ASCS: Tom Bridgman: The Solar Capacitor Model I, II
© Lloyd

This hasn't been posted on the TB forum yet.

Electric Cosmos: The Solar Capacitor Model. I
Sunday, November 30, 2008

Electric Cosmos: The Solar Capacitor Model. II
Thursday, December 18, 2008
- ... let's compare these predictions of the Solar Capacitor model with some actual observations.
- Plenty of satellites patrol the region between the Earth and the Sun: SOHO, ACE, Wind, more recently, STEREO A and B.
- They measure solar wind speed, composition, magnetic field, even electron energies.
- The data are all public. You can find some of it in places like the online archives and virtual observatories
    Virtual Space Physics Observatory
    ACE Real Time Solar Wind Data
    Solar Wind Data Server
- The solar capacitor model requires a flux of about 1e9 protons/cm^2/s at energies of around 990 MeV, nearly a billion times larger than the measured flux at this energy!
- Next, we look at the low-energy (velocity) protons, such as from Table 1 of "Space Weather: The Solar Perspective" by Rainer Schwenn
- Here, we see that the low energy flux is very large, but those don't help Scott's model.
- For solar wind electrons, we can check "Kinetic Physics of the Solar Corona and Solar Wind" by Eckart Marsch:
*- which shows that the electron velocity distribution in the solar wind at 1AU is not towards the Sun, as required in the solar capacitor model, but largely isotropic (with some enhancement along the magnetic field direction).
- We can also examine the 27-day history of the solar electron flux at geostationary orbit http://sd-www.jhuapl.edu/UPOS/MEV/index.html
- Here, we see that in situ measurements demonstrate there is no stream of electrons inbound towards the Sun, contrary to the predictions of the solar capacitor model!
- Next, let's compare the particle flux predicted for the Solar capacitor to the flux of particles trapped in the Earth's radiation belts
- What do we find? Protons with an energy greater than 10MeV have a flux of 1e5/cm^2/s.
- Electrons with energies greater than 1MeV have a flux of 1e6/cm^2/s
- These peak fluxes at these energies are fatal doses for astronauts!
- They also do a notorious amount of damage to spacecraft electronics unless the electronics are radiation hardened.
- Yet the steady electron and proton fluxes of the solar wind predicted to power the solar capacitor model is 100-1000 times higher than the flux in the Earth's radiation belts!
- Note that this is the STEADY flux in the solar capacitor model.
- What does it imply about the radiation shielding required to protect astronauts for interplanetary travel?
- There are a number of interesting predictions by this model [that] the EU advocates don't talk about.
- What is the total energy in the outbound proton flow?

Electric Cosmos: The Solar Capacitor Model. III
Sunday, April 26, 2009
- I've encountered a few of complaints from EU advocates
- 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.
- 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.
- Anonymous said...
- "Black box" analysis of the Thornhill Electric Sun model (per pln2bz' post), part 1.
- The key output is, of course, the energy output of the Sun. This is ~3.8 x 10^26 J; expressed as power, ~3.8 x 10^26 W.
- If the speed of the electrons is 1 m/s (metre per second), then each electron has a kinetic energy of 5 x 10^-31 J (kinetic energy is 1/2 mass times speed squared, and as long as the electrons are not 'relativistic', the mass is the rest mass).
- How many electrons, moving at 1 m/s, do we need to generate 3.8 x 10^26 J?
    8 x 10^56 (= 3.8 x 10^26 / 5 x 10^-31).
- What, then, is the density of electrons at the heliosphere?
    3 x 10^29 per cubic metre (m^-3) (the number that pass through each square metre divided by their speed; the electrons all move in only one direction, towards the Sun).
- What is the estimated density of electrons, at the heliosphere, as determined by astronomical observations?
- it is between 1,000 and 10 million (electrons per cubic metre).
- [That's between 10^3 and 10^7, or at least 22 orders of magnitude {greater} than EU theory predicts.]

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