Classical waves do not have a diverging vacuum polarization?

On November 4, 2025 we wrote our most detailed analysis of QED vacuum polarization divergence so far. We brought up a hypothesis that destructive interference wipes out the large 4-momenta |k| in the vacuum polarization loop below.

    mildly relativistic

    e- • --------------------------------------

                          | 

                          | q virtual photon

                          |

                        /  \ k + q virtual electron-

                        \  / positron pair e- e+

                          |  

                          | q virtual photon

                          |

       ● ---------------------------------------

   e+ static

Let us analyze purely classical waves which interact. The electron and the positron are presented as waves in the Dirac field. These waves meet each other.

There is an interaction between the electromagnetic field and the Dirac field.

The waves disturb the electromagnetic field, causing waves there. The waves in the electromagnetic field, in turn, disturb the Dirac field.

The vacuum polarization loop is the disturbance in the Dirac field.

If these all are classical waves, then there cannot be any divergence. A divergence would break conservation of energy.

Deep question. Do the quantum waves in the physical world, also in a Feynman diagram, behave like classical waves, or can they behave like in the Feynman integrals, where divergences occur?

If the answer to the question is that the waves must behave like classical waves, then we have a solution to the divergence problem of QED: there is no divergence if the calculations are done in the classical way.

Photon versus a laser beam. A laser beam is a classical wave. A photon behaves just like a laser beam until we measure the photon, and the wave "collapses". That is, a photon must be described as a classical wave until the measurement. Does the same hold for any intermediate state of a Feynman diagram, until the outgoing particles are measured?

Divergences are a result of "breaking into more degrees of freedom"

If one interprets the diagram above as a Feynman diagram, then the phase of the outgoing waves is not affected by k at all. There is a constructive interference for all k. This leads to the notorious divergence problem.

The divergence of the vacuum polarization loop happens when the system "breaks into more degrees of freedom". The value of the the 4-momentum k can be chosen freely – it is a new degree of freedom.

Classical waves also can break into more degrees of freedom. But for them, this does not create any divergences. Why?

The obvious answer is that destructive interference wipes out any classical waves which have high momenta k (short wavelength). If a wave is wiped out, then it cannot pass a disturbance forward.

A classical analogue of the vacuum polarization loop

Let us have two elastic metal plates. They correspond to the electron and the positron in the loop.

Let us model the electromagnetic wave with  a wave propagating in a rubber membrane. The rubber membrane is somehow loosely attached to the plates, maybe via very elastic rubber blocks. This constitutes the interaction.

                 metal plate (e+)

                ------------------

                 interaction

          /\/\/\/\/\/\/\/\/\/\/\   rubber membrane   

                 interaction

                ------------------ (e-)

                 metal plate

   

           --> wave propagation direction

As the rubber wave meets the plates, it interacts with them and creates waves into the plates. Later, these created waves can be absorbed back into the rubber membrane.

Does this mean that the rubber wave "breaks into more degrees of freedom"?

We can model the interaction by assuming that each small area element of the rubber membrane hits with a "sharp hammer" both metal plates. This is the Green's function approach to analyze the process.

The impulse from the sharp hammer generates waves of varying momenta k. There is no limit on how large |k| can be.

Let us analyze different momenta. Does it make sense that the e+ plate receives k and the e- plate -k? If the rubber wave pushes the plates apart (like an electric field pushes e+ and e- apart), then this assumption is reasonable. An area element of the rubber membrane hits both plates with a sharp hammer: one upward and the other downward.

Later, the process can happen time-reversed: the rubber membrane absorbs some waves from the plates.

We seem to have a process analogous to what happens in the Feynman diagram vacuum polarization loop. We can freely choose k, and it will contribute to the waves absorbed back into the rubber membrane.

Why does this not lead to a divergence in the classical system?

One aspect is conservation of energy. Hitting with an infinitely sharp hammer would consume an infinite amount of energy? Then all hits must be done with a "blunt hammer". Destructive interference wipes away all high |k|.

Why should Feynman diagrams allow a sharp hammer? Why not require a blunt hammer?

The vertex correction: the infrared divergence with small |k|

Another loop in a Feynman diagram appears in the vertex correction of QED. Again, the system "breaks into more degrees of freedom", because of the loop. Let us try to find a classical counterpart.

         ^

           \  p + q

             \

               \

              |  \

              |    \   p - k + q

          k  |      |-----------------● Z+ nucleus     

              |    /               q

              |  /   p - k

               /

             /

           /

          e-

          p

The electron e- bumps from the nucleus Z+. The momentum exchange is q. The electron sends to itself a virtual photon whose 4-momentum is k.

We can choose k freely. This causes a divergence in the Feynman integral for small |k|.

There is a direct classical analogue of this process. If we model the electric field of the electron with an elastic block of rubber attached to the electron, then the momentum exchange q makes the rubber to "wobble". The electron gives some momentum k and kinetic energy to the rubber block, and absorbs some of k back when the electron leaves the nucleus.

We already analyzed this in our post on September 24, 2025. The Feynman integral method does not understand the fact that an infinite number of "soft" real photons, or virtual photons with small |k|, are produced when the classical electron passes once by the classical nucleus. The integral thinks that these are all separate cases. The integral diverges to infinity because it thinks that the cases are separate, while they are not.

We see that the "break into more degrees of freedom" is the infrared case causes a divergence problem which is quite different from the ultraviolet case (vacuum polarization).

Further analysis of the classical vacuum polarization analogue

The metal plate analogue above is awkward. Very different from any "polarization". What about assuming a polarizable medium? When a classical electromagnetic field meets that medium, it produces a "polarization wave".

 

             /\/\/\/\/\                   ############

                --->

     electromagnetic     polarizable medium

              wave

We might have two charged fields. One would have a uniform positive charge density, and the other the canceling negative charge density. These two classical fields would correspond to the Dirac field.

The classical analysis is similar to our rubber membrane - metal plate model. Since the incoming wave is smooth, destructive interference cancels all high momenta |k|.

Quantum mechanics: the interference pattern of two created particles

We have tackled with this question many times in the past few years. Suppose that two waves are born simultaneously: e+ and e-.

For classical fields, it is enough that destructive interference cancels the waves e+ and e- separately. But in quantum mechanics, the waves are entangled. How does destructive interference act on them?

***  WORK IN PROGRESS  ***

Neutrino oscillations

Even though neutrinos in the Sun are mostly created as the "electron flavor", νe, when they travel 150 million kilometers to Earth, roughly a half of those neutrinos here are observed to have the muon or tau flavor: νμ, ντ. This is the "solar neutrino problem" which catalyzed a lot of neutrino research in the past 60 years.

