Lorentz covariance is a property of a system, not of a single electron?

UPDATE Nov 13, 2018: See our new post on the correct way to measure V in the new energy-momentum relation. That should resolve the mystery of the double inertial mass.

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Our previous blog post showed that treating potential energy as rest mass, or rest mass as potential energy, makes sense. The Schrödinger equation becomes more logical when rest mass and the potential energy are identified. The hamiltonian operator H then includes all energy of the particle, also the energy in the rest mass.

This link contains an open-access .pdf file of Dirac's original 1928 paper:

http://rspa.royalsocietypublishing.org/content/117/778/610

The Dirac equation has been called the "square root" of the Klein-Gordon equation. The 1+1-dimensional Dirac equation which we treated in our blog post about zitterbewegung uses the second component of the wave function to store the current value of p / (E + m).

We know that if |p| is small, then the Schrödinger equation is the correct one.

Make Ψ_2 = dΨ_1/dx

Another approach to remove a second derivative is to introduce a new variable which contains the first spatial derivative of the wave function. We can make two differential equations which are formally first order, though, in reality, second order.

The two equations for the Schrödinger equation would be:

      -i * dΨ_2/dx +  (m + V) Ψ_1 = i dΨ_1/dt
      -i * dΨ_1/dx - 2(m + V) Ψ_2 = 0.

The Dirac two equations:

     -i * dΨ_2/dx + (m + V) Ψ_1 = i dΨ_1/dt
     -i * dΨ_1/dx -  (m + V) Ψ_2 = i dΨ_2/dt.

Let us try to write Schrödinger's two equations in a form which is Lorentz covariant:

(S1) -i * dΨ_2/dx +  (m + V) Ψ_1       
         = i dΨ_1/dt
(S2) -i * dΨ_1/dx + (E - 2(m + V)) Ψ_2
         = i dΨ_2/dt.

These are the same equations as Dirac's, but there is an extra E_kin * Ψ_2 in the lower equation. We denote E_kin = E - (m + V). E is the total energy of the particle.

If V gives a preferred coordinate system, like in the hydrogen atom, can we define that E is the energy in that coordinate system? Then E stays constant in the Lorentz transformation. If E is not constant, it may spoil Lorentz covariance.

If the two equations are symmetric in dx and dt, they maybe are Lorentz covariant, because we know that the similar Dirac equations are.

Lorentz transformation of the Dirac equations

Convention 1. Throughout this article we assume that v is small and that we can ignore the gamma factor of a Lorentz transformation, and also all terms of the form v^2.

Suppose that we have two wave functions Ψ_1 and Ψ_2 and there is no potential. An inertial observer A measures the function values at every spacetime point.

     > tilt

   ^ t

   |

   |

    ------------> x    ^ tilt

    -------------- string 1

    -------------- string 2

The wave functions may, for example, be wave heights in two strings which are stacked on top of each other. That is a very concrete example.

Let us have an observer B who is moving at a constant speed v relative to A.

If v is small, the coordinate system of B has the coordinate axis x' tilted upward the angle v relative to x and the axis t' is tilted right the angle v relative to t.

The Lorentz transformation is:

      γ  = 1 / sqrt(1 - v^2)
      x' = γ (x - vt)
      t'  = γ (t - vx).

If v is small, we may assume γ = 1.

Let us solve the above for t' = 0, x' = 1. We get:
       1 = x - vt
       0 = t - vx
      =>
       x = 1
       t  = v,

when v is small. That is, the axes of B are tilted by the angle v but they are not contracted.

If A has an equation on the directional derivatives of the wave functions along the axes of A, we may ask if the same equations hold for the directional derivatives along the axes of B. If they do, then the equations are Lorentz covariant.

Let us try to prove the Lorentz covariance of the Dirac equations in the case V = 0. Dirac equations according to observer A are schematically:

       -x_2 + P1 = t_1
       -x_1  - P2 = t_2.

B measures the directional derivatives in a different way if his coordinate axes are tilted by the small angle v:

       -x_2 - v t_2 + P1 = t_1 + v x_1
       -x_1 - v t_1  - P2 = t_2 + v x_2.

We should show that the two latter equations hold. We get:

       -x_2 + P1 + v P2 = t_1
       -x_1 -  P2 -  v P1 = t_2.

The equations are different from the first ones. The wave functions in the frame of B have to differ from the ones in the frame of A. This is ok if all observable physical quantities are the same.

The waves of the electron need to be Lorentz transformed along with t and x. This is like the electromagnetic field (E, B) which needs to be transformed when we move to new coordinates. If there were an electric potential in our equations, the potential should be transformed.

In our string example, it would not be ok to transform the waves because the displacement of a string is an observable quantity.

Let us study the Dirac covariance in more detail.

Hsin-Chia Cheng from UC Davis:

http://cheng.physics.ucdavis.edu

has a field theory course online. The link:

http://cheng.physics.ucdavis.edu/teaching/230A-s07/rqm4_rev.pdf

contains a derivation of the transformation S for the wave function. For a boost in the x direction in the 1+3-dimensional case:

       S = I cosh(η / 2) - α_1 sinh(η / 2),

where η = -v for small v. The Lorentz transformation matrix is

       cosh(v)    sinh(v)

       sinh(v)    cosh(v).

We can try our own α matrix for the α_1 in the transformation S:

        0        1

        1        0

Let us try a transformation:

        P1' = P1 - v/2 P2
        P2' = P2 - v/2 P1.

We need to calculate the directional derivatives again. We can ignore terms with v^2 in them. Schematically, we get:

  -x_2 - v/2 t_2 + P1 - v/2 P2  = t_1 + v/2 x_1

  -x_1 - v/2 t_1  - P2 - v/2 P1  = t_2 + v/2 x_2.

An easy manipulation shows that the above is equivalent to the untransformed equations. We showed that, assuming v^2 is very small and γ = 1, the Dirac equation in 1+1 dimensions has a natural-looking Lorentz transformation for the wave function.

We should still show that the transformed function is physically sensible. Let us check that the momentum of the electron in the transformed wave function makes sense.

The standard plane wave of an electron in the laboratory frame of A is

       (1,  p / (E + m)) * exp(-i (E t - p x)).

The frame of B is moving with a speed v. We denote the speed of the electron by v_e in the frame of A. Let us assume that p and v are positive and v is small:

       v_e = p / E
       v_e' = (v_e - v) / (1 - v v_e)
               = (v_e - v) (1 + v v_e)
               = v_e - v (1 - v_e^2)
       v_e'^2 = v_e^2 - v * 2 (v_e - v_e^3)

       E'^2 = v_e'^2 E'^2 + m^2
       E'^2 = m^2 / (1 - v_e'^2)
                = m^2 / (1 - v_e^2
                               + v * 2(v_e - v_e^3))
                = E^2 (1 - v * 2(v_e - v_e^3)
                                    / (1 - v_e^2))
                = E^2 (1 - 2 v v_e)
       E'      = E - vp
       p'      = (E - vp) (p/E - v + v p^2/E^2)
                = p - vE

We showed that energy and momentum transform like time and space.

