Explaining nova and supernova with new physics theories – 1
26 Aug 2020, 15:20:51
Over the past few months, I have been engaged in updating notions relating to the fundamentals of physics as they apply to our understanding of the universe. In brief, I have updated the Newtonian formula for universal gravitation, by introducing the number of electrons or protons in a given matter as opposed to its mass.
(The new upgraded formula for gravitational attraction for uncharged masses with N1 and N2 protons/electrons in each; where the number of protons and neutrons in the nucleus are roughly similar; and separated by distance D meters is F=2.87163669*10^-26*N1*N2/D^2 Newtons.)
This formula explains how the Sun’s gravity cannot hold down the escaping electrons and protons that form the ionic wind. With whatever knowledge we could have about the rate of loss of matter from the Sun as a result of this ionic drift into space, I calculated the possible age of the Sun, that is, when its outer atmosphere will cease to exist, and then the Sun would become ‘dead’ or rather, “dark matter” now explained as black holes, neutron stars, etc. as opposed to stars that have lost their atmosphere.
In such a calculation I did not take the loss of matter in the Sun arising from the supposed fusion process that is said to cause the Sun’s energy. I did so for the simple reason that there is no fusion in the core of the Sun, as the core of the Sun has to be very cold in order to sustain the superconducting current causing the reasonably steady magnetic field of the Sun. (As is, in a proportionately attenuated manner, the case for the Earth and the planets). The main energy comes from the Sun comes not from fusion at its core, but from two other sources: its magnetic field causing currents accelerating the ions and causing collisions providing heat; and its enormous compressing gravity forces again causing compression of the hydrogen in its atmosphere. There is also a third source, that which causes the solar flares – we will discuss this later.
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The facts about nova, from its entry in the Encyclopaedia Britannica, are given below in order to explain it in terms of the new physics theories:
“nova, plural novas or novae, any of a class of exploding stars whose luminosity temporarily increases from several thousand to as much as 100,000 times its normal level. A nova reaches maximum luminosity within hours after its outburst and may shine intensely for several days or occasionally for a few weeks, after which it slowly returns to its former level of luminosity. Stars that become novas are nearly always too faint before eruption to be seen with the naked eye. Their sudden increase in luminosity, however, is sometimes great enough to make them readily visible in the nighttime sky. To observers, such objects may appear to be new stars; hence the name nova from the Latin word for “new”.
“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; however the flow of hydrogen-rich material flows immediately and the whole process that produced the outburst repeats itself, resulting in another explosion about 1000 to 10000 years later. Recent research, however, suggests that such outbursts may recur at much longer intervals – every 100,000 years or so. It is explained that a nova eruption separates the members of the binary system, interrupting the transfer of matter until the two stars move closer together again after a considerable length of time.”
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About supernova now, from the same source as above:
Any of a class of violently exploding stars whose luminosity after eruption suddenly increases millions or billions of times its normal level. Supernovas resemble novas in several respects. Both are characterised by a tremendous, rapid brightening lasting for a few weeks, followed by a slow dimming. Spectroscopically, they show blue-shifted emission lines, which imply that hot gases are blown outward. But a supernova explosion, unlike a nova outburst, is a cataclysmic event for the star, one that essentially ends its active (ie energy-generating) lifetime. When a star goes “supernova” considerable amounts of its matter, equalling the material of several Suns, may be blasted into space with such a burst of energy as to enable the exploding star to outshine its entire home galaxy.
Historically, only seven supernovas are known to have been recorded, with the most famous occurring in the year 1054 AD. The remnants of this explosion are visible today as the Crab Nebula, which is composed of a glowing ejecta of gases flying outward in an irregular manner and a rapidly spinning, pulsating neutron star, called a pulsar. Other prominent supernovas are known to have been observed from Earth in AD 185, 393, 1006, 1181, 1572 and 1604.
The closest and the most easily observed supernova since the supernova sighting of 1604 was first sighted on the night of Feb. 24. 1987, by the Canadian astronomer Ian K. Shelton while working at the Las Campanas Observatory in Chile. Designated SN1987A, the formerly extremely faint object attained a magnitude of 4.5 within just a few hours, thus becoming visible to the naked eye. The newly appearing supernova was located in the Large Magellanic Cloud at a distance of about 163,000 light years. Its brightness peaked in May with a magnitude of about 3 and slowly declined in the following months.
Supernovas may be divided into two broad classes, Type 1 and Type 2, according to the way they detonate. Type 1 supernovas are brighter than Type 2.
(to be continued)
Explaining nova and supernova with new physics theories – 2
26 Aug 2020, 15:20:51
In the preceding section (#1) I have described the nature of the nova and the supernova. In this section I will quote from the Encyclopaedia Britannica about the reasons for these phenomena.
The existing reasons for the nova and supernova are given out below in some detail, for the purpose of my providing alternative explanations for the cause of the nova and supernova. Those explanations will be provided in later sections. They will be based upon my insights and discoveries relating to the fundamentals of physics.
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“The so-called classic explosion, associated with Type2 supernovas, is to have as progenitor a very massive star of at least eight solar masses that is at the end of its active lifetime. Until this stage of its evolution, the star has shone by means of the nuclear energy released at and near its core in the process of squeezing and heating lighter elements such as hydrogen and helium into successively heavier elements – ie the process of nuclear fusion. Forming elements heavier than iron absorbs rather than produces energy, however, and since energy is no longer available, an iron core is built up at the very centre of the aging, heavyweight star. 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.
“In some cases, the core collapse may be too great to produce a supernova and the imploding star is compressed into an even smaller and denser body than a neutron star – namely, a black hole. Infalling material disappears into the black hole, the gravitational field of which is so intense that not even light can escape. The entire star is not taken in by the black hole, since much of the falling envelope of the star either rebounds from the temporary formation of the spinning neutron core or misses passing through the very centre of the core and is spun off instead.