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

The standard theoretical model is to assume that there are three different neutrinos with different rest masses in the range 0.05 eV ... 0.5 eV: ν₁, ν₂, ν₃.

Let us have a reaction which creates an electron neutrino νe. The neutrino can be any of mass states ν₁, ν₂, ν₃. We do not know which it is. Various rest masses have various probability amplitudes.

Let α = e. The probability of observing the neutrino as the flavor β at the distance L from its creation is:

where mj are the various neutrino rest masses, and Uαj are constant components from the Pontecorvo et al. 3 × 3 matrix.

The formula on the right sums the waves of different particles: neutrinos with different rest masses m₁, m₂, m₃? That is strange. Is that allowed in quantum mechanics?

In quantum electrodynamics, the electron Dirac field is classical, and can describe an electron or a positron. There we have another case in which we sum waves of different particles: the electron and the positron.

Suppose that a neutrino with a rest mass m flies in vacuum. Obviously, the rest mass of the neutrino cannot change. There cannot be an oscillation of neutrinos if we look at their rest mass.

Anca Tureanu (2025) says that we cannot use the interference of the wave functions of the neutrino wave functions when the neutrinos have different masses

https://helda.helsinki.fi/bitstreams/9d374eb1-c3b2-4637-a770-43534a4d6e8d/download

https://quantumzeitgeist.com/university-helsinki-tureanu-challenges-decades-old/

Professor Anca Tureanu from the University of Helsinki says that, in quantum field theory, one is not allowed to calculate interference of wave functions of neutrinos of different rest masses.

A principle in quantum field theory, and in quantum mechanics, is that we can only calculate an interference of two waves if we cannot know which of the two wave histories happened. In the double-slit experiment, we cannot know if the photon passed the left or the right slit. We are allowed to calculate the interference pattern of the two different paths of the photon. But if we place a detector which can determine the path of the photon through the slits, the interference pattern disappears.

In the Nuclear Physics, Section B 1018 (2025) paper, Tureanu writes:

She claims that the end state of a reaction in quantum field theory is a mixed state. A mixed state has classical probabilities. There is no interference of classical probabilities.

But this is a strange claim. In quantum mechanics a "pure" state, that is, a superposition state, stays pure until it is measured. When we measure the outgoing particles, we obtain classical probabilities. The state is no longer pure. The state "decoheres".

Note that the pure state includes ALL outgoing particles. We cannot speak separately of the neutrino.

If we know the location of the reaction well, we do not know the mass state of the created neutrino

Suppose that we prepare the incoming particles in such a way that their momentum (and energy) are known extremely precisely. We measure the energies and the momenta of outgoing particles, except of the neutrino. That way we will know the energy and momentum of the created neutrino, and can determine its rest mass m.

The uncertainty principle says that we then cannot know the location of the reaction precisely:

       Δp Δx  ≥  ħ / 2.

The flavor probability formula has the parameter L which tells the distance to the reaction.

The energy-momentum relation is

       E²  =  p²  +  m².

Let E and |p| be roughly 1, and m ≈ 10⁻⁶. To determine a rough value of m from the energy-momentum relation, we have to know p to the precision better than Δp = 1/2 * 10⁻¹². We use natural units, so that ħ = 1. Then

      Δx  >  10¹².

We see that the formula

      exp( -i mj² L / (2 E) )

in the flavor probability equation above then has a relatively large uncertainty because of L:

       (10⁻⁶)² * 10¹² / 2 = 0.5.

That is, we know the rest mass of the neutrino, but do not know the neutrino oscillation phase too well. The uncertainty, actually, is quite a lot larger than 0.5 because we must measure momenta of several particles. Also, we chose Δp too optimistically above. A better choice might be 1/3 of the value we chose. Then the uncertainty 0.5 easily becomes 3 or more.

That is, we cannot know the phase of the neutrino oscillation. The flavor probability formula in this case gives the average over a whole cycle of the oscillation. The average is a constant and does not depend on the value of L. This is reasonable.

If we know L relatively precisely, then we cannot know the rest mass state of the neutrino. It makes sense to calculate the sum (interference) of different rest mass waves.

What is the created neutrino? Which state does it have?

If we know the location of the reaction relatively well, then we cannot know the rest mass of the created neutrino.

The created neutrino can then be seen as a "malformed" classical neutrino field wave, which does not have a definite rest mass until we measure the rest mass in some way.

There is only one neutrino particle which has 3 different states? In the model above, there is a single neutrino field which has (stable?) states of three different rest masses. Should we say that the neutrino is just one particle, which appears in three stable states?

An analogous question: are the electron and the positron a single particle which has two states?

Can a neutrino decay to a lower rest mass state? Could it emit a photon?

A particle reaction emits waves, not particles?

In bremsstrahlung, an electron can emit photon(s) of various energies and momenta.

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

Classically, the electron emits an electromagnetic wave of a complicated form, as the electron e passes the nucleus marked with +. For high energies of the photon, the quantum mechanical calculation yields probability amplitudes which differ surprisingly much from the probabilities one could naively assume from the Fourier decomposition of the classical treatment.

A basic principle of quantum mechanics is that you must treat everything as waves until a measurement is made. You cannot assume that a wave "has collapsed" until it is measured. A typical example of this principle is the Bell inequality: if you assume that there are "hidden variables" which determine the state into which a quantum wave will collapse, then you will calculate results which differ from quantum mechanical results.

This basic principle of quantum mechanics suggests that in a reaction, we must treat outgoing particles as waves, until the particle is measured in a detector.

However, we have a problem here. What is the shape of the bremsstrahlung wave? It cannot be the classical shape. From Feynman diagrams, we can calculate the probability amplitudes of various photons with a momentum p. Could it be that we ger the wave in the momentum space simply by superimposing the waves with different p, multiplied by its probability amplitude? The phase of an individual wave we may get from the position and the momentum change of the electron as it passes the nucleus.

It looks like the model of neutrino oscillation in Wikipedia is compatible with quantum mechanics.

Are the electron, the muon, and the tau particle states of the same particle with different rest masses? What about antiparticles?

The Dirac equation governs these particles, as well as their antiparticles. 

Do we know anything about interference of electron and positron waves? What about interference of electron and muon waves?

An electron colliding with a nucleus

The bremsstrahlung reaction creates photons. Let us forget photons for a moment, and look at the wave function of the electron colliding with the nucleus.

We may assume that the nucleus is moving at a velocity v, so that, depending on its path, the electron may gain or lose some kinetic energy.

                              -------------

                           /    v            \

     e-       --------      <-- ● Z       ----------->  p'

     p        ---------------------------------------->  p''

The Schrödinger equation describes the behavior of the electron e-. The initial momentum of the electron is p.

