginary mass and move faster then light.

That isn't really solved by calling it a "virtual particle" meaning to say that it isn't really a particle. It causes more confusion with non-specialist for instance thinking that "virtual particles are made out of virtual matter" because "only virtual matter can move faster than light".

It is like calling a horse a car, but then, since it isn't a real car, calling it a "virtual car" just to show that you realize that it isn't a real car. OK but it will keep confusing people anyway and you'll get stories about how the Wild West was conquered by people driving mystical virtual cars that could jump over fences and that new virtual cars didn't come from the assembly plant but out of the virtual car's own trunk...

Yes, I get that … you're saying that virtual particles aren't particles …

I made it clear I understood you were saying that in my last post …

Are you saying only that they're not particles, or are you saying that they're not particles and they're not real?

… but I still don't get whether you're saying that virtual particles (or virtual horses, or whatever you want to call them) are real.

This whole thread started (obviously ) with the title … "What are space-like and time-like virtual photons?".

ok, we've established many things that they're not, but answering what they are assumes the pre-requisite that they "are" …

if they're not real, in other words if they're just a mathematical trick, then the answer is that from a physical, or reality, point of view, they don't exist (alternatively, from a mathematical point of view, they exist because we create them in the maths, but they don't model anything real).

 

[tiny-tim]

Hi Hans!

ok, this is on the issue of whether virtual particles are real, which I think you say they are …

The QFT Hamiltonian density is an operator.

A genuine operator, not just a complex number.

As such, it is made up of the operators peculiar to QFT … creation and annihilation operators of particles (typically, electrons and photons). It is an infinite sum of n products of "triples" of three co-happening creation and annihilation operators (each with every possible momentum for that particle, subject only to conservation of 3-momentum), but different triples happening at n separate space-time events randomly dispersed throughout space-time, and "mating" with each other in every possible permutation (… pause to draw breath …), and all this in every possible way and for every possible value of n!

In the "position-space" representation, these creation and annihilation operators represent "on-shell" particles, and 3-momentum (but not energy) has to be conserved.

The "position-space" representation is not Lorentz-covariant, and so far as I know, there is no Lorentz-covariant theory which does involve operators.

The "momentum-space" representation is Lorentz-covariant, but it does not involve operators (obvious, since they've already dropped out, but I'll just add that all those "virtual particles" with 4-momentum (E,p) and E² ≠ p² + m² can't have an operator).

I maintain that a theory without operators does not even purport to model reality, and so cannot be described as real. "Momentum-space" QFT has no operators, so it isn't even a candidate for reality, and nor are artefacts which only appear in "momentum-space" QFT (of course, it, and they, are perfectly valid mathematically, as an aid to solving the problems of "position-space" QFT).

A theory which does involve operators is a candidate for being described as real … it models a "reality" in which the creation and annihilation operators correspond to actual events. But if that "reality" does not correspond to our reality, then it is a candidate which fails the test.

I maintain that, even if "position-space" QFT passes the "reality" test, its artefacts, the position-space "virtual particles", fail the test because the theory doesn't even purport to say when and where, or in what numbers, they are created-and-annihilated …

they are created-and-annihilated at every event in space-time (including long before and after the "collision", and a long way from it), an electron positron and photon (for example) created at event x₁ may be annihilated at quite separate events x₅ x₉₉ and x_31,773, and all these interactions happen in every possible way (involving every possible number of particles), concurrently!

(this is totally unlike, for example, epicycles, which at least have definite identities, ie positions and numbers, and could well be real if it wasn't for the fact that we can see they're not there; and totally unlike the aether, which there's only one of! )

And how would you describe this reality anyway? … wouldn't you have to start by saying that the QFT field contains an infinitely dense sea of "vertices", of three creation or annihilation operators, a sea of "events waiting to happen", which then "mate" …

no, seriously, if it's reality, you should be able to describe it …

so how would you do so? ​

 

[Frame Dragger]

no, seriously, if it's reality, you should be able to describe it …

Why? Let me be clear, I'm not arguing for the existence of virtual particles outside of the math, but why do you think we should be able to describe reality? I'm not going mystical on you either, I mean this in a very real sense. It's Anthropic thinking to assume that reality needs to make sense to us, or be configured in such a way that even at extremes beyond our daily experience (QM-Microscopic/(GR-Macroscopic) we could describe it.

I don't think we've reached a fundamental blockade in our understanding of physics, but it's good to remember that increasingly distant and probabilistic approximations may be the best description of reality we can ever produce. In such a theory, there will be artifacts that are reflected only in the theory, but which can never be eliminated from the theory without destroying its utility as a predictive tool. This is already happening in QM, but I believe that is a matter of the state of the science and not a fundamental "wall" produced by the nature of our existence.

