________
/ \
+ ●-------------● -
\_________/
Let us assume that the energy density of an electric field is
ε_0 / 2 * E^2.
Let us visualize the field with the lines of force. The density of lines is proportional to the strength |E| of the electric field.
We get two rules of thumb:
1. Electric field lines repel each other with a "potential" proportional to 1 / r^2, where r is the distance between the lines.
2. There is "tension" in a field line. The line wants to become shorter. The tension is proportional to E^2.
z
^ E
| | | | |
| | | | |
--------------------------|--->
0 x_0
Rule 1 follows from the energy density ~ E^2. Suppose that we have a uniform electric field E to the direction of the z axis between 0 and x_0 in the diagram above.
If we squeeze the entire field between 0 and x_0 / 2, then the electric field strength doubles, as well as the density of field lines. The energy density of the field grows 4-fold. The volume containing the field is only half after the squeezing. Thus, the total energy of the entire field doubled.
The density of field lines grows by a factor of 2 if we reduce their distance by a factor of 1 / sqrt(2). The "potential energy" stored in the repulsion of the field lines is thus proportional to 1 / r^2, where r is the distance between the lines.
Rule 2 follows immediately from the field energy density, which is ~ E^2.
Note that the tension in a field line is not constant along the line. It depends on E^2 at the location.
Let us look at the diagram at the top of this page and analyze the attractive force. The repulsion of field lines prevents them from all gathering between the charges.
If we move the charges closer, the field lines become shorter. That is the origin of the attractive force. The tension in the straight field line between the charges is much larger that the curved ones, because the field strength E is greatest between the charges.
In the link, Benjamin Crowell derives the electromagnetic field tensor. There is negative pressure to the direction of a field line, but positive pressure to the normal directions. Our model casts light on this fact.
What are real photons in the field line model?
^ charge was suddenly moved up
|
● ------------
\ electric field line
-----------------------------------
wrinkle
Recall how Edward M. Purcell calculates the energy of electromagnetic radiation. Look at the diagram at the top of the linked page (by Daniel V. Schroeder, 1999).
A real photon, or an electromagnetic wave, is a "wrinkle" in field lines. The wrinkle moves at the speed of light. A bent electric field line requires a time-dependent magnetic field to be present, to bend the field line. Electric field lines are denser at the wrinkle, which implies that some extra energy E^2 is concentrated there.
The energy density of the symmetric static Coulomb field is proportional to 1 / L^4, where L is the distance. The energy density in the wrinkle is much greater than in the surrounding field. This explains how an electromagnetic wave is able to carry energy to across great distances.
A wrinkle is always transverse (the oscillation is normal to the direction of the wave movement). There exist no longitudinal electromagnetic waves.
What is a virtual (off-shell) photon, which carries only spatial momentum, in the field line model?
Suppose that we have two charges attached to a frame so that the charges are static. The Coulomb force between them is tension in the field lines.
Thus, a virtual photon which contains just spatial momentum, no energy, is longitudinal tension of field lines.
^
t / (new frame v)
^
| | | | |
| | | | | crests of the
| | | | | Fourier component
------------------------------> x
In a Feynman diagram, such a photon is a Fourier component of the static Coulomb field. Why is the photon a sine wave electric potential in a Feynman diagram, but the photon is longitudinal tension in the field line model? These descriptions look very different graphically. What is the connection?
What happens if we change to a different inertial frame?
If we do a Lorentz transformation to a different frame which moves up right at the speed v, in the diagram above, then the wave potential in the moving frame no longer is of the form sin(p x). It will be something like
sin(E t - p x).
In the new frame, our virtual photon does carry some energy E, as well as spatial momentum p.
In the field line model, the a moving observer sees the field line diagram Lorentz-contracted in the direction of v. Field lines look distorted in a parallelogram fashion to a moving observer.
The moving observer sees a time-dependent magnetic field along with the electric field. He may interpret that the magnetic field has distorted the electric field lines.
How should we interpret the energy that seems to be flowing in the Feynman style diagram? If a charge pushes another, and the other charge moves to the direction of the force, then the first charge is doing work. This is the energy which flows in the Feynman diagram.
A static system of charges is easiest to study in the static frame. There is a preferred frame. Switching to a moving frame just complicates things.
How are longitudinal forces of field lines (virtual photons) converted to electromagnetic waves (real photons)?
Our rubber plate model tells us that by waving the plate up and down, we can produce transverse waves (real photons).
When we wave the rubber plate, the system is described as longitudinal forces as well as deformation of the rubber plate. How do these become converted to beautiful sine waves which escape to infinity?
It is obviously a complicated process which should be calculated in a computer simulation, or tested with a real rubber plate.
The same probably holds for the field line model. When we wave a charge, the longitudinal tension in the field lines feeds energy to form beautiful sine waves, but the process is quite complicated.
In a Feynman diagram, the production of real photons is described in a miraculously simple way. We just need to make sure that spatial momentum and energy are conserved, and we get a correct numerical prediction!
So, what is the connection between virtual photons and real photons in a Feynman diagram? These photons are quite different things. A Lorentz transformation always keeps virtual photons virtual and real photons real. There is a complicated process which can produce real photons from virtual photons.
Z+ -------------------------
| virtual photon
e- -------------------------
\
~~~~~~~~~~~~~
real photon
However, in a Feynman diagram, we draw the charge directly producing the outgoing real photon. The diagram above depicts bremsstrahlung.
What is the role of creation and annihilation operators of canonical quantization? They are supposed to create and destroy real photons.
Not much. In our example, it is fuzzy where a real photon is actually "created". The converse process would be annihilation.
The use of creation and annihilation operators confuses us, if we want to model a collision in a detailed way. These operators are only suitable for describing the measured end result of the whole process: real photons were created somewhere in the process, and observed.
What are the virtual photons in the field line model, in a dynamical encounter of two charges?
Z+ ●
^
|
|
e-
Suppose that an electron e- flies past a nucleus Z+.
The electron will first accelerate and then decelerate. Its electric field lines will "lag" in the acceleration phase and slow it down, but will "pull on the electron" to speed it up in the deceleration phase.
We may interpret this that the magnetic field of the moving electron first resists the acceleration of the electron, and then resists the deceleration.
In the approach, the electron moves spatial momentum p to its field, and gets some of p back in the receding phase. The electron moves considerable momentum to itself, via its own the field. This is the vertex correction of Feynman diagrams.
In the field line model, in the dynamic case, bending, stretching, and tension of field lines is "virtual photons". There can be large momentum p stored in these deformations of the field lines. The deformations do not obey the energy-momentum relation
E / c = p
of real photons.
Conclusions
"Virtual photons" exist as a phenomenon of the classical electromagnetic field. There is (probably) no quantum aspect at all in them. Quanta are only observed when we measure real photons after a collision experiment.
A photon with momentum q, which is registered in the measurement, means that the classical field has "collapsed" to indicate one Fourier component of it: the component which describes an outgoing photon with the momentum q.
In the Copenhagen interpretation we speak about a collapse of the "wave function". With electromagnetic fields and macroscopic charges, we can take the wave function to be the classical electromagnetic wave. What about microscopic charges?
No comments:
Post a Comment