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

[jeebs]

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.

 

[humanino]

A spacetime interval is called timelike if there is a rest frame in which the interval separates events by only time, that is in this rest frame there is no space in the interval. In this case, the 4-length squared (which does not depend on the frame) $\Delta s^2 = \Delta t^2 - \Delta x^2$ is positive (with a choice of units such as the speed of light c=1 and the choice I made for the signature, time comes with a plus sign and space with a minus sign). A timelike interval can belong to the trajectory of massive particle (for the above rest frame I was referring to). The interval is called spacelike if its squared is negative, or equivalently if it can not be in the trajectory of a massive particle. The limiting case where the 4-length squared vanishes is called lightlike because it belongs to a lightcone, that is it could be the trajectory of a real photon.

If you annihilate an electron-positron pair into a virtual photon, the virtual photon will be timelike (there is a referential in which the electron and positron have opposite momenta, and this is the rest frame of the timelike photon). If you scatter two electrons off each other and they exchange a virtual photon, the virtual photon will be spacelike.

 

[humanino]

If you scatter two electrons off each other and they exchange a virtual photon, the virtual photon will be spacelike.

Can you find a simple argument to show that ?

 

[Frame Dragger]

Can you find a simple argument to show that ?

As far as I know the only instance in which a "virtual" pair can produce "real" photon would be in the form of Hawking Radiation. and the other photon that is lost in the BH could be described in terms of the past worldline of the "real" photon that escapes. It's past the event horizon after all. Obviously Humanino deleted his post, but in terms of what he said, I don't believe that is the most accurate intrepation of that interaction.

 

[humanino]

Obviously Humanino deleted his post

I did not delete anything.

One can find a simple argument by thinking of the mass of the exchanged photon in Moller scattering in the frame where one of the electrons is at rest.

 

[Frame Dragger]

I did not delete anything.

One can find a simple argument by thinking of the mass of the exchanged photon in Moller scattering in the frame where one of the electrons is at rest.

My bad, I was thrown by you quoting your previous post. Sorry!

I can see where your argument leads, but does it actually happen that way in nature? I don't know, so this isn't a snarky or leading question. I didn't think that virtual pairs were terribly conducive to producing a "real" (space-like) particle under anything approaching normal circumstances. That said, what you said seems to make sense. I'm confused! (obviously)

 

[tiny-tim]

Hi jeebs!

… virtual photons… 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.

No … virtual particles aren't real (the clue's in the name! ).

They're just a mathematical trick to help calculations.

They exist only in the mind.

(btw, virtual electrons and positrons are also needed, and since there's twice as many of them as of virtual photons in most Feynman diagrams, it's a little biased to say that it's only the virtual photons that "mediate" the electromagnetic interaction)

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?

Real photons are (obviously!) light-like, meaning that the magnitudes of their energy and momentum are the same. In other words, they move at the speed of light.

Virtual photons (just a mathematical trick, remember) in the "position representation" Feynman diagrams are light-like.

Virtual photons in the "momentum representation" Feynman diagrams are "off-shell", and so can be space-like or time-like.

 

[arivero]

I never get all this stuff on virtual photons, I think it is north-american folklore, something in the secondary school textbooks, isn't it?

In the rest of the world:
- electromagnetic interaction is mediated by the electromagnetic field.
- Photons are the quanta of the electromagnetic field.
- so photons appear in the diagrams of quantum electrodinamics.

Now,in quantum field theory, the calculation is done for all possible values of energy and momentum. It means that only the external particles being calculated are on-shell, [itex]E^2-p^2=m^2[/tex]. Particles in the diagrams of the calculation do not need to meet this equation, they are called off-shell.

This virtual particle thing... is the same that the off-shell particles?

