Ok, I'm dumb, but help me out!!!

[upriver]

In a hypothetical situation, how would I provide the "math" to someone if the question was "What is the function of the polar plumes and how much current do they carry in Alfven waves?" "What is the function of the neutral helium tubes and what is their plasma density and do they also have Alfven waves?"
These questions are perfectly in line with the standard model.
Could you please provide me with examples.

[01101001]

In a hypothetical situation, how would I provide the "math" to someone if the question was "What is the function of the polar plumes and how much current do they carry in Alfven waves?" "What is the function of the neutral helium tubes and what is their plasma density and do they also have Alfven waves?"
These questions are perfectly in line with the standard model.
Could you please provide me with examples.

I'm not sure I understand your question. I believe the answer to how is to study the subject deeply enough so that you have the expertise to provide the requested mathematics. Until then, try: "I don't know."

That usually works for me.

[Karl]

In a hypothetical situation, how would I provide the "math" to someone if the question was "What is the function of the polar plumes and how much current do they carry in Alfven waves?" "What is the function of the neutral helium tubes and what is their plasma density and do they also have Alfven waves?"
These questions are perfectly in line with the standard model.
Could you please provide me with examples.

Perhaps the best way to approach it would be to examine what the observable effects would be and explain the presence or absence of observations that support your model.

For instance, the Io flux tubes and plasma torus at Jupiter cause readily observable effects that are detectable via remote sensing at earth, and corrobarated by local measurements on spacecraft. Surely something on a much larger scale would have larger effects? The Ulysses spacecraft flys through the region you are talking about with extremely sensitive electromagnetic and magnetic sensors. What levels would you expect to see out of these sensors?

[Ricimer]

well, you could try to show how much current they could carry by using a few known numbers and observations.

Here's an example, using math, on the electric sun and power sources:

I figured this report I whipped up a few months back (and also posted on SDC) might be useful around here:

NOTE: the truely relavant section on the EU is about half-way down.

As it currently exists (and imo will continue to exist) the Standard Model of stellar evolution claims that the stars is powered by nuclear fusion of various elements (mostly hydrogen).

Many people have problems with this statement, and produce alternative models, like the "electric sun" theory (http://www.electric-cosmos.org/sun.htm). Ignoring some of the problems with the underlying premisis (that sunspots are holes to the interior for instance) there is a fundamental question which must be answered by any stellar model.

Where does the energy come from?

stars radiate energy and very high rates, for a very, very, long time. Even if you believe that the sun was much dimmer in the past, say even half as bright, you are dealing with a large amount of energy. This reduction, at most, can add only 2 orders of magnitude to any life-time estimate of a stellar model. This is because you can only reduce the luminosity by 100%, or 10^2. An order of magnitude is a power of ten. As such if a energy scheme is off by a factor greater than 100x the required value, it is not a viable energy source by itself.

As an example for such an analysis, (which I was required to write up for an advanced lab course) I give you the following (if the math gets you, skip on a bit) treatise on using gravitational contraction as an energy source. This is the same energy source that makes a brick hurt if you drop it on your foot.

The short version: The total time an object can radiate at a given luminosity, is determined by dividing the total energy provided by the method (in this case gravitational potential energy U) by the luminosity. I.e. T = U/L. For the sun, this gives a lifetime of only 30 million years (before it's gone competely) and thus any model that relies on gravitational contraction a significant source of energy, will not work.

The Long Version:

A note on Notation: As the format from word doesn't translate well, any terms that are 10 followed by a number (e.g. 1041) should be read 10^41, as the superscript may not have translated well. I tried to catch all the instances, but I may have missed one or two.

Can gravitational contraction be a viable energy source to power the sun? Ignoring the effects hydromagnetic dynamics, virial theorem, and other complicating details, (which are shown to be unnecessary in this analysis) this can be easily determined by considering the amount of energy available to the sun via contraction, and considering the rate of contraction required.

All contracting, or falling, objects are converting gravitational potential energy into kinetic. The amount of potential energy available to a spherical object is:

U ~ GM^2 / R

G is the gravitational constant (6.67 x10-11), M is the mass of the object, and R is the radius. The sun has a radius of 695,000 km (~700 million meters), and a mass of 2x1030 kg. . This is derived from the fact the sun radiates a spectrum that approximates a blackbody curve (with an excess in the infrared, and deficiency in the ultraviolet) with a shape that puts the sun’s temperature is approximately 6000o K. If luminosity equals:

L = 4 * π * σ * R^2 * T^4

Then the sun then radiates energy at ~4 x 1026 joules per second at the surface.