Let us calculate first what is the standard plane wave of the electron in the frame of B:

      (1, (p - vE) / (E - vp + m))
              * exp(-i (Et' - vpt' - px'+ vEx')).

From

       t'  = t - vx
       x' = x - vt

we get the exponent (we ignore v^2):

      Et - vEx - vpt - px + vpt + vEx
      = Et - px.

The exponent part is just like the wave of A.

The vector part is

       (1,  (p - vE) / (E - vp + m)).

The conserved current

   t
   ^
   |
   |
    --------------> x
    e-        e-8

Suppose that A sees a density of ρ of electrons moving at speed v_e. A sees an electric current of J = ρ v_e.
0
A sees successive electrons at coordinates (0, 0), (0, 1 / ρ). B will see them at coordinates (0, 0) and

       (-v / ρ, 1 / ρ).

B sees the first electron still move for a distance v v_e / ρ before B thinks its time coordinate is zero. B sees after that the first electron at coordinates

       (0, 1 / ρ + v v_e / ρ)

B sees an electron density

       ρ' = ρ / (1 + v v_e) = ρ  - v ρ v_e = ρ - v J.

A sees successive electrons at (0, 0) and (1 / (ρ v_e), 0). B sees them at (0, 0) and

     (1 / (ρ v_e) - v / (ρ v_e)).

It takes the electron time v / (ρ v_e^2) to reach x' = 0. The current which B sees is

      1 / (1 / (ρ v_e) + v / (ρ v_e^2))
      = ρ v_e / (1 + v / v_e)
      = J  - v ρ

The conserved current (ρ, J) of a Dirac wave is

       (Ψ† Ψ,     Ψ† α_1 Ψ).

Note that in our notation, α_0 is the identity matrix. Our matrices have some signs flipped relative to the standard Dirac gamma matrices.

The current for the standard plane wave of observer A is

       (1 + p^2 / (E + m)^2,
        2 p / (E + m))
    = (1 + b^2,  2b),

where we denote b = p / (E + m).

The current seen by B in his own standard plane wave is:

       (1 + (p - vE)^2 / (E + m - vp)^2,
        2 (p - vE) / (E + m - vp)).

The Lorentz transformation of the standard plane wave of A is

       (1 - v / 2 b,   b - v / 2)  *  exp(...).

Its current is

      (1 - 2 v b + b^2,
        2 b - v - v b^2).

The above is also the Lorentz transformation of the current of the standard plane wave of A. The values match, which means that the Dirac wave function transforms in a sensible way.

Can we prove the Lorentz covariance of the Schrödinger equations?

We introduced the two Schrödinger equations which only differ from the Dirac equations by the term E_kin in the lower equation.

(S1)  -i * dΨ_2/dx +  (m + V) Ψ_1      
          = i dΨ_1/dt

(S2)  -i * dΨ_1/dx + (E_kin - m - V) Ψ_2
          = i dΨ_2/dt.

We will study if we can make a Lorentz transformation work with these equations.

Also, we need to study how to Lorentz transform the electric potential V. We conjecture that the rest mass m is electric potential energy, too. How to transform m?

Our two Schrödinger equations have the obvious error that for the standard plane wave solution they claim:

       E = p^2 / (2m) + m
      <=>
       2 mE - m^2 = p^2 + m^2.

That is not the correct energy-momentum relation for large |p|:

       E^2 = p^2 + m^2.

The electromagnetic four-potential (ϕ, A) is the correct Lorentz covariant way to treat a potential:

https://en.m.wikipedia.org/wiki/Electromagnetic_four-potential

If we identify the rest mass with the scalar potential, we get an energy-momentum relation

       E^2 = (p + A)^2 + ϕ^2,

where ϕ includes the rest mass m. The first term in the right side is the energy of the magnetic field of the electron and the second term is the energy of the electric field of the electron.

The Lorentz transformation for the potentials is:

      ϕ = ϕ - v A

      A = A - v ϕ.

In 1+1 dimensions we cannot really treat the rest mass as potential energy because the electric force in one spatial dimension is constant. We need to move to 1+3 dimensions.

Actually, if rest mass is electric potential energy, then it is doubtful that we can describe a single electron in a Lorentz covariant way. Suppose that we have a universe with just one electron and one positron. The rest mass of the electron is potential energy in the electric field of the positron. The positron sets a preferred frame of reference. Lorentz covariance bans preferred frames.

The system the electron & the positron is Lorentz covariant, but individually, the particles are not.

A correct Lorentz transformation must act on all particles in the system.

A big question is that since the electron can be annihilated by any positron, what particles should we consider when we describe an electron? If we have two positrons flying to opposite directions, what is the preferred frame where we should describe the electron?

Energy-momentum relation for an electron-positron pair

We need to understand the Lorentz transformation properly before we can do quantum mechanics. Let us consider an electron-positron pair which is static in the laboratory frame symmetrically on the y axis.

                         ^ x

                         |

                         |

                         |

       ----------------------------> y

             e+                    e+

When the distance between the particles was just 10^-15 m, the total energy of the system was zero. The whole rest mass of the particles can be defined to be potential energy in their mutual electric attraction.

The energy-momentum relation for the static particle is

       E^2 = ϕ^2 = m^2,

where m denotes what we traditionally regard as the rest mass.

Let the electron move to the positive x direction at a slow speed v_e. The magnetic fields cause an additional attraction between the particles.

The energy-momentum relation, obtained through a Lorentz transformation with v_e, for the electron is

       E^2 = (v_e E + v_e ϕ)^2 + ϕ^2
               = 4 (m v_e)^2 + ϕ^2
               = 4 p^2 + ϕ^2.

That is, the inertial mass seems to be twice the rest mass of the electron. That is strange. Could this have something to do with the fact that the magnetic field generated by an electric current tends to keep the current going? It seems to store momentum which it will give back to the current if the current decreases.

Let us think. The potential energy of the system the electron & the positron is 1.022 MeV. We showed in an earlier blog post that the division of the inertial mass on the parts of the system depends on the configuration. In our example, all inertial mass was on the electron.

If the positron is at a potential which makes its total energy zero, then the whole inertial mass of 1.022 MeV is on the electron.

The Lorentz transformation might not be the right tool.

The new positron equations

Dirac's astonishing find "explained" why the electron has a spin, and why the positron exists.

The Dirac equations in our 1+1 case do not have a spin, but they still predict the existence of the positron as the negative energy solution.

Do our new Schrödinger equations have negative energy solutions? The basic positive energy solution of a free particle is

       Ψ = (1,  p / (2m))  *  exp(-i (E t - p x)).

From (S1) and (S2) we get:

       p^2/(2m) + m = E
       p - 2m * p / (2m) = 0.

If we assume that m is negative, then E is negative. That is probably the positron equation.