If so, are you telling that in the tree-level diagram exchange of a photon, the integration limits are such that always $E^2 < p^2$? I am not claiming it to be false; it is only that I had not noticed the phenomena, can you point it more clearly by writing the equation? All this stuff of s,t,u channels always confuse me.Edit:

If you annihilate an electron-positron pair into a virtual photon, the virtual photon will be

.
Here i think you mean "annihilate an electron-positron pair into a virtual photon what disintegrates again into a electron-positron pair". The other thing is obviously timelike because it is not virtual at all.

 

[sylas]

If you annihilate an electron-positron pair into a virtual photon, the virtual photon will be timelike (there is a referential in which the electron and positron have opposite momenta, and this is the rest frame of the timelike photon).

Why virtual? When you annihilate an electron positron pair you get real photons... at least two of them. You can't get just one photon, virtual or otherwise, from this annihilation.

Cheers -- sylas

 

[Frame Dragger]

Why virtual? When you annihilate an electron positron pair you get real photons... at least two of them. You can't get just one photon, virtual or otherwise, from this annihilation.

Cheers -- sylas

As I said earlier, the only instance I know of where a "virtual pair" becomes a single particle is at the event horizon of a black hole in the process of Hawking Radiation. Even then, there is a requirement that one of the pair be lost inside the event horizon, formulated as the past worldline of the "real" photon that is emitted. Otherwise I thought virtual particle creation/annihilation was the description of the quantum vacuum, and by definition a bit like renormalization in the "trick" sense that Tiny Tim mentioned.

 

[arivero]

at least two of them.

Hmm now you mention it, I hope the discussion on timelike vs spacelike will not depend on the mass nor the spin of the carrier, will it? Photons are tricky.

 

[Frame Dragger]

Hmm now you mention it, I hope the discussion on timelike vs spacelike will not depend on the mass nor the spin of the carrier, will it? Photons are tricky.

As opposed to the REST of particle physics...?! :tongue2:

 

[humanino]

Why virtual? When you annihilate an electron positron pair you get real photons... at least two of them. You can't get just one photon, virtual or otherwise, from this annihilation.

Please keep in mind that I am only trying to help the original poster, and as in kindergarden I used only Moller and Bhabha scattering. At this level, all particles could be scalars, it would not change their invariant mass ! All I was trying to say is that this virtual photon (in the diagram, time flows in the horizontal direction, space is the vertical direction)
http://upload.wikimedia.org/wikipedia/commons/a/a4/Electron-positron-annihilation.svg
is timelike whereas this virtual photon
http://upload.wikimedia.org/wikipedia/commons/f/f5/Electron-positron-scattering.svg
is spacelike. In both cases, there are simple, almost geometrical, arguments to show what the invariant mass of the one-photon-exchanged is.If you want to annihilate an electron-positron pair into a photon pair, the tree-level diagram will have an electron exchanged. This is obviously irrelevant for the original poster.

 

[Frame Dragger]

Please keep in mind that I am only trying to help the original poster, and as in kindergarden I used only Moller and Bhabha scattering. At this level, all particles could be scalars, it would not change their invariant mass ! All I was trying to say is that this virtual photon (in the diagram, time flows in the horizontal direction, space is the vertical direction)
http://upload.wikimedia.org/wikipedia/commons/a/a4/Electron-positron-annihilation.svg
is timelike whereas this virtual photon
http://upload.wikimedia.org/wikipedia/commons/f/f5/Electron-positron-scattering.svg
is spacelike. In both cases, there are simple, almost geometrical, arguments to show what the invariant mass of the one-photon-exchanged is.


If you want to annihilate an electron-positron pair into a photon pair, the tree-level diagram will have an electron exchanged. This is obviously irrelevant for the original poster.

That's a very lucid explanation; works for me. And no, I'm not being sarcastic.

 

[jeebs]

A spacetime interval is called timelike if there is a rest frame in which the interval separates events by only time, that is in this rest frame there is no space in the interval.

I'm still not clear on this.

You were on about an electron-positron pair annihilating, and I am happy with there being a rest frame where their momenta are equal and opposite. What I don't get is what events (plural) are being separated here?
There's only one event as far as I'm aware, the particle annihilation...

