Heating up the electromagnetic field of the vacuum - the fate of a wave close to an event horizon

Let us model the electromagnetic field of the vacuum with a drum skin. We can produce an ordinary circular wave with our finger by pressing a point of the skin rhythmically.

The radial coordinate wave equation close to the finger is not like the one-dimensional standard wave equation of a string. The circular waves close to the finger will partially reflect back. A solution is a standing wave whose radial profile is a Bessel function and which does synchronized oscillation up and down.

Far from the finger, the radial wave equation is close to the equation for a string, and the waves will spread with little reflection back.

If we throw a stone into water, the center will keep oscillating up and down for quite some time while waves who got far enough from the center will start spreading somewhat like plane waves.

If the medium is not uniform, then we can describe the propagation of a plane wave as the main wave plus perturbation sources caused by the nonuniformity of the medium. The plane wave may gradually decay into a various more or less random disturbances of the field.

Does the space around an event horizon heat up?

How does a wave which is approaching an event horizon behave?

The medium close to the event horizon is not uniform, that is, it is not equivalent to the flat Minkowski space. For instance, a static observer near the event horizon will feel a strong gravitational pull, which would not happen to an inertial observer in the Minkowski space. How does the pull affect light propagation?

If we believe in the equivalence principle, we may try studying the light wave in a freely falling laboratory. There are tidal forces in the lab, but at least we get rid of the strong gravitational pull.

We may assume that a static observer drops the lab, so that the lab can for a while study a light wave propagating downward.

The frequency of the light will drop because of the Doppler effect. That is, the input to the laboratory is a down chirp.

A wave of a form

       ψ = exp(-i ω(t, x) (t - x))

does not usually satisfy the standard wave equation

       d^2/dx^2 ψ = d^2/dt^2 ψ,

if ω changes with time t and position x.

A simple formula for a chirp would be

       ω = ω_0 + ε (t - x).

Maybe the down chirp cannot be transmitted through the laboratory without producing a perturbation source?

Since a wave will cycle through an infinite number of wavelengths before reaching the event horizon, it must travel through an infinite number of laboratories. If a perturbation source is produced in each laboratory, then the wave may decay entirely into perturbations before it reaches the horizon. If this is the case, then it is hard for anything to fall through the horizon because diffusion of heat is slow compared to the propagation of a plane wave.

A wave in an infinitely long swimming pool and fishing floats

Suppose that we have an infinitely long pool of water of some fixed width. We can make a wave which travels along the pool without any reflection or disturbance.

Suppose then that we add a fishing float at every 25 meters of the pool. The float will produce a minuscule perturbation wave each time the crest of our water wave passes it. Since the pool is infinitely long, eventually the floats will absorb all the energy of the wave we made, and convert it to more or less random waves zigzagging the pool.

This example shows that a minor modification to a wave equation may cause all energy to be reflected.

If a light wave which approaches an event horizon does not obey the standard wave equation exactly under some choices of coordinates, then most probably all of its energy will eventually be reflected.

We need to write the wave equation close to the horizon in a way which contains the standard wave equation plus a small source term (= perturbation). Then we can study what is the impact of the source term.

In the Schwarzschild coordinates, where x is a short distance from the horizon, the speed of light is proportional to x.

This corresponds to a wave equation of the form

   x^2 * d^2/dx^2 ψ = d^2/dt^2 ψ.

An ingoing gaussian pulse

http://laplace.physics.ubc.ca/People/gdreid/resources/summer_2013/scalar_wave/project1WaveEquation.pdf

Graham Reid calculated in the Schwarzschild metric the fate of an ingoing spherically symmetric gaussian pulse for various widths of the pulse. The result is that the black hole absorbs between 0% and 100% of the pulse, less for narrow pulses.

Our conjecture was a 0% absorption. We need to check the assumptions in the calculation.

