Explaining the nova and supernova phenomena with new physics theories -2, by Arindam Banerjee

[Arindam Banerjee]

Explaining nova and supernova with new physics theories – 9
Solar flares and their causes - 4
The Models and Calculations of Solar Energy without fusion – 2

The calculation of the solar energy from purely gravitational effects does not have anything to do with the constitution of the core – whether it is hot, or cold. Similarly, the calculation of the solar energy from purely electromagnetic effects does not, strictly speaking, depend upon the constitution of the core. We will find it from the measurement data relating to the strength of the magnetic field in the photosphere; the density of ions and their speed.

As we will see, solar energy is mainly of electromagnetic origin. The main reason for the energy of our Sun is the Sun’s magnetic field interacting with the ions (charged particles) in the photosphere. Just as the Earth appears to have a large bar magnet inside it (with the North of this magnet near the Geographic North Pole, and the South of the magnet near the Geographic South Pole), the Sun too appears to have a huge bar magnet inside it. From these poles, magnetic lines of force spread out all over the Sun, reasonably uniformly save – very importantly – for the magnetic spot regions where they are far more intense.

We will now see what is known as magnetic field strength, also known as magnetic induction. It is a vector, which means that it has a direction as well as a magnitude, and is represented by the letter B. Consider all the magnetic lines of force (usually called magnetic flux) of the Sun, cutting into the photosphere region. These are all parallel lines, going from the Magnetic North Pole of the Sun to the Magnetic South Pole of the Sun. Any cross section of the entire photosphere region, that is, a great circle with a very large hole in it, is the surface region of the Sun’s total magnetic flux through the photosphere Call it A. Let the total magnetic flux by called, Sun-Flux. Then the average B at any point in the photosphere of the Sun is Sun-Flux divided by A; its direction is the direction of the magnetic flux at any point. Its value is given as 1 gauss; which is double that of the Earth’s magnetic field strength, or magnetic induction.

Just as an uncharged particle is attracted by a gravitational field, and accelerated by gravitational force; a charged particle (ion) is affected by a magnetic field, and accelerated by electromagnetic force, when moving with respect to the magnetic lines of force in the magnetic field. However, there is a difference. The uncharged particle moves in the direction of the gravitational field – thus a ball drops straight down to the ground by gravity, does not move sideways. But for the charged particle moving in a magnetic field, the direction of its acceleration is perpendicular both to the magnetic field and the direction of its motion. In short, it moves sideways, not down as is the case for gravity.

An ion (electron or proton) that is moving in the direction out of the Sun, will be affected by the magnetic lines of force that are running parallel to the Sun. The force upon it will act sideways – making it veer away from its original outward path by accelerating it toward a sideways, parallel path to the Sun. Thus a recently ionised hydrogen atom, an electron and proton that is, will have its two components accelerating in opposite directions. All ions moving up will behave like that; now all ions moving down will have their electrons going towards protons and protons going towards electrons. They will combine, once again to form the hydrogen atom, and in this arrest of its kinetic energy, give out radiant energy from the excited newly formed hydrogen atom. Prior to this, the ion may have collided with hydrogen atoms, causing ionisation with formation of radiant energy, and also radiant energy, part of which is visible light. If the movement of any ion is parallel to that of the magnetic induction vector, then no force will act upon the ion. But at any other angle, the force will be kept. Incidentally, this is the way the ions in the ionosphere of the Earth or Sun are kept attached – the ions that are not very fast are moved aside by the magnetic field strength, so that they remain parallel to the Earth (or Sun). However, ions (protons and electrons) do escape from the Sun, especially at the times of great solar flares when they get too much outward velocity.

By now our model for the Sun’s energy is getting clear – the magnetic field of the Sun accelerates the electrons and protons (formed by the ionisation or electron deprivation of the hydrogen atom) so that they smash against atoms causing ionisation and radiant energy, before combining again to form hydrogen atoms. This is a cyclic process – ions with increased kinetic energy smash into atoms, creating more ions, before recombining into atoms. This process would not have worked had there not been a magnetic field to accelerate the ions. No loss of mass is involved. We note that the charge on the electron is in magnitude the same as the charge of the proton. These are oppositely charged. This means that under the magnetic field they go off in opposite directions – they do not cancel each other as may appear in mathematics as they have different signs. To recombine, a proton thus created from a hydrogen atom must meet an electron created by the ionisation of another hydrogen atom.

From the physics textbook, Halliday Resnick Vol 2, Chapter 33-2 “The Defintion of B” we go straight into the formula for force upon a charge in a magnetic field. I will be quoting from the book, below.

“Let us fire a positive test charge q with arbitrary velocity v through a point P. If a sideways deflecting force F acts on it, we assert that a magnetic field is present at P and we define the magnetic induction B of this field in terms of F and other measured quantities.”
“if we vary the direction of v through point P, keeping the magnitude of v unchanged, we find, in general, that although F will remain at right angles to v its magnitude F will change. For a particular orientation of v, and also for its opposite orientation –v, the force F becomes zero. We define this direction as the direction of B.”
“The magnitude of the magnetic deflecting force F… is given by
F= q * v * B * sin(theta),
where theta is the angle between v and B.”

Let us see how the above formula can help us to show whether or not the electromagnetic force is responsible for most solar energies.

The force acting upon a proton will be thus
F= q(proton) * v * B * sin(theta).
There will be very many protons as we shall soon see, and let us assume that at the time of their ionisation there was this unknown velocity v which let us hold is the average velocity for all protons. Also there is a sin(theta) value which should be 45 degrees for the average, as half the protons will be clustered statistically around the direction of B, and the other half perpendicular to it.
There will also be the force acting upon the electron, which is
F = q(electron) * v * B * sin(theta).
Force by Newton’s second law of motion is mass * acceleration. So the acceleration of the proton due to the electromagnetic force upon it is
Acceleration of proton (Ap) = q(proton) * v * B * sin(theta)/(mass of proton), and similarly
Acceleration of electron (Ae) = q(electron) * v * B * sin(theta)/(mass of electron).
We find from above that the acceleration of the electron is many orders of magnitude higher than the acceleration of the proton, as the electron is correspondingly lighter. The electron will move much faster. If it cannot combine with a proton in the photosphere, it will escape into the Sun’s ionosphere, where, if it will meet a proton it will recombine to form a hydrogen atom and descend to the photosphere. Or else, be part of the ionic drift. The high velocity of the electron thus does not contribute to solar power as such – that is, not the way that proton-hydrogen collision does for the energy we get.
In time t, the proton under this force alone, with no consideration for any initial velocity (for conservative reasons, as we are only considering electromagnetic force impact) moves a distance Sp according to the formula
Sp = 0.5 * Ap * t^2.
In one second, the proton will move a distance Sp = 0.5 * Ap. Similarly in one second the electron will move a distance Se = 0.5 * Ae.
The energy from electromagnetics obtained by the proton in one second is the force multiplied by the distance over which the force acts. It is
Energy of proton in one second, or power from proton = (mass of proton)*Ap*Sp.
Similarly for the electron
Energy of electron in one second, or power from electron = (mass of electron)*Ae*Se. This power of the electron is very much more than the power of the proton! Our task here is to show that fusion at the core of the Sun is not responsible for the Sun’s energy. To prove our task, let us first ignore the power from the electron as the cause of the solar energy. We will, however need it to explain solar flares, novae and supernovae.
Power from the ionised hydrogen atom, neglecting the impact of the negatively charged electrons is Force multiplied by distance per second, or mass of proton multiplied by its acceleration and the distance covered in one second by that force. Call it Piah, so
Piah = (mass of proton)*Ap*Sp.
Substituting the values of Ap, Sp and simplifying
Piah = (mass of proton) * (q * v * B * sin(theta)/(mass of proton)) * 0.5 * (q * v * B * sin(theta)/(mass of proton)). or,
Piah = 0.5(q*v*B*sin(theta))^2/(mass of proton)
Now let us put in some data values here, obtained from books and internet. Mass of proton and its charge are well known. They are 1.67 * 10^-27 Kg and 1.6 * 10^-19 Coulombs respectively. With theta held at 45 deg, sin(theta) is 0.707
The value of B is 1 gauss, and in MKSA it is 10^-4 Tesla. The link for the quote is given below.
https://www.windows2universe.org/sun/sun_magnetic_field.html#:~:text=The%20Sun%20has%20a%20very,Earth%20(around%200.5%20Gauss).
The Sun has a very large and very complex magnetic field. The magnetic field at an average place on the Sun is around 1 Gauss, about twice as strong as the average field on the surface of Earth (around 0.5 Gauss).

