THE SUN: CONCLUSION & APPENDICES
Stellar Model Checker; Introduction; The Ideal Gas Laws; Compressive Ionization; Degrees of Ionization
Conclusion © Charles Chandler
The original source of the energy that went into the Sun was the momentum of the particles in the collapse of a dusty plasma. The implosion, plus the gravity field from the compressed matter, created hydrostatic pressure sufficient for electron degeneracy pressure, wherein charges are separated, thus converting the energy to electrostatic potentials. The electric force between charged double-layers pulls them together, even if electron degeneracy pressure prohibits recombination. Further compacting the matter increases the density of the gravity field. Thus the equilibrium that was finally achieved is a very dense by-product of a force feedback loop involving electric and gravity fields.
With the primary energy store being electrostatic potential, 2/3 of the power output of the Sun is the recombination of opposite charges (i.e., electrostatic discharges), while 1/3 of the power output is from nuclear fusion within the discharge channels.
The prime mover in the energy release, and a source of heat in its own right, is charge recombination due to equatorial s-waves 120 Mm below the surface. The output is constrained by positive and negative feedback loops. The release of heat at the crest of the wave pushes down the next trough, and accelerates the wave, but destructive interference attenuates the wave heights, resulting in a steady output, though the feedback loop oscillates in an 11.2 year cycle. The s-waves also create differential rotation. The main implication is that magnetic field lines close just outside the equatorial band, and with the field normal to the surface, Birkeland currents can stream outward without magnetic braking. The current density can become great enough for organized electrodynamic effects such as sunspots.
Hence a wide variety of data have been taken into account, without finding reason to abandon this mechanistic approach. After all, the Sun is a physical object, so somewhere out there, we ought to be able to find a physical description of it. This search has yielded interesting and potentially valuable results.
This is not to ignore the economy of mathematical simplifications. But math should not preclude physics. As our knowledge increases, it becomes possible to make the transition from heuristics to mechanics. In so doing, we gain the ability to anticipate new discoveries. Phenomenology is OK for assimilating existing datasets, and for presenting them in a way that seems to make sense. But when we come to understand the physical forces responsible for the phenomena, we know where to look for new types of data that might be even more valuable.
Appendices
Stellar Model Checker
Introduction © Charles Chandler
The objective of this project is to provide a calculation engine for checking the consistency of solar models, given the proposed densities, pressures, temperatures, and elements. The strategy is to use finite element analysis, mainly for its conceptual transparency. If we divide the Sun into small enough pieces, we can calculate the fundamental forces acting on each piece with simple arithmetic, and then just add it all up to get the net results. This eliminates the complexities (and the attendant possibility of undetected errors) in the relevant integral calculus formulas.
Figure 1. Distribution of points generated by the Golden Spiral method.
Figure 2. Truncated Pyramids
This engine divides the Sun into 1000 pyramids, with "bases" equal to 1/1000 the area of the solar surface (shown as inwardly pointed cones in Figure 1). The pyramids are sub-divided into 100 equal-volume truncations. (See Figure 2. Note that the pyramids all have the same volume, even though in 2D, they don't have the same area.) The truncated pyramids are the discrete parcels used in the FEA calculations.
It doesn't matter that the square bases of the pyramids do not form a regular polyhedron, as we just need equally spaced centroids for the calculation of the gravitational and electric forces between the parcels. It also doesn't matter that there is no perfect distribution of an arbitrary number of points on a sphere, as the results are averaged from all of the parcels.
With the Sun divided into 100,000 parcels, we can then assign the model densities, pressures, temperatures, and elements to the respective parcels, and check the physical properties of the model.
Is the model in equilibrium, given the gravitational, hydrostatic, and electrostatic forces?
Do the energy sources look realistic, given the model temperatures?
Will the model produce helioseismic shadows at .27 and .70 SR?
Will the model produce a granular layer, 4 Mm deep?
Will the model produce 4600 K blackbody radiation on the limb, and 6400 K BBR normal to the surface?
Within each model, more specific questions can be asked. For example, if the model states that the 4600~6400 K BBR is generated by ohmic heating in high-pressure plasma, what is the depth at which the necessary pressure is achieved? Given that pressure, what is the spacing between the atoms? Knowing the spacing, and knowing the frequency of photons, we can calculate the atomic speeds. Are those speeds realistic, for that amount of ohmic heating? What is the source of the current? Will the overlying plasma generate the necessary resistance for that much ohmic heating? And above the level at which the 4600 K BBR is being generated, will the plasma be cool enough to only produce absorption lines?