The merger of two black holes

If we cannot get any signal from behind the event horizon, how can two equal-mass nonrotating black holes merge into one spherical black hole? How do we know that the masses were combined into a spherically symmetric system?

           background stars

      ● --->                  <--- ●

    -----------------O--------------> x

                  observer

Let us imagine a head-on collision of two identical nonrotating black holes. They move towards each other on the x axis and will meet at the origin. Our observer is some distance away on the negative y axis.

Both black holes act as powerful gravitational lenses. Long before the collision, they start to distort each other's images. They start to form the Einstein ring in the lens and change shape to a half moon plus an extra image on the other side of the lens.

 

    (|  |)

The main images stretch vertically and are distorted into a crescent shape.

When the black holes are close enough, maybe the half moon images of the black holes combine into a full moon? We would need a computer simulation to calculate the image.

The observer will see many images of each object which fell into the black holes.

The spacetime geometry is determined by the path of light rays. If something looks spherical, then it has distorted the geometry like a spherical object.

Let us ignore the fact that the light coming from an event horizon dies off at a very quick rate.

Because of a mirror symmetry, when looking to the direction of the y axis, the observer will always see light which originated from the background stars. Geodesic lines dictate that at one direction, he can only see light from one object, and because of the symmetry, it cannot be an object which fell into either of the black holes.

Our observer will see the half moons of the black holes glued together but split in the middle where he will see background light arriving.

An object falling into a spherical black hole

Suppose that we have a large black hole at the origin of the x axis. Our observer throws a 1 kilogram weight straight towards it.

We assume that the black hole was built from a collapse of a shell of 1 kilogram weights.

Our observer sees the images of those very many weights floating at the event horizon.

When the extra 1 kg weight moves towards the event horizon, it acts as a gravitational lens. It magnifies the image of the entire black hole slightly. Close to the position where it will hit the horizon, it magnifies more, kind of "making room" for itself to land.

Does the weight leave a bump in the gravitational field of the black hole? Or does the optical magnification even out the apparent mass distribution on the event horizon, so that the gravitational field stays spherically symmetric?

We may imagine a spherical metal ball holding a negative charge, and adding an extra electron to it. The electron will make room for itself and even out the charge density on the surface. Does the same happen if we drop the 1 kg weight to the black hole?

The Penrose diagram for a collapsing star

Is the metric smooth at the point where the surface of the star descends below the event horizon?

Assume that we have two rays of light L1 and L2 passing through the star. L1 just manages to exit the Schwarzschild radius area before the event horizon forms.

L2 nearly misses that and starts descending towards the central singularity.

We assume that the Penrose diagram uses Kruskal-Szekeres coordinates outside the surface of the star. The metric is the standard Schwarzschild metric outside the star.

In the diagram, L1 continues at a 45 degree angle up right and also L2 continues at a 45 degree angle up right. The metric in the diagram is smooth even at the point where the event horizon is born.

Gluing together the near-Minkowski metric inside the star and the Schwarzschild metric outside

https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

In the Kruskal-Szekeres coordinates it is confusing that inside the event horizon, the coordinate t of the global Schwarzschild coordinates moves backwards as our falling object descends towards the singularity. We may have an event at (t, r) = (1, 0.5 r_s) where an object is falling into the singularity. What does that mean? If the star had not yet started collapsing when the clock of a distant observer showed time 1, it sounds nonsensical that "at the same time" an object is falling towards the singularity. The resolution to the confusion is that t inside the event horizon is a coordinate variable which is not connected to the time which the distant observer sees from his clock.

           T

           ^

            |

            |

            |          Schwarzschild metric

          t_M      in the Kruskal-Szekeres diagram

            ^                   

            |    \               / event horizon

            |       \          /

            |          \     /

            |            \ /

            |              \

            |                \   star surface

            |                  \

            |                   \

            -------> r_M ---------------------------> X

            The Minkowski coordinates with

            the Minkowski metric are inside the star

We need coordinates to specify events in our spacetime manifold. One should not assume anything about the simultaneity of events, based on coordinate values alone. Also, the proper time of an observer may flow to a direction which is opposite to the coordinate t.