According to Feynman, a positron is a "negative energy electron going backward in time". But in real life, a positron certainly is a positive energy particle going forward in time. The idea that a positron is actually an electron scattered back in time is beautiful. In annihilation, the wave function of the electron might continue (after a short superluminal trip) as the wave function of the positron. The wave of the electron rotates clockwise with time. Then the wave of the positron should rotate anticlockwise with time.

The relativistic positron equations might be (S1) and (S2) where we have inverted the sign of (m + V).

t           _ _  superluminal section
  ^       /      \
   |     /          \
   |   e-           e+
    -----------------> x

In the diagram, the electron is a wave which rotates clockwise with time and anticlockwise with space:

        Ψ_e- = (1,  p / (2m))  *  exp(-i (E t - p x))

When viewed visually, the spatial wave moves right with increasing time.

It annihilates with the positron whose wave function is

         Ψ_e+ = (1,  p / (2m))  *  exp(-i (-E t - p x)).
       
The wave rotates in the positron anticlockwise with time, and anticlockwise with space. When viewed visually, the spatial wave moves left with increasing time.

We have conjectured in our earlier blog posts that there is a short superluminal section in the path when the potential energy between the particles is < -1.022 MeV. The energy-momentum relation

        E^2 = p^2 + (m + V)^2

would have the sum m + V imaginary if p^2 > E^2. An imaginary mass corresponds to a tachyon, and a tachyon has to move superluminally.

Relativistic corrections in the hydrogen atom in the non-relativistic Schrödinger equation

The energy-momentum relation without a potential is

        E^2 = p^2 + m^2.

What is the kinetic energy if p is smallish?

        E_kin = sqrt(p^2 + m^2) - m

                   = m sqrt(p^2/m^2 + 1) - m

                   = 1/2 * p^2/m - 1/8 * p^4/m^3.

The term -1/8... is the leading relativistic correction. It reduces 1s by 9 * 10^-4 eV.

Correcting the mass for m + V should raise 1s. Is this the real reason for the Darwin term?

A very brief ballpark calculation gives that the correction raises the hydrogen energy levels by 1 meV. How precisely do we know the absolute energy -13.6 eV of the 1s-orbital? The Rydberg constant has a value with 11 decimals, but what actual measurements have shown?

Since s-orbitals have their probability amplitude largest at the proton, their energy is raised by some 10^-4 eV. This is the same order of magnitude as the Darwin term.

All rest mass is potential energy?

UPDATE Nov 15, 2018: There is something wrong with either classical electromagnetism or our proof below. Our post on Nov 15, 2018 studies that problem.


Theorem 1. The potential energy of a particle adds to its inertial mass, in other words, to the rest mass. The exact amount of inertial mass which the potential energy adds to each particle depends on the geometry of the setup.

Note that if there are unknown forces between the electron and the platform, then the inertia of the electron can be arbitrarily high. The proof below gives the minimum possible inertia. See our Nov 15, 2018 post.

Proof. Let us give a more detailed proof where the potential energy of a particle has to be added to its rest mass, if we believe the conservation law of the center of mass.

                   z                      r

                   e-  <--- F

                O/    

                 |\_______________________o m + V

                /\              ---> F

        ====================

        -     -     -    -     -    -    -     -

                  x                       y

                 -V                    +V

Let the electron have an inertial mass m when it is static at position z.

A weightless man living on a not infinite uniform negatively charged plane uses an energy store V to pull an electron from position z closer to the plane at x. The plane must not be infinite because then the potential energy of the electron would be infinite.

We assume that the plane is very massive, so that the round trip of the electron only changes the tilt of the plane a negligible amount.

Since the electron comes closer at x, the plane will be tilted by a small angle, leaning down on the right.

The magnetic force between the electron and the plane is proportional to v^2 where v is the relative speed of the electron and the plane. If the man moves the electron slowly, we can ignore the magnetic force.

The man moves the electron from position x to y. Since the electric field is not perfectly uniform close to the plane (force F), he moves the electron along a slightly curved path where the electron stays at a constant electric potential relative to the plane. He then does not need to use any energy to win horizontal forces on the electron.

The man has to win the inertia of the electron to move it. He pulls on a long weightless rod which is attached to an electrically neutral mass of m + V. We claim that m + V is the inertial mass of the electron at x.

Once the man is at y, he moves the electron to r and stores the released potential energy V in an energy store at y.

Then he moves the electron back to y. This time he uses a rod attached to a weight m to win the inertia of the electron. He moves the electron at a constant electric potential so that he does not need to win horizontal forces.

The end result of the process is that the electron is back at its original horizontal position. Horizontal forces affecting the plane were negligible. Thus, the center of mass of the plane is at the original horizontal position, though the plane did tilt slightly when the electron completed its loop.

The energy V has moved a distance s = y - x to the right.

A weight V + m moved the distance s to the left and then the weight m moved back the same distance.

If the energy stores do not have any interaction with the plane, then we can treat them as separate objects that have an inertial mass V. The center of mass of the plane stayed still. The center of mass of energy V moved the distance s to the right and it was balanced by a weight V moving s to the left.

We conclude that the center of mass of the whole system the plane & the electron & energy stores & weights stayed still.

If the inertial mass of the electron would have been some m + V' != m + V at x, then the center of mass would have moved. QED.

Location of the extra inertial mass

If we have a fully symmetric setup of two charges, then obviously we cannot add the full potential energy to the inertial mass of both particles. That would breach conservation of momentum.

The energy of an electric field is stored in a density |E|^2 where E is the electric field strength. In our plane example above, the extra energy in the electric field is mostly stored just upwards from the electron. That is the obvious reason why the inertial mass of the electron appears to grow by the full amount V, and the plane's inertial mass remains constant.

Theorem 2. The inertial mass of a particle is the same to each direction at a location x. The mass may change when the particle moves, because the geometry of the setup changes.

UPDATE Nov 18, 2018: This theorem is wrong. See our post about inertial mass Nov 17, 2018.

Proof. We only prove this for the special setup of Theorem 1. The general proof is left for future.

Theorem 1 essentially handles the case where the particle moves perpendicular to the force field.
                     
               ____________________
     V    x ________________  e-z|
   ds                                      |    | pipe
               _________________|    |
     dV  y ___________________ r|
                   \O
                     |
                    /\
    ==============
    -   -   -   -   -   -   -   -

Let us repeat the consideration when the energy stores are vertical relative to each other. The stores are rigidly attached to the plane. There is a pipe as in the diagram. The pipe extends very far from the charged plane and is rigidly attached to the plane. The plane is very massive, so that the round trip of the electron does not tilt it significantly.

The distance between x and y is small, ds.

Let the inertial mass of the electron be m at z.

The man uses the energy V to pull the electron to x.

Then he lets the (small) energy dV pull on the electron so that its force exactly cancels the electric repulsion. The man uses a rod attached to a weight m + V to win the inertia of the electron an pushes the electron down to y.

Then he lets the electron to slide away in the pipe back to r. A weightless string runs from y to the electron. The string stores the energy at y.

In the final trip r -> z the man lets the energy store pull on the string so that it exactly cancels the electric repulsion. The man uses a rod attached to a weight m to win the inertia of the electron and pushes the electron back to z.