That said... why should reality make sense? There is evidence already that from a human perspective, it doesn't.

 

[Hans de Vries]

Hi Hans!
Yes, I get that … you're saying that virtual particles aren't particles …
I made it clear I understood you were saying that in my last post … … but I still don't get whether you're saying that virtual particles (or virtual horses, or whatever you want to call them) are real.

This whole thread started (obviously ) with the title … "What are space-like and time-like virtual photons?".

ok, we've established many things that they're not, but answering what they are assumes the pre-requisite that they "are" …

if they're not real, in other words if they're just a mathematical trick, then the answer is that from a physical, or reality, point of view, they don't exist (alternatively, from a mathematical point of view, they exist because we create them in the maths, but they don't model anything real).

Hi Tiny Tim,

They are real as electromagnetic fields.

If you can call the electromagnetic field "real" then you can call what is going on
during scattering ("the exchange of a virtual photon") real as well.

QFT determines the charge/current density as

$$\bar{\psi}\gamma^\mu\psi$$

This is the charge current density in the scattering zone. This charge current density
is the source of an electromagnetic field which is calculated in just the same way as
any electromagnetic field from any charge/current density. The charge/current density
contains an interference term (the transition current) which is the source of the electro
magnetic field we are talking about. This is what the math describes.
Regards, Hans

 

[Hans de Vries]

Hi Hans!

ok, this is on the issue of whether virtual particles are real, which I think you say they are …

The QFT Hamiltonian density is an operator.

A genuine operator, not just a complex number.

As such, it is made up of the operators peculiar to QFT … creation and annihilation operators of particles (typically, electrons and photons). It is an infinite sum of n products of "triples" of three co-happening creation and annihilation operators (each with every possible momentum for that particle, subject only to conservation of 3-momentum), but different triples happening at n separate space-time events randomly dispersed throughout space-time, and "mating" with each other in every possible permutation (… pause to draw breath …), and all this in every possible way and for every possible value of n!

In the "position-space" representation, these creation and annihilation operators represent "on-shell" particles, and 3-momentum (but not energy) has to be conserved.

The "position-space" representation is not Lorentz-covariant, and so far as I know, there is no Lorentz-covariant theory which does involve operators.

The "momentum-space" representation is Lorentz-covariant, but it does not involve operators (obvious, since they've already dropped out, but I'll just add that all those "virtual particles" with 4-momentum (E,p) and E² ≠ p² + m² can't have an operator).

I maintain that a theory without operators does not even purport to model reality, and so cannot be described as real. "Momentum-space" QFT has no operators, so it isn't even a candidate for reality, and nor are artefacts which only appear in "momentum-space" QFT (of course, it, and they, are perfectly valid mathematically, as an aid to solving the problems of "position-space" QFT).

A theory which does involve operators is a candidate for being described as real … it models a "reality" in which the creation and annihilation operators correspond to actual events. But if that "reality" does not correspond to our reality, then it is a candidate which fails the test.

I maintain that, even if "position-space" QFT passes the "reality" test, its artefacts, the position-space "virtual particles", fail the test because the theory doesn't even purport to say when and where, or in what numbers, they are created-and-annihilated …

they are created-and-annihilated at every event in space-time (including long before and after the "collision", and a long way from it), an electron positron and photon (for example) created at event x₁ may be annihilated at quite separate events x₅ x₉₉ and x_31,773, and all these interactions happen in every possible way (involving every possible number of particles), concurrently!

(this is totally unlike, for example, epicycles, which at least have definite identities, ie positions and numbers, and could well be real if it wasn't for the fact that we can see they're not there; and totally unlike the aether, which there's only one of! )

And how would you describe this reality anyway? … wouldn't you have to start by saying that the QFT field contains an infinitely dense sea of "vertices", of three creation or annihilation operators, a sea of "events waiting to happen", which then "mate" …​

Hi Tiny Tim

Quantum Electro Dynamics has a number of fundamental corner stones
of which we are highly certain that they are correct.

- Lorentz Invariance
- Gauge Invariance
- Maxwell's theory of the electromagnetic field.
- Dirac's theory of the electron's field.

But there are many "axillary" theories which are not that fundamental
at all and neither are we very sure that they are correct or we even
know that they can't be correct.

Quantization of real photons is one of those "auxillary" theories.

1) It's not Lorentz invariant.
2) It's not Gauge invariant
2) It does not really explain unitarity of detection, which was its original purpose

Quantization of "virtual photons" is even worse then then quantization
of "real photons". By calling a real electromagnetic field a virtual photon
one assigns an imaginary mass and a FTL velocity to them. Nothing of
this is found back in the QED mathematics which describes the EM field
in the classical way propagating at the speed of light.

no, seriously, if it's reality, you should be able to describe it …

so how would you do so? ​

The standard theory describes it (as electromagnetic fields originating from
charge/current density fields)Regards, Hans

 

[tiny-tim]

Hi Hans!

