# Dirac Sea = Unceasing Day

1. Jun 12, 2007

If the universe is a Dirac Sea, then wouldn't the light generated from virtual particle annihilation be detectable? Not only this, but wouldn't it drown out all other light sources? I can't find any good explanations so far for why virtual particle annihilation does/doesn't produce photons or Z^0 Bosons etc like regular particle annihilation, please "enlighten" me.
As an aside, if a photon is its own antiparticle, why doesn't it annihilate itself upon existence?

2. Jun 12, 2007

### strangerep

The Dirac Sea picture is no longer necessary in modern advanced formulations
of quantum field theory. Unfortunately, it still hangs around in the more
elementary treatments of relativistic quantum mechanics. Don't waste your
time or energy trying to unravel any such puzzles it generates. Your time and
energy are better spent trying to learn some QFT.

Expressions like "an electron annihilates with a positron" are misleading.
Nothing is really lost, just converted. E.g: the initial total electric charge
was 0, and the total charge of the 2 photons emerging from the so-called
annihilation is also 0. Similarly, total linear momentum, and angular
momentum are also conserved.

I.e: the electron-positron pair are just converting into other particles
in such a way as to conserve certain quantities.

Photons do in fact interact with other photons, but the effect is
extremely weak. It's like a very slight scattering of one photon
off another. Once again the total incoming and outgoing
charge and momenta remain conserved.

- strangerep.

Last edited: Jun 12, 2007
3. Jun 12, 2007

Are you saying that virtual particles are no longer necessary or a viable approach? I thought they were still being used to explain things. If so, then why don't constant pair creation/annihilation in the vacuum of space create constant ambient light? If not, then could you reference me to an article which refutes the virtual partical theory?
Thats fascinating, is there a specific name for that phenomena which i can look up online?
thanks,

4. Jun 13, 2007

### strangerep

Virtual particles are nothing more than a name for a theoretical/mathematical
construct which arises when obtaining scattering cross-sections in quantum
field theory (QFT) by perturbation methods. They correspond to internal
lines in Feymann diagrams (which are themselves just a pictorial
representation of terms in this perturbation series).

When experts "explain" things using the concept of virtual particles,
they're either just using this as a shorthand word for such internal lines,
or else they're trying to explain QFT to a much less technical
audience who do not yet have the knowledge to tackle full-on QFT.
If you're not yet at that level, then just take the "virtual particle" thing
with a large grain of salt - until you get to the level of understanding at
least the basics of QFT.

"Virtual particles" are not a "theory". It's just a name for a particular
item in the mathematics of QFT as described above, which has no
physically-relevant significance in itself. The fact that "virtual particles"
have only mathematical, but not physical, relevance is most compellingly
demonstrated by the fact that different perturbation schemes in QFT
involve different versions of virtual particles - but they also produce
the same physically-testable cross sections relevant to scattering
his FAQ at http://www.mat.univie.ac.at/~neum/physics-faq.txt
(Section 3; in particular S3c. How real are 'virtual particles'?).

As a starting point, you could look up the closely-related
"Delbruck Scattering", which is scattering of a photon off the intense
coulomb field of a nucleus. This has definitely been observed
experimentally. Wikipedia has an entry about it. Wiki also says that
photon-photon scattering (i.e: scattering of 2 free photons) has not been
observed. However, I vaguely recall a recent experiment at SLAC which
claimed to have observed this for the first time via intense laser fields.
I don't recall the details, sorry.

- strangerep.

5. Jun 14, 2007

### Hans de Vries

You could try "light by light" scattering.

It's a contributer to the magnetic anomaly of the leptons. especially
the ones with higher mass. for instance that of the muon:

Code (Text):

--- Muon magnetic anomaly and its subcomponents -------

0.00116592080___(63)  experimental (world avg.)