The energy V moved from x to y. A weight m + V moved up ds and a weight m moved down ds. There was no net vertical force on the plane in any phase, and its center of mass did not move. The plane tilted a negligible angle.

We see that the center of mass of the whole system was conserved. If the inertial mass of the electron would have been different from m + V at x, the center of mass would have moved. QED.


An ages-old problem is where is the potential energy stored in a bound system. Suppose that we have a bound system where the particles are static and the forces between them do not change significantly if we move a particle a short distance dr to an arbitrary direction.

How is the energy divided between the particles? We cannot take the system apart and weigh each particle individually. But we can measure the inertia of each particle if we use a rod to move it a distance dr. The sum of these measurements must be equal to the inertia of the whole system when all particles are moved simultaneously. The breakdown of inertial masses is a measure of how much potential energy each individual particle possesses.

The full energy-momentum relation should obviously contain all the particles in a system as well as the energy of their fields.  When the particles move, the field energy moves. For any move dr of an individual particle, the field energy is moved some dr'.

Potential momentum

https://en.wikibooks.org/wiki/Modern_Physics/Potential_Momentum

In the link, the energy-momentum relation is given as

         (E - V)^2 = (p - q)^2 + m^2.

We assume c = 1 and just one spatial dimension. A potential momentum q is subtracted from the momentum p to make the equation Lorentz-invariant. Can this work?

     p      E
     ^     ^
     |    /
     |  /
     |/
      ------> m

If we draw m and p as orthogonal vectors, the mass-energy relation (without potential) states that E is the length of their vector sum. The formula above means that we reduce the length of the p vector by amount q (potential momentum) to make the sum vector E shorter by amount V.

 p - q        E - V
      ^     ^
      |   /
      | /
       ------> m

The procedure is equivalent to some "agent" drawing part V of the kinetic energy of the particle and storing the energy to some energy storage at location x.

Suppose that a man far away uses a rod and a weight m to move the particle to location y in the potential field.

When the particle leaves the potential, it gets the energy back from the storage.

But how can the potential field store energy at location x and give back the same amount of energy at location y? The field after the particle left is the same as before it arrived. The energy stored at x must have flowed from x to y.

What force moved the energy V from x to y? It must have been the man who used the rod. He had to overcome the inertia of energy V.

The potential momentum concept forgets that someone has to move the energy V, too. That error leads to thinking that rest mass (= inertial mass) of the particle stays the same.

The Aharonov-Bohm effect

https://en.m.wikipedia.org/wiki/Aharonov–Bohm_effect

Yakir Aharonov and David Bohm noted, among many other physicists, that a potential affects the phase of a wave function, and that happens also in a constant potential where the particle does not meet "field lines" of the force field.

This fact is, of course, evident in the Schrödinger equation. The interference pattern of a particle depends on the choice of the zero level of the potential V.

The effect has been measured for a magnetic vector potential but not for a static electric potential.

All rest mass is potential energy?

The rest mass of the electron can be considered potential energy which the particle acquired in pair creation.

The same might hold for all particles with a non-zero rest mass.

Let us look at the Schrödinger equation:

        -1 / (2m) * d^2Ψ/dx^2 + V Ψ  = i dΨ/dt.

What is the role of the rest mass m versus the role of V?

The hamiltonian operator

         H = p^2 / (2m) + V

does not include the (potential) energy in the rest mass. Why does the equation then give accurate predictions?

A more correct hamiltonian would have m + V in the place of V.

The free Schrödinger particle is

         exp(-i (E t - p x)),

where E only includes the kinetic energy. Why it does not contain the rest mass energy?

If we include the rest mass energy, the free particle would be

         exp(-i ((m + E) t - p x))

where E only contains the kinetic energy.

A "full-energy" Schrödinger equation is

       -1 / (2(m + V)) d^2Ψ / dx^2 + (m + V) Ψ
       = i dΨ / dt.

If we are looking for a solution of type

        Ψ(x) * exp(-i (m + E) t)

for the full-energy equation, we get on the right side of the equation

         (m + E) Ψ(x) exp(-i (m + E) t)

and on the left side

    -1/(2 (m + V)) d^2Ψ(x)/dx^2 * exp(-i(m+E)t)
        + (m + V) Ψ(x) exp(-i (m + E)).

We can divide by exp(...) to remove a common factor. We remove the term m Ψ... on both sides. In 1 / (2(m + V)), V is negligible. The end result is the familiar time-independent Schrödinger equation.

We conclude that in the Schrödinger equation we can treat m as potential energy.

Destructive interference does not conserve energy in a linear wave equation

Definition 1. Suppose that we have a partial differential equation which describes a wave. We say that the solutions of the equation are closed under sums if the following holds:

If Ψ_1 and Ψ_2 are arbitrary solutions of the equation, then Ψ_1 + Ψ_2 is, too.


Definition 2. Let

        |Ψ_i(x)| < k^i for all x,

for some 0 < k < 1, and let all Ψ_i be solutions. We say that the solutions of the equation are closed under convergent sums if the above implies that the sum of all Ψ_i is a solution.


Linear wave equations have the property of Definition 2.

Definition 3. A wave equation is local if we can combine into a solution Ψ two solutions, Ψ_1 to the left from spatial point x_0, and Ψ_2 to the right of x_0, provided that the combined solution is continuous and has a continuous derivative at x_0, and the other parameters of the equation, the potential, the string weight, etc. stay constant in the vicinity of x_0. Note that far away from x_0, the parameters of the equation of the solutions Ψ_1 and Ψ_2 are allowed to be different.


Locality means that we can glue together two solutions if they match in the vicinity of x_0. The different parts of the solution can only "communicate" through the wave function in the vicinity of x_0.

In our previous post about the Klein paradox, we uncovered a fundamental question in the linear wave equation formulation of quantum mechanics.

Suppose that we have a smooth potential barrier, where the potential has a continuous derivative.

         e- ------>
               ____
        ___/        \___   smooth potential
            A        B
     
        <--   <-- reflections

A flux of particles meets the barrier from the left. Conventional wisdom says that if there is destructive interference of the reflections from each end of the barrier, then the transmitted flux is 100%.

There is ample evidence that the conventional wisdom is true. The antireflective coating in optics works like this. My eyeglasses seem to reflect mostly green light, because the coating causes destructive interference for yellow light.

The solutions of the Schrödinger wave equation are closed under convergent sums.

The question is what mechanism creates the extra flux from B to the right, if there happens to be a destructive interference of reflected fluxes at A?

If we just add an ad hoc flux from B to the right, that would make the wave function discontinuous at B.

Destructive interference in a linear wave equation

                Ψ --->    Ψ_A --->   Ψ_B -->  right-moving waves

     ------------------=========XXXXXXXXXX

                            A                  B

             <--- Ψ_1  <--- Ψ_2  left-moving waves

Suppose that we have a tense string whose weight per meter differs at different sections. If we feed a wave Ψ from the left to the string, there will be reflections Ψ_1 from A and Ψ_2 from B to the left. We have marked in the diagram all waves in each section.