They are real as electromagnetic fields.

If you can call the electromagnetic field "real" then you can call what is going on
during scattering ("the exchange of a virtual photon") real as well.

(NB: I haven't read your next post yet)

I'm still not quite clear whether your saying that "virtual particles" are real.

QFT determines the charge/current density as

$$\bar{\psi}\gamma^\mu\psi$$

This is the charge current density in the scattering zone. This charge current density
is the source of an electromagnetic field which is calculated in just the same way as
any electromagnetic field from any charge/current density.

Yes, I'm happy to accept that $\psi^\ast\gamma^\mu\psi$ is a current density and therefore real.

And that the associated field is therefore real.

The charge/current density
contains an interference term (the transition current) which is the source of the electro
magnetic field we are talking about. This is what the math describes.

This is where you've lost me.

Which part of $\psi^\ast\gamma^\mu\psi$ is the transition current you're talking about?

(I wiki'ed and googled "transition current", but didn't find anything helpful, and it's not in the index of Weinberg's QTF, Vol I, which is the source of most of my knowledge. I vaguely recall seeing $\psi_f^\ast (\partial_\mu \psi_i)\ -\ (\partial_\mu \psi_f^\ast)\psi_i$ somewhere else, but I think it left out an A² term, and I don't see where it comes in the the simple derivation of the Dyson series, nor why an approximation made of an artificial combination of idealised initial and final fields should be regarded as in any way modelling the real field in the "scattering zone" of an interaction. )

 

[tiny-tim]

… I maintain that, even if "position-space" QFT passes the "reality" test, its artefacts, the position-space "virtual particles", fail the test because the theory doesn't even purport to say when and where, or in what numbers, they are created-and-annihilated …
…

And how would you describe this reality anyway? … wouldn't you have to start by saying that the QFT field contains an infinitely dense sea of "vertices", of three creation or annihilation operators, a sea of "events waiting to happen", which then "mate" …

no, seriously, if it's reality, you should be able to describe it …

so how would you do so? ​

The standard theory describes it (as electromagnetic fields originating from charge/current density fields)

But that description ("electromagnetic fields originating from charge/current density fields") doesn't contain any mention of these virtual photons and electrons.

 

[Hans de Vries]

Hi Hans!
I'm still not quite clear whether your saying that "virtual particles" are real.

Hi, Tiny Tim

The math describes EM fields. These fields follow Maxwell's laws and originate
from the charge current density in the scattering zone just as they would classically.
If I use the term "virtual photon" then it's just a nickname for the EM field.

Yes, I'm happy to accept that $\psi^\ast\gamma^\mu\psi$ is a current density and therefore real.

And that the associated field is therefore real.

ok

Which part of $\psi^\ast\gamma^\mu\psi$ is the transition current you're talking about?

(I wiki'ed and googled "transition current", but didn't find anything helpful, and it's not in the index of Weinberg's QTF, Vol I, which is the source of most of my knowledge. I vaguely recall seeing $\psi_f^\ast (\partial_\mu \psi_i)\ -\ (\partial_\mu \psi_f^\ast)\psi_i$ somewhere else, but I think it left out an A² term, and I don't see where it comes in the the simple derivation of the Dyson series, nor why an approximation made of an artificial combination of idealised initial and final fields should be regarded as in any way modelling the real field in the "scattering zone" of an interaction. )

An electron in a scattering zone can change its momentum suddenly from an initial
momentum to a final momentum. Not via a gradually changing path but under a sharp
angle. The reason that it can do so is because an electron can interfere with itself.

For some time the electron will be in a transition state going from initial to final
momentum. Its wavefunction will be a linear combination of the two states and these
states interfere with each other. Now what is this transition interference current?

$$\psi = \psi_f+\psi_i$$

Thus the charge current density is given by:

$$\bar{\psi}\gamma^\mu\psi ~=~ \Big(\bar{\psi_f}\gamma^\mu\psi_f~+~\bar{\psi_i}\gamma^\mu\psi_i \Big) ~~+~~ \Big(\bar{\psi_f}\gamma^\mu\psi_i~+~\bar{\psi_i}\gamma^\mu\psi_f\Big)$$

The last two terms (the crossterms) represent the interference current. The two
interference terms are each others complex conjugate. Each term contains all
information needed. The Feynman diagram expressions use the first of the two terms.
The basic interference pattern is the same as in the case of a Klein Gordon particle:

$$\psi_{int} ~~=~~ \exp\left\{ -i(p_f^\mu-p_i^\mu)x_\mu \right\}~+~ \exp\left\{ -i(p_f^\mu-p_i^\mu)x_\mu \right\} ~~=~~ 2\cos\left\{ (p_f^\mu-p_i^\mu)x_\mu \right\}$$

The cosine functions represents an alternating charge density pattern in case of
a Klein Gordon particle which shifts with a speed of anywhere between 0 and c.
However, we are dealing with a Dirac particle field which also has a spin density.