Or do you mean that the particles collide and annihilate, and then some time later a photon leaves that point where they collided? Is it not considered to be happening instantly?

 

[humanino]

what events (plural) are being separated here?

The creation and annihilation of the virtual photon. This is not supposed to be meaningful : as the virtual photon is in a well-defined eigenstate of momentum, it should be "everywhere".

I'm taking a classical analogy "as if" this virtual photon would correspond to a classical particle. Please keep in mind that I'm suggesting this only to explain the etymology of "spacelike" and "timelike" virtual photons.

If we really wanted to be realistic, we would have to take wavepackets for the incoming and outgoing particles. That is not what I suggest to do, as it is quite painful and not very illuminating. Yes, I do suggest to take the Feynman diagram as a "little drawing" of what is really happening, although it's only the second term (first order, the first term being at zeroth order and corresponding the identity) in a perturbative expansion of the amplitude, and by itself is meaningless.

 

[tiny-tim]

virtual = off-shell ?

Hi arivero!

Now,in quantum field theory, the calculation is done for all possible values of energy and momentum.

It means that only the external particles being calculated are on-shell, [itex]E^2-p^2=m^2[/tex].

{Internal} Particles in the diagrams of the calculation do not need to meet this equation, they are called off-shell.

This virtual particle thing... is the same that the off-shell particles?

(I added the "{Internal}")

No, the virtual particles in the position-space representation Feynman diagrams are on-shell. Only in the momentum-space representation are they off-shell.

I've looked at how a couple of books deal with this …​

Weinberg, in Quantum Theory of Fields Volume I, doesn't seem to use the word "virtual" at all, just "on the mass shell" and "off the mass shell" (or "off-shell").

V. B. Berestetskii, E. M. Lifgarbagez, and L. P. Pitaevskii, in the free online http://books.google.com/books?id=oA...d=16#v=onepage&q="virtual particles"&f=false" at p.312, §79, say "virtual states" in the position-space representation and "virtual particles" in the momentum-space representation.

Does anyone have any examples from other books?

 

[Frame Dragger]

@Tiny-Tim: I haven't, but that doesn't mean you're not right. I hope you are... it would make for better terminology than "virtual".

 

[arivero]

(EDIT: I did this post before reading #17 above (Hi!) but fortunately I kept myself in the momentum space representation)

Ok, I sort of convinced myself. I was first looking a bit at http://en.wikipedia.org/w/index.php?title=Bhabha_scattering&oldid=332559874 , a calculation for which I have sad rememberances when undergradute; but in any case it is not even needed. It is enough to understand than in a channel the 4-momentum is exchanged, in the other is accumulated. So in the former case the carries takes a 4-momentum which is a difference, in the later it carries a sum. Now it is only plain relativity: take $||E,p||=E^2-\sum p_i^2=m^2$, wash, rinse, and repeat.

To be specific, consider a pair of four-momentum vectors $(E_1,p_1), (E_2,p_2)$, with respective masses $m_1, m_2$. Both of them will fullfill the relativistic equation above. For simplicity you can choose a reference frame where $p_1=(0,0,0)$, so $E_1=m_1$. But it is not compulsory.

Now, questions are: what can we tell of the "square mass" of the sum $(E_1+E_2,p_1+p_2)$ and difference $(E_1-E_2,p_1-p_2)$? We are looking for the relativistic equivalent of "triangle inequalities". What we found just by applying the energy-momentum formula and knowing that $E_i >= m_i$, is that

$$M_+^2=|| (E_1+E_2,p_1+p_2)||^2 > (m_1+m_2)^2$$

$$M_ -^2=|| (E_1-E_2,p_1-p_2)||^2 < (m_1-m_2)^2$$

So:
- the s-channel (which is a sum) particle is always time-like.
- the t-channel (which is a difference) is mainly space-like, but it could have a small contribution from time-like exchanges. If the rest mass of the input and output particle is the same, then the channel is completely space-like.