Theorem 4. Let us assume the solutions of a wave equation are closed under convergent sums and the equation is local. Then destructive interference in reflection can break conservation of energy.

Proof. We assume that reflected waves carry much less energy than transmitted waves.

We assume that there is a destructive interference, so that Ψ_1 contains much less energy than Ψ_2.

Is it possible that the energy is reflected to the right, so that energy is conserved?

We will prove that energy is not conserved.

From the uniform part of the string there is no reflection, because there is a symmetry by translation in the space coordinate x. There is no reason why a reflected wave of a specific phase φ should originate from a uniform section of the string.

Diagram 1.
       Ψ --->        Ψ_A' ---> 

--------------------=====================  

     <--- Ψ_1'

Diagram 2.

                         Ψ_A' --->         Ψ_B' --->
======================XXXXXXXXX
                       <--- Ψ_2'

Diagram 3.
                       Ψ_A'' --->
-------------------=======================
     <--- Ψ_1''    <--- Ψ_2'

Diagram 4.

                        Ψ_A'' --->          Ψ_B'' --->
=====================XXXXXXXXXX
                         <--- Ψ_2''

We can continue drawing such diagrams where the upper one tells what wave we feed to the lower one.

Let us sum all the solutions in diagrams 1, 3, 5, 7, ..., and respectively in 2, 4, 6, ....

Odd numbers:

Ψ --->       Ψ_A --->
-------------==================
 <--- Ψ_1   <---- Ψ_2

Even numbers:

                 Ψ_A --->              Ψ_B --->
====================XXXXXXX
                <--- Ψ_2

Then we can use the locality to glue together the two diagrams and we get the first diagram of this section.

For energy streams, we get from the Even diagram

             E_B = E_A - E_2.

We assumed that there is a significant destructive interference in E_1' and E_1''. We can assume that E_1 is much smaller than E_2', and therefore smaller than E_2:

             0 > E_1 - E_2.

Diagram 1:

             E = E_A' + E_1'.

E_A'' and so on are very small compared to E_1'. Therefore,

            E > E_A.

The outgoing energy flux is

            E_B + E_1 = E_A + E_1 - E_2

which is < E. QED.


But energy is conserved in classical mechanics. Which of our assumptions was wrong? Break of locality would be spooky. We conclude that the solutions are not closed under sums. The wave Ψ_2 at A is probably affected by the incoming wave in such a fashion that Ψ_2 reflects completely back.

If a string is uniform, we believe that solutions are closed under sums, but if the weight changes at A, that probably is not true.

Path integral approach?

If solutions of quantum mechanical equations are closed under sums and equations are local, how do we salvage conservation of energy?

What if we think of the quantum mechanical experiment as a path integral? The particle may have different paths. The reflected paths end up with a probability amplitude zero. What rule then inflates the amplitude of transmitted paths? Maybe we need to add a rule of normalization:

Normalization rule 5. If the end result of a path integral would not conserve the probability of finding a particle, we have to normalize the end result by multiplying it by a suitable real number C.


In the many-worlds interpretation of quantum mechanics, some branches have a probability amplitude zero. If we claim that an observer can only exist in a non-zero branch, then the normalization results from this fact. An observer can never see a particle disappear.

The observer must choose one of the branches that have a non-zero probability amplitude. The Born rule states that the probability of each branch has a weight

            |Ψ|^2,

where the complex number Ψ is the probability amplitude of the branch.

What about a converse rule? An observer cannot exist in a branch where a particle appeared from empty space. The time-reversed process would make the particle to disappear. The probability amplitude of the history of such an observer would be zero.

Destructive interference happens only in the head of the one scientist?

How does the wave function "know" that there is a destructive interference at A, so that it knows to send more particles forward at B?

Conjecture 5 of

http://meta-phys-thoughts.blogspot.com/2018/10/huygens-principle-smoothens.html

is relevant here. When we say that there is destructive interference at A, we kind of sum path probability amplitudes before any observation is made. Conjecture 5 states that such summing only makes sense in the head of (just one) observing scientist. In his head, the wave function may "collapse" without any spooky action at a distance.

The collapse happens according to the Born rule. The destructive interference inside his head sets an amplitude zero on any observation where he would see the particle number not conserved in the process. We may think that we also normalize the wave function before applying the Born rule, so that the sum of weights is one.

The concept of a "flux" of particles is wrong, because in it we are summing intermediate probability amplitudes without an observation.

The non-conservation of the "fluxes" in destructive interference is a symptom of using a wrong concept.

Wave equations generally assume that one can sum the probability amplitudes at a spacetime point. In which cases does that work and in which not? Our treatment of regularization in Feynman diagrams suggests that in some cases intermediate sums can be infinite.

Make quantum mechanics nonlinear: parallel universes interact?

If we use a path integral approach, is there some way to restore conservation of energy without normalization?

We may assume that just one particle at a time enters the reflection experiment. If reflected back from B, it should somehow know that it must reflect from A to conserve energy. It must somehow interact with other paths.

That is a new concept. The whole path integral approach with propagators assumes that we can calculate the probability amplitudes of paths individually.

In the double slit experiment we assume that different paths interfere - we have to sum their probability amplitudes at each point on the screen. If the paths also interact with each other, that means that the same photon can communicate with its own copies in different "parallel universes".

Since different particles are typically indistinguishable, it makes sense to allow the same particle in different universes to interact with itself. Banning such interaction would be hard, except in the case where we know that there is just one particle.

We do not observe people communicating with their copies in other universes (?). Maybe decoherence makes such communication impossible for large objects?

Claim 6. Different paths in a path integral must interact, if we want to model the destructive interference at a potential step with a path integral.

Normalization does not really work

Light is also a classical wave. Energy fluxes have to be conserved at each stage. It is not enough to normalize them at the final stage.

Suppose that we have an experiment where the reflected energy stream would be 20%, but destructive interference reduces it to 10%.

Energy conservation requires that 90 % is transmitted.

On the other hand, if we normalize 10% + 80% to 100%, then transmission is only 89%.

Quantum mechanical processes should have classical physics as the limit when a large number of particles is concerned.

This shows that the normalization approach is wrong.

Nonlinear Schrödinger equation

Are there any practical experiments about the energy flux in antireflective coatings?

https://en.m.wikipedia.org/wiki/Nonlinear_Schrödinger_equation

There is ample literature on a nonlinear Schrödinger equation, but at first sight, destructive interference in reflection is not mentioned on the Internet.

A comment on the Physics Stack Exchange links an antireflective coating and destructive interference to retarded and advanced solutions of Maxwell's equations. But if Maxwell's equations are linear, we already proved that they cannot explain the phenomenon.

We solved the Klein paradox

UPDATE Nov 12, 2018: see our potential step calculations from the Nov 12, 2018 post.

...

Theorem 1. The correct energy-momentum relation under a potential is

         E^2 = p^2 + (m + V)^2.

That is, the potential energy is added to the rest mass.