“Type 1 supernovas have only recently become explicable, though not without some uncertainty, on the basis of observational data. It appears that the immediate process resulting in the Type 1 explosion – the collapsing core suddenly becoming rigid and causing the infalling material to rebound – is similar to what happens in the Type 2 supernovas. The progenitor of the Type 1 variety, however, is a lighter-weight star of only four to eight solar masses. This kind of supernova may result when the extremely massive white dwarf component in a binary star system draws so much matter from its red-giant companion that it collapses to a neutron core.
“Supernova explosions release not only tremendous amounts of electromagnetic radiation but also cosmic rays and many of the heavier elements that make up the components of the solar system, including the Earth. Recent observations have shown that the collection of such ejected matter with nebulae (clouds of gas and dust) in the vicinity can sometimes compress the nebulae sufficiently in certain regions for gravity to cause these regions to collapse and thereby form new stars and, presumably, planets. Both the nebular material and the heavy elements ejected from a supernova would be contained in these newly formed stellar and planetary bodies. Furthermore isotropic evidence in meteorites from the asteroid belt between the orbits of Mars and Jupiter indicates that the occurrence of one of more nearby supernovas about 4,600,000,000 years ago triggered the formation of the solar system.”
1 Sept 2020, 22:11:49
Explaining nova and supernova with new physics theories – 3
- A hard nut to crack!
In the preceding sections (#1 and #2) I have described the nature of the nova and the supernova, and the explanation of the supernova phenomenon from the modern physics viewpoint that is held to be true, universally.
In this section I present what I consider logical and practical fallacies relating to these standard explanations. Those comments are placed between brackets, and relate to the immediately preceding quoted text from the Encyclopaedia Britannica.
In later sections I will provide alternative explanations for the causes of the nova and supernova, which will be based upon my insights and discoveries relating to the fundamentals of physics.
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“The so-called classic explosion, associated with Type2 supernovas, is to have as progenitor a very massive star of at least eight solar masses that is at the end of its active lifetime. Until this stage of its evolution, the star has shone by means of the nuclear energy released at and near its core in the process of squeezing and heating lighter elements such as hydrogen and helium into successively heavier elements – ie the process of nuclear fusion.”
(Here Encyclopaedia Britannica – EB from now – effortlessly tells us that fusion is a reality. This is not something that is ever proven, only explained by the theory of relativity with the hydrogen bomb and solar energy as supposed proofs.)
“Forming elements heavier than iron absorbs rather than produces energy, however, and since energy is no longer available, an iron core is built up at the very centre of the aging, heavyweight star. “
(Going by this logic, that energy has been produced by loss of mass from the equation e=mc^2, it must be that the weight of the star has reduced from all that energy generation.)
“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.”
(If the core is now iron, it can no longer create fusion energy as per the earlier logic. But that is only from within the iron core. So long as there is hydrogen or lighter elements than iron outside the core, which have to be very hot as they are squeezed against the iron core, they can still be fused for energy generation until there is no hydrogen left. Or the pressure from the remaining hydrogen is not enough for fusion. In which case the star should become cold, but that never happens. All that is happening with certainty, if the logic of fusion is valid, is that with the hydrogen turning to iron, the star is becoming denser. As the mass is not increasing, its volume has to decrease. However the gravitational pull near the centre of the star remains the same, irrespective of density. It has to be less than what it was before as a result of mass loss due to fusion energy, again going by Einsteinian theory of e=mc^2. One can thus visualise the star diminishing in diameter, slowly and steadily. In which case it is difficult to see how or why there has been any sudden increase in the “tremendous pull of its own gravity” creating the supernova. One can argue that with a large iron shield around the core, the gravitational pressure from all around would be distributed and largely confined to the outer shells of the iron, thus keeping the core always free from any kind of pressure. Just as it is hard to crack a nut simply by putting pressure all around it – it always helps if the nut, like a bathysphere or submarine, has a circular cross section for distribution of forces within the shell. Indeed, mathematically it can easily be shown that the gravitational force at the centre of any star or any planet including our Earth, has to be zero. This was the understanding in the 19th century, which led to Jules Verne’s famous book, “Journey to the Centre of the Earth”.)
“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.”
(Indeed extraordinary crushing force would be required to reduce the iron core to neutrons. Where does that come from? At what place does this supposed crushing start? Does it happen at the centre of the star? From any mathematical point of view, the gravitational force at the centre of the star has to be zero. No doubt there will be terrific pressure at the layers below the surface of the iron core, just as on Earth there are the magma layers caused by pressure from the material above the magma layers. Extending this analogy to the star about to undergo supernova, the iron under the surface of the large core should become very hot and become a very high pressure gas or liquid instead of becoming neutrons. There is no case for any kind of collapse. The idea of the neutron star density, as stated in the EB, is an unconvincing conjecture, with no basis in physics from any kind of experiment or observation.)
“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.”
(Strange. When the core was collapsing, as is supposed by EB, did the gravitational forces on the material outside the contracting core suddenly cease to operate? Were they not always hugging the core, very warmly indeed, causing all the supposed fusion? Why should there be a sudden space gap causing all the rebound of the infalling gases? If the gravity was so great, how could they escape its clutches? If at all there was no space between the hard core and the material outside, there can be no shock wave so powerful as to blow away the masses.)
“In some cases, the core collapse may be too great to produce a supernova and the imploding star is compressed into an even smaller and denser body than a neutron star – namely, a black hole. Infalling material disappears into the black hole, the gravitational field of which is so intense that not even light can escape. The entire star is not taken in by the black hole, since much of the falling envelope of the star either rebounds from the temporary formation of the spinning neutron core or misses passing through the very centre of the core and is spun off instead.”
(This is pure conjecture. There can be no scientific basis behind the existence of black holes as they are so very far away if they do exist. Efforts to create black holes in the laboratory have failed.)