Piah (watts) = 0.5 * (1.6 * 10^-19 * v * 10^-4 * 0.707))^2/1.67 * 10^-27, or
Piah (watts) = 0.5 * v^2 * (1.132 * 10^-23)^2/1.67 * 10^-27 = 0.5 * v^2 * 1.28 * 10^-46/1.67 * 10^-27, or
Piah (watts) = 3.8 * 10^-20 * v^2

We now find out how many protons are there in the photosphere.
The mass of the photosphere has been found in the preceding section to be (assuming a depth of 400 Km and a density of 3 * 10^-4 Kg per cubic meter to be 3.65 * 10^20 Kgs.
Let i be the fraction of the hydrogen in the photosphere that is ionised.
Then the total mass of the ionised mass in the photosphere is i * 3.65 * 10^20 Kgs.
Going by the fundamentals of Physical Chemistry, in 1 gram of Hydrogen – a mole - there are N molecules of Hydrogen, where N is the Avogadro’s Number, 6.02 * 10^23. In one Kg of Hydrogen there are then 1000 * N atoms, or 1000*N protons when their atoms are ionised (deprived of electrons).
The total number of protons in the photosphere (Npp) is thus 1000*N*(i * 3.65 * 10^20) units. Or i * 3.65 * N * 10^23, or i * 21.93 * 10^46, rounded to
Npp = i * 2.2*10^47

Now the unknowns are i and v. Let us see what a ball-park value of i (the degree of ionisation in the solar photosphere could be). Internet search gives us this link:
https://astronomy.stackexchange.com/questions/7883/what-is-the-degree-of-ionization-is-the-solar-photosphere
where it is written:
“The book "Introduction to Stellar Astrophysics" by Boehm-Vitense, vol. 2, 1997 reprint, on page 76 claims that one out of every 10^4 hydrogen atoms are ionized, so hydrogen is mostly neutral in the photosphere;”
In the same link we have a more detailed reference
“The degree of ionization in the photosphere varies with depth of course, but overall it is small. Table 1 of the Bilderberg Continuum Atmosphere (Solar Physics, 3, 5, 1968) gives the pressure and the electron pressure at various optical depths in a comprehensive model. The ratio of the pressure gives the Ne/Ntotal. At optical depth = 1 at 5000 Angstroms, Pgas = 1.412E5 (cgs) and Pe = 6.239E1. The ratio is 4.4E-4. The hydrogen ionization, also given in the table is 4.07E-4.”

We will take, for the cause of conservativism, the lower value of 10^-4 for the value of hydrogen ionisation in the photosphere. Then the number of protons in the photosphere will be:
Npp = 2.2 * 10^43.
The total solar energy from the photosphere, caused by proton movement alone, will then be Npp * Piah = 2.2 * 10^43 * 3.8 * 10^-20 *V^2 = 8.36 * 10^23 * v^2 watts.
Now, what is left is to find the value of v, which is the velocity of the proton in the direction perpendicular to the magnetic lines of force. Its absolute velocity is not the value required. Internet searching in “Google Scholar” for the value of the velocities of the atoms in the photosphere shows us in the link https://link.springer.com/article/10.1007/BF00153898
for the paper “A physical mechanism for the production of solar flares” gives a figure of greater than1 km/second for the velocity of the gases in the photosphere. Other sources give much higher values. These values are absolute values – their component that is perpendicular to the magnetic lines of force is not known.

Taking the value of v as 1Km/second or 1000m/s and putting it in the above equation we get the value of total solar energy from the photosphere to be 8.36 * 10^29 watts.

Hey! The actual total solar energy as calculated (and verified by experiment) is only 3.864 * 10^26 watts. And of that only 3.5% is from gravitational forces. Our calculation above gives the electromagnetic sources for solar energy over 2000 times the actual solar energy. We have been very conservative – we have not taken the energy from electrons for the energy from the photosphere; we have taken a low value for the ionisation of the photosphere. Some sources say that the photosphere is only 100Km in depth, not 400Km as we have taken. Even when we take the former value of 100Km, the energy from electromagnetic forces would still be 500 times the actual energy.

This does amount to an overkill for my proposal that the solar energy is NOT from fusion at the core. Of that, there can be no doubt. So my main thesis, that the Sun has a cold iron core, where circulate huge currents from superconductivity, causing the magnetic field, which moves the ions to create the bulk of the solar energy, is validated.

The huge value of 8.36 * 10^29 arises from our taking the value of v to be 1000m/s. This v can never be known by measurement, for we cannot experiment on the Sun, but it can be calculated from measurement. What is it? Well, we have found in the earlier section that the solar energy from gravity alone is 13.7 yotta watts. The solar energy not from gravitational forces is then (386.4 – 13.7) yotta watts or 372.7 yotta watts. This then should be equal to the power from the electromagnetic forces. Taking the photosphere to be 400 Km deep, and with i=10^-4, B=10^-4 teslas we get the relation
8.36 * 10^23 *v^2 = 372.7 * 10^24.
From where v^2 = 3727/8.36 or v = 21 m/s

If the photosphere were 100 Km deep, there would be one fourth the amount of protons, in which case the value of v would be only double at the very reasonable value of 42 m/s. Thus it seems fairly certain that solar energy need have nothing whatsoever to do with very hot fusion processes in the core of the Sun. That the core of the Sun, like any other star, is very cold, and unaffected by whatever going on around it, is crucial for our understanding of novae and supernovas.
In the next section we will consider the impact of the fast moving and powerful electrons for creating solar flares.

Arindam Banerjee
Melbourne 26/09/2020

Explaining nova and supernova with new physics theories – 10
Solar flares and their causes – 5
The Models and Calculations of Solar Energy without fusion – 3