In the diagram above, we have two alternative paths for the electron. Depending on the path, the electron exits with a momentum p' or p''.

The momentum of the nucleus changes by an amount p - p', or p - p''.

In the Schrödinger equation, the waves for different paths of the electron do interfere, even though the electron in each path has a different kinetic energy and momentum.

The collision can be considered a "reaction". The particle e- exits the reaction. The electron in each path has a different "state". 

The collision is actually a "double-slit" experiment, in which the nucleus forms the obstacle between the two "slits".

We conclude that in quantum mechanics, the waves of a particle do interfere, even though the waves correspond to different kinetic states of the particle.

In the case of neutrino oscillations, the neutrino can be in different rest mass states. Could it be that having a different rest mass removes the interference? Is having a different rest mass a "kinetic state"?

A simplified model of an excited hydrogen atom

An excited hydrogen atom has a "rest mass" which is larger than of the atom in the lowest energy state. What do we know about the interference of hydrogen atoms?

https://en.wikipedia.org/wiki/Double-slit_experiment

People have been able to run the double-slit experiment with large molecules, like the "buckyball". This suggests that we should be able to see interference between a hydrogen atom in the ground state and the same atom in an excited state.

If we have no way of knowing if the atom is excited or not, why would interference disappear? We can measure the rest mass later, and we will know. But the same hold for the electron in the previous section: we can measure the momentum of the electron later – this does not prevent interference from occurring between electrons which have different momenta.

            y position of proton

            ^

            |

            |

            |

            |

             --------------------------> x position of e-

Let us consider the following simplified model of a hydrogen atom. We have one spatial dimension in the world. The y coordinate tells the position of the proton. The x coordinate is the position of the electron. That is, we model both particles with one particle whose spatial coordinates are (x, y).

The potential:

       V  =  1 / | x  -  y |.

If we fix y to some value, the wave function is somewhat like a particle in a box.

The inertia of our one particle is larger in the y direction than to the x direction. The proton is heavier.

In this simple model, there most probably is interference between different momentum states of the proton?

There is a problem: in the Schrödinger model we have hard time raising the atom to an excited state. Could that succeed by letting another electron to "collide" into the the electron?

The interference happens in the MEASURING process – there is no need for interference in the fields being measured? A macroscopic Feynman diagram

Let us assume that the neutrino particles  ν₁, ν₂, ν₃ are completely different particles. Each neutrino type is described with its own field which is not related to the field of the other neutrino type in any way. That is, the field of ν₁ is as different from the field of ν₂, as is the field of the electron e- different from the field of, say, a quark.

Our measuring process checks if a neutrino causes some reaction which is associated with the electron flavor neutrino νe. The process is sensitive to each of the three different fields, corresponding to ν₁, ν₂, ν₃.

The neutrino is produced in some process, say, beta decay, and then measured in a process which destroys the neutrino. We cannot know which type of a neutrino, ν₁, ν₂, ν₃, mediated this.

Quantum mechanics in this case says that there has to be interference in the end result of the measurement. The neutrino waves, or probability amplitudes, do not "collapse" when they mediate the process.

In Feynman diagrams, this principle is the rule that we must sum the probability amplitudes of all diagrams which produce the same output.

This argument resolves the problem. There is an interference regardless of whether the individual fields describing different neutrino masses take whatever form. The interference is in the end product of the measuring process: the outgoing particles in the reaction which we use to measure the neutrino flavor.

Below is a schematic "macroscopic Feynman diagram":

                                 --------------------------- particles

                               /

                             /   ν₁ or ν₂ or ν₃

 --- beta decay  -----------------------  measurement

                                                       \ 

                                                         \  

 ------------------------------------------------------ particles

                            <------------------->

                              macroscopic

                                 distance

It is a "macroscopic" Feynman diagram because the internal neutrino line describes an on-shell particle which travels a macroscopic distance.

A traditional Feynman diagram describes a reaction which happens in a small spatial volume in a short time. Off-shell (virtual) particles can exist for a short time. They do not need to satisfy the energy-momentum relation

       E²  =  p²  +  m².

The neutrino between the beta decay and the measurement can be of any type. We must sum the probability amplitudes for the constellations of outgoing particles in the diagram. There is always interference. We could even replace the neutrino with a totally different particle type. Also in that case, we must sum the probability amplitudes.

Can we determine the neutrino type from the outgoing particles?

In principle, yes. Let us prepare the "measurement particles" in the diagram in the way that we know their energies and momenta extremely precisely. We then only know their position very poorly. After they have reacted with the incoming neutrino, let us measure the momenta and energies of the outgoing particles very precisely. We can then deduce the neutrino type, ν₁ or ν₂ or ν₃, from the energy-momentum relation. In this case, we must not calculate any interference. We know which of the three diagrams happened.

Discussion

The problem arose when we forgot that the neutrino is an internal line in a (macroscopic) Feynman diagram, and started to think of the neutrino as a "classical particle" whose rest mass and state can be known without disturbing anything in the experiment.

If we could, magically, measure the rest mass of the neutrino in the Feynman diagram, then there would be no interference. We would know which of the Feynman diagrams, ν₁ or ν₂ or ν₃, happened, and then we are not allowed to sum the probability amplitudes.

This discussion does not settle if ν₁, ν₂, ν₃ are different states of the "same" particle. They can be the same particle, or different particles, and that does not affect the macroscopic Feynman diagram in any way.

The term "neutrino oscillation" is misleading. It implies that ν₁, ν₂, ν₃ are the "same particle" which somehow oscillates between states. We do not know if it is the same particle. Another term would be "neutrino flavor interference pattern". In the double-slit experiment we do not have "photon oscillation". We have an interference pattern.

In quantum mechanics we have a general rule: we must treat everything as a wave phenomenon, unless there is a special reason why the system has "decohered" into the realm of classical probabilities. There always is interference, if interference is not ruled out by something, e.g., a measurement.

Conclusions

The treatment of neutrino oscillations in Wikipedia is compatible with quantum mechanics. If we know fairly well the location of the reaction where the neutrino is born, then there is an uncertainty in the momentum of the neutrino, which prevents us from knowing which neutrino type it is: ν₁ or ν₂ or ν₃.

In quantum mechanics we have to treat all processes using the wave model, unless we are able to measure something, in which case the wave "collapses", and we enter the realm of classical probabilities. We cannot know or measure the rest mass of the precisely enough. We must use the wave model. In the wave model, there is an interference of waves (or probability amplitudes).

The model in Wikipedia assumes interference, which is the correct quantum mechanical treatment.