0.00116591785___(61)   total Standard Model

0.0011658471809_(23)   total QED contribution.
0.00116140972758   1st order alpha/2pi
-0.00000177230505   2nd order + VP
-0.00000185694638   2nd order without mu/mu VP
0.00000008464133   2nd order mu/mu VP only
0.00000001480420   3rd order + mu/mu VP
-0.00000000005031   4th order + mu/mu VP

0.00000590406002   mu/me 2nd order
0.00000002406858   mu/me 3rd order VP
0.00000026253510   mu/me 3rd order light by light
0.00000028660368   mu/me 3rd order total
0.00000000381334   mu/me 4th order
0.00000000004483   mu/me 5th order
0.00000000042119   mu/mτ 2nd order
-0.00000000002234   mu/mτ 3rd order VP
0.00000000002685   mu/mτ 3rd order light by light
0.00000000000451   mu/mτ 3rd order total
0.00000000000661   mu/me + mu/mτ loops 3rd order

0.00000000154___(3)    Electro Weak VP

Now, indeed, one could argue that these depend on the perturbation
scheme, as strangerep mentioned. Nevertheless, this is pretty much
the general accepted pertubation scheme, and the sums of light by
light scattering to all orders should always add to the same number
independent of the scheme.

Regards, Hans

Last edited: Jun 14, 2007
6. Jun 14, 2007

### strangerep

I'm not exactly sure what you meant in your last sentence above.
The anomalous magnetic moment is calculated theoretically by
summing many higher order diagrams involving the full standard
model (i.e: involving virtual hadrons and all the other stuff you
mentioned). But in the end, the only thing we measure physically is
the total anomalous magnetic moment. We have no means to
measure separately the contributions from (say) internal
photon-photon loops, or indeed any other piece of the total
picture.

But maybe I misunderstood you.

- strangerep.

7. Jun 15, 2007

### Hans de Vries

I just mean to say here that one might expect that some physical effects
(like for instance Vacuum Polarization) play their role independent of a
particular perturbation scheme.

Personally, I'm very interested to look beyond the standard perturbation
scheme. Very interesting are for instance the Volkov solutions which are
exact solution of the interaction of (sinusoidal) EM fields with the Dirac
electron. They show an inherent quantization independent of an imposed
quantization scheme like canonical quantization.

The solutions have components like:

$$\mbox{\huge  e^{iQ \sin(\omega t)}}\ \ =\ \sum_{k\ =-\infty}^\infty\ \mbox{\huge J}_k(Q)\ \mbox{\huge  e^{ik\omega t} }$$

This is a superposition where the Bessel coefficients can be interpreted as
amplitudes:

J0(x) = amplitude to absorb/emit 0 quanta
J1(x) = amplitude to absorb/emit 1 quanta
J2(x) = amplitude to absorb/emit 2 quanta
J3(x) = amplitude to absorb/emit 3 quanta
...............

These Bessel coefficients have the unique property that they are Unitary for
both amplitudes as well as probabilities for any value of Q:

$$\sum_{k\ =-\infty}^\infty \mbox{\huge J}_k(Q)\ = 1, \quad \ \ \sum_{k\ =-\infty}^\infty \left|\mbox{\huge J}_k(Q)\right|^2\ = 1$$

At the other side, Some effects we treat non-pertubative may actually
be the result of recursive physical processes. A nice example is:

$$\frac{1}{p^2-m^2}\ =\ \frac{1}{p^2} + \frac{m^2}{p^4} + \frac{m^4}{p^6} + ......$$

Which represents the following series in configuration space.

$$\Box^{-1}\ -\ m^2 \Box^{-2}\ +\ m^4 \Box^{-3}\ -\ ...$$

Where the first term maps a source on the light cone as for a massless particle,
the second term represents a re-transmission from the first term, on the light
cone as well. The minus sign represents an opposition to change from the
vacuum's center value. The third term represents the second retransmission.
et-cetera.