The Ideal Gas Laws © Charles Chandler
The laws of hydrostatics tell us that the density, pressure, and temperature of gases (and plasmas) are all related, and that if we know 2 of them, we can calculate the third. So this is something that we can check.
We can also check that the pressure is realistic, given the force of gravity. For a discrete celestial object such as the Sun, density and pressure are not just related by temperature. The density of the matter generates the gravitational field that supplies the pressure necessary to compress the matter. So we can step through the following process:
Given the model densities of each parcel, calculate the force of gravity, from each parcel to each parcel, and add it all up to get the total gravity acting on that parcel. This is the parcel's weight.
A parcel's weight supplies the pressure that compresses the parcel below it. So normally, pressure is calculated from the known density & temperature, but here we can calculate the pressure just from the density, given the densities of all of the other parcels, and the force of gravity.
Knowing the calculated pressure, and the model temperature, we can recalculate the density. If the new density doesn't match the model density, the model is not correct.
When all of this is done for the Dalsgaard model, asserting a 75/25 mix of hydrogen/helium, it all checks out. In other words, if those were the correct densities, pressures, and temperatures, gravity would put it all into hydrostatic equilibrium. (See Figure 1.)
Figure 1. Density per solar radius in the Dalsgaard model, based on the ideal gas laws, with gravity pulling in, and with heat from nuclear fusion in the core generating the hydrostatic pressure that pushes back out. The X axis shows the decimal of the solar radius starting from the center, and the percentage of the solar volume, starting from the surface. The Y axis shows g/cm3. The densities of liquid platinum, iron, helium, and hydrogen are shown for reference. The average density of the Sun is 100.15 = 1.408 g/cm3 = 1408 kg/m3.
But it doesn't produce any internal helioseismic boundaries, nor a distinct limb, much less the granular dynamics on the limb.
Compressive Ionization © Charles Chandler
But it was naïve of us (and Dalsgaard) to simply run out the ideal gas laws, assuming that pressure and temperature always define density. When dealing with extreme pressures, there is another factor that has to be taken into account: the Coulomb barrier. Compressing a gas (or plasma) is easy enough, until you get down to the density of a liquid, wherein all of the atoms have achieved a closest-packed arrangement. Further compression is much more difficult, as liquids are incompressible. This is because at the liquid density, the outer electron shells of neighboring atoms are in contact with each other. Further compression will cause the failure of those shells, expelling the electrons, and leaving the atoms ionized. Thereafter, electrostatic repulsion between the like-charged atoms pushes them apart. And since the electric force obeys the inverse square law, the repulsion increases exponentially with further compression. Hence we need to add the Coulomb force to the hydrostatic pressure before concluding that the model is in equilibrium.
We also need to add the electrostatic attraction between electrostatic layers. To understand why, let's do a thought experiment.
Consider that we have a box that measures 1 meter on all sides, filled with air. The air will be evenly dispersed within the box. Then we insert a divider into the box, to split the volume into two. The air is still evenly dispersed within each compartment. Suppose the divider is a perfect insulator, so no electric currents can flow through it, but it has a high permittivity, so electric fields can shine straight through it. We now take out our atomic tweezers, and one by one, we pick every electron out of the one side and put it into the other, and put every atom into the other compartment. Now we have one compartment with nothing but atomic nuclei on one side, and electrons on the other.
Now, with one compartment filled with nothing but positive ions, and the other with nothing but electrons, will the particles in each compartment be thoroughly dispersed due to electrostatic repulsion, or will they be concentrated in the center, as close as they can get to the divider, attracted to the opposite charge just on the other side?
The answer is both, but the latter is the more powerful force. The following images will help illustrate. First, we have three positive charges. Notice the collisions of lines of force between each ion. This is the source of the electrostatic repulsion between them.
Figure 1. Lines of force from three positive ions.
Now we add three complementary negative charges. Notice that most of the lines of force have been bent towards the opposite charges, and the field density between like charges is not as great. This means less electrostatic repulsion between the ions. And that means that they will be closer together. So it's not just that the ions will be attracted to the opposite charges, minus their repulsion from each other. There's less repulsion, due to the redirection of the lines of force. This allows the ions to get packed together more tightly. And the closer they get to the opposite charges, the electric force increases exponentially. So we will expect a major concentration of charges on either side of the divider, despite the repulsion of like charges within each compartment.
