The way to force a minimum frequency to oscillation is to use a harmonic oscillator:
d^2 ψ / dx^2 - m^2 ψ = d^2 ψ / dt^2
drum skin B
======================
| | | | | | |
springs that resist displacement
of the skin
The massive Klein-Gordon equation does the job. If we ignore the term d^2 ψ / dx^2, we have the differential equation for a harmonic oscillator.
B is like an array of harmonic oscillators which are coupled by the drum skin.
The impulse response of the massive Klein-Gordon equation
Let us imagine a Dirac delta term
δ(t, x)
summed to the Klein-Gordon equation at a certain point in spacetime t, x. How does the equation react to such delta source? An answer is a Green's function.
The Klein-Gordon equation can be interpreted as calculating the vertical acceleration of a a small square of the drum skin B. The term d^2 ψ / dt^2 is the acceleration. The term d^2 ψ / dx^2 is the vertical force which results from the bending of the tense skin. The term m^2 ψ is the force exerted by the spring of a harmonic oscillator.
A Dirac delta impulse is like hitting the drum skin with a sharp hammer.
Obviously, a single hit with a hammer would make B to oscillate and create sine waves, that is, real particles. A single hit is like the creation operator in quantum field theory.
Modeling particle creation in the massive Klein-Gordon equation
Let us assume some unknown coupling between drum skins B and C. In our previous blog post, the coupling was a thin rubber membrane (between strings).
In our previous blog post we found out that colliding waves in C can create particles in B. A transient standing wave in C will create sine waves in B.
Let the resonant frequency of the harmonic oscillators in B be f_B. We may imagine that f_B corresponds to 511 keV.
If colliding waves in C have the frequency f_B, they will act as an oscillating source in B. The oscillation creates waves which slowly spread to the environment. If we use the circular polarization framework in B, opposite polarizations might correspond to an electron and a positron.
| | springs
● ●
/ \ / \ drum skin
● ● ●
| | | springs
● = oscillator
If the frequencies in C are larger than f_B, then the oscillators in B may raise their resonant frequency by using the tension in the drum skin, like in the diagram above. The black dots are the oscillators. If the angles in the tense drum skin are steep, the tension in will make the oscillators to oscillate faster. These waves correspond to electrons and positrons with kinetic energy > 0 in QED.
Modeling the Coulomb force
Let us have a plane wave in B. If the wavelength is shorter in some direction, then the wave tends to turn to that direction.
We can make the wavelength shorter in some area by making the springs looser there. The wave has to keep its frequency. It has to compensate the loosening of the spring by steepening the curves in the string - the wavelength has to become shorter.
If a wave packet has positive potential energy, it is located in an area where the springs are tight: this corresponds to a large m in the Klein-Gordon equation.
Recall that we solved the Klein paradox a couple of years ago by letting potential energy add to the inertial mass of the particle.
We have a model for the Coulomb potential: a high electric potential A in some area, through some mechanism, tightens the springs in B in that area, a low potential loosens the springs.
The bare charge of the electron is positive under its electric field?
The interaction term in the QED lagrangian is
-e ψ-bar γ^μ A_μ ψ.
If we assume that the electron has a positive electric charge, then creating a potential well at the location of the electron will probably lower the total potential energy of the system. The cost is increased potential energy in the electric field component of A, but we save potential energy of the charge of the electron.
There would be no direct Coulomb force between charges. The repulsion or the attraction would be a result of the potential pit (or hill) created by the other charge.
This is like pressing a drum skin with a slippery finger to make a pit. Two fingers tend to slide toward each other because together the fingers can press the skin down more: potential energy is freed.
How does an oscillating electric charge produce electromagnetic waves?
drum skin B
======================
| | | | | | springs
------------------------------------------
very tense lightweight
drum skin C
The drum skin C takes the role of the electromagnetic field in the following. Pits and hills in C control the tightness of springs in B.
In our model, if we through some mechanism force a wave packet (charge) to move back and forth in B, it will force the "Coulomb field" in C to move accordingly. If the pit or hill in C would not move along with the wave packet in B, then the potential energy of the charge would increase unnecessarily.
It is intuitively clear that a pit moving back and forth in the drum skin C will produce sine waves far away.
The wave-particle duality: whenever we measure the relative position of particles, we have to use a particle model or a wave packet model
Collisions of particles are hard to model with plane waves. How can low-amplitude waves spread over a large volume of space cause scattering to large angles?
For example, if an electron and a positron collide, they will change their direction significantly.