The alternating spin density is the same as an alternating transverse current density
pattern according to Stokes law. You can see this in figure 1.7 in my book here.
http://physics-quest.org/Book_Chapter_EM_basic.pdf"

The left side of figure 1.7 is effectively the same as the right side according to
Stokes. If you google for "Gordon decomposition" then this is the extra term from
spin in the vector current.

The electromagnetic field from the alternating charge current pattern is derived
in the classical way and this is what is "nicknamed" the "virtual photon"


Regards, Hans

 

[Frame Dragger]

Hi, Tiny Tim

The math describes EM fields. These fields follow Maxwell's laws and originate
from the charge current density in the scattering zone just as they would classically.
If I use the term "virtual photon" then it's just a nickname for the EM field.



ok





An electron in a scattering zone can change its momentum suddenly from an initial
momentum to a final momentum. Not via a gradually changing path but under a sharp
angle. The reason that it can do so is because an electron can interfere with itself.

For some time the electron will be in a transition state going from initial to final
momentum. Its wavefunction will be a linear combination of the two states and these
states interfere with each other. Now what is this transition interference current?

$$\psi = \psi_f+\psi_i$$

Thus the charge current density is given by:

$$\bar{\psi}\gamma^\mu\psi ~=~ \Big(\bar{\psi_f}\gamma^\mu\psi_f~+~\bar{\psi_i}\gamma^\mu\psi_i \Big) ~~+~~ \Big(\bar{\psi_f}\gamma^\mu\psi_i~+~\bar{\psi_i}\gamma^\mu\psi_f\Big)$$

The last two terms (the crossterms) represent the interference current. The two
interference terms are each others complex conjugate. Each term contains all
information needed. The Feynman diagram expressions use the first of the two terms.
The basic interference pattern is the same as in the case of a Klein Gordon particle:

$$\psi_{int} ~~=~~ \exp\left\{ -i(p_f^\mu-p_i^\mu)x_\mu \right\}~+~ \exp\left\{ -i(p_f^\mu-p_i^\mu)x_\mu \right\} ~~=~~ 2\cos\left\{ (p_f^\mu-p_i^\mu)x_\mu \right\}$$

The cosine functions represents an alternating charge density pattern in case of
a Klein Gordon particle which shifts with a speed of anywhere between 0 and c.
However, we are dealing with a Dirac particle field which also has a spin density.

The alternating spin density is the same as an alternating transverse current density
pattern according to Stokes law. You can see this in figure 1.7 in my book here.
http://physics-quest.org/Book_Chapter_EM_basic.pdf"

The left side of figure 1.7 is effectively the same as the right side according to
Stokes. If you google for "Gordon decomposition" then this is the extra term from
spin in the vector current.

The electromagnetic field from the alternating charge current pattern is derived
in the classical way and this is what is "nicknamed" the "virtual photon"


Regards, Hans

This would seem to predict the absence of the observed Casimir Effect.

 

[Hans de Vries]

This would seem to predict the absence of the observed Casimir Effect.

Why?

Regards, Hans

 

[Frame Dragger]

Why?

Regards, Hans

You say that 'virtual photon' is just a nickname essentially, for the classical EM field. If that's the case, explain the Casmimir Effect within that framework. I don't see how you could. Expand that to vacuum polarization and virtual photon to 'virtual any-particle' and I don't see how your argument is consistant.

 

[Hans de Vries]

You say that 'virtual photon' is just a nickname essentially, for the classical EM field. If that's the case, explain the Casmimir Effect within that framework. I don't see how you could. Expand that to vacuum polarization and virtual photon to 'virtual any-particle' and I don't see how your argument is consistant.

You would first have to give a definition of a "virtual photon". Many different things get
all called "virtual photon" causing a lot of confusion.

I'm describing the literal physical meaning of the Dirac algebra in the case of a "virtual
photon" exchange in a Feynman diagram.

In case of the Casimir force the EM field is quantized as a result of the boundary conditions
represented by the flat plates. These are very different processes.Regards, Hans

 

[tiny-tim]

what is the issue?

Hi Hans!

The math describes EM fields. These fields follow Maxwell's laws and originate
from the charge current density in the scattering zone just as they would classically.
If I use the term "virtual photon" then it's just a nickname for the EM field.

Now I'm completely confused as to what we're talking about.

I was hoping we were discussing what other people mean by "virtual particles".