Once we have a logical explanation for solar flares, it will be more easy to understand what causes novae and supernovas.
Quoted from the Wikipedia, information about solar flares: https://en.wikipedia.org/wiki/Solar_flare
“A solar flare is a sudden flash of increased brightness on the Sun, usually observed near its surface and in close proximity to a sunspot group. Powerful flares are often, but not always, accompanied by a coronal mass ejection. Even the most powerful flares are barely detectable.”
“Solar flares… an energy release of typically 10^20joules of energy suffices to produce a clearly observable event, while a major event can emit up to 10^25 joules.[2]
“Flares occur when accelerated charged particles, mainly electrons, interact with the plasma medium.
“Flares occur in active regions around sunspots, where intense magnetic fields penetrate the photosphere to link the corona to the solar interior. Flares are powered by the sudden (timescales of minutes to tens of minutes) release of magnetic energy stored in the corona.
“Although there is a general agreement on the source of a flare's energy, the mechanisms involved are still not well understood. It's not clear how the magnetic energy is transformed into the kinetic energy of the particles, nor is it known how some particles can be accelerated to the GeV range (109 electron volt) and beyond.”
We will offer explanations about the above facts relating to solar flares, with our new approaches in physics.
As per our new model of the visible Sun, there is within it an Earth-like non-gas sphere of radius around 433,802 Km encased within a largely Hydrogen atmosphere of depth 261,197 km. Let us call this non-gas sphere SE.
Let us now picture what the SE surface may be like. As we assume it to be Earth like, it should be mostly rocky with other minerals in it, like iron. Under such an extraordinarily high pressure from layers of hydrogen 261,197 Km deep, it should be very hot, but not gaseous as the enormous pressure will keep it in a molten state, as a thick liquid. It could be like sand on the desert shifted by winds to form large dunes. These dunes could be very high, and its position not permanent.
In the context of solar flares, it is likely that such dune-like structures, of enormous size, should contain disproportionate amounts of magnetic material like iron. The Sun’s magnetic field, which should be uniform as on Earth, as it is effectively formed by a huge bar magnet with North and South poles, is thus skewed by the magnetic material concentration on the surface of SE. This skewing – just as for the ferrite antenna in a mobile phone - causes concentrations of magnetic field strength over extensive areas. Wherever there are excessive concentrations of magnetic field strength, we have sunspots – such is their characteristic. It is only when we consider as is now done that the Sun to be homogenously filled with hydrogen, mainly, and not any ferrous material, that we have difficulty in explaining what causes sunspots.
Sunspots are areas of the sun where the Sun’s radiant energy intensity is relatively low. It is held that this is so because of the very strong magnetic fields there, acting as a blank for the radiation. Earlier we had shown that the Sun’s energy comes from its magnetic field accelerating the ions in the photosphere. It would appear that there is a contradiction – if the Sun’s magnetic field causes the Sun’s energy, then why is there less energy radiating from the sunspots? Why do solar flares happen at the edge of the sunspots, and not within the sunspots?
The answer could be that with the development of the strong magnetic field (with the shifting hills of magnetic material on the SE below) there is an ion depletion region in the photosphere above. The strong magnetic fields, exerting relatively more force, drive out most of the ions outside the sunspot, more readily than elsewhere; so the normal electromagnetic forces accelerating the protons, as happens elsewhere in the photosphere, does not happen to that extent in the sunspot. This causes the cooling effect on the sunspots.
Solar flares often occur at the edges of the sunspots. At the edges, the magnetic field strength from the Sun’s core is still strong but not strong enough to drive away most of the ions, as in the sunspot. The presences of a still strong magnetic field, and sufficient ions, are the underlying causes of solar flares. These are directly caused, as mentioned earlier, by very fast electrons. In the earlier section we had seen how the protons, accelerated by the Sun’s magnetic field, are mainly responsible for the energy radiated from the Sun. Now we will see the role of electrons in creating solar flares, among other issues.
We had earlier derived, with the notation given earlier, that
Acceleration of electron (Ae) = q(electron) * v * B * sin(theta)/(mass of electron).
This is the acceleration of an electron at the edge of the sunspot, when it has just been ionised and is moving at a speed that has a component v perpendicular to the magnetic lines of force. V had been estimated to be between 20-40m/s, let us take it as 30m/s. B here is much higher than normal; let us say it is 100 times (this is erring on the side of caution, it should be much more, check out https://iopscience.iop.org/article/10.3847/2041-8213/aaa3d8 “Here we report clear evidence of the magnetic field of 6250 G”) in which case B will be 10^-2 teslas. With theta equal to 45 deg, the acceleration of the electron Aes near the edge of the sunspot will be in the region of:
Aes = 1.6 * 10^-19 * 30 * 10^-2 * 0.707 / 9.1 * 10^-31 = 3.7 * 10^10 m/s/s.
This is an extremely high acceleration! Normally the acceleration Ae is of the order of 3.7*10^8 m/s/s which is very high. With this latter value of acceleration and assuming the average photosphere depth to be 200Km, the electron would go through the photosphere in a very short time as compared to the proton. It would reach the Sun’s outer layers, the chromosphere and the corona. Those regions would be overpopulated with electrons, and thus create a potential difference between them and the photosphere. This potential difference would have a dampening effect upon the electron’s velocity, which would slow down on its outward path. By getting continually deflected tangentially by the Sun’s magnetic field it would remain bound to the Sun unless the initial velocity was too high, in which case there would be ionic drift. This potential difference, creating a strong electric field, would drag up the loose protons in the photosphere, to form hydrogen atoms which would then drop to the photosphere with the Sun’s gravity, if not sent out as ionic wind (positive this time) if there is no such recombination.
Let us investigate this issue of normal solar energy a bit further, to show the cyclic nature of the mass-energy process involved, which has no need for any kind of nuclear reaction, fission or the supposed fusion, to happen.
The normal average electron to go through 200 Km of photosphere would take from the formula S=0.5att, or t=sqrt(S/0.5a)=sqrt(2*10^5/(0.5*3.7*10^8))=sqrt(1.08*10^-3)=sqrt(0.00108)=0.032seconds.
The normal average proton (where B = 1 gauss) would take by the same analysis:
Ap = 1.6 * 10^-19 * 30 * 10^-4 * 0.707 / 1.67 * 10^-27 = 20.3*10^4m/s/s and t=sqrt(S/0.5a)=sqrt(200000/(0.5*20.3*10^4)) = sqrt(20/10.15) = 1.4 second.
Since the electron stays in the photosphere a lot less time than the proton, 1.4s as compared to 0.032s, the chances of its hitting another proton or hydrogen atom is correspondingly less. Thus, as shown in the earlier section, the main solar energy comes from protons hitting hydrogen atoms in the photosphere, creating radiant energy from their kinetic energy - much as light bulbs do in our homes.
Returning now to the issue of solar flares resulting from highly energetic electrons - those electrons moved by the powerful magnetic fields do not last long in the photosphere, as shown by the above calculations. As they are very fast, if they do strike any hydrogen atom they can cause ionisation. In the strong magnetic field, the electrons thus generated will move very fast and create more ionisation; thus a chain reaction may be created. There will be many high energy electrons created, and that is one characteristic of the solar flare. Those electrons should have a net direction, away from the sunspot. This lack of randomness, thus, in their movement, will lead to the formation of currents, in turn creating the strong magnetic fields associated with sunspots. Again, these magnetic fields will accelerate the ions, creating more collisions and more current.
The above activities do not explain the great energies associated with solar flares, and the radiation. For that, a conceptual leap is required. Simply put, a solar flare is a long and powerful hydrogen bomb that is happening on the surface of the Sun.
This assertion (while obvious to any beholder with a telescope) goes against the current theory, which holds that fusion of causes hydrogen bombs, only the core of the Sun is hot enough (through the pressure from surrounding masses) to create the conditions for fusion. Nuclear fusion, converting deuterium hydrogen into helium, cannot happen on the surface of the Sun, as the surface is not hot enough. As I have shown earlier in this series of articles, this view, though current, becomes unacceptable when it is held that the core of the Sun (as also Earth) is very cold; and that the hydrogen bomb created on Earth works not with the fusion of hydrogen to helium, but with the splitting of the deuterium atoms to two protons and an electron. The protons then repel each other with great force; gamma rays are created with the sudden increase of the electric field at very close distances; and high energy electrons are released. With the creation of gamma rays and high energy electrons – many of which escape to outer space along with the protons they drag up with their powerful electric field – a chain reaction is started, which will spread all over the contiguous areas.
On the Sun, the conditions relating to very fast electrons are similar to what is done with fission atomic bombs. It is known that gamma rays are responsible for the action of the hydrogen bomb; but the beta rays from radioactivity are also very powerful. The very fast electrons, caused by the intense magnetic field to begin with, can break up the deuterium isotope into two protons and an electron, just as the hydrogen bomb on Earth; and initiate the chain reaction in the photosphere. It is important to note that only very powerful electrons can break up the deuterium nucleus to two protons and an electron. Less powerful electrons can only break it up into a neutron (that is, a proton-electron closely bound set) and a proton. This sort of fission merely soaks up energy and does not create the great energy caused by proton to proton repulsion. (Note: I am indebted to Michael Moroney of sci.physics for this information.) Laboratory experiments on Earth that break up the deuterium into a proton and a neutron cannot replicate what is happening on the Sun; so the notion that the fission of deuterium soaks energy instead of creating energy is prevalent as of date.
One would then ask, why should not the whole photosphere blow up? The answer is important, for our understanding of novae, where much of the photosphere does blow up – for whatever reason! The entire photosphere, as also much of what lies below, blows up in a supernova leaving only a core behind. (That core is currently considered a neutron star, or a black hole.)
The continuation of the hydrogen bomb explosion depends primarily upon the availability of deuterium isotope. When that supply is depleted locally, the explosion has to come to a stop. The second issue is the strength of the magnetic field. When moving away from the sunspot region, the magnetic field strength will decrease - so the initiating condition for the fission of deuterium, as discussed earlier, will no longer remain. The weaker electrons can only ionise, or break up the deuterium nucleus into a proton and a neutron.
To solidify the case that a solar flare is a hydrogen bomb in action, let us now consider some figures mentioned earlier. The maximum power for the solar flare is around 10^25 joules, a typical solar flare is 10^20 joules, while the power of a one megaton hydrogen bomb is about 10^15. We have worked out that this bomb requires 100 Kg of heavy water where all the hydrogen is deuterium.
100 Kg of heavy water will contain 20 Kg of deuterium hydrogen. Since the power of the typical solar flare (barely visible) is 100,000 times more powerful, it will involve 2 million Kg of deuterium hydrogen. Now the density of the photosphere is 0.2 gm/cubic meter, or 2*10^-4 Kg/cubic meter. 2 million Kgs of deuterium will occupy a volume of mass/density = 10^10 cubic meter. On Earth the percentage of heavy water is .02%, so assuming it the same on Sun’s photosphere, the solar flare should occupy a core volume of 10^10/.0002 or 5*10^13, about a cubic volume of side about 37 Km in length. Just as a hydrogen bomb explosion on Earth is not confined to its bomb dimensions, it is clear that the overall explosion will cover a far vaster volume, and so be observed from Earth with a telescope.
In the next section we will apply the new theory that solar flares are hydrogen bomb explosions to the understanding of the nova phenomena, as observed.
Arindam Banerjee
Melbourne, 6 Oct 2020