A possible solution is to treat the electron and the positron as a single particle moving in 6-dimensional space, and treat their interaction potential as an external potential which affects the single 6D particle. But then it is complicated to explain electromagnetic radiation sent by the particles.
When particles scatter from each other, we gain precise information about their relative position. That suggests that we should use a particle model or a sharp wave packet model to model the process.
It is like a photon hitting the image sensor of a digital camera. We have to treat a photon as a particle to explain why it hit just a single pixel cell.
When a hydrogen atom decays to a lower energy state, it for a short time acts like a dipole antenna. When the atom is excited to a higher energy state, the converse process happens. In this case, it is natural to threat the photon as a wave packet.
In our mechanical drum skin model this means that we have to model scattering events using sharp wave packets. If we try to model scattering with waves spread over a large area, the mechanical model fails.
What about the double-slit experiment where the distance of the slits is L? Can we use sharp wave packets to model it? Not very sharp. The packets have to be of the size ~L to form an interference pattern on the screen.
How do colliding waves in the electromagnetic field create waves in the Dirac field?
This is a difficult question. In our previous string model we assumed that there is an ad hoc rubber membrane which couples the two strings. We cannot resort to ad hoc mechanisms now, but have to derive, in some way, the behavior from the QED lagrangian.
Since the collision of two photons measures their relative position, the previous section of this blog post suggests that we should model the photons with sharp wave packets or particles.
photon
~~~~~ --------------------- e-
| virtual
| electron
~~~~~ --------------------- e+
photon
In a Feynman diagram, a photon scatters a positron which "travels backward in time", into a virtual electron. Another photon scatters that virtual electron "forward in time" into a real electron.
The diagram is full of mysteries:
(a) What is the energy and the charge of the virtual electron?
(b) Do the electron and the positron attract each other? If yes, when does the attraction start?
(c) From where does the virtual electron appear? Should we take seriously the claim that the the positron arrives from the future?
(d) We can create a 5 million volt potential difference with a van de Graaff generator. Electron-positron pairs do not form spontaneously to neutralize the charges. Why not? Is there some maximum distance that the virtual electron can travel?
An oscillating dipole model of the photon: we can make the spatial dimensions much smaller than the wavelength
Should we model the photons with sharp wave packets whose size is of the order of their wavelength, or 10^-12 m? The wave packet is too large and its energy way too diluted in space.
Compare this to a hydrogen atom whose radius is 5 * 10^-11 m and which absorbs a photon whose wavelength is 122 nm. The wavelength is 1000X the atom diameter. The way to model the photon is some kind of a hybrid particle-wave model, where the oscillation is restricted to an area much smaller than the wavelength of the photon.
e+
| ^
| | oscillation
| v
e-
A possible way to model a photon is to use an imagined electric dipole which has an electron and a positron at a close distance. The dipole can be, say, 10^-15 m long, but represent a wave whose wavelength is 10^-12 m. The dipole oscillates much slower than the speed of light.
A linearly oscillating dipole represents a linearly polarized photon. A rotating dipole models a circularly polarized photon.
A drum skin wave model of pair production
When our two photons collide, they form a standing wave where a strong electric field oscillates, e.g., up and down.
^
| e+ created
|
| e- pair
electric
field of
photons
Everything happens in an area whose size is 1 / 1000 of the photon wavelength, or less than 10^-15 m.
Apparently, the electric field in that tiny area is strong enough to produce a new electron-positron pair.
What is the role of the virtual electron? It is a wave for which the energy E is typically small or zero, and the momentum p is substantial. It is an off-shell electron.
Let us look at a wave model.
An analogous construct in a drum skin is a pit which we make by pressing the skin with an elongated object.
object
____ ##### ____ skin
| \________/ |
| | pit | | springs
As long as we hold the object still, there is no oscillation relative to time. The "energy" E in an oscillator is zero:
exp(-i (E t - p x)).
But the skin is deformed: the "momentum" p is not zero.
There certainly is energy in the pit: we had to deform the drum skin to make the pit. The E in the above formula is not the true energy of the deformation.
The elongated object is really the interaction of the electromagnetic field with the Dirac field. The interaction deforms the Dirac field.
Once we remove the elongated object (or the interaction), we get two sine waves which start to spread to every direction in the drum skin. These are analogous to the created real electron and the real positron.
Why would Nature create these particles? Nature tries to minimize the potential energy in the hamiltonian of the system. When the photons are very close to each other, creating a pair apparently helps to reduce potential energy.