If we are collapsing a thin shell of dust, then the metric inside the shell is Minkowski. We must glue together the Minkowski metric and the Schwarzschild metric at the shell surface.

In the Kruskal-Szekeres diagram, null geodesics, that is, the paths of rays of light, run at a 45 degree angle up, leaning either left or right. We must glue together the metrics in a smooth way. That means that in the diagram, null geodesics must enter or leave the Minkowski metric area at 45 degree angles. It seems easy to draw Minkowski coordinates t_M and r_M in such a way that this is true.

In the diagram, the Minkowski coordinates inside the star are independent from the Schwarzschild or Kruskal-Szekeres coordinates outside the star - but the metrics must agree at the star surface. We may have that in the Minkowski coordinates, at (1, 0.5 r_s) the star had not yet started collapsing, while at the Schwarzschild coordinates (1, 0.5 r_s) the surface of the star is already below that point.

Does an outside observer get "signals" about the mass movements inside the event horizon?

Suppose that we collapse a dust shell which is not spherically symmetric, but which anyway results in a non-rotating black hole. Is the gravitational field of the end result spherically symmetric?

The no-hair hypothesis of black holes claims that the gravitational field will very rapidly settle to a spherically symmetric state. Does that mean that we are getting signals from mass movements inside the event horizon?

In a spherically symmetric collapse, in the Kruskal-Szekeres diagram, an outside observer at a late Schwarzschild time t will be located very close to the event horizon line which runs at a 45 degree angle up right. In the past light cone of the observer, the surface of the star is very close to the event horizon. We may interpret that the gravitational field which the observer feels, comes from the mass which he sees falling towards the horizon, or essentially floating at the horizon.

The outside observer does not see anything which resides inside the event horizon.

What the observer will see at a late time depends on the last light rays which could reach him from the falling mass. Suppose that the black hole was born from a perfectly symmetric dust shell. What if we drop a 1 kilogram weight on one side of the event horizon? Does the outside observer see a permanent deformation of the gravitational field at that point?

Are the black hole singularity models of Penrose and Hawking correct?

In the 1960s papers and in the 1970 paper Roger Penrose and Stephen Hawking present proofs that singularities will inevitably form in a star collapse into a black hole.

That is at odds with our previous blog post where a collapse will slow down in the global Schwarzschild coordinates, and a "frozen star" will be the end result.

The mathematical question is if there exists a manifold with a reasonable metric which, furthermore, fulfills the Einstein equation.

If the star would continue collapsing after an infinite time of the global Schwarzschild time has passed, is it really possible to extend the spacetime manifold in such a way? Does the manifold after that just contain the internals of the star, or could it also contain the exterior space? What would it mean that the exterior, asymptotically a Minkowski space, would exist after an infinite time has passed?

This reminds us of nonstandard models of arithmetic, where an outside observer sees the ordering of natural numbers as N + Z + Z ..., where N denotes the usual ordering and Z denotes the ordering of negative and positive integers. For an outside observer, natural numbers continue after an infinite number of them have been counted.

If the manifold after an infinite global time just contains the internals of the star, is that a reasonable solution? Why did the internals of the star survive an infinite global time but the surrounding Minkowski space did not?

An astronaut falling into the forming black hole will pass the event horizon at the infinite global Schwarzschild time. What if, at the same time, a ray of light was moving outward from inside the star, and the astronaut meets that ray of light at the infinite time? What is the path of the light after that? It cannot go to the non-existent surrounding space. Does it turn back? But would the sharp turn violate the principle that the metric has to be smooth?

There is no principle in general relativity which forces the proper time of a particle to move forward infinitely long. The proper time of a photon does not progress at all. Why should we require that the proper time of a falling astronaut should extend past the moment where he meets the event horizon? Penrose and Hawking should present grounds for extending the proper time of the astronaut.