Explaining nova and supernova with new physics theories – 11
The cause of novae - 1

A fair amount of background material has been covered in the earlier sections, in order to understand the alternative theories to explain the phenomenon of nova and supernova. The new model of any star is a Earth-type sphere contained within a thick atmosphere of hydrogen, with a certain percentage of it being deuterium – that is the isotope of hydrogen where there is an extra neutron added to the proton. The surface of this sphere is thick silicon based liquid, continually shifting and reforming under the extreme atmospheric pressures. Below this liquid, which may be as deep as a few hundreds of thousands of kilometres, there are more solid rocky layers, which get increasingly cooler as they near the core. These layers insulate the heat from above, and act as a shell to absorb the pressures from around; so the core is cool and may be very cold if the star has a strong magnetic field, caused by constantly circulating currents. The main energy from a star radiating from the photosphere comes from the magnetic field accelerating the positive and negative ions. The positive ion, the proton, moves much more slowly than the electron, and collides with the hydrogen atoms in the star’s photosphere, causing radiant or electromagnetic energy from its kinetic energy. The much faster moving electrons usually escape into the chromosphere or the corona, where they form a negative sheet creating a powerful electric field, which pulls up the protons to form hydrogen atoms, or neutrons when there is a direct bonding with a straight collision. These uncharged hydrogen atoms and neutrons fall back into the photosphere, to collide with other hydrogen atoms, again creating radiant energy, this time from gravity.

The notion of “strong nuclear force”, in this alternate model for the deuterium atom, has been replaced by electrostatic force bonding of two protons with a single electron; it is implicit that a neutron is a tight bonding of an electron with a proton, thus presenting a zero net charge. High-energy electrons and gamma rays as well, such as coming from an atomic bomb explosion, are in this model capable of breaking this bonding electron away from both the protons. The repulsive force between the two liberated protons is immense, contributing to the great energies associated with the explosion of the hydrogen bomb. Solar flares are such hydrogen bomb explosions, where local magnetic fields of much higher intensity than usual, accelerate the electrons to such velocities associated with the atomic bomb explosions that initiate the hydrogen bomb explosion. A chain reaction is set off among the deuterium nuclei, which continues till there is depletion of the deuterium locally.

The electrons, being light and very fast, are more likely to leave the star as a result of the acceleration from the magnetic field. They do get trapped by the magnetic field, which forces them to move in a tangential direction. When solar flares happen, the velocities involved are far larger than usual, so even the magnetic field is not enough for restraint. While positive ions are also discharged into outer space, it seems reasonable to conclude that with time the positive ion concentration (protons) in the photosphere of the star can only increase, given that under normal circumstances it is mainly the electrons that are ejected out of the photosphere.

The neutrons and the hydrogen atom combine to form those isotopes, when the electron in the neutron attracts the proton in the hydrogen atom. The increase of the proton concentration of the photosphere, inevitably leads to the formation of more deuterium isotopes of hydrogen, leading to its increased concentration in the photosphere; and from there to the gaseous layers below the photosphere. This concentration increases with loss of mass from the photosphere, caused by normal statistical processes relating to ionic velocities, and solar flares causing direct eviction.

With the proton concentration for an aged star being higher than normal, the energy from it is likely to come mostly from electromagnetism, as opposed to gravity; with the magnetic field causing the protons to collide with the hydrogen atoms. While in our Sun, the concentration of the proton in the photosphere was one in ten thousand, in an aged star that could be a few orders of magnitude more. The result for that is that a star about to go nova should be very bright. Indeed observation shows just that. As quoted from the Britannica in the very first section, stars that go nova are white dwarfs, barely visible to the naked eye.

Going by the data given in the Britannica, we take for our nova as having intensity value 10,000 times the normal, lasting for “several days or a few weeks”. (The intensity factor mentioned in the Britannica was “several thousand to 100,000). Using this information, and taking our Sun as reference for the “normal” source of stellar power, with the nova lasting 30 days, let us see how our new theories in physics explain the nova phenomenon.

The total energy from the nova, from the above data is (time in seconds * power) that is,
10000 * 30 * 24 * 60 * 60 * power normally radiated by the Sun or 3.85*10^26 watts,
or 9979.2*10^33 or around 10^37 joules.
We have shown in an earlier section that 100Kg of deuterium creates the energy, from its fission making it a hydrogen bomb, of magnitude 1.7*10^15 joules; thus one kilogram of deuterium would create energy of 1.7*10^13 joules.
Thus the amount of deuterium required for the nova explosion is 10^37/(1.7*10^13) = 6*10^23 Kg. The percentage of deuterium, if say 0.333%, means that the gases blown out will be some 333 times the mass of the deuterium, so the loss of gaseous mass due to the nova will be say 2*10^26 Kg.
The mass of the Sun is roughly 2*10^30 Kg, so from this calculation we find that 2*10^26/2^10^30 or 1/10000 of the material of the Sun has been ejected by the nova.
There is, then, found a match between our calculations based upon our new theories in physics with observed reality, when we go back to the first section of this series where this line is quoted from the Encyclopaedia Britannica, about nova “causing the ejection of hot surfaces gases on the order of 1/10000 the amount of material in the Sun”.
********************
It is practically important to internalise the knowledge that deuterium fission, and not the supposed deuterium fusion, is the cause of novae and supernovae. Research for the last 70 years directed to get energy from deuterium has failed to provide positive results. Deuterium is abundant in the oceans. It is a great source of potential energy, which can be harnessed for power. Plants were powerful x rays and electron beams break up the deuterium nucleus as is probably happening in the stars, will create controlled power when the protons strike other water molecules creating steam – and steam can be converted to electricity using steam turbines.