Arindam Banerjee
Explaining nova and supernova with new physics theories – 4 30 Sept 2020, 12:12:20
- The notion of fusion has an alternative, that of fission of deuterium
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To understand the principle of nova and supernova, it is essential to have a clear idea about the basic process relating to the creation of energy. It has long been supposed that fusion of the deuterium isotope of hydrogen with an ordinary atom of hydrogen creates a great deal of energy. This is counter-intuitive. Anything that has to be fused (joined together, like a broken plate or electric junction requiring soldering) requires energy, and is far from generating energy. How can joining two atoms create more energy? On the other hand, when something break, it splits apart into fractions that have a fair bit of kinetic energy. So fission undoubtedly causes the generation of energy. Could it be that the X rays from the fission atomic bomb is breaking up the deuterium nucleus (proton plus neutron or looked another way, two protons joined by an electron) by snapping the electron link causes the energy by making the two protons fly away from each other with great speed? This seems far more logical to me. With this background, we will go into some detail into the phenomena of cold fusion and the hydrogen bomb. In the later sections, we will see how that applies to an alternative explanation of the nova and supernova.
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"Cold fusion", an apparently observed exothermic phenomenon is very probably deuterium fission
It is a fact well-known that deuterium (hydrogen nucleus has an extra neutron which can be seen as a neutrally charged proton-electron mass) occurs plentifully in sea-water, naturally. Thus, ever since the forties, efforts have been made to use this deuterium to provide unlimited energy. The reason for this lies in the success of the hydrogen bomb. The extraordinary power there came first from the fission of an atomic bomb which created the temperatures necessary to cause what was supposed to be fusion. Why was it supposed to be fusion? The answer lies in the textbook "nuclear engineering" by Irving Kaplan of MIT (Add. Wes. 1977). On page 668 we have
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Atomic mass of 4 hydrogen atoms = 4.13258 amu
Atomic mass of 1 helium atom = 4.00387 amu
Difference in mass = 0.02871 amu
= 26.7 MeV
= 42.7 x 10^-6 et
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To put it in words, when as a result of intense temperature (as in the atom bomb or at the core of the Sun due to intense pressure of gravity) four hydrogen atoms merge to form a helium atom. This process involves loss in mass as shown. From the law of conservation of mass and energy which mathematically shows that mass and energy are interchangeable concepts, from e=mcc, that mass is energy released when destroyed as what must happen when four hydrogen atoms turn into one helium atom. This was taken as gospel truth by all the scientists who have for decades tried to create the temperatures required to turn hydrogen into helium in the lab. They have failed so far, despite billions of dollars spent and over 70 years of work.
They have failed so far with "hot fusion" following the solar model. Which incidentally has to be false, for he core of the Sun or any large body like the Earth has to be very cold, if there is a magnetic field around them. But this is another story.
Now in the 80s a phenomenon called "cold fusion" was announced by Fleishman and Pons. These were serious researchers whose work created tremendous attention. However these days one does not hear about them. I wonder why. A senior physicist I knew said that they were honest and diligent people, but they did not get the credit they deserved for their discovery. Most likely such was so as their work did not satisfy the "hot fusion" researchers who failed to see how on Earth repelling protons could fuse together in normal lab conditions.
In the below article, I have shown how a 1 Megaton TNT hydrogen bomb explosion can be explained not by employing the law of conservation of mass and energy, but by pure and simple electrostatic repulsion of protons, that caused by breaking the nuclear bonding of the deuterium nucleus seen as two protons held by a single electron. The analysis shows how it is unnecessary to implement the law of conservation of mass and energy in the explanation for the hydrogen bomb phenomenon.
To turn to cold fusion, now. If indeed there is no fusion of hydrogen to helium, simple disintegration of the deuterium nucleus instead, then such a phenomenon need not be dependent purely on a preceding nuclear blast causing beta and gamma rays. The disintegration of deuterium to form heat could well be done in the lab under certain phenomenon. That is most probably what happened in the experiments of Fleishman and Pons, but as fusion was mentioned, no one thought of deuterium fission. Till your present writer.
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Analysis of the hydrogen bomb mathematically using internal forces instead of binding energies involving e=mc^2
The hydrogen bomb is supposedly the most powerful illustration of e=mcc, as the binding energies are involved somehow in fusing hydrogen and deuterium nuciei into helium.
Let us suppose that is not true, holding e=mcc to be false to begin with.
Further, let us hold that the deuterium nuclei is two protons held together by one electron which acts as a superglue joining two positive charges just about strong enough to make it NOT fly away when the nucleus is hit with a high energy electron along with gamma ray involving strong vibration - the latter typically caused by the normal atom bomb fissioning with strong induced radioactivity.
Let us now see if we can explain the energy of a 1 Megaton TNT H-bomb using known electrostatic forces only, and not binding energies or strong force assumptions.
The assertion now is that the deuterium atom, instead of fusing with a proton to form a tritium atom releasing extraordinary energy (the standard position now held, presented in all books, accepted by all institutions) what is really happening is that the electron can no longer hold the two protons together as a result of collision from a high energy electron (beta ray) along with gamma ray vibrations. In which case the two protons fly away from each other, repelled by their mutual positive charge.
Let us see if this model works to explain the energies involved for a 1 Megaton TNT H-bomb explosion. As per google search, 1 Kg of TNT releases 1.67 x 10^9 joules. 1 Megaton (a million tons) would release 1.67 x 10^15 joules.
Can simple formulas, known for decades, relating to electrostatics be used to show that fission as stated above will create such an energy?
Let us get the basic data first, that is publicly known and found via google searching.
mass of proton 1.67 x 10^-29 Kg
charge of proton 1.602 x 10^-19 coulombs
diameter of proton 10^-15 meter
k value is 8.98 x 10^9 MKAS units
Using the formula for electrostatic attraction and repulsion, and assuming that at rest in the deuterim atom the protons were separated by two proton diameters, we present the age old formula
F = k x q1 x q2/r^2 and putting in the values all in MKAS units
F = 8.98 x 10^9 x 1.602 x 10^-19 x 1.602 x 10^-19/(2 X 10^-15 x 2 x 10^-15)
or
F = 8.98 x 1.602^2 x 10^-29/(4 x 10^-30)
or
F1 = 57.6 Newtons
So the fissioning of a deuterium nucleus leads to at the initial position a repulsive force of 57.6 newtons. This is very important.