The word "neutrino oscillation" is somewhat misleading. We cannot directly observe an interference of the neutrino types ν₁, ν₂, ν₃. We can only observe the reaction which they cause in other particles. It is better to say that the interference is in the end result of the experiment. Then we do not need to claim that there is an "oscillation" in the neutrino itself.

Anca Tureanu (2025) says that quantum mechanics does not allow an interference to happen between particles which have different rest masses. Our analysis above shows that we need not, and maybe should not, assume an interference of the neutrino types ν₁, ν₂, ν₃.

Anca Tureanu (2025) further writes that in a quantum field theory reaction, the outgoing particles should be assumed to have classical probabilities, that they should not be treated as waves (or probability amplitudes). That is an incorrect claim. We must use the wave description unless the outgoing particles have been measured, so that the wave "collapsed" into a classical probability.

Another way to characterize a quantum field theory reaction: the ingoing particles are in a "pure" state (synonyms for this are a superposition state or a coherent state). Unless something is measured, the outgoing particles also form a pure state, not a "mixed" state (in a mixed state we would use classical probabilities). A measurement makes the pure state to "decohere", so that it becomes mixed.

Solving the Hubble tension: slow down the expansion of the universe during the first 500 million years

Supernovae of type Ia, and Cepheids suggest that the universe is right now expanding at a rate 73 km/s / Mpc, and, furthermore, the expansion rate is accelerating (dark energy).

But the mapping the cosmic microwave background, and combining it to a ΛCDM model says that the expansion rate is only 67 km/s / Mpc.

This is called the Hubble tension. In our previous blog post we suggested a simple solution to the Hubble tension: slow down the expansion rate of the universe by a factor about 1/2X, when the universe is 150 million ... 500 million years old.

We do not know why the expansion of the universe started to speed up some 5 billion years ago. We cannot take for granted that the universe has obeyed the ΛCDM formula during the previous 8.7 billion years, either.

Let us investigate how people derive the figure 67 km/s / Mpc.

What we know of the cosmic microwave background

Literature seems to be certain that the "last scattering" temperature of the CMB was 3,000 K, at a precision +-0.5%. Since the temperature today is 2.73 K, we know that the redshift since the last scattering is

       z  =  3,000 K / 2.73 K

           = 1,100.

The spatial size of the sound horizon in comoving coordinates

https://www.mv.helsinki.fi/home/hkurkisu/Cosm9.pdf

In the link there is a textbook by Dr. Hannu Kurki-Suonio, in which the spatial size of the sound horizon is calculated:

where cs(t) is the sound speed (about 60% of c), and a(t) is the scale factor. The integral formula assumes a standard ΛCDM model. The speed of sound before the recombination (last scattering, or decoupling) is:

where ωb is the baryon density and ωγ is the radiation density. We know their approximate values by observing the current universe, and from knowing that the current temperature of the CMB is 2.725 K.

The expansion speed of the universe before recombination follows the formula:

There Ωr is the share of the radiation density of the critical density and Ωm is the share of matter of the critical density. Since |a| is very small, we can ignore the a² and a⁴ terms.

The value of the integral is:

The sound horizon size r* depends on the density of baryons and matter in the current universe. Its dependence on the Hubble constant H₀ in the current universe is not significant: if we calculate time backwards, the final collapse speed of the universe does not depend much on the collapse speed right now.

Apparently, the baryon density and the matter density in the current universe are known quite well. The radiation density we know very well, too. The redshift at the decoupling, zdec, is known well. Thus, if we assume the ΛCDM, we know the sound horizon size r* quite precisely.

The Hubble tension

Our best knowledge of the expansion speed of the current universe come from type Ia supernovae and Cepheids. The data from supernovae and Cepheids agrees about the value 73 km/s / Mpc.

If we assume the ΛCDM model, and calculate backward from the current universe, the cosmic microwave backround rises to the temperature 3,000 K at the redshift 1,100, when the scale factor a of the universe is 1/1,100th of its current value.

The last scattering / recombination / decoupling / end of baryon drag almost certainly happens at that temperature.

The Hubble H₀ constant now has the value 73 km/s / Mpc. The matter density of the current universe is Ωm = 0.31 times the critical density. We can proceed to calculate backward ΛCDM, and can determine the distance D(1,100) in comoving coordinates to the last scattering. That is, a photon scattered then moves the distance D(1,100) in comoving coordinates, before hitting our eyes.

   

                            /|

                         /   |

                      /      |

                   /         |

                / θ*       | r*

          o  -------------    last scattering

          |    D(1,100)                       

         /\                       comoving coordinates

An observer on Earth should see the sound horizon length in the cosmic microwave background to subtend an angle

       θ*  =  r* / D(1,100).

Recall that r* is given in comoving coordinates, too.

We can analyze the CMB very precisely from the Planck data, and obtain a value for θ* which is significantly smaller than the value which we calculated above. This is the Hubble tension. "Features" in the CMB look smaller in the sky than they should be.

The Hubble tension. The tension is between the assumption:

- we have measured H₀ correctly in the current universe, as well as other relevant parameters, and ΛCDM is correct,

and the fact:

- features in the CMB look smaller in the sky than what we calculated.

It is an oversimplification to say that the tension is just in the value of H₀.

A simple way to remove the tension is to assume that the universe expanded slower between the last scattering and the current time than assumed by ΛCDM. If the universe expands slower, then a photon sent from the last scattering has more time to travel in comoving coordinates, and the distance D(1,100) is longer. Consequently, the calculated angle θ* is smaller. We can make the angle so small that it agrees with the angle which we see in the sky.

Dark energy already showed us that the universe may expand surprisingly fast. It is not a stretch to claim that it can also expand surprisingly slowly.

The Hubble tension really is a tension in the angle θ* calculated from the current universe versus what we see in the CMB in the sky?

Let us measure the mass density Ωm in the current universe, as well as the value of the Hubble constant H₀ from type Ia supernovae.

Assuming ΛCDM, we can then calculate how the uneven distribution of the current galaxy clusters should show up in the CMB.

Let us check if galaxy surveys, combined with H₀ = 73 km/s / Mpc, really conflict with the CMB we see in the sky.

https://arxiv.org/html/2505.01661v1

Zhiwei Yang et al. (2025) derive H₀ from DESI, H0LiCOW, and Pantheon supernovae. DESI has measured the redshifts of millions of galaxies and built a 3D model of the current universe (actually, it is the universe in the past 10 billion years, since the redshifts z are up to 2 in the survey).

https://shsuyu.github.io/H0LiCOW/site/

H0LiCOW measures gravitational lensing of quasars and derives a value for H₀ from them.

https://arxiv.org/abs/2112.03863

Pantheon+ contains the light curves of 1,550 type Ia supernovae. They can be used to determine the distances to galaxies, as well as Ωm.