A collapsing star and the Kruskal-Szekeres coordinates

https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

Let us use the Kruskal-Szekeres coordinates which have been extended to include the interior of an event horizon. The coordinates come with the Schwarzschild metric outside the star. Inside the star, we have a metric which is between a flat Minkowski space and the Schwarzschild metric.

Let us assume that a spherically symmetric star collapses. Let us assume that the last layer of atoms passes the Schwarzschild radius at an event (spacetime point) x and no interior layer of atoms has yet passed its respective Schwarzschild radius. That is, the event horizon forms at x.

Suppose that a ray of light is rising upwards from the star and comes to x. Since the Schwarzschild metric inside the event horizon dictates that every ray of light goes towards the center, our ray must turn abruptly and start descending back into the collapsing star.

Does that mean the metric is not smooth at x? In a smooth metric, all rays of light should go along smooth paths and there should be no abrupt turns.

Is it possible to glue together a Schwarzscild metric and the interior metric without introducing singularities?

Folklore says that, using the coordinates of a freely falling observer, the spacetime is smooth at the event horizon. Suppose that x happens at the center of a falling spaceship. People in the spaceship will not observe any sharp turn of the ray of light. Rather, they will think the ray is still going upward after x.

An astronaut falling into a black hole never reaches the event horizon - there is no information paradox or a firewall?

It is a popular belief that an astronaut falling towards a black hole will pass through the event horizon at the speed of light. This view has been advocated in many popular science books as well as in scientific papers.

https://en.wikipedia.org/wiki/Black_hole_complementarity

https://arxiv.org/abs/1207.3123#

The firewall paradox of black holes talks about an observer falling through the event horizon and still performing observations after passing the horizon.

https://www.sciencemag.org/news/2007/06/no-more-black-holes

But in the Schwarzschild solution it is easy to calculate that in the global Schwarzschild coordinate time it takes an infinite time for the astronaut to reach the event horizon. Lawrence Krauss in the above link talks about this.

Do we have any reason to assume that the fall of the astronaut continues longer than the infinity of the global Schwarzschild time?

We do not see any reason for such an assumption. The astronaut will "freeze" just above horizon. His proper time will not continue past the finite proper time it takes for him to reach the horizon.

https://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

The popular belief that the astronaut would be able to pass the horizon is based on the Gullstrand-Painleve coordinates where the flow of time is the proper time of the astronaut. Does it make sense to extend the proper time of the infalling observer past the infinite time of the global Schwarzschild coordinates? Suppose that the universe will end by some mechanism before an infinite time has passed. Then the astronaut would be wiped out before he reaches the event horizon.

Gullstrand and Painleve introduced their coordinates to argue that the Einstein theory is nonsensical because the "singularity" of the Schwarzschild coordinates at the event horizon can be removed with their coordinates. But their coordinate system does not make sense if the age of any part of the universe cannot be "greater than" infinity.

Note that since no true event horizon ever forms, there is no genuine singularity in the Schwarzschild coordinates at the event horizon.

Later, the Gullstrand-Painleve coordinates were used in popular science books to introduce the model where an observer can fall past the event horizon.

There is no firewall paradox or information loss paradox

Since nothing will ever go past the event horizon, there cannot be any firewall paradox. There is no duplication of quantum information at the horizon.

The information loss paradox of black holes does not exist either. The paradox was how the information devoured by a black hole can be returned in seemingly random Hawking radiation. In this blog we have argued that Hawking radiation probably does not exist because it would break basic principles of quantum physics. Now we have another argument which removes the information loss paradox: the black hole never "devours" anything. The information is preserved in the infalling matter which never reaches the horizon.

Several authors (e.g., Lawrence Krauss in the sciencemag.org link above) have previously noted that the hypothetical Hawking radiation would make the black hole to evaporate before anything can reach the event horizon.

There is no singularity inside the event horizon

If we let a dust sphere collapse on a star, and there is enough mass that the star and the sphere would form a black hole, then the proper time inside the forming horizon will essentially stay still once the dust sphere is close to the horizon. No singularity of infinite density will ever form. The star inside the horizon is frozen at its place.

