## Sunday, April 26, 2009

### Electric Cosmos: The Solar Capacitor Model. III

I've encountered a few of complaints from EU advocates on my analysis of their models, i.e.

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:
1. you can't treat it electrostatically (actually, I haven't - it's a straight energy transfer problem, conserving matter and energy)
2. you haven't included the nonlinearities (have yet to hear these nonlinearities actually specified)
3. whatever new excuse du jour you have
Unless you can show, or point me to where someone has actually DEMONSTRATED mathematically, that these solve the ES problems.

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.

Anonymous said...

A "black box" analysis of Thornhill's Electric Sun/Star model (if that's what it can be called), as presented by pln2bz, is in the comments on the "Hubble's Universe Unfiltered: A Flash of Brilliance (VIDEO)" story in Digg (I can't include a link, it seems).

I'll try copying Part 1 here (broken into pieces, to avoid the character limit).
----------------------------------
"Black box" analysis of the Thornhill Electric Sun model (per pln2bz' post), part 1.

In a series of posts, I'll present some analyses of the Thornhill Electric Sun model, as given by pln2bz (above). I intend to start simple, and with components that are easy to understand, and relatively easy to constrain.

The approach I'll be using is pretty much a "black box" one, in that I'll be looking at the inputs and outputs, and making no comments on what goes on in between.

I'll also be somewhat sloppy with the calculations, working with no more than two significant figures, and often round up, or down, sometimes to the nearest order of magnitude (OOM). I'll do all the calculations in SI units, a.k.a. metres, kilograms, seconds; energy in these units is joules (J), and power watts (W).

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. This output is pretty constant (it varies by less than 1%, over timescales ranging from microseconds to decades, possibly millennia), so I won't be looking at any time-varying phenomena, or mechanisms.

In the model, the charge carriers are electrons, and their source is the heliosphere; the electron has a (rest) mass of 9.1 x 10^-31 kg; the heliosphere I will assume, for this first analysis, to be a sphere of radius 100 au (~1.5 x 10^13 m) centred on the Sun.

In this model, the Sun gets its energy from the kinetic energy of the electrons (which come from the heliosphere), delivered to the Sun's photosphere (roughly, the Sun's surface); as this is a black box approach, I do not consider how the electron's kinetic energy is converted to photons (which is how the Sun's energy is lost to it).

The surface area of the heliosphere is 2.8 x 10^27 square metres (4pi r^2).

Keeping things very simple, let's see what happens if the speed of the electrons when they leave the heliosphere is the same as the speed of the (same) electrons when they arrive at the photosphere (all the electrons move in only one direction, towards the centre of the Sun).

Let's do a calculation, to get a feel for the numbers ...

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).

How many electrons, then, must leave the heliosphere, every second?
8 x 10^56 (electrons are not created or destroyed between heliosphere and photosphere; 1 second at the photosphere is 1 second at the heliosphere).

How many electrons, then, must pass through each square metre of the heliosphere's surface, every second?
3 x 10^29 (= 8 x 10^56 / 2.8 x 10^27).

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?
I'll leave the answer for a later post ... except to say that it is between 1,000 and 10 million (electrons per cubic metre).

Obviously, this particular example (my first, getting-a-feel-for-the-numbers, toy) is wildly inconsistent with reality.
----------------------------------
APODNereid
PS feedback welcome!

Anonymous said...

Last bit of Part 1 (and let me try to put in a link: http://digg.com/space/Hubble_s_Universe_Unfiltered_A_Flash_of_Brilliance_VIDEO)
-------------------------------------
In my next post I'll ask the question "what speed must the electrons have, in order for them to deliver 3.8 x 10^26 J to the photosphere AND for the density of electrons at the heliosphere to be between 1,000 and 10 million per cubic metre?" (within the bounds of the very simple, toy, model presented in this post).

Things to consider later:
* additional sources of electrons, between the heliosphere and the Sun's photosphere
* sinks for electrons, between the heliosphere and the Sun's photosphere
* a bigger heliosphere; a smaller heliosphere
* how electrons may deliver energy to the Sun's photosphere, other than by their kinetic energy
* incorporating relativity
-------------------------------------

APODNereid

Anonymous said...

"Black box" analysis of the Thornhill Electric Sun model (per pln2bz' post), part 2.

What speed must the electrons have, in order for them to deliver 3.8 x 10^26 J to the photosphere AND for the density of electrons at the heliosphere to be between 1,000 and 10 million per cubic metre (within the bounds of the very simple, toy, model presented in the post above)?

Recall that the toy model has the electrons leaving the heliosphere with the same speed as the electrons arriving at the Sun's photosphere, so the number passing through each square meter of heliosphere surface, in one second, will be between 1,000 x electron speed and 10 million x electron speed.

The energy each electron delivers is 1/2 the mass of the electron times its speed, squared (assuming the electrons are not moving at relativistic speeds).

Doing some simple algebra, the speed the electrons must have is the cube root of ((2 x Sun's energy output) divided by (electron density at heliosphere x mass of electron x area of heliosphere)); and the values for each bound are as follows (please check my arithmetic):
1,000: 700 million m/s
10 million: 30 million m/s

Now the speed of light is only 300 million m/s, so clearly the electrons must be highly relativistic of the heliosphere's electron density is only 1,000 per cubic metre!

In a later post I'll re-do the calculations, to take account of the fact that the electrons are relativistic.

If the heliosphere's electron density is 10 million per cubic metre, the electrons still need to move at a fast clip.

How fast? Let's compare it to the average speed of electrons in a plasma, in equilibrium ... and let's use 1 million K, which is one published estimate of the electron temperature in the heliosphere:
http://physicsworld.com/cws/article/news/34895

The mean speed (one kind of average speed) is the square root of (8 x k x T/(pi x mass of electron)), where k is Boltzmann's constant, and T the temperature. So, plugging in the numbers, we find that the estimated mean speed of the electrons in the heliosphere is 6 million m/s.

Houston, we have a problem.

But the model is still only a toy; perhaps the electrons could leave the heliosphere at a more leisurely pace, and arrive at the photosphere moving much faster?

Perhaps the current picks up additional electrons, as it passes through the space between the heliosphere and the Sun?

I'll look into these questions in a later post, as well as:
* a bigger heliosphere; a smaller heliosphere
* how electrons may deliver energy to the Sun's photosphere, other than by their kinetic energy
* incorporating relativity.

Request to [any interested reader]: please check everything, and let me know if you see any flaws.

(Part 3 to follow)

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