The cathode part is really a given.
Curious, then, that the word appears only twice on your website, and neither instance is in the phrase "
cathode solar model" (or "
"cathode" solar model").
That's what makes the solar wind work DRD. Since you *REFUSE* to accept "current flow" in any form, it's still a great "mystery" to you, even though Birkeland "predicted" it 100 years ago. He simulated it too. You've yet to reproduce a single of his experiments with "magnetic reconnection" and you never will because that is pure "pseudoscience" according to the author the the MHD theory that you're kludging all to hell in an effort to avoid that dreaded "electricity'' word.
Would you like to count how many times the word "current" is used in the papers I've referenced?
Goodness, in the title alone I count four (out of six)!
But let's get back to your "
"cathode" solar model" shall we?
Start with "
from the sun": let's assume it's the cathode, i.e. the place to which electrons flow; let's also assume that the actual cathode is the photosphere, which is 700,000 km from the centre of the Sun (we can adjust the numbers a bit later).
Next, "
to the heliosphere": let's assume this is the surface of a sphere, whose radius is 100 au, with the Sun at its centre.
Now in all "
"electric universe" concepts" that I have seen, out there on the world-wide internet, the Sun is powered by a "
current flow", so in this first "
"cathode solar model"", the current must produce ~3.8 x 10^26 W, which is the observed power (energy per second) of the Sun.
How does the "
current flow" produce this power? Let's assume it does so by converting the kinetic energy of the electrons into electromagnetic radiation, on (or at) the photosphere. We won't worry ourselves about how this happens ("
electrical processes" perhaps), for now.
Let's keep it simple and assume that electrons arrive at the photosphere at the same rate as they leave the heliosphere - x electrons leave the heliosphere in one second, and x electrons arrive at the photosphere in one second. In other words, electrons are neither created nor destroyed between the heliosphere and photosphere, and that the current flow is a steady one.
So, how many electrons leave the heliosphere every second? Well, the electron density there is 10 million per cubic metre, and the electrons are moving at 6 million metres per second (again, we can adjust the numbers later), so across each square metre of heliosphere surface there will be 60 trillion electrons crossing every second. Now the heliosphere's surface is ~3 x 10^27 square metres, so 1.8 x 10^41 electrons depart for the photosphere every second.
How fast are these electrons moving when they reach the photosphere? Well, let's keep it very simple and use the physics of Birkeland's day; specifically the part of Newtonian physics which says that the kinetic energy of a body moving at speed v is half its mass times v squared. Now the mass of an electron is 9.1 x 10^-31 kg, and every second 1.8 x 10^41 electrons give up their kinetic energy for light. So we have a simple equation (Ne is the number of electrons, m the mass of an electron, and v its speed):
1/2 mv^2 * Ne = 3.8 x 10^26
which, when we plug in the numbers, gives us the speed of the electrons as 700 million metres per second.
Before I proceed to check this model against empirical reality (i.e. results of experiments in the lab),
I'd like MM to check my model for any flaws, errors, shortcomings, etc.
Of course, every other JREF member reading this is more than welcome to do so too, but if you could, please limit your comments to any mistakes I may have made in my math.
I should add that I have developed this model using only the simplest formulae/math I could; in fact there's little here beyond arithmetic; some extremely simple algebra; the standard definitions of things like energy, power, and density; and the formula for kinetic energy.
Waiting, waiting, waiting ...