If energy radiated by the sun is due entirely to the gravitational potential energy, then the luminosity of the sun (the energy radiated per unit time) is given by:

L = dE/dt = dU/dt

By the chain rule we can put this in terms of the radius, allowing us to consider the change of radius due to the energy loss.

L = dU/dt = dU/dR * dR/dt

Substituting the spherical potential energy in for U, we obtain:

L = - d/dt[ -GM2/R ] = - GM2/R2 dR/dt

This equates the change per second of the potential energy to the change per second of the radius.

dR/dt = L / GM^2/R^2 = L*R / U

The value of U, obtained from the figures above, is 3.82 x 10^41 J. Thus dR/dt is:

4 x 1026 W * 7 x 108 meters / 3.82 x 1041 J = 7.3 x 10-7 meters per second

This equates to 23 meters a year. At that rate, the sun shrinks from 7 x 10^8 meters to zero in ~30 million years. While this is easily long enough to sustain the sun for recorded history, it does not match the radioactive dating of the earth to ~4.5 billion years. It is an easy premise to accept that the earth must have formed during, or after the sun’s formation. Especially if one considers the earth’s nearly circular orbit about the sun, good evidence it formed in place as opposed to being a captured object (which should result in much more elliptical orbits) to talkorigins.org, thirty million years ago primitive monkeys didn’t exist, let alone the older (and often extinct) species. The K-T extinction event (thought to be the demise of the dinosaurs) is dated to 65 million years ago (sdnhm.org). Under gravitational contraction the sun wouldn’t exist that long. In order for the sun to radiate at it’s current luminosity for the age of the earth, it would have to have a radius of ~10^11 meters. As the earth is only 150 x10^9 meters from the sun, the earth would have been enveloped by the sun in the beginning, most likely rendering earth’s formation impossible.

An even shorter analysis is merely comparing the observed sun’s luminosity, to the potential. U/L gives ~10^14 seconds, or 30 million years (as expected). A similar analysis for Sirius (U/L) produces an even shorter time period of ~10^13 seconds, or 3 million years. Sirius is larger (R= 2*Rsun) and more massive (M = 2*Msun), giving it a higher store of potential energy, but it has a much higher luminosity as well, at 23 Lsun. This higher mass, larger radius trend, but exponentially higher luminosity, is typical of ninety percent of the stars in existence. The other ten percent are stars that are bloated, having a much higher radius, and luminosity for a given mass. The higher luminosity is due to a greater surface area, as the temperature is lower. A typical example of such bloated red giants is Betelguese. With a mass roughly equal to 15 solar masses, radius of about 650 solar radii, and a luminosity ofapproximately 70,000 times that of the sun (uiuc.edu). Under those conditions, Betelguese lasts for a mere 5 x 10^9 seconds, or about 150 years.
These time scales do not suffice for the full life time of the star, though they do apply to the dynamics of star formation as it coalesces out of interstellar gas and dust (which is a gravitational contraction phase). The time scales here approximately match those sited at southalabama.edu (except for betelguese, which is off by 10^3 for some reason).

Any mechanism for the power of the sun must explain the initial origin of the released energy. The Electric sun model must supply an energy source to separate the charges and induce the “lightning” that is the sun. The first response is usually to claim that currents of plasma are carrying charge, however something needs to drive these currents. Convection requires a temperature differential, which requires and energy source. Gravitational contraction cannot supply the energy needs of the sun for the duration required (4.5 billion years) at the current luminosity.

According to nasa.gov the sun’s luminosity, after protostar formation, was about 50% of the current output. Even reducing the luminosity of the sun for the entire duration to this new, lower, output will not suffice, as our timescale for gravitational contraction is orders of magnitude to short.