This is for a vertex. As one interaction tree has at least two vertex, I guess that there are really two bounds in each channel, but on the other hand there is more kinematic in game.

Finally: why do we call "virtual" to the particles here? Because $M_\pm^2$ is fixed from the values of external 4-momentum (remember we are calculating the probability of having such and such outputs with such and such inputs, then all the external are given as premises of the calculation). So they do not coincide with the mass "m0" of the interaction carrier (zero mass for the photon, 91 GeV for a Z, etc). In some cases we can still think that each value of M can "live" during a time h/(M-m0)c^2, but it is more precise to use the propagator formula. Furthermore, one always want to take into account the interference between all the possible diagrams.

 

[Frame Dragger]

(EDIT: I did this post before reading #17 above (Hi!) but fortunately I kept myself in the momentum space representation)

Ok, I sort of convinced myself. I was first looking a bit at http://en.wikipedia.org/w/index.php?title=Bhabha_scattering&oldid=332559874 , a calculation for which I have sad rememberances when undergradute; but in any case it is not even needed. It is enough to understand than in a channel the 4-momentum is exchanged, in the other is accumulated. So in the former case the carries takes a 4-momentum which is a difference, in the later it carries a sum. Now it is only plain relativity: take $||E,p||=E^2-\sum p_i^2=m^2$, wash, rinse, and repeat.

To be specific, consider a pair of four-momentum vectors $(E_1,p_1), (E_2,p_2)$, with respective masses $m_1, m_2$. Both of them will fullfill the relativistic equation above. For simplicity you can choose a reference frame where $p_1=(0,0,0)$, so $E_1=m_1$. But it is not compulsory.

Now, questions are: what can we tell of the "square mass" of the sum $(E_1+E_2,p_1+p_2)$ and difference $(E_1-E_2,p_1-p_2)$? We are looking for the relativistic equivalent of "triangle inequalities". What we found just by applying the energy-momentum formula and knowing that $E_i >= m_i$, is that

$$M_+^2=|| (E_1+E_2,p_1+p_2)||^2 > (m_1+m_2)^2$$

$$M_ -^2=|| (E_1-E_2,p_1-p_2)||^2 < (m_1-m_2)^2$$

So:
- the s-channel (which is a sum) particle is always time-like.
- the t-channel (which is a difference) is mainly space-like, but it could have a small contribution from time-like exchanges. If the rest mass of the input and output particle is the same, then the channel is completely space-like.

This is for a vertex. As one interaction tree has at least two vertex, I guess that there are really two bounds in each channel, but on the other hand there is more kinematic in game.

Finally: why do we call "virtual" to the particles here? Because $M_\pm^2$ is fixed from the values of external 4-momentum (remember we are calculating the probability of having such and such outputs with such and such inputs, then all the external are given as premises of the calculation). So they do not coincide with the mass "m0" of the interaction carrier (zero mass for the photon, 91 GeV for a Z, etc). In some cases we can still think that each value of M can "live" during a time h/(M-m0)c^2, but it is more precise to use the propagator formula. Furthermore, one always want to take into account the interference between all the possible diagrams.

Hmmmm... ok, I see where you're going. That does make the "virtual" title more sensible.

 

[Hans de Vries]

Unfortunately,

All this naive pop-sci talk about "virtual", "space-like = faster than light" particles is only
aiding people like Tom van Flandern in their "faster than light" propagation theories.. :uhh:
Working out the real (Dirac) math shows that what is going on is nothing mystical but
rather real, down to Earth physics. The only thing which differs from classical electro-
magnetism is that the electron can interfere with itself during its transition from initial
to final state.

This interference (transition) current is a sinusoidal pattern moving at a speed slower
than c as it should be. The electromagnetic field (classical and qed) from this alternating
source field is shifting with the same speed lower than c. This is just standard electro-
magnetism as guaranteed via the propagator 1/q^2

But here we go...