Proof. See our example at the end of our previous blog post, of a man living on a plane of negative charge. If the electron in his hand would not behave like an object with a rest mass m + V, then the conservation of the center of mass would be broken. If we assume that very basic conservation law of newtonian mechanics, the potential energy has to be added to the rest mass.

Also, the rest mass of the electron is actually potential energy that it acquired in pair creation. QED.


The Dirac equation which we introduced in the previous blog post should give reasonable results when a flux of electrons hits a step potential. For now, we omit the term V Ψ. We calculate the momentum after a potential step with the energy-momentum relation and do not use the wave equation in that.

Let us use the 1+1 dimensional equation. Let there be a step potential which is 0 if x < 0 and V if x > 0. Let a flux of electrons hit the step from the left. The rest mass of the electron is m.

A solution to the Dirac equation is

      Ψ(p, m) = (1, p / (E + m)) exp(-i (E t - p x)).

Let us then solve the step potential exactly. We assume that the wave function at x < 0 is a sum

        A Ψ(p, m) + B Ψ(-p, m),

where A and B are complex numbers. The incoming flux is A and the reflected flux is B.

The transmitted flux is

       C Ψ(q, m + V),

where C is complex and q may be real or imaginary.

The energy-momentum relation for the incoming flux is

        E^2 = p^2 + m^2,

and for the transmitted flux

        E^2 = q^2 + (m + V)^2.

We may assume A = 1. The continuity of component 1 of the wave function at 0 requires:

       1 + B = C,

and component 2:

      p / (E + m) - B p / (E + m)
      = C q / (E + m + V)
      =>
      1 - B = C q (E + m) / (p (E + m + V)).

Let us denote C = 1 + C'. We have:

       1 + B = 1 + C'

       1 - B = (1 + C') (q / p) (E + m) / (E + m + V).

Let us denote
 
       1 - 2 R = (q / p) (E + m) / (E + m + V).

R comes from "Reflection". If V is small and positive, then R is small and positive.

We have  B = C' and

        1 - C' = (1 + C') (1 - 2 R)
        =>
        2 = (1 + C') (2 - 2 R)
        =>
        1 + C' = 1 / (1 - R)
        =>
        C' = R / (1 - R).

To summarize, the reflected flux is

        B = R / (1 - R),

and the transmitted flux is

        C =1 + R / (1 - R).

What if we require dΨ/dx to be continuous?

If the potential V(x) is continuous, then it might make sense to require our wave function to have a continuous derivative in space. Let us see what equations we get for A, B, C, where A = 1.

Component 1 gives:

          p (1 - B) = C q.

Component 2 gives:

          p^2 (1 + B) / (E + m)
                              = C q^2 / (E + m + dV),

where we assume that dV is close to zero. Let us again denote C = 1 + C'.

          p (1 - B) = q (1 + C'),

          (1 + B) p^2 / (E + m)
       = (1 + C') q^2 / (E + m + dV).

Continuity of the wave function earlier produced equations:

         1 + B = 1 + C',

         1 - B = (1 + C')
                    * (q / p) (E + m) / (E + m + dV).

Thus, B = C'. Let us see if the equations yield anything sensible.

         q = p (1 - B) / (1 + B),

         q^2 = p^2  (1 - dV / (E + m)),

         q = p (1 - B) / (1 + B).

In the two last equations we used the fact that dV is close to zero. The energy-momentum relation gives:

         q^2 = p^2 + m^2 - (m + dV)^2
                = p^2 (1 - 2m dV)

The energy-momentum relation, of course, gives a sensible result. On the other hand, the requirement that the derivative of the second component of the wave function be continuous produces a nonsensical result. If we used the old Dirac equation, then this continuity requirement would imply q = p, which does not make sense, either.

Destructive interference in wave equations

Suppose that the potential step is rectangular:
             ____
   _____|       |____

If the width of the step is suitable, then the reflections from each end of the rectangle will have a destructive interference. Then the reflected flux is zero, and transmission is 100%. This phenomenon is used in antireflective coatings of eyeglasses.

But how do we describe this in a linear wave equation? What mechanism causes the transmitted flux to increase if there is a destructive interference at the other end of the barrier? The next blog post concerns this fundamental problem.

The Klein paradox

We also need to study what happens when V goes to infinity. The energy-momentum relation gives then

         q = i V.

A wide rectangular barrier will let through an infinitesimal flux because q is imaginary. That is a consequence of our new energy-momentum relation where we add the potential energy to the rest mass.

The physical setup of a discontinuous potential is not realistic. We should calculate the result with a steep continuous potential.

https://www.hep.phy.cam.ac.uk/theory/webber/GFT/gft_handout2_06.pdf

The Klein paradox for the Klein-Gordon equation matches the wave function value and the spatial derivative at the barrier borders.

Our new energy-momentum relation does remove the Klein paradox there, because the momentum inside barrier is imaginary, while it is real for the traditional energy-momentum relation.

Old "solutions" to the Klein paradox claimed that a pair is created at the other end of the wall and that a positron mysteriously finds the incoming electron and annihilates it. The positron is bound inside the strong field of the barrier. Why and how would it find an incoming electron?

Our solution says that the transmission is, indeed, very small. The Dirac equation and the Klein-Gordon equation do not describe pair creation. There should be no tunneling by pair creation if we solve the problem using those equations.

A new Dirac equation with a potential

We uncovered a new problem in our study of zitterbewegung. See our blog post on Oct 14, 2018.

The problem is a fundamental one and deserves a blog post of its own: how to add a potential term to the Dirac equation?

http://rspa.royalsocietypublishing.org/content/117/778/610

Dirac himself, in his famous 1928 article, added the potential energy of the electron as a simple term V Ψ to his equation. His choice is, at least in some cases, equivalent to subtracting the potential energy from the total energy of the electron. But that is a strange choice because the potential energy "moves" along with the electron, and should be added to the rest mass of the electron.

Intuitively, an electron under a constant potential should act like a free electron.

How is the matter in the Schrödinger equation? A particle which moves to a lower potential zone gains kinetic energy and acts like a free particle with more kinetic energy. Since the Schrödinger equation is nonrelativistic, we do not need to add the (small) potential energy to the rest mass of the particle.

In the Dirac equation, we may have something like

        dE / dx = -dV(x) / dx,

        d/dx sqrt(p(x)^2 + m^2) = -dV(x) / dx,

        2 p(x) dp(x) / dx
        * 1/2 * 1/sqrt(...) = -dV(x) / dx.

We get a complicated formula for d^2 Ψ / dx^2, that is, dp(x) / dx.

        dp(x) / dx = -dV(x) / dx
                            * sqrt(p(x)^2 + m^2)
                            / p(x).

If m = 0, we get simply dp(x) / dx = -dV(x) / dx, which implies p(x) = -V(x) + C.

The massless case m = 0

The Dirac equation with m = 0 is

        α * -i dΨ/dx = i dΨ/dt,

or in the component form:

       -i dΨ_2/dx = i dΨ_1/dt
       -i dΨ_1/dx = i dΨ_2/dt.