Electronic References

http://image.gsfc.nasa.gov/poetry/ask/a11496.html

http://www.southalabama.edu/physics/...1/starform.htm

http://www.astro.uiuc.edu/~kaler/sow/betelgeuse.html

http://sdnhm.org/fieldguide/fossils/timeline.html



Here's the really Relevant part

A quick analysis of the electric sun theory, as presented at it's webpage (shown in first post):

The Basic Electric Sun Theory, as outlined at it’s own web-page:

http://www.electric-cosmos.org/sun.htmf

this Electric Sun model are as follows:

• Most of the space within our galaxy is occupied by plasma (rarefied ionized gas) containing electrons (negative charges) and ionized atoms (positive charges). Every charged particle in the plasma has an electric potential energy (voltage) just as every pebble on a mountain has a mechanical potential energy with respect to sea level.

• The Sun is at a more positive electrical potential (voltage) than is the space plasma surrounding it - probably in the order of 10 billion volts.

• The Sun is powered, not from within itself, but from outside, by the electric (Birkeland) currents that flow in our arm of our galaxy as they do in all galaxies. In the Plasma Universe model, these currents create the galaxies and the stars within those galaxies by the electromagnetic z-pinch effect. It is only a small extrapolation to propose that these currents also power those stars. Galactic currents are of low current density, but, because the size of the stars are large, the total current (Amperage) is high. The Sun's radiated power at any instant is due to the energy imparted by a combination of incoming cosmic electrons and outgoing +ions. As the Sun moves around the galactic center it may come into regions of higher or lower total current and so its output may vary both periodically and randomly.

• Positive ions leave the Sun and cosmic electrons enter the Sun. Both of these flows add to form a net positive current leaving the Sun. This constitutes a plasma discharge analogous in every way (except size) to those that have been observed in electrical laboratories for decades.

• Because of the Sun's positive charge (voltage), it acts as the anode in a plasma discharge. As such, it exhibits many of the phenomena observed in earthbound plasma laboratories, such as anode tufting. The granules observed on the surface of the photosphere are anode tufts (plasma in the arc mode).

So the full version doesn’t claim gravitational contraction at all (though the version I was familiar with did), but an external voltage difference of 1010 volts between space and the sun. In order to supply as much energy as gravitational contraction did, at this voltage, the sun must have a charge of ~1031 coulombs, giving a surface charge density (as the sun should be a conductor) of 1031 coulombs / 1.5 x 1018 m2 = ~6 x 1013 coulombs per m2. That’s a pretty big charge. I wonder if this is large enough to discharge lightning onto mercury, or how fast the solar wind should be. The energy available to accelerate the electron is m*v2 ~ Q*V

v2 = Q*V / Me, where V is 1010 volts (as claimed), Q is ~1 x1019, and Me is 9 x 10-31

This gives a velocity, excluding relativistic corrections, of 10^11 meters per second. Clearly exceeding C, and indicating a highly relativistic velocity. Even protons are ejected at relativistic velocities (without relativistic corrections the velocity is ~3x10^9 m/s). Observed solar wind velocity is on the order of 106 m/s, fast but hardly relativistic.

My charge calculations are based of the interpretation that the energy is comming from the instantaneous charge seperation between the plasma stream and the sun. The amount of charge, by this arrangement, is only enough to match gravitational contraction (which isn't enough either). Even combined this won't cut it (gets us 60 million years! yay!)

If those 10^31 coulombs of charge are instead distributed through the entire volume, we can see if the system is stable.

an electric charge is ~10^-19 coulombs, which means we require 10^50 charged particles to get the required 10^31 coulombs of charge.

The sun has 10^30 kg, and an atom weights ~10^-27kg, giving us 10^57 particles total.

That's one charged particle out of every 10^7. The density of the sun is such that atoms are typically seperated by one width, so each charge is encapsulated by a box 200 particles on a side, for a net # of charges of ~10^7

Now, lets look at the total potential energies between these two figures.

The gravitational potential is G*(10^-27kg*10^7)^2 / R giving 10^-46 / R

The Electrical potential is K*(10^-1^2 / R giving 10^-40 / R.

The electrical potential of the system, that is the energy required to assemble and hold this much charge, exceeds that of gravity (the attracting force) by 10^6 times! A million times!

If this is the charge distribution, the sun blows up. And as stated, this isn't enough charge.

[upriver]

This is not supposed to be an ES thread. I'll get in trouble. SO I have to build from standard observations strictly.

I understand what your saying, I think, Ricimer.