A singularity in Einstein's equations would be ugly because the equations are not defined at such a point. Fortunately, no singularity ever forms.

Did Einstein write anything about this? He tried to show that black holes cannot form in astrophysics, but did he comment on the Gullstrand-Painleve coordinates?

An ever-slowing computer

Imagine a computer whose clock speed is decreasing in a way that it will only complete a finite number N of cycles during the infinite age of the universe. We may define the proper time of the computer to be the number of the completed cycles.

Does it make sense to build a physical model of the computer past the infinite age of the universe? In the model, the computer would keep crunching numbers at cycles N + 1, N + 2, and so on. The model does look nice. But the model does not make sense physically, if we assume that there is no universe "after" an infinite time has passed.

Why would we then use the Gullstrand-Painleve coordinates to extend time past the infinite time of the global Schwarzschild coordinates?

What happens to infalling matter according to optical gravity?

In this blog we have developed a new interpretation of general relativity where we assume that spacetime itself is a flat Minkowski space, but the strength of forces and the inertial mass of energy change according to the location in (Schwarzschild) coordinates.

Specifically, the speed of light slows down close to the event horizon and would be zero at the horizon if the horizon would form.

A zero speed of light in optics means an infinite optical density. All light would in this analogy model be reflected from the horizon.

But what happens to particles with a non-zero rest mass? They do not have enough energy to escape. Maybe those particles will keep bouncing off from the horizon, at the same time losing their kinetic energy to radiation.

If the local conditions close to the horizon have a great density of hot particles, then some unknown mechanism might be able to convert all particles to massless ones, and eventually all mass-energy would be reflected from the event horizon. The forming black hole would evaporate through a mechanism which is different from the hypothetical Hawking radiation. Information would be preserved. There is no information loss paradox in this scheme.

Kerr black holes

https://en.wikipedia.org/wiki/Kerr_metric

A rotating mass solution of the Einstein equations is much more complicated than the Schwarzschild solution.

Wikipedia mentions that the stability of the "internal" Kerr solution is not known. But if proper time is almost frozen inside some horizon of the Kerr solution, maybe there is no problem in this?

Last year in this blog we raised the question how a black hole merger happens if the proper time is almost frozen inside the two event horizons. How does the mass distribution settle down in a merger? Does the mass distribution preserve a cigar-like shape even after the merger has happened?

Implications for quantum gravity

The Gullstrand-Painleve coordinates show that,  classically, there is no "drama" at the event horizon. The astronaut would not meet any high-energy phenomena as he approaches the event horizon. If we believe this classical scheme, then we can model everything with our current combination general relativity and quantum mechanics. There is no need for quantum gravity which would conceptually unify the two theories.

But the wave nature of matter suggests that the curved geometry close to the event horizon will make particles reflect at essentially the speed of light back from the horizon area. The astronaut will encounter particles of extremely high energies.

Also, the reflection process of quantum mechanical waves from curved spacetime requires us to develop a combined theory of gravitation and quantum mechanics, at least a semi-classical one.

Our blog post in the last fall showed that the reflection process of light waves from material of variable optical density (e.g., glass) is poorly understood, even though we can make empirical tests in the laboratory. In the case of curved spacetime, we only have empirical observations from the bending of light in the gravitational field of galaxies and stars. Reflection in these astronomical setups is negligible. In the case of black holes, reflection might produce effects that could be visible in the accretion disk. We need to develop models to estimate matter and light behavior close to the event horizon.

https://arxiv.org/abs/1612.00266

Niayesh Afshordi and others claim to have observed gravitational wave reflection from the the event horizon. Further events from the LIGO will reveal if the correlation they observed was real.

We believe that curved spacetime at the event horizon will cause reflection of gravitational waves as well as any other waves, but we do not have a mathematical model yet to calculate what the reflections should look like.

The Dirac equation follows from "curvature" of space under a magnetic field?

Let us start from the energy-momentum relation of special relativity:

       E^2 = p^2 + m^2.