Let us try a solution where Ψ_1 = exp(-i (E t - p x))

                    dΨ_2/dx = i E exp(-i (E t - p x))
  -i p exp(-i (E t - p x)) = dΨ_2/dt.

We get

         Ψ_2 = i E / (i p) exp(-i (E t - p x))
         Ψ_2 = i p / (i E) exp(-i (E t - p x)).

This implies E^2 = p^2 => E = +-p.

Thus, the solution is:

          Ψ_1 = Ψ_2 = exp(-i E (t - x))
or
                 
          Ψ_1 = exp(-i E (t + x))
          Ψ_2 = -exp(-i E (t + x)).

For a light-speed particle, E = |p|, assuming c = 1.

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

The Wikipedia article uses these two solutions to calculate the reflection and the transmission at a potential step. In the massless case, the equation

             α p Ψ + VΨ = i dΨ/dt

does yield a somewhat reasonable solution, but it produces the Klein paradox.

Should we rotate the phase of the wave function?

Suppose that a single electron approaches a step potential at x = x_0. If we represent the electron as a wave of a type

              exp(-i (E t - p x)),

there is no way we can make a continuous wave function by gluing together that wave with

             exp(-i (E' t - p' x))

at x_0. Should we allow a discontinuous wave function? Or is there a total reflection of the wave?

Since the phase of the wave function is not an observable, it does little harm to have a discontinuity of the phase at x_0. Suppose that we rotate the phase of the wave function in the way that the phase is always 0 at x = 0, regardless of t. Then we have a function

           exp(-i p x),

which we can easily glue together with a function

           exp(-i p' x)

at x_0, and we obtain a continuous wave function. But does this make sense? Let us consider a more realistic setup.

A path integral approach

     e- ----->
                    \     |
                      \   |
proton O           x
                      /   |
      e- ------>   
                           screen

The 1+1-dimensional case is too restricted. Let us look at a 1+2-dimensional case. Suppose that there is a static 1 / r potential of a proton. A flux of electrons hits it and forms an interference pattern on a screen. We should calculate the interference pattern.

If the flux consists of electrons of a fixed momentum p, then there are just 2 paths through which an electron can end up at a point x on the screen. We can calculate the phase of each of these 2 paths and their interference.

From this case we see that we are not allowed to rotate the phase of the incoming flux arbitrarily at different spacetime points. It would spoil the interference pattern.

Does the above diagram give us a clue how to add a potential to the Dirac equation?

The two paths to the point x are classical in the sense that we know the initial momentum p of the electron exactly and we can measure its position at x with an arbitrary precision. It is like a classical particle with a precisely defined rotating phase hitting the screen. Is there any sense in trying to derive a differential equation which describes the phase at x? The phase is a function of two classical paths. It does not matter what happens outside those paths, as long as no new paths to x appear.

Wave equations describe a smooth process where what happens at x is determined in a smooth way from what happens in its surroundings.

We may imagine that the flux which hits the proton is just a single electron, whose momentum we know exactly, but whose position has a huge uncertainty. Then it is obvious that the different paths of the electron are independent. The paths do not interact in any way. This suggests that there is really no sense in describing the process as a wave equation. Why does the Dirac equation work then?

A potential step

Let us consider in 1+1 dimensions the case where a flux of electrons of momentum p hits a low potential step at x =0, of height V.

 ^ t           X

 |          |  /  

 |          | /

            |/

           /|

         /  |

       /    |

     /      |

   e-

            0             ----> x       

            step         

                        

The flux of free electrons, or a single free electron whose position we do not know, is described by the 2-vector

            (1, p / (E + m)) * exp(-i (E t - p x)),

where E^2 = p^2 + m^2. How does the electron behave once it has crossed the potential step? Its kinetic energy decreases by V. A crucial question is what happens to E?

Only one path leads to X. If we translate the path upward by Δt, the phase of the start of the path advances by E Δt. The end of the path is a spacetime point X'. How much does the phase advance in X' relative to X?

If we assume that the phase of the electron rotates at a fixed speed per distance after crossing 0, then the phase in X' relative to X has advanced the same E Δt as at the start of the path.

Even though p^2 + m^2 is reduced after the potential step, the potential energy V seems to make for it. Thus, we may write:

           E = sqrt(p^2 + m^2) + V.

What is the phase of the electron with respect to x, if x > 0? In the Schrödinger equation, it is determined by p x, where p is the momentum. We may guess that the same holds for the Dirac case.

The phase in the area x > 0 would then be:

           exp(-i (E t - p x)),

where

           p^2 = (E - V)^2 - m^2
                  = E^2 - m^2 - 2VE + V^2.

Let us try to add the potential term to the Dirac equation in the same way as in the Schrödinger equation:

          α * -i dΨ/dx + β m Ψ + VΨ = i dΨ/dt.

Let us try a solution exp(-i (E t - p x)). From our zitterbewegung blog post, we get the formula

         p^2 / (E + m) + m + V = E,

         p^2 + m^2 + VE + Vm = E^2,

         p^2 = E^2 - m^2 - VE + Vm.
 
The formula does not look right. The relation between E, m, p, V is complicated.

Has anyone found a natural way to add V to the Dirac equation? Maybe we should look at the paper by Oskar Klein from 1929, which introduces the Klein paradox.

Add the potential energy to the rest mass of the electron?

                e-

            O/
           /|
            /\
  =======================
   -  -  -  -   -  -   -   -  -   -  -  -  -  -
           x                        y

Suppose that we have a fixed planar negative charge. An inhabitant living on the plane pulls an electron close to the plane. The electron acquires a potential energy V. Does it then behave like a particle whose rest mass is m + V, where m is the rest mass of the electron in empty space?

In the pulling, the inhabitant can use an energy store of size V at location x. The inhabitant carries the electron to location y and fills another energy store there with energy V by letting the electron move away.

In the process, energy m + V moved from x to y.

The conservation of the center of mass requires that the electron has to behave just like a particle with a rest mass of m + V. Otherwise the center of mass of the system the plane & the electron would change.

This suggests that we should define

          m(x) = m + V(x).

The Dirac equation would then be

        α * -i dΨ/dx + β (m + V(x)) Ψ = i dΨ/dt.

That would make a lot of sense, because if we move the electron under a potential V > 0, we move an inertial mass which would imply a higher rest mass.

We dropped the term V(x) Ψ because the purpose of that term was to subtract the potential energy from the total energy. We add the potential energy to the rest mass, and the total energy remains the same.

The rest mass of the electron itself can be considered as potential energy which it gains when it in pair production is separated from the positron.

But what if the electron descends into a negative potential whose absolute value exceeds its rest mass? If the electron would be able to give up its kinetic energy in that case, there would be a breach of energy conservation.

In the plane example, the inhabitant can let a positron approach the plane and fill an energy store V at location x. We assume V > m. Then he carries the positron to y where he uses an energy store V to push the positron away. Energy V moved y -> x and energy m moved x -> y. The positron should have a negative rest mass in this process, which would be an utterly strange experience for the inhabitant.