Are you now saying that your use of the phrase "virtual particle" or "virtual photon" is only your use? If so, of course you're perfectly entitled to have a personal use of language, but it doesn't answer the OP's question …

"What are space-like and time-like virtual photons?"​

… since (I assume) he meant, what do people generally mean by "virtual photons"?

I'll restate (or state? in case there's any doubt) my position, which is:

People generally fall into two categories:
i] either they regard virtual particles as simply a mathematical trick (in which case i agree with them),
ii]or they regard virtual particles as real things which are capable of having various properties such as position, time of creation/annihilation, momentum, etc (in which case, I think they're wrong because I see neither experimental evidence nor theoretical prediction of things with such properties).​

Or are you saying that using "virtual photon" simply as another name for the EM field is a common usage (at least among a substantial proportion of authors and/or "laity")?

If so, I doubt that the OP was referring to that usage (and btw I've certainly not come across it) when he asked specifically about time-like and space-like virtual photons, and about how long they exist and can they be directly observed …

Hi,
I'm trying to get my story straight about virtual photons. So far, my understanding of them is that they are the particles that mediate the electromagnetic interaction, and they only exist a short time due to the Heisenberg uncertainty principle meaning that they can't be directly observed.
I've heard of time-like and space-like virtual photons though, and google as I might, I just can't find out what these terms mean. Can anyone explain them to me?
Thanks.

hmm … I'm tempted to suggest wrapping it up like this …

are you saying: "The 'virtual photons' the OP was talking about aren't real, and the term should be used only as a nickname for the EM field" ?

but unfortunately I have a completely different difficulty …

when you say "it's just a nickname for the EM field", I'm not entirely following what you mean by "field" …

i] the electromagnetic (EM) field itself has nothing to do with quantum field theory, or indeed quantum theory of any sort … the EM field was well known to Maxwell, and most of its aspects can be understood perfectly well by high school students …

ii] the field in Quantum Field Theory (QFT) refers to a specific technique, of representing a specific particle (with a specific momentum etc) by a field composed of operators representing every possible momentum etc …

so if you mean i], I don't see how "virtual photons", which are a QFT concept, can be regarded by people generally as referring to a nineteenth-century pre-quantum-theory field

and if you mean ii], calling that field the "EM field" seems to me misleading: in QFT, isn't it the particles (electrons, photons, etc) that have fields, while the force (or interaction) just has a Hamiltonian (composed, admittedly, of fields of particles)?

 

[Frame Dragger]

I'm with Tim on this one. I appreciate what you're saying, and maybe you've had terrible experiences with people inserting virtual particles into basic EM theory, but this doesn't fit the definition of virtual that I know. Virtual photons are mathematical artifacts used in Perturbation Theory... you're talking about something interesting, but different.

 

[Hans de Vries]

Hi, Tiny Tim

I see that you are using Weinberg's The Quantum Theory of Fields vol.I for studying QFT.

Well, It's a very valuable reference work but you'll only understand what he writes
it if you already know about, and understand, the subjects he is talking about...

For understanding Feynman diagrams the most effective road is using David Griffith's
introduction to elementary particles, especially chapters 6 and 7.

Then, for the most important stuff missing in Griffith's, for example the Poincaré group,
the theory of gauge invariance, the Lagrangian formulation, use Lewis H. Ryder's book
Quantum Field Theory.

To see how the path integral formalism can be used to generate series of Feynman
diagrams, take Anthony Zee's book: Quantum Field Theory in a nutshell which starts
of directly with the path integral formalism.

Hi Hans!


Now I'm completely confused as to what we're talking about.

I was hoping we were discussing what other people mean by "virtual particles".

Are you now saying that your use of the phrase "virtual particle" or "virtual photon" is only your use? If so, of course you're perfectly entitled to have a personal use of language, but it doesn't answer the OP's question …

"What are space-like and time-like virtual photons?"​

… since (I assume) he meant, what do people generally mean by "virtual photons"?

I'll restate (or state? in case there's any doubt) my position, which is:

People generally fall into two categories:
i] either they regard virtual particles as simply a mathematical trick (in which case i agree with them),
ii]or they regard virtual particles as real things which are capable of having various properties such as position, time of creation/annihilation, momentum, etc (in which case, I think they're wrong because I see neither experimental evidence nor theoretical prediction of things with such properties).​

Or are you saying that using "virtual photon" simply as another name for the EM field is a common usage (at least among a substantial proportion of authors and/or "laity")?

If so, I doubt that the OP was referring to that usage (and btw I've certainly not come across it) when he asked specifically about time-like and space-like virtual photons, and about how long they exist and can they be directly observed …

I think the best way to answer the OP's question …
What are space-like and time-like virtual photons?"

is to discuss the meaning of the actual mathematics which is going into
the Feynman diagrams. This is what I'm trying to do in my posts..