Arindam Banerjee
Melbourne 14/10/2020

Explaining nova and supernova with new physics theories – 12
The cause of novae – 2

Out of the many trillions of stars that we observe, only a few stars go to the nova phase at a given time. Most do not do so. As shown earlier, they simply lose the hydrogen around the “Sun-Earth sphere” or SE, and become what is known as dark matter, that does not emit light. The hydrogen as ionic particles is ejected out as periodic solar flares, or through the statistics underlying the electromagnetic solar energy formation processes, as shown earlier - where mostly electrons leave the Sun, thus keeping the photosphere more proton-rich causing greater brightness from proton to atom collisions.

From our point of view, a nebula is composed of the ions (electrons and protons) and hydrogen atoms naturally issuing out of the stars from solar flares, novae and supernovae. The protons and electrons thus ejected unite to form hydrogen, with their mutual electrostatic attraction. Gravitational force draws the hydrogen together from the isolated reaches to form the large nebula.

When a SE enters a nebula, its pull of gravity collects the hydrogen in the nebula to form the hydrogen atmosphere once again, to become incandescent yet again, over a multi-trillion year cycle. From this viewpoint, it is wrong to think that any star is almost purely composed of hydrogen. It is currently supposed to be so from the notion that nebulae, out of gravity, condense to become stars.

What, then, is different about the star that does go nova? Usually it is a small but very bright star, not different from the rest superficially. To understand this, it is necessary to understand why most stars do not go nova.

As our Sun shows, solar flares do happen, and that regularly. Our theory has it that these are large hydrogen bombs, caused by fission of deuterium into protons. The protons could leave the Sun, or combine with electrons to form neutrons or hydrogen atoms with just the proton nucleus. Because the deuterium in the Sun is thus depleted, not increased, there is no chance for the Sun to go nova as that would need the percentage of deuterium in the Sun’s gases to keep on increasing till a critical percentage would be reached, causing the nova.

The likelihood, then, is that only those stars that cannot shed off the deuterium in their gaseous atmosphere can go nova. As has been discussed earlier, as per our new theory the solar flare is caused by the concentration of magnetic lines of force on the Sun, in certain areas. It was also asserted that this difference in concentration is caused by the concentration of magnetic material in the SE. If there was no such concentration of magnetic material causing concentrations of magnetic field strength in some areas, then the magnetic lines of force would be uniform all over the surface of the star. In which case, there would be no solar flares; thus the deuterium content in the gases of the star would keep on building up, till the star went nova.

After the nova, the star resumes its original brilliance. This can be explained by the lower layers of the star, initially below the blown-off photosphere and including perhaps some non-incandescent layers below it, have now become the photosphere, with the same energy creation method as used to be before. As the star retains its uniform magnetic field characteristic, the nova should repeat, for that very star. From Encyclopaedia Britannica: “the whole process that produced the outburst repeats itself, resulting in another explosion about 1000 to 10000 years later”.

As for the the possible cause of novae, the Britannica holds: “Most novas are thought to occur in double-star systems in which members revolve closely around each other. Both members of such a system, commonly called a close binary star, are aged: one is a red giant and the other a white dwarf. In certain cases, the red giant expands into the gravitational domain of its companion. The gravitational field of the white dwarf is so strong that hydrogen rich material from the outer atmosphere of the red giant is pulled into the smaller star. When a sizable quantity of this material accumulates on the surface of the white dwarf, a nuclear explosion occurs there, causing the ejection of hot surfaces gases on the order of 1/10000 the amount of material in the Sun. According to the prevailing theory, the white dwarf settles down after the explosion.”

These observed facts are consistent with our more detailed explanation for the nova, which was done without taking the second star in the binary system, the red giant, into consideration. Obviously, the gaseous matter lost in the nova explosion is replenished by the gas from the red giant. The process for the nova is more easily continued with such replenishment.

Arindam Banerjee
Melbourne 17/10/2020

Explaining nova and supernova with new physics theories – 13
The Cause of the Supernova – 1

The supernova is by far the grandest phenomenon in the universe. It is the death of a star, usually a large one, very bright. It reaches a brightness billions of times greater than normal during the time it stays in the supernova state; its brightness may outshine that of the entire galaxy! Unlike the star that went nova, the supernova undergoing star reduces to a neutron star, or a “black hole”, according to the current scientific consensus. This object has a very strong magnetic field, supposed to be a trillion gauss! Apart from whatever material may have gone into the making of this reduced object, the rest of the star is cast into outer space. It is thought that other stars, planets, nebulas, along with comets, meteors, asteroids, stray satellites may have originated from the supernova. Our own solar system could have been formed as the result of a supernova. As seen in earlier sections, supernovas are very rare, considering the billions of trillions of stars that we may observe. In which case there has to be something particularly special about the star that goes to supernova state.

In the earlier sections (Explaining nova and supernova with new physics theories – 1 and 2), I had quoted extensively from the Encyclopaedia Britannica about the purported causes of the supernova, along with my doubts regarding their tenability.

Before proceeding to offer our own explanation of the supernova with our new physics theories, let us briefly recapitulate the current scientific position about the cause of new physics. An easy method is look up Wikipedia, from where we find:
“Theoretical studies indicate that most supernovae are triggered by one of two basic mechanisms: the sudden re-ignition of nuclear fusion in a degenerate star such as a white dwarf, or the sudden gravitational collapse of a massive star's core. In the first class of events, the object's temperature is raised enough to trigger runaway nuclear fusion, completely disrupting the star. Possible causes are an accumulation of material from a binary companion through accretion, or a stellar merger. In the massive star case, the core of a massive star may undergo sudden collapse, releasing gravitational potential energy as a supernova. While some observed supernovae are more complex than these two simplified theories, the astrophysical mechanics have been established and accepted by most astronomers for some time.”

The words “theoretical studies” must be noted in the above quote. These words indicate that certain assumptions, based upon theory or conjecture, have been made to come to conclusions that are confidently asserted as facts. One such assumption is nuclear fusion. In earlier sections, we have shown that the energy from a star need not come from fusion happening in the core of the star; and on the other hand, it is impossible to create the conditions for the pressure and temperature supposedly necessary for fusion. It is difficult to see why, given no external cause, there should be sudden change in the physical situation of the star - even supposing there has been nuclear fusion in the core. That would be changing the core density and hence volume from the conversion of hydrogen to ultimately iron. This would be a slow steady process, going by what may be going on in every single star. Exactly what would create so much energy so suddenly? It is not as if the gravitational force has suddenly increased – going by the logic of nuclear fusion involving the loss of mass with energy, the mass of the star should be decreasing, not increasing. A “stellar merger” – suggested as a possible cause for the supernova, seems more likely. Certainly collision between two stars would create a great deal of energy. But why should most of the masses fly outward with such great speed if it was collision? Going by appearances from telescopes, a vast amount of internal force is expended for the supernova.

In our model, which does not allow for the logically inexplicable collapse of the core to become neutron stars or black holes, the core of the star is untouched, even when there is a supernova. Let us now offer alternative explanations for the cause of supernovae.

The magnetic field of our Sun is not uniform. Within the sunspots the magnetic field strength is nearly 9000 times the normal intensity, which is 1 Gauss. This difference is vast, and one explanation is to conjecture that below the Sun there is a planet like sphere, rather like our own planet Earth. On this sphere, called SE in earlier sections, there is a difference in mineral content spatially, in the magnetic sense. Where on the SE there is more concentration of magnetic material, there the magnetic lines of force gets concentrated and skewed much more; and thus we have such a difference in magnetic field strength intensity on the surface of the Sun.

