Electric Cosmos: The Solar Resistor Model
Electric Cosmos: The Solar Capacitor Model. I.
Electric Cosmos: The Solar Capacitor Model. II.
The major complaint seems to be that I haven't included the claim that the electrons are actually moving at very slow velocities in these models. They claim that these electrons will not be relativistic.
But they never answer the issue of how electrons traveling through a 1e9 (1 billion) volt potential drop, in free space, do not become relativistic!
If the electrons don't carry the full energy (all 4e17 amps worth) when they strike the solar surface, then you don't have enough energy to explain the solar luminosity:
1e9 volts* 4e17 amps = 4e26 watts.
The electrons in this model are only carrying kinetic energy to the solar surface.
If you still want to power the Sun with electrons striking the surface at a few cm per second (I believe that was the approximate value from an earlier message), then you have a host of other problems.
The “Drift electrons”
Let the electron velocity be 10 cm/sec = 0.1 m/s. Then each electron has a kinetic energy of
KE=0.5*(9.11e-31 kg)*(0.1 m/s)^2 = 4.56e-33 joules.
which it can release when it hits the solar surface.
To explain the solar luminosity of about 4e26 watts, you now need a current of
4e26 watts/4.56e-33 joules = 8.78e58 electrons/s = 1.40e40 amps
This is 3.5e22 (over a billion billion) times more electrons than you started with at the heliopause, as computed in Electric Cosmos: The Solar Capacitor Model. I.!
Where do all these extra electrons come from?
Your only other choice would be to install another 1e9 volt drop in potential between the orbit of the Earth and the solar photosphere. If so, where is it?
Problems keeping charge neutrality
Do you want to keep the heliospheric region electrically neutral?
To cancel the charge of the inbound electrons, you need the same number, 8.78e58, of protons passing through the same region at the same speed in the opposite direction. Therefore, at the orbit of the Earth, these 8.78e58 protons are spread out in a spherical shell 150e9 meters in radius and a thickness of 0.1m. This a volume of
4*pi*(150e9 m)^2 * 0.1m = 2.83e22 m^3 = 2.83 e28 cm^3.
This corresponds to a proton density of
8.58e58 protons/2.83e28cm^3 = 3.03e30 protons/cm^3.
The measured particle density of the solar wind is a few protons/cm^3 (see left banner on http://spaceweather.com/). You clearly don't match the observations.
In addition, 3.03e30 protons/cm^3 has a mass density of
(3.03e30 protons/cm^3)*(1.67e-24gm) = 5.06e6 gm/cm^3.
Note that the density of lead = 11.35 gm/cm^3! So your solar wind is thousands of time denser than lead!
Okay, let's not keep the solar wind electrically neutral...
Don't want to insist on charge neutralization? You've still got 8.78e58 electrons per second building up on the surface of the sun. Shall I compute how much energy it will take to keep them there? After all, I've yet to see the circuit complete on this Electric Sun model to take them away.
Most of the math errors I found didn't improve things for ES. I suspect any remaining ones won't help that much either.
So which law of physics do the EU advocates insist on violating? Conservation of charge or conservation of mass/energy?
And don't hide behind the standard whines:
- you can't treat it electrostatically (actually, I haven't - it's a straight energy transfer problem, conserving matter and energy)
- you haven't included the nonlinearities (have yet to hear these nonlinearities actually specified)
- whatever new excuse du jour you have
And if you can't ensure that your models generate reliable predictions for the Earth-Sun particle environment, how do you know how safe astronauts will be on their way to the Moon or Mars? Even the worst estimates for astronaut radiation exposure pales to the particle environment predicted by ES.