© Science AdminsThis was on thunderbolts, in the Star Models thread.
Postby Lloyd » Sat Jan 25, 2014 4:02 pm
Charles, how does your solar model account for the steady stream of ions (mostly electrons I think) leaving the Sun as the solar wind? Mathis' model seems to account for that as due to the steady flow of photons into and out of the Sun, knocking electrons out of atoms as well as reforming new electrons there. I'm not as clear on how your model accounts for the rather steady outflow of the solar wind. I can see how compressive ionization could squeeze out electrons, but it looks like it would not be a continuous process on a large scale. I think you credit S-waves with mixing up the Sun's double layers a little and reversing the compressive ionization process briefly etc, but I question how much heat and light that could produce, as well as escaped electrons. Do you know how to calculate that?
Postby CharlesChandler » Sat Jan 25, 2014 5:45 pm
Yes. In my model, there are current-free double-layers, where the charge separation mechanism is pressure. All other factors being the same, these double-layers would truly be current-free, with absolutely no discharges. If that were the case, the Sun would be a dark star. Ah, but all other factors are not the same, and these double-layers can get disrupted, resulting in electrostatic discharges. As concerns the heat and light that are released from the surface of the Sun, the near-surface disruptions of significance are solar flares. In addition[] to an instantaneous flash of light, these are also responsible for the sustained electrostatic discharge from the main body of the Sun throughout the quiet periods. The reason is that they expel matter from the Sun (i.e., in CMEs), which disrupts the charge balance, thereby driving electric currents. I can demonstrate that the top layer of the Sun is positively charged, which clings tightly to a negative layer beneath it. When a CME expels matter from the top layer, it leaves the Sun with a net deficiency of protons — a.k.a., a net surplus of electrons. The electrons will flow out of the Sun, in pursuit of the positive ions that were expelled. How do we calculate that current? If we know the mass expelled by CMEs, and we assume that it is mainly positive charge, we know that the electric current that will flow out of the Sun will have the same amount of charge. So that will tell us the amps.
The number of observed CMEs per day ranges from .2 at the minimum to 3.5 at the maximum, for an average of 1.85 per day.
The average mass per CME is 1.6×10^12 kg.
The mass loss to CMEs is:
(1.85 CME/day) × (1.6×10^12 kg/CME) = 2.96×10^12 kg/day = 3.43×10^7 kg/s.
Assuming that this is all hydrogen, and given that a kg of hydrogen contains 5.35×10^26 atoms, the number of expelled hydrogen atoms is:
(3.43×10^7 kg/s) × (5.35×10^26 atoms/kg) = 1.83×10^34 atoms/second.
The charge of a proton is 1.6×10^−19 Coulombs.
The "current" (in positive ions) due to CMEs is:
(1.83×10^34 atoms/second) × (1.6×10^−19 Coulombs/atom) = 2.93×10^15 Coulombs/second.
Since an amp equals 1 Coulomb per second, that's 2.93×10^15 A.
Now, if we just knew the volts, we'd be able to get the watts. Alfvén (in "Remarks on the Rotation of a Magnetized Sphere with Application to Solar Radiation. Arkiv för Matematik, Astronomi och Fysik, 28A (6): 1-9, 1941") said the volts are 1.7×10^9. I think that this was on the basis of the measured Stark effect, but he didn't say. Anyway, he certainly couldn't have anticipated the following calcs, and he didn't think that the Sun is electrically powered, so we can take his number at face value.
Volts = 1.7×10^9.
Watts = Amps × Volts = (2.93×10^15 A) × (1.7×10^9 V) = 4.99×10^24 W.
The known power output of the Sun is 3.8×10^26 W. Assuming that 1/3 of the power is coming from nuclear fusion, and that 1/6 is coming from arc discharges deep inside the Sun (that drive the supergranules), that leaves 1/2 of the power unattributed, or 2.35×10^25 W. Ohmic heating, as just calculated, accounts for 4.99×10^24 W, so we're less than an order of magnitude off, out of 25. If we then acknowledge that the CME counts are conservative, as they do not include events on the opposite side of the Sun, we can conclude that these numbers are within range.