We now see what energies are involved as a result of this electrostaic repulsion force.
Let us consider the repulsive for the proton at a distance of 20 proton diameters, that is, when a single proton has moved away by 10 proton diameters or 10^-14m
It is F = 8.98 x 1.602^2 x 10^-29/(20 x 10^-15 * 20 x 10^-15)
or
F2 = 0.576 Newtons
Very roughly, the average force acting on the proton as it moves a distance of 10 proton diameters or 10^-14 m is the mean of these two values.
The F = (F1 + F2)/2 = 29.088 Newtons
The average acceleration of the proton from rest over the distance 10^-14m can be calculated from the formula F = ma or a = F/m,
or
a = 29.088/(1.67 x 10^-27)
or
a = 17.418 x 10^27 or 1.7418 x 10^28 meters per second square.
The velocity of the proton after it has travelled 10^-14 m is obtained from the formula:
V^2 = 2 x a x s where V is the velocity, s is the distance travelled under average acceleration a and the initial velocity is zero.
or
V^2 = 2 x 1.7418 x 10^28 x 10^-14
or
V^2 = 3.4836 x 10^14
and
V = 1.866 x 10^7 meters/second
Now the kinetic energy of the single proton after it has travelled 10^-14m after fission is
0.5 x m x V^2
or
0.5 x 1.67 x 10^-27 x 3.4836 x 10^14
or
2.9088 x 10^-13 joules.
As there are two protons involved, the kinetic energy from a deuterium fission would be
5.8 x 10^-13 joules
Let us consider that in our bomb we have used M=100 Kg of heavy water.
Now 20 gms of deuterium water contain 6.023 x 10^23 molecules of heavy water as per Avogadro's formula. (H2O has 2*2+16=20 atomic weights for heavy water, assuming both the hydrogen atoms are deuterium isotopes.)
So 100Kg will contain 100*1000/20 x 6.023 x 10^23 fissionable atoms or 3.01 x 10^27 atoms.
If all these atoms fission in the hydrogen bomb explosion, then the total energy from just that will be:
5.8 x 10^-13 x 3.01 x 10^27
= 17.46 x 10^14
= 1.746 x 10^15 joules.
Now let us go back to see what is the energy yield from a 1 Megaton TNT H bomb.
Its value is
1.67 x 10^15 joules
which is quite close to the value we have got of 1.838 x 10^15 joules and that, let me remind, is obtained by pure electrostatic internal forces caused by the repelled fissioned deuterium protons
Cheers,
Arindam Banerjee
Melbourne, 23/03/2020, updated for the calculation on 30/09/2020
Explaining nova and supernova with new physics theories – 5 7 Sept 2020, 16:50:23
The Big Bang Theory, based upon Einsteinian relativity, conjectures that the Universe started with a big bang and has been expanding ever since. When it all happened, is a somewhat elastic proposition. Some thirty years ago the figure mentioned was 8 billion years. Now that has changed to around 14 billion years.
In practice, NASA makes powerful telescopes to know about the numbers of stars in the observable universe. It comes to 10^21 or so – ten billion galaxies with an average of a hundred billion stars each.
There are about 10 billion galaxies in the observable universe! The number of stars in a galaxy varies, but assuming an average of 100 billion stars per galaxy means that there are about 1,000,000,000,000,000,000,000 (that's 1 billion trillion) stars in the observable universe!
Now we know about novas and supernovas, that do happen. There is also dark matter – apparently there is about 27% as much matter out there that do not shed light, as the matter that do. Dark matter is mysterious apparently – some details are at
https://earthsky.org/astronomy-essentials/definition-what-is-dark-matter
In these series of articles, I am trying to find the cause of nova and supernova, and from that, the nature of the universe, how it all works, how we have come to this stage!
Given the vast number of stars in the universe, 10^21, and the fact that they are born and then they die, it is necessary to know the life time of a star.
With my new formula for gravitation, and discounting the role of fusion causing mass loss, the only way the star can lose mass is through ionisation, with ionic matter streaming away from the star for ions are not affected by gravitational force. In the following article I have worked out that the life of our Sun will be about 3.6 trillion years. This is much more than the Big Bang Creation time which is 13.6 billion years. Since supernovas (meaning death leading to birth of stars in some cases) happen pretty regularly, it would appear at first sight that the Big Bang Theory does not stand scrutiny time-span wise. Among trillions of stars in the nearby galaxies only a handful have gone supernova. This makes the figure of 3.6 trillion years as the life-time of our Sun to be plausible.
The following article also discusses the structure of the Sun – it asserts that the Sun is like a very large planet, with an iron core, perhaps a silicon cover and with a thick hydrogen atmosphere stretching out very far (enough to bend the light from stars during the solar eclipse and thus fortifying the cult of relativity where it is assumed that gravity causes spacewarp leading to stellar displacement). The photograph of dark matter as shown blocking the star is the physical evidence of as star that has lost its hydrogen atmosphere.
Task: How long will it take the Sun to die from loss of matter from ionisation (solar wind)
Data:
The Sun Profile
diameter: 1,390,000 km; radius = 6.95 * 10^10 cm; volume = 4/3 * pi * radius^3 = 1.406*10^33 cc
mass: 1.989e30 kg or 1.989*10^33 gm
temperature: 5800 K (surface) 15,600,000 K (core)
Note: The core temperature of the Sun, in my model of the Sun (and all heavenly bodies with a steady magnetic field) is nearly 0 deg K, or enough to set up the superconducting current that forms the strong magnetic field, causing force upon the ions, and sending them into outer space as solar wind. The ions are caused by the immense pressure, from collisions.
further data: density of hydrogen at stp is Density (at STP) 0.08988 g/L
average density of Sun 1.408 gm/cc
average density of Earth 5.51 gm/cc
This is all the data I need to roughly calculate the length of the life cycle of the Sun.