The magnetic poles of the Sun turn around every 11 years or so, regularly. One plausible reason, once we accept the existence of the SE containing the magnetic iron core, is that the SE is rotating within the Sun. Unaffected by unusual external forces, the currents within this core creating the Sun’s magnetism will suitably tilt with respect to the ecliptic, till they reverse in direction with the full 180 degree swing or half rotation, causing reversion of polarity. The change of the Sun’s magnetic poles, as a constant feature, is a further indication of the existence of the SE. In the next section we will show how the SE is necessary for the re-formation of a dark matter star (comprising of just the SE) when it enters a nebula; and how it is impossible for a star to become a star, from just the hydrogen in the nebula.


Let us now try to explain why some stars (blue stars) are hotter and smaller, and some other stars (red stars) are less hot and larger. In earlier sections we had shown that the light from the star, reflecting its energy output, in the normal case depends upon two forces – gravitational and electromagnetic. In our Sun’s case (our Sun is a “medium” white star) most of the energy (around 96%) comes from electromagnetic forces, with a concentration of one ion (proton) in 10,000 hydrogen atoms. Let us say that the concentration of ions was much lesser say 1 in a million. In that case, the solar energy would have to come from its gravitational forces, from relatively slower collisions between the falling atoms, not providing that much energy, and that too at the lower frequencies meaning more red light. The star would not be bright, but it would be large as a lot of hydrogen would be required to produce that energy. It would also be young, for it is with time that the proton concentration increases in a star. In contrast the blue star gets most of its energy from electromagnetism, with a highly increased concentration of protons causing powerful collisions more plentifully, releasing more energy at the higher frequencies causing the blue colour.

In the earlier section it has been proposed that for most stars, the deuterium content in the hydrogen photosphere is kept down by solar flares. Solar flares happen in the border regions of sunspots with the rest of the photosphere, where the magnetic field intensity is still extremely strong, causing very fast acceleration of the electrons. Such very fast electrons, it is conjectured, break down with kinetic impact the deuterium nucleus into two protons and another electron, with gamma rays as well. A successful chain reaction will become the equivalent of a very powerful hydrogen bomb. It will continue to explode so long as the deuterium is available in sufficient concentrations.

The cause for the supernova is similar to the cause of the nova, as mentioned in the earlier section. Over time, the electrons in the star’s photosphere and its below regions are boiled off by the electromagnetic forces creating ionisation, along with the usual gravitational forces. There is a surplus of protons, which keep on getting accelerated by the magnetic field creating more and more ions. Some of the electrons created by ionisation fuse two protons together to form a deuterium ion. With a uniform magnetic field, this process of deuterium buildup is uniform. With none or comparatively less solar flares to burn off the deuterium by fission, the buildup of deuterium goes on. With the deuterium content thus built up, the star should go nova when the criticality condition for explosion is reached.

For the supernova to happen, the star has to be big; it should not have burnt off the deuterium with solar flares or minor novae. With time this concentration builds up, for the star, to criticality levels. The deuterium component is then more or less uniformly present in all the layers of the huge star. The supernova happens with the triggering done by excess deuterium concentration in an area where the magnetic field somehow gets increased. The cause for that could be random movement in the SE causing unusual magnetic mineral concentration.

As the word suggests, the supernova is indeed a super nova, much more bright and powerful. The entire star is destroyed. In our model, it means that the entire gaseous part of the star ceased to exist – it gets blown out – by the fission happening in the gaseous outer layers of the star. However, this fission process, in our model, does not explain the “solid” materials that are ejected in the supernova. If only hydrogen is involved, then why are the heavier elements found in the supernova explosion?

The cause for the heavier elements that are found in the supernova is explained, from the current standpoint, as the result of nuclear fusion. Nuclear fusion creates energy, from the lighter elements, till the hydrogen becomes iron. Beyond iron, the heavier elements for their creation require energy. Thus, all elements are formed before and after the supernova – prior to the explosion, the hydrogen had turned to iron; and after the great energy of the supernova, elements heavier than iron were formed from the great energy. About the reason for the supernova energy, we once again repeat the quote from the Encylopaedia Britannica presented in the earlier sections of this series.
“When the iron core becomes too massive, its ability to support itself by means of the outward explosive thrust of internal fusion reactions fails to counteract the tremendous pull of its own gravity. Consequently, the core collapses until it reaches a point at which its constituent nuclei and free electrons are crushed together into a hard, rapidly spinning core. This core consists almost entirely of neutrons, which are composed in a volume only 10km across but whose combined weight equals that of several Suns. A teaspoon of this extraordinarily dense material would weigh 50,000,000,000 tons on Earth.
“The supernova detonation occurs when material falls in from the outer layers of the star and then rebounds off the core, which has stopped collapsing and suddenly presents a hard surface to the infalling gases. The shock wave that is generated by this collision propagates outward and blows off the star’s outer gaseous layers. The amount of material blasted outward depends on the star’s original mass.“

We now offer an alternative explanation for the supernova.

There is no question about the facts – definitely heavy elements are ejected from the supernova. On the face of it, this seems to kill our solid cold core theory which does not allow the fusion process.

Let us consider our stellar model yet again – below the very hot, dense, high pressure gases the surface of the SE is immensely pressurised liquid. It is under pressure from the cold iro magnetic core below, and the gases above, under normal circumstances.

When the supernova happens, unlike the nova, all the gases blow out into outer space, in their entirety. This happens due to the deuterium fission process. This time the impact is far more powerful and all-pervasive, as the deuterium has become critical in all the gaseous layers. What is described as the “gravitational potential energy” is indeed a potential energy of a kind – it is the potential energy of the deuterium nucleus becoming kinetic energy due to the electronic band splitting, and releasing the two protons to go very fast by mutual repulsion. The “velocity addition” effect I had discovered back in 1999, which describes the great power behind any explosion from tight packing causing fast and many outward particle collisions, has a multiplicative effect upon the energy produced from the abovementioned internal force producing process.

The consequence for that sudden sharp, blow-out of the gases from the star is a very sharp drop in pressure upon the surface of the SE.

The heavier-than-hydrogen liquid matter on the SE surface, along with the heavier gases nearer the surface, will blow out into outer space as a consequence of this drop in pressure. There could be fusion processes going on when all this matter, bursting out, meets with the very hot ions and atoms. However, this fusion will only soak up the energy, not create any energy. The energy is from deuterium fission with its after-effects relating to “velocity addition”, adding to the release of the gravitational energy upon the pressurised liquid masses below the gaseous layers; the strong magnetic field will also impact upon the ions, giving them increased kinetic energy.

Only the comparatively small iron core, that is solid, will remain of the star that has gone supernova. Apart from that iron, all the matter that was in the star will be blown out into outer space.

The above is the alternative explanation for the supernova. We will now try to find out what could be the size of the iron core that is left, and supposed to be a neutron star or black hole. Going to the top of this article, we find that it has a very powerful magnetic field of a trillion gauss. This is remarkable if indeed it is a neutron star which has no charge to create the current for the magnetic field. Of course there are no bets as what to the theoretical black hole may be all about in the field of reality.