I do work this out in great detail in my book which contains expressions
which describe the transition interference charge/current density:


$$\begin{array}{rcrr} & & \textbf{\mbox{\textsf{vector transition current}}} ~~~~~~~ & \\ \frac{1}{2m} J^0_{Vfi}~~= & + & & ~~~\gamma\cos(\omega) \\ & + & \hat{\phi} \cdot\hat{s}~ & ~~i\gamma\sin(\omega) \\ &&&\\ \frac{1}{2m} \vec{J}_{Vfi}~~= & + & (~\hat{\beta}~+~i\hat{\beta}\times\hat{s}) & ~\beta\gamma\cos(\omega) \\ & + & (~(\hat{\phi}\cdot\hat{\beta})\hat{s}+i\hat{\phi}\times\hat{\beta}-(\hat{\phi}\times\hat{\beta})\times\hat{s}) & i\beta\gamma\sin(\omega) \\&&&\\&&&\\ & & \textbf{\mbox{\textsf{axial transition current}}} ~~~~~~~~ & \\ \frac{1}{2m} J^0_{Afi}~~= & + & \hat{\beta}\cdot\hat{s}~ & ~\beta\gamma\cos(\omega) \\ & + & (\hat{\phi}\cdot\hat{\beta}+i(\hat{\phi}\times\hat{\beta})\cdot\hat{s}) & i\beta\gamma\sin(\omega) \\ &&&\\ \frac{1}{2m} \vec{J}_{Afi}~~= & + & \hat{s}~ & ~~~\gamma\cos(\omega) \\ & + & (~\hat{\phi }~+~i\hat{\phi }\times\hat{s}) & ~~i\gamma\sin(\omega) \\ &&&\\ \end{array}$$

The above expressions are for an electron initially at rest with spin s
and a final state corresponding with a boost to a velocity beta and
a rotation of the spin pointer over an angle w around the axis phi.

The above expression has to be multiplied with the interference
cosine function of my previous post. I intent to post the chapter
online when it is ready.

hmm … I'm tempted to suggest wrapping it up like this …

are you saying: "The 'virtual photons' the OP was talking about aren't real, and the term should be used only as a nickname for the EM field" ?

but unfortunately I have a completely different difficulty …

when you say "it's just a nickname for the EM field", I'm not entirely following what you mean by "field" …

i] the electromagnetic (EM) field itself has nothing to do with quantum field theory, or indeed quantum theory of any sort … the EM field was well known to Maxwell, and most of its aspects can be understood perfectly well by high school students …

Fortunately, that's not true at all . The electromagnetic potential field A
becomes the "wave-function" of the photon in Quantum Electro Dynamics and
the classical Lagrangian density is the same one as the one used in QED

The EM potential field A is related to the charge/current density j via the
classical EM relation. This is the same relation as the one which is used in
the Feynman diagrams

$$\Box A^\mu =\ \mu_o j^\mu$$

$$\begin{center} \mbox{d'Alembertian operator: }~~~~ \Box\ =\ \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2$$


In momentum space this operator is q^2 and its inverse operator is 1/q^2
Multiplying the transition current in the momentum domain with 1/q^2 means
that we are deriving the EM field A in momentum space which is emitted
from the transition charge/current density.

The final step in the Feynman diagram expression for (for instance) electron-
electron scattering is the absorption of this field by transition current of the
other electron. This requires the transition currents to be equal but opposite.

Everything in the Feynman diagrams can be physically understood. It is
just a question of working it out mathematically.


Regards, Hans

 

[tiny-tim]

Hi Hans!

Before I answer the rest of your post, may I just ask for an example that may clear up my misunderstanding: you say …

Everything in the Feynman diagrams can be physically understood. It is just a question of working it out mathematically.

… ok then … what is the physical understanding of an internal electron line in an EM Feynman diagram?

oh, and just a quick question about terminology that I may be misunderstanding (since Weinberg's book makes me used to deriving the 1/(q² + m²) factor without any differentiating in sight ): you say (of the d'Alembertian differential operator) …

In momentum space this operator is q^2 and its inverse operator is 1/q^2

… am I right in thinking that the same operator is more generally 1/(q² + m²), operating equally on both "virtual photons" and "virtual electrons" in the same momentum space?

 

[Hans de Vries]

Hi Hans!

Before I answer the rest of your post, may I just ask for an example that may clear up my misunderstanding: you say …… ok then … what is the physical understanding of an internal electron line in an EM Feynman diagram?