Zhiwei Yang et al. calculated H₀ = 73 km/s / Mpc and r* = 138 Mpc. The size of the sound horizon r* is given in the current universe.

The sound horizon size derived from the first 380,000 years of the universe is 147 Mpc. Why the difference?

Let us check what values for r* have other authors derived from DESI and other galaxy surveys.

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

The Sloan Digital Sky Survey team got a value r* = 150 Mpc. The value differs a lot from the Yang et al. value.

https://arxiv.org/abs/2510.19149

E. A. Zaborowski et al. (2025) derive the value H₀ = 70 km/s / Mpc from DESI.

https://arxiv.org/abs/2406.18298

Tonghua Liu et al. (2024) list values obtained for r* in various studies: 143 Mpc, 137 Mpc, 144 Mpc. Their own result is 140 Mpc.

There seems to be too much variation in the estimates of the sound horizon size r* in the current universe. We cannot draw any conclusions. The Hubble constant H₀ derived in these calculations varies a lot, too. It is not clear if a Hubble tension exists in the calculated values.

Thus, currently, the tension is between the Hubble value 73 km/s / Mpc derived from Cepheids and supernovae versus the Hubble value calculated from the first 380,000 years of the universe plus a ΛCDM model.

What would the sound horizon angular size θ* be in ΛCMD if we tune the universe expansion rate?

In our blog we have brought forward an idea that the universe on the average follows a FLRW model, but there are "bounces" which make it to expand slower or faster. The current epoch of dark energy is a bounce to a faster expansion.

Let us check if there would be a Hubble tension in a CDM model where Λ would be zero "on the average", when the universe expands.

How much has dark energy recently accelerated the expansion relative to CDM?

We can model the expansion with newtonian gravity.

https://astronomy.stackexchange.com/questions/2194/what-was-the-age-of-the-universe-when-the-average-density-was-one-atmosphere

If we assume that H₀ = 67.8 km/s / Mpc, then the critical density is now

       ρc  =  8.4 * 10⁻²⁷ kg/m³.

The ratio of matter and dark matter to the critical density is believed to be

       Ωm  =  0.31.

Note that the estimate 0.31 depends on the cosmological model that we use. It is based on how much we believe that the gravity of the mass has slowed down the expansion of the universe since the Big Bang.

Let us assume that the expansion of the universe has slightly accelerated in the past 5 billion years. We calculate the amount of energy involved.

Let us have a spherical volume of the current universe. Let the radius be R.

If the mass density of the universe would be 1/0.31 = 3.2-fold, then newtonian gravity would asymptotically stop the expansion. Let us have a test mass m on the surface of the sphere R. The negative potential energy of m is linearly proportional to the mass M of R, or the density of R.

                     ×                   • ---> v

                               R        m

We conclude that the negative potential energy of m is -0.31 times its kinetic energy E = 1/2 m v².

Suppose that the velocity v of m has stayed the same for the past 9 billion years. Then dark energy has done ~ 0.31 * E of work on m, against the gravity potential. Quite a lot of work!

Let us use the "epoch" scale from the previous blog post. The scale factor is a ~ 1 / 2ⁿ during epoch n.

The time lengths of various epochs:

   t₀ : 9 billion years

   t₁ : 3 billion years

   t₂ : 1 billion years

   t₃ : 350 million years

   t₄ : 100 million years

   ...

   t₁₀: 140,000 years.

The contribution to the "angular distance" D of each epoch – this measures how much a photon propagates, in comoving spatial coordinates, during each epoch:

  D₀ : 1

  D₁ : 0.7

  D₂ : 0.5

  D₃ : 0.35

  D₄ : 0.25

  ...

  D₁₀: 0.03.

The sum D = 3.3.

Let us calculate the kinetic energy of our test mass m during each epoch. Recall that the current gravity potential of m is -0.31 * E, where E is the current kinetic energy of m. When a < 1, the gravity potential energy has added to the kinetic energy:

  0 : E

  1 : 1.31 E

  2 : 2 E

  3 : 3 E

  4 : 6 E

  5 : 11 E

  ...

  10 : 300 E.

If we want to make the angular distance D 8% larger, so that the Hubble tension would be removed, we have to reduce the kinetic energy of m a lot during epochs 2, 3, and 4. The expansion rate should only be 75%. That is, the kinetic energy must be reduced by E during epoch 2, and by 3 E during epoch 4.

Let us think. What does this mean? Initially, during epoch 10, the test mass m has a lot of kinetic energy. As m climbs up from the gravity potential well, it loses kinetic energy. The energy goes to the energy of the gravity field. In earlier blog posts we have suggested that retardation of gravity stores extra energy into the gravity field. Could that extra energy be so large that it could be 3 E during epoch 4?

Note that we have a fine-tuning problem here. How did nature "know" that the kinetic energy of m during epoch 10 must be set almost exactly to the negative gravity potential -300 E? As if nature would be fine-tuning the process, so that it is almost exactly a collapse of a cloud run backwards.

Entropy of the universe as a fine-tuning problem

The universe had a low entropy in the Big Bang. A low entropy is unlikely for a random system. Here we have another fine-tuning problem.

The entropy of the universe grows after the Big Bang, as the universe expands. Suppose that the universe follows an FLRW model, and starts to contract again. Suppose that energy is conserved. Then the macroscopic state of the universe in a Big Crunch should be very similar to the corresponding state at the Big Bang. The state is very hot, and particles have random motion. That is, the entropy has to start decreasing at some point before the Big Crunch. This would break the rule that entropy has to increase. What is going on?

Let us imagine an explosion in almost-newtonian mechanics and newtonian gravity. Let the explosion start from a very small volume and be almost uniform. The entropy of the system starts growing. In particular, photons will carry away a large amount of entropy. Let us assume that gravity will pull back most of the mass which exploded. Gravity will compress it back into a small volume. The end result might be a crystal which has a very low entropy. All entropy created after the explosion is carried away by photons.

Photons cannot escape from an FLRW model because there is nothing outside the model. Thus, an FLRW model leads us to a paradox: entropy should magically start decreasing at some point.

Let us assume that black holes do not exist in the FLRW model. It is not clear how they would behave in a Big Crunch.

Entropy bans a Big Crunch? Maybe a Big Crunch is not possible at all? Since FLRW models do allow a Big Crunch, we here have an argument against the correctness of FLRW models: they break the laws of entropy.

Suppose that we have an isolated physical system whose volume first increases and then decreases. A big volume contains more "degrees of freedom". Entropy can then increase. But if some strange law of physics (like an FLRW model) makes the volume to decrease, then there will be fewer degrees of freedom, and entropy may be forced to decrease.