We now try to explain the powerful magnetic field of the stellar core remains of the supernova; we will check if it could be an iron ball at very low temperatures, sustaining superconducting currents to cause the magnetic field, the way it always did before the supernova. We know from the magnetic field strengths near the sunspots on our Sun, that a magnetic field strength of more than 9000 Gauss could set off a solar flare. From this it is plausible to note that the magnetic field strength of the star that went supernova, could be around 9000 Gauss all over. The magnetic field strength of the core is one trillion Gauss. How can that be? The answer is simple – when the star loses all its matter as a result of the supernova then the distances involved get reduced correspondingly. Assuming that there was no change at all in the iron core of the star, meaning its magnetism did not change, let us see what could be the size of the iron core. As this star is very large, say about ten times the size of our sun, its radius would be seven million kilometres. The magnetic field strength is inversely proportional to the square of the distance from the generating currents. Going by this inverse square law, the most fundamental in electromagnetism, the cold iron core radius has to be related by the inverse square law basis to the star radius. One trillion Gauss divided by 9,000 Gauss is about hundred and eleven million; the square root of this hundred and eleven million, or 10,535, is the ratio of the unexploded star radius to the core radius. This makes the core radius at seven million divided by 10,535 or about 664 kilometers.

Thus from a radius of 7,000,000 Km the star shrinks to an iron core of radius 664 Km, as per our new explanation for the supernova. This size is extremely small by astronomical standards, so the explanations given for its existence, as a neutron star or as a black hole, may have seemed plausible so far.

Arindam Banerjee
Melbourne, 26 Oct 2020

Explaining nova and supernova with new physics theories – 14
The Cause of the Supernova – 2; Rebirth of the star

Our Sun is not likely to go to nova, let alone supernova. It will continue to shine as it does for many billions of years, while slowly losing its hydrogen atmosphere, till it becomes dark matter. Most stars are like our Sun; when not shining they are like very large planets.

In this section we shall investigate how the dark star, which has lost its mainly hydrogen atmosphere, again becomes bright. In the earlier section we had seen how the supernova scatters the stellar material all over the galaxy, creating a nebula.

We assert that such a dark star, upon entering a nebula, becomes bright again by pulling in the hydrogen (and whatever other matter exists) by the force of gravity. Let us try to get some rough idea about how long it would take for the dark star to become bright again.

Assume a dark star, with the same mass of our sun, enters such a nebula – let us assume that it is purely composed of hydrogen. We want to have an idea about how long it would take to double its mass.

The mass of the sun is 2*10^30 Kg; so to double its size it needs to acquire the same mass from the nebula. Information from the Internet would have it that in a nebula, the concentration of matter is 500 to 10000 particles per cubic centimetre. Taking a value of 5000 particles per cc, for the concentration, and knowing that the mass of a proton is 1.67*10^-27 Kg, we have in one cubic meter the average nebula mass to be 8.35*10^-18 Kg. Thus the volume of nebula required for doubling the mass of the sun would be 2*10^30/8.35*10^-18 = 0.24*10^48 cubic meters. Its radius would be the cubic root of 3*0.24*10^48/(4*3.142) = (0.38)*10^16 meters.

Consider a hydrogen atom at the distance of 0.38*10^16 meters away from the Sun. How long will it take for it to reach the Sun and thus be a part of it? The gravitational force upon that atom by the law of universal gravitation will be G*mass of hydrogen atom* mass of Sun divided by the square of the distance which is R1= 0.38*10^16 meters, where G is the gravitational constant. The acceleration upon it would be G*mass of Sun/R1^2, and that would work out as 6.67*10^-11 * 2*10^30/(0.38*10^16)^2 = 92*10^-13 or 9.2*10^-12 m/sec/sec. Let us try to work out, roughly, the time it would take to travel 10% of the distance to the Sun. That distance is 0.1 * 0.38*10^16 or 3.8*10^14 m. The force of acceleration at 90% of the radius will be, at the distance of 0.9 * 0.38 *10^16 will be by the same method, equal to 11.35*10-12 m/s/s. The average acceleration a, roughly, over the distance S=3.8*10^14m would then be 10.27m/s/s. Starting from zero velocity, the time t it would take the hydrogen atom to cover that distance would be from the formula S=1/2(att),
t = 0.86*10^13 seconds or 272, 704 years.
The acceleration would increase after the first 10% distance covered. Without going into more complex mathematics, the example above indicates that within a million years the size of our sun would double when placed in a nebula, given the kind of concentration we have assumed it possesses. (The outermost hydrogen atom will take that much time to reach the sun.) It will be easy to show, going by the above method, that any mass will double in size in that nebula in the same time span (a million years).

When a dark star, or for that matter any star, falls into a nebula it will keep on soaking up the hydrogen and other matter there, till the nebula is no more. If it was dark, it becomes bright; if it was small, it becomes bigger. As we have seen, nebulas form from supernovas, and collection from the ionic winds from the living stars. It is one grand, never-ending cycle.

We now deal with a key issue about stars – certainly the most fundamental issue in our alternative explanation for solar energy and novae and supernovae. It is about discussing for viability, the formation of the hydrogen core for the star, which is the current scientific notion. How is that even possible? We shall examine this with some minuteness in the next section.

Cheers,
Arindam Banerjee
Melbourne 27/10/2020

Explaining nova and supernova with new physics theories – 15
The Cause of the Supernova – 3; The core of stars


The current scientific understanding about the supernova’s impact upon the creation of the stars and planets is that the material ejected from the dying star contains elements heavier than hydrogen, going up to Uranium. These are supposed to be formed from the fusion process underlying the energy of the supernova. The star was composed of hydrogen initially; by the internal fusion processes happening at the core, starting from hydrogen to helium, from helium to heavier elements, ultimately to iron, makes the core fully iron; somehow that iron core collapses due to gravity to form a neutron star or a black hole. The resulting gap in space caused by this sharp collapse releases the gravitational potential energy of the star. In other words, the material on top of the core falls down to the core. With this fall, the gravitational potential energy gets transformed to kinetic energy, with all the outer material bouncing back from the neutron star or black hole. This kinetic energy seeds the fusion process for the hydrogen to helium and from there to all the other elements in the periodic table, creating the stupendous energy from fusion till the level of iron is reached; from there, fusion absorbs energy, but with so much energy around the mutation to elements with higher atomic numbers is possible. All this matter is cast out into space, to form a nebula which is mostly composed of hydrogen.

Again, according to the current scientific understanding, it is only the hydrogen in the nebula that forms a star. When sufficient hydrogen is compacted into a certain volume, by gravity, then the fusion process starts again, through the conversion of hydrogen to helium in the very hot core, providing energy and brightness.

In the earlier sections I have provided objections to the abovementioned model. I have shown that the core cannot be hot, so fusion energy at the core is not possible. I have asserted that under the outer gaseous layers of the Sun, there exists a large planet like sphere with a solid iron core, very cold, and thus supports the superconducting currents creating the Sun’s magnetic field.

Whether or not either theory is correct, what is beyond dispute is that besides hydrogen, matter of heavier atomic numbers, from Helium to Uranium, are ejected into outer space as a result of the supernova, to form a nebula.

We will now try to see how exactly the hydrogen in the nebula can compress to form a star. It has to start at the core, and for that, two hydrogen atoms must join together to form a molecule. That molecule will draw other molecules, and so on.

In the nebula, the hydrogen atoms are by no means still. They are all going off in different directions, much as a gas on our Earth. The force of gravity is simply not strong enough to hold together two hydrogen atoms going off on different directions and at different speeds. They must have the same velocity exactly (meaning same speed and direction) for any chance of unity.

Let us now consider two hydrogen atoms separated by one meter, at the same velocity. What is the time required for them to combine? Only gravitational forces exist, so the gravitational force between them will be, by the method done in the earlier section, equal to the product of the gravitational constant G, their masses m, divided by the square of the distance R. Which will work out to be Gmm/R^2 = 6.67*10^-11*1.67*10^-27*1.67*10^-27/1 = 18.6*10^-65 newtons. The acceleration upon each will be mG/R^2 or 11.13*10^-38 metres/s/s. To travel 5 cm (meaning reducing distance to 95cm) at a slightly higher average acceleration a, say 13*10^-38 m/s/s will take time t from the formula s=att/2 or .05 = .5*13*10^-38tt, or t=square root of 0.0077*10^38 or .088*10^19 seconds or 8.8*10^17 seconds or more than 27 billion years, twice as much the time of our universe as per theories such as Big Bang.