Hi Tiny Tim,

The internal electron line is most easily thought of as the propagation from
the interaction between the incoming electron and photon. Starting from the
interactive Dirac equation.

$$\gamma^\mu \left\{~ i\,\mbox{\Large \partial}_\mu ~-~ e\mbox{\Large A}_\mu ~\right\}\psi ~=~ m\,\psi$$

We can isolate the interaction term at the left hand side.

$$\Big(\gamma^\mu e\mbox{\Large A}_\mu \Big)\psi ~~=~~ \Big(~ i\gamma^\mu\,\mbox{\Large \partial}_\mu ~-~ m ~\Big) \psi$$

The propagator for the internal electron is the inverse of the operator on the
right hand site. The operator's Fourier transform in the momentum domain is.

$$\gamma^\mu\,p_\mu - m$$

and so the inverse operator is

$$\frac{1}{\gamma^\mu\,p_\mu - m} ~~=~~ \frac{\gamma^\mu\,p_\mu ~+~ m}{p^2-m^2}$$

oh, and just a quick question about terminology that I may be misunderstanding (since Weinberg's book makes me used to deriving the 1/(q² + m²) factor without any differentiating in sight ): you say (of the d'Alembertian differential operator) …… am I right in thinking that the same operator is more generally 1/(q² + m²), operating equally on both "virtual photons" and "virtual electrons" in the same momentum space?

The operators (propagators) always use the invariant mass of the particle.
The off-the-shell "masses" are not really masses. The virtual particles are
always interactions between the fermion field and the electromagnetic field.
The two different types of field always propagate with their own invariant
mass.Regards, Hans

 

[Frame Dragger]

The operators (propagators) always use the invariant mass of the particle.
The off-the-shell "masses" are not really masses. The virtual particles are
always interactions between the fermion field and the electromagnetic field.
The two different types of field always propagate with their own invariant
mass.

Is this not equivalent to saying that virtual particles are a mathematical trick to describe/calculate those interactions within Perturbation Theory!?

 

[tiny-tim]

Hi Hans!

… ok then … what is the physical understanding of an internal electron line in an EM Feynman diagram?

The internal electron line is most easily thought of as the propagation from the interaction between the incoming electron and photon. Starting from the interactive Dirac equation …
…
We can isolate the interaction term at the left hand side.
…
The propagator for the internal electron is the inverse of the operator on the
right hand site. The operator's Fourier transform in the momentum domain is.
…
and so the inverse operator is …

That's the mathematical process that the line represents.

A Feynman diagram is just a diagram. It, and any line in it, does no more than represent a mathematical process. And that mathematical process may or may not represent a physical process.

I'm asking, what is the physical understanding of an internal electron line?

d'Alembertian operator: …

In momentum space this operator is q^2 and its inverse operator is 1/q^2

The operators (propagators) always use the invariant mass of the particle.
The off-the-shell "masses" are not really masses. The virtual particles are
always interactions between the fermion field and the electromagnetic field.
The two different types of field always propagate with their own invariant
mass.

I'm really asking whether it's the same d'Alembertian operator (in coordinate space).

Differential operators as a way of getting to Feynman diagrams are completely foreign to me.

I'm just asking whether there are two d'Alembertian operators, d'A_e for an electron, and d'A_φ for a photon, or is there just one operator, d'A ?

 

[Hans de Vries]

The operators (propagators) always use the invariant mass of the particle.
The off-the-shell "masses" are not really masses. The virtual particles are
always interactions between the fermion field and the electromagnetic field.
The two different types of field always propagate with their own invariant
mass.

Is this not equivalent to saying that virtual particles are a mathematical trick to describe/calculate those interactions within Perturbation Theory!?

I would says that it shows that using the term "virtual particle" is a bad
and confusing habbit. I do it, and everybody does it but it doesn't
behave as such.

A "space-like virtual particle" would propagate with an imaginary mass
and FTL. In reality there is (in QED) only the massless photon propagator
and the lepton propagator its invariant mass.


Regards, Hans

 

[Frame Dragger]

I would says that it shows that using the term "virtual particle" is a bad
and confusing habbit. I do it, and everybody does it but it doesn't
behave as such.

A "space-like virtual particle" would propagate with an imaginary mass
and FTL. In reality there is (in QED) only the massless photon propagator
and the lepton propagator its invariant mass.


Regards, Hans

I agree that a better term should be used, and in general I think the lexicon of SR/GR/QM needs to be revised and codified. That said, it seems that you're agreeing with me, which would disagree with your previous position and therefore agreeing with Tiny-Tim. I'm confused.

 

[Hans de Vries]

Hi Hans!

That's the mathematical process that the line represents.

A Feynman diagram is just a diagram. It, and any line in it, does no more than represent a mathematical process. And that mathematical process may or may not represent a physical process.

I'm asking, what is the physical understanding of an internal electron line?


I'm really asking whether it's the same d'Alembertian operator (in coordinate space).