Figure 2. Lines of force with three positive and three negative charges.
Once the initial charge separation has been set up, we can then expect a third layer to form. The next image shows a solid conductor that has been added on the right side. Though the conductor would otherwise be neutral, it attracts the lines of force from the negative charges, and it becomes positively ionized, with an electric force pulling it toward the negative charges.
Figure 3. Solid conductor attracts lines of force.
So if the walls of our 1 meter box were conductors, and if they could move, they would get drawn inward, and it would no longer be 1 cubic meter of air — it would be a lot less, due to the electrostatic potential consolidating the matter. This is interesting because if the topic is the Sun, and if compressive ionization has set up charged double-layers, the compaction of the plasma will increase the density of the gravitational field. Furthermore, if gravity is the force responsible for the compressive ionization, and if such ionization makes the gravitational field more dense, then this constitutes a force feedback loop. The net result is that the plasma will be much more dense than the ideal gas laws would have predicted.
Degrees of Ionization © Charles Chandler
Determining the degree of ionization due to pressure will be a non-trivial undertaking, as it gets deep into supercritical theory, and the laboratory confirmations are sparse at best, as it is extremely different to generate such pressures, and measuring the conditions inside such matter is not possible.
Simplistically, we might think that we could just look up the radii of the electron shells for the various elements. (There isn't a fixed radius for each shell, because the number of protons in the nucleus varies, and this produces an attractive force that varies. See http://chemistry.osu.edu/~woodward/ch121/ch7_radius.htm.) An element should be compressible with a force predicted by the ideal gas laws until the outer electron shells overlap. Further compression requires the expulsion of the electrons, producing a repulsion calculable by the Coulomb force between the ions, given their distance, and the net charges. So the compressive force needs to fare from the ideal gas laws to that plus the Coulomb force as the distance between atoms crosses this threshold. For atoms heavier than helium, with more than one electron shell, there will be additional threshold, as each shell is evacuated. For instance, if hydrogen atoms are pressed together until the k-shells overlap, the atoms will be 1.0584e-10 m apart. Up to that density, we have only the ideal gas laws. Past that density, it's that plus the Coulomb forces.
But that implies that the k-shell radius determines the liquid density, and the numbers don't match up. The mass of a hydrogen atom is 1.67e-24 g. At the 1.0584e-10 m spacing, we'd get a density of 1.4085328418683 g/cm3. Aspden and others have concluded that the Sun is all hydrogen compressed to the k-shell radius, which seems convincing as it's a heckuva coincidence that those numbers match the density of the Sun so precisely. And yet the known density of liquid hydrogen is 0.07085 g/cm3, or 1/20 of the density defined by the k-shell radius. So some other force kicks in, long before the electrons are getting expelled.
Some have suggested that it's magnetic pressure between conflicting electron spins. This would seem to make sense, in that it explains why the density within each elemental period varies. It density was just a function of the radius of the outermost shell, all of the elements in each period would have the same liquid density. But the density is greater in the middle of the period. This is because the atomic mass is increasing, but once the outer shell is halfway populated, you start getting spin conflicts, and the density starts going down, even as the atomic mass increases.
Whether or not spin conflicts could expel electrons is unknown. It's possible that this simply puts the electrons into higher energy states, and the back-pressure is the difference between the lower and higher states. So perhaps we could use the ideal gas laws down to the liquid density. From there, we could fare from no additional force, to the full Coulomb force between ions once past the shell radii, totally ignoring the nature of the repulsion that emerges between the liquid density and the shell conflict density, and simply interpolating the unknown force.
But this would oversimplify more than just that. If the temperatures are great enough that there are no bound electrons, the shells just aren't there, and the physics gets a lot more complicated. There will still be ionization due to gravity, but it will be more a matter of gravity acting 1836+ times more forcefully on the atomic nuclei than on the electrons. The ions still repel each other, as do the free electrons. But the ions are subject to 1836+ more gravitational force, and therefore, there will be more of them nearer the center of gravity.
The last complexity that needs to be considered is that while hydrogen and helium might become fully ionized, and thereafter behave as so many disassociated atoms and electrons, the heavier elements are not likely to become fully ionized, no matter the temperature and pressure. So it will be a mixed environment, in which some electrons have been liberated by temperature, and some by pressure. The problem becomes fully non-linear when the temperature cannot be known until the liquid lines have been set, and the electrostatic potentials estimated, which requires that the densities and pressures be known.