In my model for the Sun and stars that shine, it is a large Earth type structure with thick hydrogen clouds, and some helium also. The gravitational and electromagnetic forces squeeze the clouds, causing heat. The immense pressure upon the surface melts the surface of the non-gas inner sphere of the Sun. So unlike the Earth, which has a crust, the Sun has no crust under the hydrogen. Just magma. The magma layers are thousands of kilometers deep but below them there is solid rock which insulates the core, which is iron and very cold, like that of Earth. Any heavenly body with a magnetic field must have a cold iron core. It could even be hollow and contain elements other than iron. Whatever heat reaches the core from the magma layers is converted into electricity. Thus the whole system is self-sustaning for all time.
An algebraic formula to calculate the time for the Sun to die from losing all its hydrogen is derived below.
Let the radius of the Sun be Rs and that of the molten/solid Sun beneath the hydrogen be Re. Then let us define x = Rs/Re.
We have to know how much hydrogen is there on the sun, its mass, and the rate for its loss from the sun.
Let the rate of loss of hydrogen from the Sun at present be Hloss.
It will be zero when the hydrogen is all gone.
So the average loss rate is 0.5Hloss.
The mass of solar hydrogen is say Smh.
The mass of solar non-hydrogen Earth-type matter is say Sme.
Then Smh = S - Sme, where S is mass of Sun.
And the time for the Sun to die, call it Tdeath, will be
Tdeath = Smh/(0.5Hloss) in seconds if the unit for Hloss is in mass lost per second. (A)
As the sun is round, and it is reasonably to say that the shape of the solar non-hydrogen Earth type matter is also round, we have
Sme = 4/3 * pi * (Re)^3 * De, following the equation of the volume of a sphere. De is the average density of the non-hydrogen Earth-type matter under the hydrogen.
S = 4/3 * pi * (Rs)^3 * D, where Rs is the radius of the Sun and D the average density of the Sun. Both of these are known, along with S.
Smh = 4/3 * pi * (Rs)^3 * D - 4/3 * pi * (Re)^3 * De (1)
Now Smh is also equal to the volume of the hydrogen atmosphere multiplied by the average density of the hydrogen atmosphere, call it Dh.
So Smh = (4/3 * pi * (Rs)^3 - 4/3 * pi * (Re)^3) * Dh (2)
Equating equations 1 and 2 above
4/3 * pi * (Rs)^3 * D - 4/3 * pi * (Re)^3 * De = (4/3 * pi * (Rs)^3 - 4/3 * pi * (Re)^3) * Dh
Cancelling common terms,
(Rs)^3 * D - (Re)^3 * De = (Rs^3 - Re^3) * Dh
Substituting Re = Rs/x, we have
(Rs)^3 * D - (Rs/x)^3 * De = (Rs^3 - (Rs/x)^3) * Dh
Cancelling common term Rs^3
D - x^(-1/3) * De = (1 - x^(-1/3) * Dh
Solving for x,
x^(-1/3)(De - Dh) = D - Dh, or
x^3 = (De - Dh)/(D - Dh) or
x = ((De - Dh)/(D - Dh))^(1/3) (3)
Substituting (3) in (1) using Re = Rs/x, we get
Smh = 4/3 * pi * (Rs)^3 * D - (4/3 * pi * (Rs)^3 * ((D - Dh)/(De - Dh)) * De)
or,
Smh = 4/3* pi * (Rs)^3 (D - De * (D - Dh)/(De - Dh)) - (4)
Substituting (4) in (A), we have the time of the death of the Sun from now in seconds as
Tdeath (seconds) = 2 * 4/3* pi * (Rs)^3 (D - De * (D - Dh)/(De - Dh))/Hloss (seconds) (B)
In the above equation, we know pi, Rs, D exactly.
De we take as same as earth so 5.51 g/cc
Hloss we find from the Internet is 6 * 10^12 gm/second. True this is supposed to be from e=mC^2 but as we do not think fusion is at all involved in the Sun's loss of mass, this is the loss from solar wind.
Dh figure is educated guess: If the pressure on the average is 1000 times more than STP, then the average density Dh will be say 0.9 g/cc. The analogy is that the average pressure of the Earth's atmosphere is 1000 times more than the outermost layers. However, this term is not particularly sensitive in the equation; certainly not as much as Hloss, the figure given in the Internet.
Substituting these values in B, we get
Tdeath (seconds) = 2 * 4/3 * 3.142 * (6.95 * 10^10)^3 * ((1.408 - .09)/(5.51 - .09))/(6 * 10^12)
or
Tdeath = 10^18*(8/3*3.142*335.7*0.24317)/6 = 113.994 * 10^18 = 1.14 * 10^20 seconds = 1.14 * 10^20/3.1536*10^7 = 3.6 * 10^12 years.
So the time of life for our Sun is 3.6 million million years, or 3.6 trillion years, going by the data put into the equation.
(Even if Hloss were a 100 times more than what is stated in Internet, 6*10^14gm/second that is, the Sun would live for 36 billion years. So, no worries. Within that much times even humans will learn not to use rockets for space travel.)
*****************************************************************************
More about the structure of the Sun...
x = ((De - Dh)/(D - Dh))^(1/3) = ((5.51 - 0.09)/(1.408 - .09))^(1/3) = 1.602
Thus radius of the molten Earth-like non-Hydrogen "dark' Sun = 695000/1.602 = 433,802 Km
The thickness of the Sun's atmopshere is 695000 - 433802 = 261197 Km.
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
12 Sept 2020, 22:02:50
to
Explaining nova and supernova with new physics theories – 6
Solar flares and their causes - 1
Solar flares are crucial for the understanding of novas and supernovas. Most obviously, to me at any rate - a nova is a huge, continuous, prolonged solar flare; a supernova even more so. This should become clear when we shall consider the descriptions and the current reasons for solar flares, from the Wikipedia topic on solar flares.
https://en.wikipedia.org/wiki/Solar_flare
The following line lifted from the above site are of importance:
“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 in the total solar irradiance (the "solar constant")“
“an energy release of typically 10^20 joules of energy suffices to produce a clearly observable event, while a major event can emit up to 10^25 joules.“
“The plasma medium is heated to tens of millions of kelvins, while electrons, protons, and heavier ions are accelerated to near the speed of light. Flares produce electromagnetic radiation across the electromagnetic spectrum at all wavelengths, from radio waves to gamma rays. Most of the energy is spread over frequencies outside the visual range and so the majority of the flares are not visible to the naked eye and must be observed with special instruments. 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. The same energy releases may produce coronal mass ejections (CMEs), although the relationship between CMEs and flares is still not well understood.”