The usual interpretations of the corresponding operators are from the Schrödinger equation:

       E = i d/dt
       p = -i ∇
       p_x = -i d/dx
       p_y = -i d/dy
       p_z = -i d/dz,

where ∇ is the gradient, that is, the vector (d/dx, d/dy, d/dz).

The wave operator in the Klein-Gordon equation

We obtain the Klein-Gordon equation by interpreting the symbols in the energy-momentum relation as operators:

       (-d^2/dt^2 + ∇^2) ψ = m^2 ψ

Let us call -d^2/dt^2 + ∇^2 the wave operator of the Klein-Gordon equation. The solutions of the Klein-Gordon equation are then eigenfunctions of the wave operator and the eigenvalues are squares of the rest mass m of the particle.

The wave operator determines the "behavior" of the wave ψ in spacetime. It ties the partial derivatives to x, y, z directions to the partial derivative in the t direction.

In the Schrödinger equation, the wave operator can be split to the kinetic energy term p^2/(2m) and the time evolution term i d/dt.

Orthogonal commuting momentum operators

For a free particle, the momentum operators E, p_x, p_y, p_z commute, and are orthogonal in the sense that they share a common eigenbase, and each operator has an eigenfunction whose eigenvalue is non-zero, but whose eigenvalues for the other operators are zero.

The common eigenbase of these operators is the plane waves

       exp(-i (Et - p ⋅ r)).

Postulate 1. If we are working in empty space with no external fields, we postulate that the correct way to describe a quantum mechanical particle is to make the wave operator a sum of (+, -) squares of momentum operators, such that E^2 has a different sign from p_x^2 and other spatial operators.

The Klein-Gordon equation follows from this recipe. The intuitive reason for forming such a wave operator is that empty space is a flat Minkowski space, and the operators p_x, p_y, p_z are symmetric and orthogonal. It is a good guess that the wave operator should look much like the Pythagorean theorem.

A free electron or a positron is described by the Klein-Gordon equation. In the plane wave, a positive energy

       E = sqrt(p^2 + m^2)

describes an electron, and a negative energy

       E = -sqrt(p^2 + m^2)

describes a positron.

Non-commuting, non-orthogonal momentum operators

Let us add an external magnetic field with a vector potential A to our spacetime. The minimal coupling replaces p with p - qA, where q is the charge of the particle.

But

       p_x - qA_x, p_y - qA_y, p_z - qA_z

do not commute if A is non-zero. Since they do not commute, they are not orthogonal either. What should the wave operator look like then?

There should probably be some "cross terms" of partial derivatives.

In the case of a free particle, the wave operator was a sum of (+-) squares of momentum operators.

Postulate 2. We postulate that the wave operator should be a square of another operator, and if the magnetic field is zero, the wave operator should be "reduced", in some sense, to the wave operator of the free particle.


https://en.wikipedia.org/wiki/Dirac_operator

Let us denote by KG the Klein-Gordon wave operator. The corresponding Dirac operator D is such that

       KG I = D^2,

where I is the identity matrix for some dimension n. We have to let n be at least 4, to find a solution for D.

https://en.wikipedia.org/wiki/Dirac_equation

The solutions for D are obtained through the well-known trick of Paul Dirac, where we add "degrees of freedom" to the equation by making the wave function 4-place, and take a "square root" of the Klein-Gordon operator by using the gamma matrices as coefficients.

The standard Dirac equation is

       D ψ = m ψ,

where ψ is a 4-place complex-valued wave function. If D satisfies KG I = D^2, then also -D satisfies it. We get the second Dirac equation

       D ψ = -m ψ.

For now, we do not add an electric potential to the equation. The 4-place vector is called a Dirac spinor.

http://physics.gu.se/~tfkhj/TOPO/DiracEquation.pdf

Let the particle have a momentum along the z axis. The eigenfunctions of the standard Dirac equation are schematically of the form

     / electron spin_z up    \
     | electron spin_z down |
     | positron spin_z down |  exp(-i (Et - p ⋅ r)).
     \ positron spin_z up     /

The eigenvalues are 4-component spinors. For the positron solutions, E is negative.