This series leads to the Bessel function representing the propagator in
configuration space. Note that all propagation is on the light cone which
is highly desirable for any underlaying theory.

Regards, Hans.

Last edited: Jun 15, 2007
8. Jun 16, 2007

### Anonym

Please provide the ref where that point is discussed in details (in addition to rel QM by E.M.Lifschitz et al: my teacher said “don’t read that book”).

Regards, Dany.

Last edited: Jun 16, 2007
9. Jun 16, 2007

### Iamu

Excuse me... I think I've heard the term "virtual particles" being used a different way, to describe particles that can be sustained by a vacuum of a particular volume.

Take two reflective plates x distance apart, and another two metal plates x/2 apart, and the vacuum energy between the first set of plates will be greater than between the second set. The first set can sustain "virtual particle" wavelengths up to a length of x, while the second set can only sustain wavelengths of up to x/2.

I've only heard the term used this way in pop physics books. Is this effect due to virtual particles?

10. Jun 18, 2007

### strangerep

I think you're talking about (a version of) the Casimir effect. This is usually
discussed in textbooks in the context of the photon field: more modes are
possible in one region than another, and that difference leads to a difference
in energy in the two regions, resulting in a net force which can be
detected experimentally.

Although the Casimir effect is often "explained" using the concept of
virtual particles, the details are tricky. When one says "vacuum", one
normally means a state of lowest energy, usually taken as 0. But
implicitly then, one is also talking about an unconstrained vacuum
occupying an infinite amount of space. Only in this case does it make
sense to say that the vacuum has exactly zero energy and momentum.
That's because of the Heisenberg uncertainty principle: if the vacuum
state is somehow restricted to a finite region of space, its momentum
becomes indeterminate. Since we're talking about the vacuum state
of the photon field, that also means its energy is also uncertain. Such
conditions are sometimes called "squeezed states". The effects of
such energy-momentum uncertainty are actually observable in
quantum optics experiments, (iirc). But all that's happened is that
by constraining your vacuum spatially, you've forced it to become
a superposition of momentum eigenstates insteading of being
a state of exactly zero energy-momentum.

- strangerep.

11. Jun 18, 2007

Modes of what- photons? This seems odd... if there is an ever-present ambient field of energy, wouldn't it be evenly distributed? This explanation of the Casmir effect looks similar to diffusion, except with supposedly ambient energy.
How do they know the effect isn't caused by minute vibrations in the laboratory, or some residual charge of molecules on the metal's surface? How do they know it isn't caused by the change in light from wave to particle when they observe(like in the double slot experiment)?
So a vacuum has momentum- or only as much as it has energy- right? Would you extrapolate from the evidences that space IS energy? Or would it be more correct to infer that energy is "birthed" from space?

thanks,

12. Jul 10, 2007

### Hans de Vries

I found a nice overview online here: (in chapter 2) Discussing the
interaction of Dirac electrons with an EM plane wave (laser beam)
in great detail, including the exact Volkov solutions.

http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-r-666.pdf

This also clarified some of the obscure notation used in the book you
mentioned above (in paragraph 40)

Regards, Hans

13. Jul 10, 2007

### meopemuk

In the standard paradigm of QFT, virtual particles are not observable. They are unobservable not because they are difficult to see and one needs a powerful instrument to see them. No. They are not observable *in principle*. So, they are not a part of physical measurable reality. They are a part of a model, which theorists use to describe physical measurable reality. So, when people say that some effect is attributed to virtual particles, you should understand that these people have used a model that involves virtual particles and they were able to describe this effect in this particular model.

There is only one physical reality, but there could be many different theoretical models. So far the model based on virtual particles worked well. However, since virtual particles cannot be directly observed, there is always a chance that a different model can be invented, which doesn't use the idea of virtual particles and describes experiments equally well or even better. So, agreement with experiment is not a proof that virtual particles *really* exist.

14. Jul 11, 2007

### Anonym

Thank you very much

Regards, Dany.