The above quotes, that are based upon observation, are sufficient for a build-up, using my new physics theories, for a logical explanation of solar flares.
What are the causes of solar flares as per current thinking? The same site provides some insights:
“Flares occur when accelerated charged particles, mainly electrons, interact with the plasma medium. Evidence suggests that the phenomenon of magnetic reconnection leads to this copious acceleration of charged particles.[5] On the Sun, magnetic reconnection may happen on solar arcades – a series of closely occurring loops following magnetic lines of force. These lines of force quickly reconnect into a lower arcade of loops leaving a helix of magnetic field unconnected to the rest of the arcade. The sudden release of energy in this reconnection is the origin of the particle acceleration. The unconnected magnetic helical field and the material that it contains may violently expand outwards forming a coronal mass ejection.[6] This also explains why solar flares typically erupt from active regions on the Sun where magnetic fields are much stronger.
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 (10^9 electron volt) and beyond. There are also some inconsistencies regarding the total number of accelerated particles, which sometimes seems to be greater than the total number in the coronal loop. Scientists are unable to forecast flares.”
In simpler terms, the cause of the solar flare is largely unknown. What seems to cause it is very powerful magnetic field causes the acceleration of charged particles, mainly electrons. The suddenness and power of the energy release causes mass ejection from the Sun. How the magnetic energy converts to the kinetic energy of the particles is not understood.
To provide a scientific reason for the solar flare, based upon the undeniable facts, let us first consider the energies involved. In the earlier section, I have shown that the energy from the hydrogen bomb is around 2*10^15 joules. This energy according to my new physics arises from the immense repelling forces caused by breaking the bond holding the two protons together by a single electron in the hydrogen-deuterium nucleus. With continuous collisions this energy may well be amplified, following the alternative formula for the kinetic mass-energy relationship that I derived back in 1999. e=0.5mVVN(N-k) – check out the publication at
https://www.outlookindia.com/website/story/newt-is-old-hat/220623
The solar flare, from the above Wikipedia reference, has an energy of 10^20 joules to 10^25 joules, which is thus about 100,000 to 10,000,000,000 times that of the hydrogen bomb.
Thus a hydrogen bomb, which is no doubt very powerful on the Earth, is not even a pinprick, energy-wise, on the Sun’s surface. Indeed the cause of the solar flare is a mystery, as stated in the Wikipedia.
The current thinking is that the solar energy comes from fusion of the hydrogen into helium, that happening constantly at the very core of the Sun, where the temperature due to intense pressure is such that the hydrogen nuclei fuse into helium nuclei. Apparently there is evidence to show that helium nuclei are lighter than the hydrogen nuclei used for fusion. And somehow, by Einstein’s famous equation, e=mc^2, that mass difference converts into heat energy which radiates out to the Sun’s surface, making it so hot. In short, the Sun is a huge and constantly functioning hydrogen bomb working on relativistic principles.
https://phys.org/news/2015-12-sun-energy.html
Further exposition of the current theory for solar energy is at:
https://en.wikipedia.org/wiki/Solar_mass
“The Sun is losing mass because of fusion reactions occurring within its core, leading to the emission of electromagnetic energy, and by the ejection of matter with the solar wind.”
The “proof” on Earth for this comes from the fact of the all-powerful hydrogen bomb! Such is the power of the bomb, that no one even dares to dispute the validity of Einstein’s equation, e=mc^2. One by-product of such esteem is that physicists can get away with anything chanting the e=mc^2 formula and covering it up with impenetrable mathematics justifying theories dead against common sense. Such is the state of the modern world.
From
https://ag.tennessee.edu/solar/Pages/What%20Is%20Solar%20Energy/Sun%27s%20Energy.aspx we find that
The total power from the Sun is calculated as: 384.6 yotta watts, or 3.846*10^26 watts. This is supposed to come from fusion at the core of the Sun, from mass to energy conversion. We will explore if this power could come from any alternative source, later in this section.
Explaining nova and supernova with new physics theories – 7
Solar flares and their causes - 2
The Reasons for Solar Energy without fusion – 1
With my model of the Sun, as a large planet with a very cold iron core where circulates a strong superconducting current creating the enormous magnetic field, there is obviously no scope for nuclear fusion due to extremely high pressure and temperature at the core, as is currently envisaged. The current model is that the Sun is a fully gaseous mass, with increasing density towards the core ultimately leading to the kind of pressure and temperature just mentioned. And that extreme pressure and temperature fuses the hydrogen into helium, with great release of energy that is convected to the surface of the Sun, causing its brightness.
Obviously, these are two very different models. They also represent two entirely different philosophical approaches. With the fusion scenario, where matter is constantly destroyed and converted to energy which being conserved has to go somewhere – and that goes to increase the entropy, or the state of disorder in the universe. With the cold core scenario, there is no loss of matter – nothing can affect the core of the Sun. It will take trillions of years for the hydrogen cover to evaporate via ionisation, and a few more trillion years to regain that lost hydrogen if the core manages to enter a nebula.
My approach would not have seemed so strange in the nineteenth century. The idea of a very hot core for the Earth was not there. The Earth could have a hollow in the core. In Jules Verne’s novel, “Journey to the Centre of the Earth”, such a hollow could be reached by going down an extinct volcano, the walls of which could protect from the hot magma. While this seems fanciful in these days where we know of continental drift, which has to happen when the crusts are floating on a sea of impassable magma, Verne’s fictional notion had the true characteristics of the best science fiction, in that it was plausible scientifically, and could be realised in the future. Just as his science fiction related to the nuclear submarine (Nautilus, from the book “Twenty Thousand Leagues under the sea”, and powerful guns (rails guns now as opposed to gunpowder guns) hurling objects into space “From the Earth to the Moon”.