A phase velocity lower than c corresponds with with a "space-like" particle, so by
interpreting an electromagnetic field which is shifting along with its source at a
physical speed lower than c, as if it were a particle with a deBroglie wavelength,
the naive suggestion is raised that something moves faster than light... Regards, Hans

 

[Frame Dragger]

Unfortunately,

All this naive pop-sci talk about "virtual", "space-like = faster than light" particles is only
aiding people like Tom van Flandern in their "faster than light" propagation theories.. :uhh:



Working out the real (Dirac) math shows that what is going on is nothing mystical but
rather real, down to Earth physics. The only thing which differs from classical electro-
magnetism is that the electron can interfere with itself during its transition from initial
to final state.

This interference (transition) current is a sinusoidal pattern moving at a speed slower
than c as it should be. The electromagnetic field (classical and qed) from this alternating
source field is shifting with the same speed lower than c. This is just standard electro-
magnetism as guaranteed via the propagator 1/q^2

But here we go...

A phase velocity lower than c corresponds with with a "space-like" particle, so by
interpreting an electromagnetic field which is shifting along with its source at a
physical speed lower than c, as if it were a particle with a deBroglie wavelength,
the naive suggestion is raised that something moves faster than light...


Regards, Hans

Did someone here imply that these particles are FTL? I missed that...

 

[Hans de Vries]

Did someone here imply that these particles are FTL? I missed that...

Hi, Frame Dragger

I wasn't talking about anyone on this thread but more about the use of the terms
"virtual, space-like particles" which suggests some form faster than light propagation.

Regards, Hans

 

[arivero]

Did someone here imply that these particles are FTL? I missed that...

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.

 

[Frame Dragger]

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.

Ugh... Doesn't the term "VIRTUAL" give it away?! This is like Catherine Asaro talking about adding an "imaginary portion" to a ship's velocity to "rotate partially out of normal space". I guess some people think that a mathematical approximation is somehow mystical. How... dissapointing.

Isn't fusion, and antimatter, or the implications of a possibly valid MWI enough for sci-fi? Leave the magic to the Fantasy genre :)

 

[Hans de Vries]

Ugh... Doesn't the term "VIRTUAL" give it away?! This is like Catherine Asaro talking about adding an "imaginary portion" to a ship's velocity to "rotate partially out of normal space". I guess some people think that a mathematical approximation is somehow mystical. How... dissapointing.

Isn't fusion, and antimatter, or the implications of a possibly valid MWI enough for sci-fi? Leave the magic to the Fantasy genre :)

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.

Regards, Hans

 

[Raghnar]

As far as I know the only instance in which a "virtual" pair can produce "real" photon would be in the form of Hawking Radiation.

In nuclear physics there are some processes that many try to interprete as the virtual phonons becoming real...

For me the "Virtual" as "The one that helps us do the math" part is veeery very tricky and maybe misleading. I don't want to go Off Topic but almost erverything you talk about in physics can be read "things that helps us do the math", the math is our way to interprete the physics reality for what we can understand about it they are real just as gravitational force.

If you can explain giromagnetic moment of electron way more precisely adding virtual contribution (that arent nothing other then the contribute of the electron with itself and with vacuum pairs Heisenberg-induced) than it is as real as you explain gravitational lensing with space-time wells.
Phyics is just about the Occam's Razor: If it is the most elementary working explanation, must be the truth.

 

[Frame Dragger]

In nuclear physics there are some processes that many try to interprete as the virtual phonons becoming real...

For me the "Virtual" as "The one that helps us do the math" part is veeery very tricky and maybe misleading. I don't want to go Off Topic but almost erverything you talk about in physics can be read "things that helps us do the math", the math is our way to interprete the physics reality for what we can understand about it they are real just as gravitational force.