How do we avoid the paradox of decreasing entropy? Either the system is:

1. not isolated, or

2. the system will expand forever.

Bounce-back (cyclic) models of the universe require that entropy has to decrease at some point. If one believes that entropy cannot decrease, then bounce-back models are ruled out.

Conclusions

We presented various thoughts about the Hubble tension and dark energy, but were unable to reach any definitive results. We will continue our investigation later.

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

Next we will look at neutrino physics.

Dark energy: a balloon model

In December 2025 we had problems finding a reasonable mathematical model for retarded gravitation in an asymptotic Minkowski space. Let us try it again, this time for a spherical balloon model. The spacetime is then topologically different from a plane.

Picture publicdomainpictures.net

Can the equation of spacetime be local for a balloon model?

Let us think of the Schwarzschild solution. The solution extends through the entire asymptotically Minkowski space. The central mass M affects locations very far away.

For a spherical rubber balloon, the equation governing its stretching is local, if we assume that the pressure inside is uniform.

Friedmann equations in general relativity look local, but that is only because we assume a uniform mass distribution. Any retardation effects are masked.

A rubber string and a weight

                 rubber string

           ●----------------------------●

      point                            point

Let us draw points uniformly around a sphere, and connect them with tense rubber strings which run on the surface of the sphere. The rubber strings simulate the gravity between the mass points. If the balloon is static, or inflates or deflates at a constant rate, this model looks like an ordinary balloon: the tension in the strings is uniform along the string.

A rubber string and a weight which the string slows down

Let us consider the simplest possible model of a tense rubber string plus weight:

          

                                              force

                                         F  <------ 

         ==| ~~~~~~~~~~~~~~ ● ---> v

    wall         rubber string            M weight

We are interested in what kind of longitudinal waves form in the string when the weight moves to the right while the tension in the string pulls the weight to the left.

Free waves in the string are sine waves. Since the speed of the waves is finite, there is "retardation".

As the weight slows down, it creates new waves. The weight perturbs the string. Calculating the precise form of the wave probably requires a computer. But we are only interested in very crude estimates.

Do binary pulsars prove that there are no longitudinal gravitational waves? Are longitudinal waves confined between masses?

Binary pulsars have confirmed that transverse waves and linearized Einstein field equations explain the energy loss of a binary pulsar up to a precision of 0.1%.

In an earlier blog post we suggested that longitudinal waves must be "absorbed into" matter quickly, because they cannot propagate in empty space. This hypothesis would explain why binary pulsars match the transverse wave model so well.

Another hypothesis: longitudinal wave effects can only exist in the field between two masses M₁ and M₂ and can never escape to empty space.

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

Longitudinal electromagnetic waves exist in plasma, but they do not exist in empty space. Could this be analogous to our model of retarded gravity? Langmuir waves exist in a plasma which is, on the average, neutral. Since gravity charges always are positive and gravity always pulls, Langmuir waves cannot exist in a gravitating system.

A model with solid rubber inside

Let us have two masses M₁ and M₂ on the surface of the balloon. Visualizing a rubber string between the masses, and running on the surface of the balloon, is somewhat hard. Let us imagine that the rubber string runs inside the balloon.

The simplest case is when M₁ and M₂ are on the opposite sides of the balloon. A rubber string runs through the center of the balloon and connects them.

The balloon expands. The rubber string is not immediately aware of the expansion slowing down. This will cause oscillation in the expansion rate.

Photo publicdomainpictures.net

Let us have a solid rubber ball. The rubber will resist the expansion in most cases. But if there is a shock wave moving from the center toward the surface, the expansion of the rubber ball may actually accelerate at a certain time. "Dark energy" in this case is the energy of the shock wave.

This model predicts that the expansion rate of the universe will oscillate, while it, on the average, is quite similar to an FLRW model.

But we should find a mathematical formula, so that we could compare it to the observed dark energy.

The strength of dark energy seems to be weakening right now

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

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

The DESI project measures the 3-dimensional mass density variations of the universe right now, and compares it to the baryon acoustic oscillations after the Big Bang. The results of DESI, when combined to other observations, suggest that dark energy is weakening right now. Wikipedia says that dark energy density right now might be 10% less than 4.5 billion years ago.

This is consistent with our own rubber ball model: on the average, the expansion should be like an FLRW model, but there should be "bounces" from the retardation of gravity.

https://www.emergentmind.com/topics/chevallier-polarski-linder-cpl

In the link we have the definition for the CPL parametrization of the dark energy equation of state, w(a), where a is the universe scale factor. The current value of a is set to 1.

https://arxiv.org/abs/2503.14738

Abdul-Karim et al. (2025) write that the energy density of dark energy may have increased in the past. This is called a "phantom crossing". In our rubber ball model, a phantom crossing is expected: at some points the bouncing slows down the expansion, at other points speeds up the expansion.

The Hubble tension

Let us measure the baryon sound horizon size in the cosmic microwave background, and match it to the measured density fluctuations in the current universe. Assuming, for example, the standard ΛCDM model, we end up with a Hubble "constant" current value of

       67 km/s / Mpc.

But a direct measurement, based on Cepheids and type Ia supernovae yields a value

       73 km/s / Mpc.

The values differ too much to be just a statistical fluctuation.

https://arxiv.org/abs/2309.06795

Marco Raveri (2023) tries to use dark energy to make these values compatible. Let us check what he suggests. His conclusion is that dark energy modifications at "late" times, 0.01 < z < 3 cannot solve the Hubble tension.

Early dark energy (EDE)

https://academic.oup.com/mnras/article/533/4/3923/7750120

https://en.wikipedia.org/wiki/Recombination_(cosmology)

Xuejian Shen et al. (2024) write about an early dark energy (EDE) model, where an unknown field boosts the expansion of the universe just before the recombination (last scattering) which happens when the universe is 380,000 years old in a standard ΛCDM model.

Shen et al. say that this model can solve the Hubble tension. Is it so?

Let us start from the observed uneven matter distribution in the recent universe, and match it to the the uneven matter distribution in the cosmic microwave background (CMB). The baryon acoustic oscillations (BAO) make bumps into the uneven distribution. Let us match the bumps in recent matter and in the CMB.

Let us calculate time backward from the present time, and model matter distribution at earlier times. We use a standard ΛCMB model. When the universe has shrunk to one 1,100th of its size, the CMB has a temperature of 3,000 K, and the last scattering occurs.

The Hubble tension is that the BAO bumps in this calculation have a larger angular scale than in the measured and mapped CMB in the sky. That is, the "features" in the actual CMB map are smaller than our calculation backward in time predicts.