In the nineteenth century they had no clue about any reason for the Sun’s energy. This matter I will discuss in detail later in this section. Then two important advances were made in the late nineteenth and early twentieth century. Radioactivity was discovered as a phenomenon that generated energy (alpha rays or helium nuclei, beta rays or fast electrons and gamma rays which are high frequency X-rays) with no apparent source for same,
The other advance was Einstein’s theory of relativity, which related the loss of mass with production of energy on the basis of the formula e=mc^2. It was assumed that the energy from radioactivity came with the loss of mass involved in nuclear reactions. From this assumption, it was possible to imagine that the source of the Sun’s energy could be radioactive in origin. When they discovered that two deuterium atoms (with a proton and neutron forming the nucleus as opposed to a single proton for the normal hydrogen atom) had more mass than a single helium atom, then the logic underlying e=mc^2 became paramount – mass was getting destroyed when two deuterium atoms united to form a helium. This process was opposed to the normal radioactive process of fission, where a heavy nucleus fissions, or breaks up, into a smaller nucleus while emitting radiation.
Einstein’s theories seemed to be most triumphantly – and very cruelly – vindicated by the atom bombs exploded in Hiroshima and Nagasaki. There seemed no other explanation for such awesome power – later enhanced by the far more powerful hydrogen bomb. It seemed obvious that the Sun’s energy had to come from some realisation of his formula e=mc^2. There was simply no other explanation – suddenly the scientific world changed its opinion about relativity. From being a “Jewish science”, ridiculed and despised by the Nazis in particular, it became the very height of wisdom. The supposed absurdities of this theory (matter becoming infinitely heavy and thin at the speed of light, time dilating, space-time warping, etc.) became serious matters of fact.
Because of all these convincing factors, no one at the present holds the idea that the solar energy could come from sources different from such radioactivity – and that the solar flares could result from the fission of the deuterium isotope. Instead, it became necessary to hold that the core of the Sun has to be very hot as a result of the Sun’s gravity, caused by immense pressure. It is such temperature and pressure that force the hydrogen nuclei to fuse to become helium nuclei. And the loss of mass as a result of such fusion creates a lot of energy - the kind of energy found in hydrogen bombs. This energy, as heat, is transmitted by convection from the core to the surface, making it radiant. This is the current standpoint.
In defence of the current standpoint, which is not satisfactory to me at least, it must be admitted that it is difficult to explain the Sun’s great energy without thinking it to be a huge and continuous hydrogen bomb, as above explained. What other reason could there be? Given the scientific mind is bound by what is considered the laws of Nature, without the mass to energy fusion process, there can be no other explanation that would not violate the laws of thermodynamics – that is, heat is a form of energy and being so, from the law of conservation of energy, heat cannot be created or destroyed. How is all that heat energy created, if there is no fusion? There HAS to be fusion or else the laws of thermodynamics will have to be abandoned!
Logically, then, if there is no fusion for solar energy, then the laws of thermodynamics as currently understood will have to be abandoned. Energy is continuously getting created and destroyed in an infinite universe – at any given point in it we are aware of its constant flow at that point by radiation. In deep space, that energy has a certain level, known as background noise, or static, so common in the early analogue radio receiver sets. This leaves us with the original 19 th century question – what creates solar energy? What is energy? Energy is commercially speaking power multiplied by time; in physics it is equivalent to the work it does, and work is force multiplied by the distance over which it acts. We also know it as kinetic energy or energy from the mass’s motion; or potential energy based upon the work the mass can do because of its position, or the mass’s state relating to its chemistry or radioactivity.
Considering no radioactive forces for solar energy, and discounting chemical energy, we are still left with two kinds of forces – gravitational and electro-magnetic. We will consider the role of gravity for solar energy, in the following paragraphs.
The photosphere of the Sun, the outermost layer of the visible Sun, gives us all the solar energy due to the high excitement of the atoms there, caused by – what? If it does not from the heat energy coming from the core, then, from where is it coming? To answer this, let us take a new view of the Sun’s interior, based upon its having a cold core. As found out in the earlier section, under the photosphere there is an increasingly dense hydrogen “atmosphere” of depth 261197 Km under which there is an Earth-like sphere (call it SE) of radius 433802 Km. The enormous pressure on the surface of the Earth-like sphere, SE, and the heat, is likely to make its surface a sticky, viscous liquid of Silicon and related minerals – the high pressure will increase the boiling point so it cannot be gaseous. The pressure and the heat will only increase as we go deeper into this SE, because of the increased gravity. It may peak at around 100,000 to 200,000 Km below the surface of SE, going – proportionately - by the distance of the magma layers below the Earth’s surface (2900 Km).
The Earth’s magma layers are caused by gravity alone, from the pressure of the rocks above the layers causing heat enough to melt the rocks. Unlike the current case for the Sun’s energy from fusion, there can be no such case for the Earth’s internal energy. That is caused by gravity. The heat from the magma layers, seeks release into the cold outer space, so it flows outwards from the magma layers to the surface. Which is why, all over the planet, there is pretty even temperature (around 24 deg C) some 10m below the surface; there being no question of fusion in the Earth’s core causing convection currents through the gases to heat, this warmth comes from the heat conducted out of the hot magma layers deep below, escaping via the surface to the outer space. No matter is lost in this process, which is everlasting. The Earth’s gravity is principally responsible for the Earth’s internal energy, which is evidently dissipated into space.
Similarly, with the the great heat of the SE’s magma layers, coming out from the surface, is conveyed by convection to the gaseous layers above the SE and from there to the photosphere. This heat energy is entirely from the Sun’s gravity; which causes not only heat in the magma layers, but on the surface of the SE and the dense gases at the bottom of the Sun’s atmosphere. All this heat has to escape – what is the direction? The obvious direction is towards the outer space, as it is very cold in outer space and so the temperature gradient is sharpest towards outer space. Now if we argue for a cold core, then the question comes why should not heat the core as well. Again, the answer is the same as for our own Earth. Between the magma layers of our Earth, and the iron cold core, there is some 3000 Km of solid rock which will act as insulation, keeping out the heat almost completely: the very little heat energy that does enter the core in converts to the superconducting currents creating the magnetic field, possible with piezo-electric effects. For the SE, its iron core is separated from the magma layers by about 200,000 Km of rocky insulation, and that is how the iron core remains cold for all time! Nothing, absolutely nothing can change this eternal situation.