Local conservation of energy, momentum, and charge requires that a Dirac wave cannot suddenly appear or disappear in empty space where there are no other fields that interact with the Dirac field.
The role of the virtual electron line might be to ensure some kind of "continuity" between the created real electron and the real positron. Feynman speaks about the electron "scattering backward in time or forward in time". Some kind of continuity of waves is required by all wave equations.
1. The created sine waves carry some momentum to every direction in the skin when the elongated object is removed. The pit is kind of a "spring" which pushes the waves and gives them the momentum. This is just like the virtual electron line in the Feynman diagram. The line transfers momentum between the created particles.
2. The energy as well as the momentum in the created sine waves completely come from the deformation of the drum skin and the springs in the pit.
3. The virtual electron in the Feynman diagram "moves" faster than light. It has more momentum p relative to the energy E than the energy-momentum relation allows. The created electron and the positron do not initially see each other.
4. The virtual electron is not a solution of the Dirac equation because it does not honor the energy-momentum relation. That is no surprise: the virtual electron is created by the interaction between the Dirac field and the electromagnetic field. It cannot be a solution of the free Dirac equation. Rather, it is a solution of the Dirac equation with a source - the impulse which the interaction gives.
5. Why created pairs do not neutralize the charge in a 5 megavolt van de Graaff generator? The elongated object in our diagram would in that case be very long. Apparently, the energy to make the very long pit to the drum skin would be too large: we would not recover that energy from the reduction of the electric field.
6. The virtual electron (the pit) may at some point of time contain all the energy to create the pair. The charge of the virtual electron is not really relevant because it very quickly decays into a real electron and a real positron, or is a combination of them all the time. We could say that the "charge" of the virtual electron is an electric dipole.
7. The virtual electron, the electron, and the positron are all created from the interaction of the electromagnetic field with the Dirac field. There is no positron which "arrives from the future".
A particle model of pair production
Imagine an electron and a positron placed at a 1.4 * 10^-15 m distance from each other. The sum of the masses of the particles is 1.022 MeV, but the electric potential is -1.022 MeV. The total energy of the system is zero.
The electric field of the electron at the distance 1.4 * 10^-15 m is
E = k e / r^2 = 7 * 10^20 V/m.
If the colliding photons (modeled with dipoles) produce an electric field which is stronger, it is able to pull the pair apart.
If the colliding photons have > 1.022 MeV energy, they might produce a pair.
The Schwinger limit for pair production is 1.3 * 10^18 V/m. We need to check why it differs from E calculated above.
Is this a perturbative model?
If we treat the positron as a particle traveling backward in time, then we could say that it absorbs a photon and is scattered into a virtual electron which moves "sideways" in time, or superluminally.
One can say that this is a perturbation: the photon only scatters a small portion of positrons traveling backward in time. Most positrons continue undisturbed.
If we look forward in time, most photons will not collide. Only a very small number will produce pairs since the cross section is very small.
However, if we assume that the two photons will collide, then the process definitely is not just a perturbation: it is a completely different process and not some fine-tuning.
How to reconcile different length scales?
Almost all the energy in the electric field of the electron is contained within a 10^-14 m radius from the center. If we want to use the electron to "fill" a pit in the electric potential, the pit has to be of this size.
However, the Compton wavelength of the electron is 2 * 10^-12 m, much larger. If we want to use a wave packet model, we cannot make the electron smaller than that.
We encounter the same length scale problem in Coulomb scattering. A relativistic electron and a positron have to pass within 10^-14 m for the momentum exchange to be substantial. How can we use wave packets to model that?
In the case of an electron scattering from a nucleus, we may assume the nucleus to be a classical point charge, and be static. Then we can model the electron as a wave. The large wavelength of the electron is not a problem in that case, because the nucleus is a particle.
It looks like in a close encounter, we have to treat one or all the particles using a particle model. Our drum skin model cannot handle such cases well, as the model is a pure wave model.
The Schwinger limit of 1.3 * 10^-18 V/m seems to assume that a pair can easily form at the distance of the Compton wavelength 2 * 10^-12 m from each other. If that were the case, would the process create electrons close to atomic nuclei, and make nuclei to radiate positrons? We need to check how Sauter and Schwinger calculated.
The length scale problem is glaring when a photon excites a hydrogen atom. We need to check if anyone has comments on this.
The famous wave-particle duality is central in analyzing particle collisions. Feynman diagrams use plane waves - they are a wave model.
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