September 18, 2026: Classical vertex correction in QED determines the Planck constant? – The physics blog of Heikki Tuuri.
Monday, November 9, 2020
An improved perpetuum mobile for general relativity
The speed of sound inside a neutron star CAN exceed the vacuum speed of light
https://www.nature.com/articles/s41567-020-0914-9
Evidence for quark-matter cores in massive neutron stars by Eemeli Annala, Aleksi Vuorinen et al. in Nature Physics June 1, 2020 studies the mass distribution of a neutron star, assuming an arbitrary function f:
pressure = f (energy density of matter).
The authors mention that most hadronic models predict that the speed of sound squared, c_s^2, is equal to 0.5 or larger for high densities (we use natural units where c = 1).
Is it possible for the speed of sound to exceed the vacuum speed of light?
The answer is definitely yes, if we define the speed of sound as the phase speed of a well-formed periodic pressure wave (a "sine" wave). That is, as the speed of the crest of wave. A crest of an infinite sine wave does not transmit any information from a point A to a point B. The speed of the crest is not constrained by the universal signal speed limit, that is, the vacuum speed of light c.
https://motls.blogspot.com/2020/10/a-fun-calculation-of-maximum-speed-of.html
https://arxiv.org/abs/gr-qc/0703121
Lubos Motl in his blog post (2020), as well as George Ellis et al. in their arxiv paper (2007) claim that the "causal limit" for the speed of sound is the speed of light, c. But they fail to define what they exactly mean by the speed of sound.
When considering the stiffness properties of matter, the natural definition for the speed of sound is the phase speed, not the signal speed. The phase speed depends on the stiffness, and can exceed the speed of light.
string
wall |--------------------------| wall
As a practical example, consider a tense string which is attached to walls at its endpoints. If we pluck the string, we can create a standing wave into it. There is no speed limit for the phase speeds of the two sine wave components of the standing wave. The standing wave does not transport any information and is not constrained by the speed of light.
Monday, December 9, 2019
Why do we only sum probability amplitudes of Feynman diagrams, never subtract?
https://en.wikipedia.org/wiki/Møller_scattering
^ ^
\ /
\ /
|~~|
/ \
/ \
e- e-
If the energy of the collision is less than 1.022 MeV, then no new electron-positron pair can be produced.
If the energy is larger, then new pairs will be created.
Intuitively, the production of a pair should reduce the probability amplitude of an elastic collision where the electrons exchange a large amount p of 4-momentum.
But the possibility of pair production is not explicitly apparent in the Feynman formula which sums the t- and u-channels of elastic scattering.
Could it be that the Feynman formula for elastic scattering in some implicit way "knows" about the possibility of pair production? That is probably not true. We may imagine new physics where a lighter variant of electron exists and can produce a large number of new pairs. How could the Feynman formula for ordinary electrons be aware of such new physics?
Furthermore, the electron-electron collision may produce a photon. That is, the collision is not elastic. How could the elastic collision formula be aware of the (complex) process of photon radiation in the collision?
The simplest pair production diagram in Møller scattering is the following:
e- e+ e- e+
^ ^ ^ ^
\ \ / /
\ \ / /
|~~\____/~~|
/ \
/ \
e- e-
The diagram contains four photon-electron vertices:
|~~
while the elastic scattering only has two such vertices. If the integral formula for the pair production has a much smaller probability amplitude (or cross section) than the formula for elastic scattering, then in the first approximation we can ignore the effect which pair production has on the elastic probability amplitude.
We have not yet found the pair production cross section from literature.
A semiclassical treatment of pair production
Monday, December 2, 2019
Why destructive interference does not cancel high 4-momentum in the vacuum polarization loop?
In the wave interpretation, p is associated with a wavelength λ = 1 / |p| in a spacetime diagram.
time
^ wavelength λ
| \ \
| \ \ -----> 4-momentum p
| \ \
------------------------------> space
We can develop the wave forward in time and space in the diagram through the Huygens principle: let each point act as a point source of new waves, and calculate the interference of the new waves at a new point.
Let
D(ψ) = 0
be a wave equation. If the right side is strictly zero at all spacetime points, we say that it is a homogeneous equation.
If the right side is not zero everywhere, then we call the non-zero part a source.