If p is zero, then the solution for the spin up electron is

       /1 \
      | 0  |
      | 0  |   exp(-i Et).                                            (1)
      \ 0 /

For the second Dirac equation, we may define that the positron with the spin down is the same formula (1) as above, and E is negative.

Question 3. The second Dirac equation above produces a second set of solutions for an electron spin up, and so on. Why is there such a degeneracy of the solutions? Is it certain that the second solutions represent the same particles as the first solutions?


If we think of the positron as an electron traveling "backward in time", and the electron may scatter so that it turns back in time, then it is natural to represent both the electron and the positron with the formula (1) above, so that we do not need to switch the spinor

       (1, 0, 0, 0)

suddenly to the spinor

       (0, 0, 0, 1).

The positron is then a solution of the second Dirac equation.

If the magnetic field is zero (A = 0), then the Dirac equation does, in a sense, reduce to the Klein-Gordon equation. The exponent part of a Dirac eigenfunction is a plane wave and is an eigenfunction of the Klein-Gordon equation. We can discard the spinor part because for a free particle there is no way to interact with the spin of the particle. With no interaction, the spinor stays the same at all times and we can ignore that degree of freedom.

The electron is fundamentally a wave phenomenon and cannot be explained as a classical point particle or a billiard ball

We wrote in our previous blog post that our four months of hard thinking did not produce a model where the electron would be described, in some sense, as a classical point particle or a classical billiard ball, and the wave associated with the electron would be derived from a path integral on some "lagrangian density".

The electron does certainly exhibit some features of a classical billiard ball or a point particle. We can measure its location with a great precision. The electron carries angular momentum in its spin, like a rotating ball.

We hypothesize that these classical features emerge from the electron wave, and the wave is the fundamental "structure" of the electron. Saying that the electron is a "point particle" is wrong, because that conveys the impression that it is a classical particle. It is better to say that the electron is a wave.

Question 4. Why does the electron wave exhibit classical features, like collapsing to a point when its location is measured, or carrying angular momentum?

Question 5. The electron cannot be described through a path integral. Does this have any bearing on Feynman diagrams, where we do calculate path integrals?

The electron does not have an intuitive physical model at all?

For the past four months we have been trying to find an "intuitive" physical model for the Pauli and Dirac equations, but we have failed. The biggest obstacle is the gyromagnetic ratio g = 2. Any classical point charge which moves in a circular orbit will have the gyromagnetic ratio g = 1.

What does an "intuitive" physical model mean? We have concentrated on studying models which contain one or more point particles. The wave of a classical point particle would be determined using a path integral approach: the lagrangian density over a path would determine the phase of the wave.

The electron as a classical point particle

Maybe the hypothesis that the electron is, in some way, a classical point particle is wrong? We can certainly measure the position of an electron with a very great precision. There are claims on the Internet that experiments set the limit of the electron radius at most to 10^-22 m.

In an earlier blog post we calculated that a point-like electron must move at the speed of light in a circle of radius 2 * 10^-13 m, to explain its spin angular momentum.

The electron in some aspects looks point-like, but its spin and magnetic moment suggest that it should quite a large sphere, or a point particle moving around quite a large circle, to explain its spin and magnetic moment.

A classical point charge involves the paradox of the infinite energy of its electric field. The electric field of the electron contains its mass 511 keV of energy outside the classical radius of the electron 3 * 10^-15 m.

Maybe it is best to reject the idea that the electron is a classical point particle, in any sense.

What is the correct kinetic energy term in the Pauli equation?

The kinetic energy operator in the Pauli equation

     (1/ (2m) (σ ∙ (p - qA))^2, + qϕ) ψ
     = (i h-bar ∂ / ∂t) ψ

is

       (σ ∙ (p - qA))^2,

where σ is the vector of the three Pauli matrices, and the exponent 2 means that we take the product of 2 × 2 matrices, that is, the composite mapping.