If you can explain giromagnetic moment of electron way more precisely adding virtual contribution (that arent nothing other then the contribute of the electron with itself and with vacuum pairs Heisenberg-induced) than it is as real as you explain gravitational lensing with space-time wells.
Phyics is just about the Occam's Razor: If it is the most elementary working explanation, must be the truth.

This is QM... I'm used to Renormalization. Virtual pair creation and dectruiction as a description of a field quantized at EVERY point... is ok. It's weird, and clearly "tricky", but it works for now. Gravitational "force" is at least something which can be observed. By they're nature a Virtual Photon can't actually be EXPERIENCED at anything like firsthand, unless perhaps you count The Casimir Effect, and I don't.

EDIT: While I've been arguing Occam's Razor vs. dBB lately, I have to say that is not ALL of phyics. That is one particularly popular view of how physics should be approached. Simplicity doesn't HAVE to equal truth, it just avoids the creation of needless and arbitrary hypothesis and conjectures.

 

[Raghnar]

This is QM... I'm used to Renormalization. Virtual pair creation and dectruiction as a description of a field quantized at EVERY point... is ok. It's weird, and clearly "tricky", but it works for now. Gravitational "force" is at least something which can be observed. By they're nature a Virtual Photon can't actually be EXPERIENCED at anything like firsthand, unless perhaps you count The Casimir Effect, and I don't.

It's only a technical matter, not a theoretical one.
If you believe that the renormalization process explain better so you have to believe that is as real as more intuitive things. "Direct" is also misleading, if you measure the orbit of Mercury and deduce that adding general relativity contribution explain better the position of its perielium so you deduce that the general relativity is "Real".
Same thing with renormalization: if you measure gyromagnetic moment of electron or the shell of a nucleus or the lamb shift in the atom (this one is pretty "Direct") and renormalizing with energy contribution explain better the value that you masuread, so you must deduce that virtual photons and phonons are "Real".

"Direct" is a mean of technics, it can be wrong because every experiment can falsify a theory but for now you have no epistemological reason to consider virtual photon "a mathematical trick".

EDIT: While I've been arguing Occam's Razor vs. dBB lately, I have to say that is not ALL of phyics. That is one particularly popular view of how physics should be approached. Simplicity doesn't HAVE to equal truth, it just avoids the creation of needless and arbitrary hypothesis and conjectures.

What is dBB? ^^
It's only on what definition of simplicty. No doubt that things can get more complicated, but if you involve lesser entities (in form of physical entities or mathematical hypothesis and laws) and explain the more phoenomena for now it is considered to be the scientifical thruth for what I know.
Maybe is not the best possible way to understand reality, but is the best we've figured for now (to my knowledge).

 

[tiny-tim]

Hi Raghnar!

… if you measure the orbit of Mercury and deduce that adding general relativity contribution explain better the position of its perielium so you deduce that the general relativity is "Real".

I don't get that …

space can be "real", aether can be "real", quarks can be "real", but eg a theory or law of quarks can't be "real", it's only a concept.

If a quark is real, you can point to it (or identify it in some other way) … you can't point to a concept such as general relativity.

Same thing with renormalization: if you measure gyromagnetic moment of electron or the shell of a nucleus or the lamb shift in the atom (this one is pretty "Direct") and renormalizing with energy contribution explain better the value that you masuread, so you must deduce that virtual photons and phonons are "Real".

You might as well say that epicycles are real …

epicyles of course are the wheels upon wheels upon wheels … upon which a planet rotates.

At one time they may have been thought of as real.

They're still perfectly good maths … any ellipse can be approximated by epicycles (a sort of Fourier series ) … we could still use them to calculate planetary orbits.

But that doesn't mean the epicycles are real.

ok, we need infinitely many (smaller and smaller) epicycles for a completely accurate treatment … but it's the same for QFT and renormalisation … we need infinitely many virtual particles for a completely accurate treatment.