Differential operators as a way of getting to Feynman diagrams are completely foreign to me.

I'm just asking whether there are two d'Alembertian operators, d'A_e for an electron, and d'A_φ for a photon, or is there just one operator, d'A ?

Hi, Tiny Tim

The Dirac algebra describes real physics, Just like the equations of the electromagnetic
field, and the mathematical symbol handling in Feynman diagrams can be understood
in terms of real physics, or at least in geometrical terms with vectors, axial vectors
and so on. But indeed, the textbooks usually don't get beyond the purely abstract
mathematical symbol handling, which is really sad.
To gain understanding one can to do the following.

-Start with the observables.
These are things which can be visualized: The charge/ current density, the axial
current density, the magnetization/polarization tensor.- Gain an understanding what propagators are in position space,
how the mass less 1/q propagator spreads a signal on the light cone, and how other
propagators can always be derived from (interacting) mass less propagators.-Study how the Weyl bispinor consist out of two interacting Chiral components
(the mass less left and right handed chiral components. How the direction of each
component is both a momentum direction and a spin direction, one with a left handed
chiralty and one with a right handed chiralty.-Understand what the pauli matrices do geometrically,
and what the i does geometrically:

Multiplying by -$i\sigma^i$ rotates the spinor by $180^o$ around the $x^i$-axis
Multiplying by $i$ rotates the spinor by -$180^o$ around its own axis

This means that a multiplication of an x-up-spinor by the pauli-x-matrix is rotated
by 0 degrees while the same multiplication rotates an x-down-spinor by 360 degrees,
(which amounts to a multiplication by -1). This provides us with a way to obtain
the x-component of the spin vector of the spinor. The spin-vector is consequently
given by:

$$\vec{(\xi)} ~=~ \left(\begin{array}{c} \xi^*\sigma^x\xi \\ \xi^*\sigma^y\xi \\ \xi^*\sigma^z\xi \end{array}\right)$$

This vector represents both a current as well as a spin direction. It means that we
can express the total vector-current, as well as the total axial current, with the
mass-less spin vectors as:

$$\begin{array}{ll} \mbox{vector current:} ~~ & =~~ +\vec{(\xi)}_R - \vec{(\xi)}_L \\ \mbox{axial current:} ~~ & =~~ -\vec{(\xi)}_R - \vec{(\xi)}_L \\ \end{array}$$

So, now we have already derived these two basic observables in a geometrical
way which we can visualize with pictures representing the underlying physics.Regards, Hans

 

[tiny-tim]

Hi Hans!

Yes, I understand all of that.

(though I don't see what propagators in position space have to do with virtual particles which can only exist in momentum space )

But i don't understand, and you're still not answering, …

what is the physical understanding of an internal electron line? ​

 

[Hans de Vries]

Hi Hans!

Yes, I understand all of that.

(though I don't see what propagators in position space have to do with virtual particles which can only exist in momentum space )

But i don't understand, and you're still not answering, …

what is the physical understanding of an internal electron line? ​

It's an electron propagating under the influence of an electromagnetic field.

An internal electron line originates from the original electron but has a changed
momentum so that the interference charge/current density between the two
generates an electromagnetic field which negates (absorbs) the em field of the
incoming photon.

Regards, Hans

 

[Frame Dragger]

It's an electron propagating under the influence of an electromagnetic field.

An internal electron line originates from the original electron but has a changed
momentum so that the interference charge/current density between the two
generates an electromagnetic field which negates (absorbs) the em field of the
incoming photon.

Regards, Hans

That happens, but that's not a virtual photon.

 

[tiny-tim]

virtual electrons?

Everything in the Feynman diagrams can be physically understood. It is just a question of working it out mathematically.

… ok then … what is the physical understanding of an internal electron line in an EM Feynman diagram?

It's an electron propagating under the influence of an electromagnetic field.

An internal electron line originates from the original electron but has a changed momentum so that the interference charge/current density between the two generates an electromagnetic field which negates (absorbs) the em field of the incoming photon.

(Do you mean "an internal electron originates from the original electron but has a changed momentum {etc}"?)

I don't understand the physical meaning of "originates from the original electron".

"originates" how? where? and how/where does it de-originate?

And are you saying then that there is a physical internal electron for each internal electron line in each Feynman diagram (of which of course there are infinitely many) for the particular process?

And are you saying that there is a separate physical interference charge/current density between the original electron and each separate internal electron? And is there similarly a physical interference charge/current density between each pair of internal electrons?

Finally, what "incoming photon" (whose em field is to be negated)?

I didn't specify either an incoming or an outgoing photon in the interaction, and there doesn't have to be one!​

 

[Hans de Vries]

That happens, but that's not a virtual photon.

They call it a "virtual electron" instead...