The energy-momentum relation suggests that the mass becomes imaginary if the kinetic energy exceeds the rest mass of a particle. The particle is then a tachyon.

The proton decays because positive frequencies slowly turn to negative

Our blog post about charge nonconservation predicts proton decay. A quark will eventually turn into an antiquark, because its positive frequencies slowly turn to negative. There are 3 quarks in a proton plus a number of gluons, and the interaction is not strictly between two particles. There are random accelerations, which necessarily produce negative frequencies.

The radius of the proton is 9 * 10^-16 m.

Quark masses are only 2 or 5 MeV, while the mass of the proton is 938 MeV.

The Compton wavelength of a quark is around 2 * 10^-13 m. Most of the proton mass comes from gluons. If we assign the proton mass evenly to the three quarks, the Compton wavelength of a quark is 4 * 10^-15 m.

Since a quark is contained within 2 * 10^-15 m, its wavelength inside proton potential well has to be around 2 * 10^-15 m or less.

A quark will bump into the potential wall surrounding the proton at most some 10^24 times per second if it moves at the speed of light.

If the gluon cloud inside the proton is essentially independent of the position of an individual quark, then the quark will move in an essentially static potential and will not get much negative frequencies per bump.

Can we somehow estimate the production rate of negative frequencies?

Let us first calculate the ballpark figure what magnitude of acceleration "significantly" deforms a single wavelength of a reflected wave. Then there should be a significant negative frequency component in the reflection.

The acceleration must be such that it will move the mirror almost a wavelength during one cycle of the wave.

                    1/2 a t^2 = λ,

where t = λ / c. We get

                    a = 2 c^2 / λ.

Conversely, we get λ =  2 c^2 / a. Let us calculate the black body temperature associated with acceleration a. Wien's displacement law gives

                     b / T = 2 c^2 / a

                     T = 1/2 b a / c^2 = 1.5 * 10^-20 a

kelvins. The Unruh temperature is 0.4 *10^-21 a kelvins. The coefficient is in the same ballpark as our 1.5 * 10^-20.

What is the acceleration when a quark bounces back? There is asymptotic freedom inside the proton, and the bounce happens from a steep potential wall. The process will take around 10^-24 seconds and the speed change is 6 * 10^8 m/s. The acceleration is 6 * 10^32 m/s^2. The Unruh temperature is T = 2 * 10^11 K, which corresponds to a wavelength of 10^-14 m.

Unruh calculated that the (nonexistent) Unruh radiation has approximately the black body spectrum. Let us make an educated guess and conjecture that the portion of negative frequencies produced in a reflection goes like the black body spectrum.

If the proton would consist just of three quarks, then each reflection of a quark would produce a significant portion of negative frequencies, because there would be a significant amount of random acceleration a in each reflection. But since the mass of an individual quark is just 1 / 200 of the proton mass, the random acceleration is just 1 / 200 of a. The temperature is thus 1 / 200 of the temperature.

Planck's law says that black body radiation is proportional to

            f^3 / (exp(h f / (k T)) - 1),

where f is the frequency, h is the Planck constant and k is the Boltzmann constant. The maximum is when h f / k T is roughly 5. In our quark case, the frequency of the quark was 5X the frequency of the maximum Unruh effect. Thus, h f / (k T) might be 25. When we drop T by a factor 200, the ratio is 5,000. We conclude that the flux of negative frequencies is not significant by a factor 10^2,500. The proton lifetime is in the ballpark of 10^2,500 years.

How can the Compton wavelength of a quark be much bigger than the proton radius?

There is a problem in the proton model. If the Compton wavelength of a quark, or 1 / 3 of the proton mass, is much larger than the proton radius, how can it be localized inside the 9 *10^-14 m radius of the proton? Maybe the particle is in a potential well, and its wavelength inside that well is shorter than it would be as a free particle. Localization inside 10^-15 m suggests that the energy of each particle with respect to the floor of the potential well is actually in the GeV range.

If we try to pull a quark out of a proton, then according to the rubber band model of QCD, the rubber band which pulls the quark to the proton will eventually break, and two hadrons will be formed. This means that each quark inside the proton really is sitting in a potential well whose depth is in the GeV range, if we set the zero potential at the rubber band break position. But why should we set the zero level there?

An explanation for the matter-antimatter asymmetry in the visible universe

Our previous blog post raised the question of baryogenesis where matter for an unknown reason is more abundant than antimatter. How is this possible if the laws of nature are perfectly symmetric for matter and antimatter?

Theorem 1. Assume that a cooling primordial soup after the Big Bang favors creation of new baryons, that is, baryons become more abundant when the soup cools.

Assume that there exists an unknown attractive force which pulls together matter baryons, and pulls together also antimatter baryons.

Then there will form areas where matter baryons dominate and areas where antimatter baryons dominate. This is spontaneous symmetry breaking, or in a sense, crystallization of the soup.

Proof. Because of the unknown attractive force, distributions where matter baryons and antimatter baryons are separated have a lower potential than mixed distributions. When the soup cools it favors a lower potential.

It is like cooling of water where two types of ice can form, but the types cannot coexist close to each other. When the water cools, ice must form. Since the ice cannot be mixed, the two types of ice have to appear as separate "crystals". QED.


We do not know how to calculate the crystallization process. We seem to live inside one huge crystal.

Crystallization takes a long time. Is this compatible with a fast Big Bang? If we assume extremely heavy particles which can decay into either matter or antimatter, the crystallization might happen very quickly. The visible universe is then the decay product of one or more such superheavy particles.

Actually, we do not need Theorem 1 at all if we assume a superheavy particle which can decay into an entire visible universe.

https://en.m.wikipedia.org/wiki/Baryogenesis

Andrei Sakharov in 1967 wrote three "necessary" conditions for baryogenesis. It turns out the first two conditions are not necessary. There is no need for baryon number violation or any symmetry violation. The crystallization process causes apparent symmetry breaking.

The third condition of Sakharov is that interactions are out of thermal equilibrium. The third condition is trivial: if something is "born", then the system must have been out of equilibrium.

Lumping together even when there is no attraction

Even if there is no unknown attraction between matter baryons, there will be some lumping together of matter. The following toy model clarifies the thing.

Suppose that we have n baskets and a random flux of baryons and antibaryons which fall in them. We assume that in each basket, antiparticles immediately annihilate. In each basket, the number

       N_baryons - N_antibaryons

will perform a random walk. After n particles, its distribution is

       2 * B(n, 1/2) - n,

where B is the binomial distribution.

The typical number of particles in a basket grows like sqrt(n) after n particles have fallen into a basket.

We had to assume that individual particles choose their basket at random. If a produced pair baryon-antibaryon would always fall in the same basket, then there would be no lumping together. Since antiparticles attract each other, the cross section for annihilation is bigger for slow particles. If the particles in a pair are born with a high kinetic energy, they will often fall in different baskets.

Lumping together is reduced by the motion of particles from a basket to an adjacent basket.