Thus it is difficult to see how a hydrogen core for any star or for that matter any celestial body can be formed.

In contrast, let us consider two pieces of iron, each 10 Kg in weight, that have been flung out in the supernova. The value of 10Kg is taken as meteors – that contain iron - have such weight. Let them be separated by 100 kilometre, or 10^5 meters. The time for them to come at 90% of that distance would be found, by the above method, to be around 15,000 years; and to join, say roughly 100,000 years.

Thus the iron around any such point would be collected up within a few million years, along with other solid remains of the supernova. The iron being heavy and also magnetic attract, and would settle in the core, pushing the lighter silicon and other mineral material around it. It would also attract the hydrogen and other gases which from density issues would surround the solid beneath. Thus the star from the nebula would be formed, when the matter becomes sufficiently large. Otherwise, planets like Earth would be formed. It is interesting to note that it has been found that so-called gas giant planets, such as Neptune, also have a magnetic field indicating that they too have an iron core.

Arindam Banerjee
Melbourne, 28 Oct 2020

Explaining nova and supernova with new physics theories – 16
The Cause of the Supernova – 4; The rings of Saturn

The rings of Saturn are very beautiful. Their formation may be further explained by our new insights into the structure of the solar system, and its workings, that have been elaborated upon in the earlier sections. Before doing that, let us first investigate the current explanations about the formation of Saturn’s rings. The following quotes are from Wikipedia https://en.wikipedia.org/wiki/Rings_of_Saturn

One theory, originally proposed by Édouard Roche in the 19th century, is that the rings were once a moon of Saturn (named Veritas, after a Roman goddess who hid in a well) whose orbit decayed until it came close enough to be ripped apart by tidal forces (see Roche limit).[46] A variation on this theory is that this moon disintegrated after being struck by a large comet or asteroid.[47] The second theory is that the rings were never part of a moon, but are instead left over from the original nebular material from which Saturn formed.[

A more traditional version of the disrupted-moon theory is that the rings are composed of debris from a moon 400 to 600 km in diameter, slightly larger than Mimas. The last time there were collisions large enough to be likely to disrupt a moon that large was during the Late Heavy Bombardment some four billion years ago.[48]
A more recent variant of this type of theory by R. M. Canup is that the rings could represent part of the remains of the icy mantle of a much larger, Titan-sized, differentiated moon that was stripped of its outer layer as it spiraled into the planet during the formative period when Saturn was still surrounded by a gaseous nebula.[49][50] This would explain the scarcity of rocky material within the rings. The rings would initially have been much more massive (≈1,000 times) and broader than at present; material in the outer portions of the rings would have coalesced into the moons of Saturn out to Tethys, also explaining the lack of rocky material in the composition of most of these moons.[50] Subsequent collisional or cryovolcanic evolution of Enceladus might then have caused selective loss of ice from this moon, raising its density to its current value of 1.61 g/cm3, compared to values of 1.15 for Mimas and 0.97 for Tethys.[50]

The above explanations do not clarify the fine band structure of Saturn’s rings, but the theory of a moon going inside Saturn does indicate how the rings could have been formed over time, with the steady accumulation of heavy matter going to the core of the planet. That is in line with the formation of the stars in a nebula, that was detailed in the earlier section.

The supernova from which formed the nebula out of which our solar system was born, let fly out all the elements and compounds that had formed the star that went supernova. The iron being the most massive united by gravitational forces to form the cores of the sun and the planets. The other minerals and dust were soaked up, and the hydrogen as well, in the regions around the sun. The outer planets are much bigger as they soaked up the scattered nebula material in the vaster regions away from the sun. Being nearer the hydrogen, helium, methane, etc. from the nebula were thus more attracted to the distant planets – Jupiter, Saturn, Uranus and Neptune.

Jupiter, Uranus and Neptune also have rings. The rings of Jupiter are not made of ice, but dust. Since they are made of dust, it is likely that they are formed from moons that have been crushed somehow, maybe by impact from some other body, or they were not particularly dense in the first place, so they got gradually disintegrated. Lack of knowledge prohibits from discussing the moons of Uranus and Neptune.

Back to Saturn’s rings, now: definitely they are formed from ice, and that too very pure ice. It is thus impossible to think of them as many dusty satellite moons that have come apart, but they could be break-ups of icy moons, with the heavier parts falling inside Saturn. It could be that there was more oxygen pick up by Saturn creating more water with combination from hydrogen; thus originally Saturn could have been more watery than say Neptune or Uranus where carbon was more abundant (more methane there).

Now if over time the heavy, dense, dusty and possibly iron-cored moon of Saturn fall into the core of the planet itself in the formative stages - leaving the water content in the outer regions of the planet, later to rotate around Saturn in icy rings - then there will be a side effect. Its density (considering that the planet is mainly gaseous) will increase with more heavy mass in the core, making its gravitational force stronger. This increased gravitational force will act upon the outer layers of the planet, drawing them in, and thus reducing the diameter of the planet. Most of the ice will by gravity be pulled into the planet. The ice existing just around the equator will form a band, with electrostatic forces holding the material.

With increasing contraction of the planet, with more and more dense supernova material (iron, silicon, other metals and minerals ) pulled into it by gravity, the distance between the icy ring and the planet’s outer surface will keep on increasing. More the matter coming in from the supernova and other space debris, more the bands formed, and greater the distance from the outermost icy band. At some stage, more than gravitational force, electrostatic forces in the icy rings will draw in the remaining ice in the neighbourhood from the supernova explosion, with increase in size. A sudden contraction, caused say by swallowing a huge moon or comet, will lead to a sudden decrease in the diameter of Saturn, thus creating the bands between the rings.

The bands could also be formed as a break-up and reformation of the ring that was broken up, and a new ring formed below it. This break up could be caused by a passing asteroid or comet breaking up the cohesion of the original ring. Observation of the rings shows their dynamic nature.

The rings of Saturn then, are huge hula-hoops around Saturn! Now a hula-hoop has to rotate for stability, while constantly nudged upwards. In the case of Saturn’s rings, there may be wonder. Why do the ice particles not fall down, straight? Why are they around the equator? The obvious reason is the tangential velocity of the ice particles, causing the so-called centrifugal force which works against the centripetal force of gravity. The other reason is the electrostatic forces among the ice particles, which holds them all together cohesively and thus distribute the gravitational pressure acting downwards, in the sideways direction along the hula-hoop, for stability. That would make it work like a very big arch, or rather two opposing arches joined together. The rings are effectively solid. They have to be, and are, just over the centre of gravity of Saturn. Thus there could never be a ring over any other latitude, for the sideways force would shift it right over the equator of the planet.

In this section I have not presented anything startlingly revolutionary. But I did quote the fact from the Wikipedia that Saturn’s moons may have entered Saturn, leaving the icy parts outside to form the rings. This is revealing. If Moons and small planets and asteroids and comets, let alone hordes of meteors and cosmic dust, etc. can get swallowed up by Saturn, then the core of the planet Saturn is not likely to be composed of just gas. It could be like that of Earth. There could be a lot of iron in it, accounting for the magnetic field.

This is then further indication of the theory that the Sun contains within itself a very large planet, Earth-like, which I have called the Sun-Earth sphere, or SE, in earlier articles. If small planets, etc. can be said to get into Saturn in its initial days, then there is no reason why they should not enter the Sun to form the SE.
Arindam Banerjee
2 Nov 2020