The Green's function method calculates the response of a wave equation to a Dirac delta source in the equation. That is, we assume that the wave equation
D(ψ) = 0
has on the right side, instead of 0, a Dirac delta term at a certain spacetime point x. The Green's function is the solution of the new equation. We say that it is the response to an impulse source of the wave equation.
If we have a tense string, then pressing the string briefly at a certain point with a finger with a force F, is an impulse F × Δt, and the resulting wave is the response to an impulse source.
If we have a source which is not concentrated to one spacetime point, but is continuous, we can build an approximate solution by summing the response to an impulse source at each each spacetime point x.
Suppose that the source is cyclic and has a certain wavelength λ in the spacetime diagram.
The response to a Dirac delta impulse contains waves for all kinds of 4-momenta p.
It is obvious that there tends to be a destructive interference for all waves where p does not match the cycle (wavelength λ) of the source.
In particular, all high 4-momenta p will have a total destructive interference.
This is the reason why tree-like Feynman diagrams have strictly restricted 4-momenta p at every part of the diagram.
However, if we allow an imagined wave, as in the previous blog post, to have any 4-momentum p, then the imagined wave introduces an arbitrarily high p, or an arbitrarily short wavelength λ, to the diagram. Destructive interference does not cancel it.
Diverging of the vacuum polarization loop integral
q + p -->
~~~O~~~~~~~~~~~~
q --> <-- p q -->
The vacuum polarization loop carries the photon 4-momentum q, as well as an arbitrary 4-momentum p which circles around the loop.
The impulse response to a Dirac delta impulse at a spacetime point x contains a certain spectrum (= propagator) of various 4-momenta. The intensity depends on the sum q + p. That is, the probability amplitude of the diagram above depends on both q and p.
If we allow any p, then there exists no sensible probability distribution for p. The integral of probability amplitudes over all p diverges, or alternatively, we may say that the integral is not defined.
The causality of a Feynman diagram
Saturday, November 30, 2019
The vacuum polarization loop
What if the waves were classical waves?
https://en.wikipedia.org/wiki/Münchhausen_trilemma
Baron Münchhausen told the story where he pulls himself out of a swamp by his own pigtail.
The diverging of the Feynman integral over a loop
But does that restrict Feynman diagrams too much, so that they would no longer agree with empirical data?
Why does Feynman use a Green's function to describe the electric field of an electron?
Feynman assumes that the distribution of various 4-momenta in the photon is the Green's function for the massless Klein-Gordon equation. Why?
Let us consider the drum skin analogy of the static electric field of the electron. If I press the drum skin with my finger, it creates a depression into the skin. That depression is analogous to the static electric field of a particle.
We may imagine that instead of pressing with a constant force F, I keep tapping the skin with my finger at a very rapid pace.
The tapping creates a depression. A single tap is equivalent to applying an "impulse source" to the wave equation of the drum skin. The Green's function for the skin wave equation, by definition, is the response of the skin to that impulse.
That is, we may imagine that the static electric field of a particle consists of a very rapid pace of Green's functions emanating from the particle. The electric field does not carry energy away. There has to be a total destructive interference for the "on-shell" waves in the decomposition of the Green's function.
On the other hand, waves carrying just linear momentum p, can progress. Those waves apparently are responsible for the static depression in the drum skin or the static electric field of a particle.
The decomposition for the various p obeys the decomposition of the Green's function.
If there is a planar wave describing another electron nearby, the photon waves for various p disturb the free Dirac equation of that other electron. That is, the equation no longer is equal to zero, but a (small) source term appears.
Each wave p creates a source term. If we perturb the planar wave solution to find a more accurate solution for the source term associated with p, then another wave appears. That wave is interpreted as the wave of an electron which absorbed the photon with a momentum p.
Relationship to the classical scattering from a static Coulomb potential
Sunday, November 3, 2019
If a photon is an orbiting virtual electron-positron pair, does that explain Compton scattering?
The history of the Klein-Nishina formula
https://arxiv.org/abs/1501.06838
Waller and Tamm (1930), and in unpublished notes, Ettore Majorana, modified the Klein-Nishina semiclassical approach to a quantum field theoretical framework. It turned out that the electron goes through intermediate states. Thomson scattering is produced by negative-energy, that is, positron, intermediate states.
We need to compare the ideas of Waller, Tamm, Majorana, and Feynman.