The early dark energy (EDE) hypothesis talks about what happened before the last scattering. The hypothesis does not help in any way in solving the Hubble tension. We have to check the literature. Can the EDE hypothesis be so much wrong?

https://www.researchsquare.com/article/rs-100387/latest.pdf

Karsten Jedamzik et al. (2020) write that EDE cannot resolve the Hubble tension.

 

Above, θ* is the angular size of the sound horizon, that is, the angular size of "features" in the CMB. We use coordinates comoving with matter. Then r* is the spatial size of the sound horizon (in those coordinates), and D(z*) is the spatial distance to the last scattering.

Jedamzik et al. say that only the quantity

       Ωm h²

affects the spatial distance to the last scattering, in a flat ΛCDM model. There, Ωm is the matter fraction of the critical density, and h is the value of the Hubble constant now.

The authors state that the Dark Energy Survey and the Kilo-Degree Survey have set strict limits on Ωm h², so that any attempt to solve the Hubble tension through manipulating Ωm h² will fail.

Let us check what they say about manipulating r*. Suppose that we calculate the development of a ΛCDM model backward in time. The matter density variations of galaxy clusters eventually turn into variations of the CMB as the temperature rises.

Solving the Hubble tension with a very slow expansion during the first billion years of the universe: also the James Webb paradox of too mature galaxies is solved?

The James Webb telescope has observed galaxies which in a standard ΛCDM model are only 500 million years old, but look much more mature, as if they were 1.5 billion years old.

Let us assume that the expansion during the first billion years of the universe was much slower than the ΛCDM says. We add an "extra" 1 billion years to the age of the universe, to the early phases of the universe. Adding that extra means that the universe expands much slower than in the ΛCDM model, during the first 1 billion years or so.

Then galaxies have much more time to mature. This would solve the James Webb paradox.

During that slow expansion during the first 1 billion years, photons from the last scattering had time to move a "longer" path than in the ΛCDM model. Let us explain what "longer" here means. If we model the expanding universe with coordinates comoving with matter, then the ruler of the spatial metric (say, a ruler 1 meter long) shrinks as time progresses. The "length" of the path of the photons is measured in these comoving coordinates.

Since the photons moved a longer path before coming to our eyes, the features in CMB will have smaller angular diameters in our eyes. This resolves the Hubble tension.

A quantitative calculation of making the universe to expand slower during the first 1 billion years

Let us make a quantitative calculation. We assume a matter-dominated universe with a zero pressure. Then we can use newtonian mechanics to calculate the development of the universe backwards from the present time.

              "feature" size

               <----->

              ●        ●        ●       ●

              ●        ●        ●       ●

                             o ~~~~~  last scattering

                             |     D

                            /\ 

                      observer       c ~ 1 / a

We set a = 1 as the current scale factor of the universe. In comoving coordinates, the speed of light

       c  ~  1 / a.

Let us look at eleven "epochs" where

       a  ≈  1 / 2ⁿ,

0 ≤ n ≤ 10.

The epoch 10 corresponds to the time of the last scattering, that is z ≈ 1,100.

In ΛCDM, the kinetic energy of a particle m in static (not co-moving coordinates) is very roughly

       1/2 m v²  ~  1 / a  =  2ⁿ.

That is, v ~ sqrt(2ⁿ).

This implies that the time the model spends in an epoch n is

       tₙ  ~  (1 / 2ⁿ)  /  sqrt(2ⁿ)

              = 1 / 2^(1.5 n)

              = 1 / 2.8ⁿ

              = 0.35ⁿ.

The sum of tₙ is t ~ 1 / (1 - 0.35) = 1.5.

The time lengths of various epochs:

   t₀ : 9 billion years

   t₁ : 3 billion years

   t₂ : 1 billion years

   t₃ : 350 million years

   t₄ : 100 million years

   ...

   t₁₀: 140,000 years.

The speed of light c ~ 1 / a = 2ⁿ. Light in that time propagates a distance in comoving coordinates:

       Dₙ  ~  2ⁿ  *  1 / 2^(1.5 n)

              = 1 / sqrt(2ⁿ)

              = 1 / 1.4ⁿ

              = 0.7ⁿ.

We are interested in the distance

               10

       D  =  ∑ Dₙ

              n = 0

(in comoving coordinates), which light travels from the last scattering, into the eye of the observer. If the contribution of epoch D₀ is 1, then the sum of the series is

       D  =  1 / (1 - 0.7) = 3.3.

The contributions of various epochs:

  D₀ : 1

  D₁ : 0.7

  D₂ : 0.5

  D₃ : 0.35

  D₄ : 0.25

  ...

  D₁₀: 0.03.

In particular, the contribution of epoch D₁₀ is only 0.03. We see that we have to modify the physics drastically during epoch 10, if we want to increase D by 8%. This shows that solving the Hubble tension by tampering only the first 400,000 years of the universe is an implausible solution.

The first one billion years of the universe corresponds to epochs 2, ..., 10. The sum of D₂, ..., D₁₀ is 1.6.

We see that if we make the universe to expand 16% slower during epochs 2, ..., 10, we can increase D by 8%. Then the angular size of "features" will be properly smaller to match how Planck and other probes saw the CMB.

But now we realize that we only have to add 160 million years to the early history of the universe. That would not explain James Webb observations.

If we double the time length of epoch 3 to 700 million years, we increase D by 11%, and add 350 million years to the early history of the universe. Could this be enough to solve the James Webb paradox?

Dark energy means that the universe has been expanding surprisingly fast during the past 7 billion years. Maybe the expansion rate fluctuates, and was surprisingly slow during the first 500 million years of the universe?

Conclusions

We still do not have a mathematical model which we could use to calculate the effects of retarded gravity. Intuitively, retardation should cause the universe to expand surprisingly slowly at certain times, and surprisingly fast at other times. That is, like a solid ball whose elastic interior can contract and expand.

We will write a new blog post about the Hubble tension. Literature strangely ignores solutions where the expansion of the universe is slowed down during the first 350 million years. Literature seems only to talk about modifying physics at z < 1 (the last 7 billion years), or at z ~ 3,000 (the first 100,000 years).

Why does literature ignore the most obvious solution? Slowing down the first 350 million years could solve also the James Webb paradox of "too" mature galaxies at z ~ 14.

de Broglie and Bohm: photon creation from a radio transmitter – the problem of new particles

Let us consider a simple semiclassical process which creates photons. Let us have two charged particles (charge Q) whose mass M is macroscopic. We let the particles pass close to each other, so that they send a radio wave with many photons.

                                                                    screen

                                                                          |

                                                                          |

                                                                          |

                              )   )   )   )   )  radio wave

           ● ---->              

          M, Q                 <---- ●

                                         M, Q

The radio wave will hit a screen far away, and we can determine the positions of individual photons.