One may ask why the operator should not be

       (p - qA)^2,

where the exponent 2 means taking the inner product of two vectors? If the vector potential A is zero, then the kinetic energy operator does, indeed, take a simpler form, the Laplace operator,

       p^2,

where the exponent 2 means taking the inner product of the vector p with itself.

An inner product p^2 in a 3-dimensional space is simple in the sense that "cross terms" of p_x, p_y of the vector p components in the x, y, z directions do not affect the value of the product. It is just the sum

       p_x^2 + p_y^2 + p_z^2

which matters.

However, when A is not zero, the magnetic field determines a preferred coordinate system in space, and it is not at all obvious that the simple inner product form of the kinetic term is the right one then.

In classical electrodynamics with the electron a point particle, in the Hamiltonian formulation, the kinetic term is an inner product also when A is not zero. But in wave mechanics, it might not be the right kinetic term.

The Cauchy stress tensor in classical mechanics is a full 3 × 3 matrix. Elastic energy can be stored in the normal stresses of a solid object, but shear stresses also contribute to elastic energy. That is, "cross terms" in that case are important in the determination of energy of the system.

We conclude that maybe the right kinetic operator in the Pauli equation should also involve cross terms, in one way or another.

If Nature has decided that the kinetic term still should have the form of being a "square" of something, then the "factorization" of p^2 with the Pauli matrices is a good candidate as the kinetic term.

Adding cross terms takes us away from a classic point particle model.

http://hitoshi.berkeley.edu/221a/landau.pdf

When A is not zero, then operators p_x - qA_x and p_y - qA_y, in general, do not commute. This might be an intuitive reason why the inner product of vectors no longer is the correct kinetic energy operator. In free space, with A zero, operators p_x, p_y, p_z are orthogonal in the sense that they have a common eigenbase, and there is an eigenfunction for which p_x is zero and p_y non-zero, and so on. Maybe it only makes sense to calculate the kinetic energy as a simple inner product if the operators involved are orthogonal? The metric in the underlying space is flat in that case?

If the electron has no point particle model, there are problems for hidden variable interpretations

The de Broglie-Bohm interpretation assumes that particles at all times have a precise position and that they move according to what the "pilot wave" tells them to do. The position of the particle is a hidden variable of the system.

But if the electron does not have an intuitive model where it is a point particle with a precise position, how can we make the de Broglie-Bohm interpretation to work? What are the hidden variables then?

https://en.wikipedia.org/wiki/De_Broglie–Bohm_theory

The Bohm model in the non-relativistic case is simply that we imagine the point particle to be a water molecule which travels along the probability current of the wave function.

According to Wikipedia, the relativistic case is problematic because then in spacetime there are no fixed planes of equal time. What does it then mean that the probability is conserved?

Hrvoje Nicolic and others have developed Bohmian models that handle the Dirac equation.

We may study probability conservation in a future blog post. Probability conservation has further problems in curved spacetime.

What is the correct kinetic energy term in the Dirac equation?

The Pauli equation comes from a non-relativistic energy momentum relation:

       E = p^2 / (2m) + m.

The energy E can be divided in a simple way into the energy of the rest mass m, and the kinetic energy term p^2 / (2m).

The relativistic energy-momentum relation

       E^2 = p^2 + m^2

does not admit such a simple formula for the kinetic energy. The formula for E would involve a square root, which is awkward.

The quantum mechanical operator for E involves a partial derivative on time, and p involves partial derivatives on spatial coordinates. A general principle of special relativity is that time and space should be treated in a unified way. That suggest that the temporal and spatial operators should be combined somehow:

       E^2 - p^2 = m^2.

https://en.wikipedia.org/wiki/Laplace_operator

The operator on the left side of the equation is a d'Alembert operator or a wave operator. In the case of the Pauli equation, we can split the wave operator into the kinetic energy part (the Laplace operator) and the partial derivative on time. In the relativistic case, such a split may not be possible.