And at least our epicycles are definite … at any instant of time, for a particular planet, we know exactly where they are, and we can draw them (not infinitely many, of course, we don't have the time, but to any order we choose) …

but the virtual particles, for an interaction between particular "real" particles, are not definite … they could be anywhere, at any time, and in any numbers.

Nobody nowadays would say that epicycles are real, so why would you say that virtual particles are real?​

If you still say they're real, then can you tell us:

how many are there (and where are they) for the simple case of two electrons approaching each other and being (electromagnetically) deflected?
​

"Direct" is a mean of technics, it can be wrong because every experiment can falsify a theory but for now you have no epistemological reason to consider virtual photon "a mathematical trick".

Yes we have … they're totally unquantifiable … if they're not a trick, the theory should at least enable us to count or locate them, at least within the theory if not in practice.

 

[Frame Dragger]

Hi Raghnar!


I don't get that …

space can be "real", aether can be "real", quarks can be "real", but eg a theory or law of quarks can't be "real", it's only a concept.

If a quark is real, you can point to it (or identify it in some other way) … you can't point to a concept such as general relativity.


You might as well say that epicycles are real …

epicyles of course are the wheels upon wheels upon wheels … upon which a planet rotates.

At one time they may have been thought of as real.

They're still perfectly good maths … any ellipse can be approximated by epicycles (a sort of Fourier series ) … we could still use them to calculate planetary orbits.

But that doesn't mean the epicycles are real.

ok, we need infinitely many (smaller and smaller) epicycles for a completely accurate treatment … but it's the same for QFT and renormalisation … we need infinitely many virtual particles for a completely accurate treatment.

And at least our epicycles are definite … at any instant of time, for a particular planet, we know exactly where they are, and we can draw them (not infinitely many, of course, we don't have the time, but to any order we choose) …

but the virtual particles, for an interaction between particular "real" particles, are not definite … they could be anywhere, at any time, and in any numbers.

Nobody nowadays would say that epicycles are real, so why would you say that virtual particles are real?​

If you still say they're real, then can you tell us:

how many are there (and where are they) for the simple case of two electrons approaching each other and being (electromagnetically) deflected?
​

Yes we have … they're totally unquantifiable … if they're not a trick, the theory should at least enable us to count or locate them, at least within the theory if not in practice.

I can't believe I'm uttering this phrase, but um, "God bless you Tiny Tim!" Seriously, thanks. By the way, can Phlogiston be real... if I really TRULY believe?! ;)

 

[Frame Dragger]

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.

Regards, Hans

Fair enough, but we know better, hence the name.

 

[Hans de Vries]

Fair enough, but we know better, hence the name.

It is a real physical process and it is so following the basic definitions of
the standard theory:

The emitting charge/current density in the scattering zone is.

$$\overline{(\phi_i+\phi_f)}\gamma^\mu(\phi_i+\phi_f)$$

The emitting interference charge/current density in the scattering zone is
given by the cross terms:

$$\bar{\phi_i}\gamma^\mu\phi_f~+~\bar{\phi_f}\gamma^\mu\phi_i~~~~$$

(both complementary terms contain the same information) Applying the
operator 1/q^2 gives the classical EM potential field $$A^\mu$$ which is emitted.

The emitted EM field is instantaneously absorbed by the other electron
field existing in the scattering zone. the instantaneous absorption highly
simplifies the process.

There is no net EM field because there are two EM fields which cancel
each other out. We therefor don't have to deal with mathematically more
complex interactions such as those giving rise to the Volkov solutions.

This is (except for the interference) just a classical EM process but it
describes a process in a scattering zone of two overlapping fields, which
something very different as the naive pop-sci picture of two electrons
throwing virtual photons to each other in order to repel. The interaction
between two electrons with non-overlapping fields is an entirely different
process.Regards, Hans

 

[tiny-tim]

It is a real physical process …

?? 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?

 

[Frame Dragger]

?? 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?

I have no idea. His last paragraph would seem to agree, but the rest seems only tangentially relevant.