We can calculate the shape of the radio wave classically, but we cannot calculate where the individual photons will hit the screen.

The process creates new particles. Is there a way to extend the de Broglie-Bohm model to this case?

Could this work: the photons "pre-exist" in the electric fields of the charges Q, and are "freed" in the scattering process? Then all the particles could have a hidden variable already at the outset of the experiment, and would sail in a deterministic way to the screen – just as in the double-slit experiment.

The outgoing photons have an uncountable number of possible momenta p. Even if we would have pre-existing photons in the field of Q, how would these obtain the required p?

Time asymmetry in the process

The initial configuration of two charges Q flying at each other is quite different from the final configuration where we have the wave function for the two charges Q plus many alternative configurations of photons. The entropy of the system grows, and there are more degrees of freedom in the final configuration.

What if there would be no growth of entropy? Then the initial and the final configurations should look similar. Maybe the number of the degrees of freedom does not grow any more?

What about inverting the direction of time? Then the degrees of freedom will shrink as time progresses.

Absorption of a photon by a hydrogen atom

An excited hydrogen atom can decay and send an individual photon. The atom can absorb the required momentum from the reaction as it launches the photon.

What about the reverse process? A wave function of the photon contains many cases in which the photon does not hit the atom at all. The photon passes the atom.

Can we somehow add more "worlds" in such a way that the the graph of possible worlds would look more time-symmetric?

In the case of a hydrogen atom absorbing a single photon, the degrees of freedom only shrink in the case of the absorption. In other cases, there is no change. A "wave function collapse" only happens in one branch of many-worlds.

Ignore photons and only consider (charged) particles which existed at the outset?

The de Broglie-Bohm approach might work if we only had undestructible particles. Let us have N such particles. Let us have a foliation of spacetime such that spacelike folios define "simultaneousness". Then we could replace the N particles with one particle moving in a space with 3 N spatial dimensions.

But we have a problem already with electrons: pair production and pair annihilation. There are no undestructible particles.

Give up determinism?

Albert Einstein said that God does not play dice. That is, the "next", measured, state of a physical system should follow deterministically from the "previous" state.

Since simultaneousness is not well defined in relativity, we maybe should say this: what happens at some point in time and space is determined by the light cone which precedes that point.

Why not require determinism also in the "initial" states of the particles? Why can the initial states be random? Does God play dice, after all?

Why is the time dimension different from spatial dimensions?

Defining a "branch" in many-worlds

Let us have a human living in spacetime. Most of the massive elementary particles in his brain survive for a long time. They are not annihilated. However, quite a few particles are replaced with other similar particles over his lifetime. How could we define the "branch" in many-worlds, in which he is living?

The brain consists mainly of electrons and of nuclei. These particles are, in most cases, conserved over 100 years.

Then we can apply the de Broglie-Bohm model to these particles, and obtain well defined "branches" of their configurations.

We do not try to define a branch in a more general sense: we do not define, e.g., which photons on Mars belong to his branch.

What if two photons collide and form an electron and a positron, and the electron subsequently is used as a new building block in his brain?

Could we simply accept a model in which the branch of this brain chooses "by random" the initial state of the electron, as it is integrated into the brain?

A new problem: which paths of separate particles are in the "same" branch of many-worlds?

Suppose that two electrons are created in pair productions, and these two electrons are integrated to the brain of the observer.

We can use the de Broglie-Bohm model to describe the paths of the two electrons. But what paths are in the "same" branch of many-worlds?

This might be the solution: we simply choose one de Broglie-Bohm path for each electron. Then the combination of those paths is taken to be in the "same" branch as the brain of the observer.

When a new particle is included into the brain, the brain performs a "quantum jump" to include a certain path of that particle.

Do we need precisely defined branches at all?

Definition of a branch does not change anything in the wave function. In the de Broglie-Bohm model, the path of a particle does not affect the wave function in any way.

However, picking a certain path may define the "real world".

Let us have an observer. Maybe he lives in the "real world", and the copies of him in other branches are just "empty" wave functions?

This does not sound natural. Why should a conscious observer always exist in the "real world" and not in the "empty" branches?

In the de Broglie-Bohm model, the empty branches do physically exist, too. They are just different branches of the wave function.

Maybe it is necessary to drop the concept of the "real world" and treat all the branches equal. This means that Schrödinger's cat is both dead and alive at the same time.

Do we have a reasonable definition of a "branch" now? A robot

Instead of an observer we may consider a computer or a robot which possesses a camera and other sensors. It also has arms for manipulating external objects.

Above we suggested that we can use the de Broglie-Bohm model to define a "branch" of many-worlds, such that an instance of the robot lives in that branch.

The robot will encounter various random (or pseudorandom) quantum phenomena, and adjust the contents of its memory accordingly. Its camera may register random photons hitting its CCD sensor. 

The robot is an almost classical machine. It probably has relatively exactly definable states, and it in most cases functions deterministically.

Is the de Broglie-Bohm model an awkward way to define a branch in this case? Why not use a traditional Copenhagen model where the next state of the machine depends probabilistically on the previous state?

Assigning hidden variables is awkward?

The observing subject

Suppose that we have a reasonable way to define a branch for the robot. The computer in the robot, in that branch, will have a similar world-view as a human observer. It will remember its history. In almost every branch, the robot notices that physical processes in its history behaved according to the probabilistic rules of quantum mechanics.

According to Rene Descartes, cogito ergo sum. That is, I as an observing subject do exist.

My subject might be attached to one branch of a robot, my physical body, or it might "jump" from branch to branch. I would not notice anything if my subject jumps around the branches of many worlds.

It sounds strange if I alternate between different branches. I very much feel that I am in just one branch. This paradox is similar to another paradox about time. In what sense is now different from moments in the past or in the future?

Conclusions

It would be nice if we could define a single "branch" in the many-worlds interpretation. Then we could imagine that a human observer lives in that branch.

The de Broglie-Bohm model gives us a way to define a branch for a robot. Simply plot the paths of the nuclei and the electrons in the mechanism of the robot.

However, that model is not very natural. Why should the branch exactly follow the strange paths of de Broglie-Bohm model?

Maybe a more natural way to define a branch is more fuzzy: it is the collection of those states of the robot which are "similar". For example, we may require that the bits in the memory of the robot have identical values in the collection of the states. This would be a "fuzzy branch".

We still have the question if an "observing subject" is somehow attached to a certain branch. This is a metaphysical problem which is hard to clarify and solve. What makes a certain branch "real" for the subject? How would the subject know if it jumps from branch to branch?

A similar metaphysical problem is in what sense is the world "now" more real than the past or the future. We have no answer.