Can we calculate the Sun’s energy from gravity alone? One of the main (and good) reasons for the Sun’s hot-core fusion theory, not mentioned earlier in this section, is that gravitational forces by themselves cannot account for the solar energy – the energy output from the Sun found from experiments relating to the amount of energy captured per square meter in outer space, is just too much! There must be some other source. Yes, electromagnetic forces are said to be responsible for the solar flares, it is now held, but how much electromagnetic forces are responsible for the solar energy has not been properly discussed. In any case, finding formulas for solar energy, both from gravity or electromagnetics, seem a daunting task!
In the next section, I will derive some straightforward methods from first principles, using publicly available information obtained from the Internet, to come up some figures about the energy of the Sun from gravitational and electromagnetic forces. These require a very minimum understanding of mathematics, mainly arithmetic. I will explain the physics involved, as simply as possible.
Yet simple as it will be, I expect that the maths and more importantly their background reasoning may be beyond the scope of all my readers. I will not keep them in suspense. My results show that gravitational forces alone do not explain the solar energy that is measured. It is at best about 3 to 4 percent! I believe, from remembered reading, that this sort of conclusion had been reached by the physicists decades ago; and had been a factor for the present hot-fusion theory for solar energy.
Considering electromagnetic forces – that is, the Sun’s magnetic field acting upon the plasma in the photosphere, thus creating currents, causing collisions producing heat – now that explains the solar energy in full; and gives the clue to the cause of solar flares, novae and supernovas.
Cheers,
Arindam Banerjee
Melbourne, 20/9/2020
Explaining nova and supernova with new physics theories – 8
Solar flares and their causes - 3
The Models and Calculations of Solar Energy without fusion – 1
The photosphere is the outer region that radiates into outer space the heat caused by the internal energies of the Sun. This section will deal with the energy radiated away by purely gravitational forces.
The photosphere is composed mainly of hydrogen. The hydrogen atom that is still, or moving tangentially to the surface of the Sun, in the photosphere, will be pulled down by the Sun’s gravity. As it accelerates down, it acquires velocity, and has kinetic energy. When this atom impacts hits another hydrogen atom – or any other atom or molecule in the photosphere - some or all of that kinetic energy gets converted into radiant energy, part of which is visible light. This is how gravity alone causes solar energy, in the photosphere.
Half the mass of the photosphere is ascending up, from the pressure exerted by the denser and hotter lower layers beneath the photosphere. Half of the mass of the photosphere is going down. Now, this is a very simplified picture – of course the ions, atoms and molecules are going in various directions! Computer simulation methods are required to provide the full picture – but so many are the individual atoms involved in the energy generation process, that even the most powerful computers will be unable to model the situation rigorously. Nevertheless, back of the envelope calculation will provide rough and ready results, which we will check with whatever facts are at hand.
The mass of the photosphere is its volume multiplied by its density. The volume is the surface area of the Sun multiplied by the depth of the photosphere. The values of surface area (6.09*10^18 sq meter), density (3 * 10^-4 Kg per cubic meter) and depth (400 Km) are obtained by elementary Internet searching. They are only indicative, and a sort of average taken of whatever information I could get. The Sun’s force of gravity in the photosphere is 274 m/sec^2.
Thus half the mass of the photosphere, descending with the Sun’s gravity, is
0.5*(6.09*10^18 * 400*10^3 *3 *10^-4 = 3.65*10^20 Kilograms. All units are SI, or MKSA (meter kilogram second ampere).
The total gravitational force F upon that mass, causing it to accelerate downwards, is its product multiplied by the Sun’s “g” value; or F = 3.65*10^20 * 274 = 1.0 * 10^23 Newtons.
Using the formula S = 0.5*g(sun)*t^2, which is the formula for the distance S covered by a body accelerating at g over time t, we have the distance our mass falls towards the Sun per second as
S = 0.5 * 274 *1^2 = 0.5 * 274 = 137 meters. Here the implicit assumption is that this is the distance every atom on the average travels under gravity, in the photosphere, unless it gets checked by other atoms and thus emit light.
Thus the potential energy getting converted into kinetic energy (with gravity) and then into radiant energy per second, or the power generated by the Sun due to its gravity alone, is
Solar power (energy generated per second) due to Sun’s gravity is gravitational force multiplied by the distance over which it acts, or
1.0*10^23 newtons * 137 meters = 1.37 * 10^25 watts.
Or 13.7 yotta watts.
Now, if we remember, going back a bit, the Sun’s total power output has been calculated to be 384.6 yotta watts.
Is this a reasonable value? Let us make a quick and rough check. The distance of the Sun from us is on the average 1.4*10^11 meters. The 384.6 yotta watts has to radiate through a sphere of that much radius, with an area of 4*pi*r^2 or 4 * 3.14 * (1.4*10^11)^2 = 2.46*10^23 square meters.
Thus, the power incident per square meter at Earth is 3846*10^23/2.46*10^23 = 3846/2.46 = 1563 watts, or about 1.5 KW per square meter.
This looks right, given our experience with solar cells!
So, we conclude, that even with our rough and ready approach which gives us a high value for the gravitational energy of the Sun, given the kind of assumptions made, yet, this solar energy from pure gravitation is only 13.7/384.6 = 3.5% of the total solar energy.
This calculation accords with the thinking of the 19th century physicists who could not consider gravity as the sole cause for the Sun’s energy. This induced them to consider radioactivity as the main cause.
In the next section I will take about the role of electromagnetic forces for the Sun’s power.
Cheers,
Arindam Banerjee
Melbourne 21/09/2020
(to be continued)