When do electrical currents (and flares) occur in the Sun's atmosphere?
NASA press release 2005/08/16

Strong electrical currents in the Sun's corona form when new field, itself likely carrying currents, emerges into existing active regions out of alignment with existing field (see this study published in the Astrophysical Journal (ApJ 628, 501; July 2005).
http://vestige.lmsal.com/TRACE/Publi.../TRACEpod.html
Scroll down, check out the movies!!

My charge calculations are based of the interpretation that the energy is comming from the instantaneous charge seperation between the plasma stream and the sun.

What if the charges are already seperated(negative and positive)? I think seperated means ions and electrons. Negative and positive recombine at the surface.

if those 10^31 coulombs of charge are instead distributed through the entire volume, we can see if the system is stable.

If they are distributed on the surface of the sun then that problem is negated. And there is evidence for a surface at iron wavelengths. See the surface of the sun. Heated iron has the interesting property of causing hydrogen recombination.

YOU CAN SEE RIDGES AND GLOWING CRACKS. Notice how the loops cross over the ridges.
http://soi.stanford.edu/results/SolP...doverview.html

A jet of hydrogen gas is dissociated as it passes through an electric arc. H2 > H + H - 422 kJ. An endothermic reaction, with the intensely hot plasma core of the arc providing the dissociation energy. The atomic hydrogen produced soon recombines; and this recombination is the source of such high temperatures
The average temperature of the flame is approximately 4000 deg. C., which is higher than the maximum temperature of any other flame. The heat is concentrated chiefly at the point of recombination of the atoms, and this recombination is accelerated catalytically by contact with the surface of the metal being welded. Thus an intense flame is obtained at the point of welding.
The molecular hydrogen burns off in the atmosphere, contributing little to the heat output.


The most important thing is that the electric sun is driven by the potential
"across" it. So they only way to know is to measure the potential across the sun or calculate it from the dimensions and properties of the helium tubes that Ulysses discovered.
Thats what my question is about. What is the potential of the helium tubes that cross our solar system, at the same place that the polar plumes extend to? Is that sufficent to produce the power output observed?
You can see by this bottom picture, that the tube is bigger than the sun, so the logical conclusion is that if it does carry current that it would carry enough to power the sun.


So how do I calculate the power density of the above helium flux tube, assuming it transfers charge from the interstellar medium?

[Ricimer]

By instantaneous charge seperation, I mean it's not a charge due to current.

The sun just "has" this charge.

As for surface vs volume locations for the charge, both are problematic. The volume distribution puts the least amount of stress on the system (charges are as far apart as possible)...but still blow the sun up, without having enough charge to power the sun.

If it's on the surface, you still blow the sun up, but you also accelerate particles off at incredibly high (and most importantly unobserved) speeds.

So the sun can't be powered by a "static" (i.e. unchanging) charge.

So your idea of "current" is really the only way to go. To calculate power, you need to know the charge flux. How much net charge is passing through any given cross-section of the tube. Find that, and you've got a chance.

BTW, I know you (and the BA) don't want this to be an "Electric Sun" thread, and by addressing only specific aspects (current in plasma tubes) I think you'll be fine.

I brought up my Electric Sun post merely because it was the best example on hand of a quick and mathematical analysis of a problem.


This section of your post below is an area where more explaination of physical properties and math to show that it can and does work that way would be useful.

A jet of hydrogen gas is dissociated as it passes through an electric arc. H2 > H + H - 422 kJ. An endothermic reaction, with the intensely hot plasma core of the arc providing the dissociation energy. The atomic hydrogen produced soon recombines; and this recombination is the source of such high temperatures
The average temperature of the flame is approximately 4000 deg. C., which is higher than the maximum temperature of any other flame. The heat is concentrated chiefly at the point of recombination of the atoms, and this recombination is accelerated catalytically by contact with the surface of the metal being welded. Thus an intense flame is obtained at the point of welding.
The molecular hydrogen burns off in the atmosphere, contributing little to the heat output.

[Mosheh Thezion]

its all makes sense, if there is a dwarf mass at the center of the sun, and that it has a very large positive charge.

then.. i imagine you wish to calculate charges and fields across the galaxy, in which case, we need data about ion density per cubic meter in space, and be able to measure potential enrgy and flow patterns at some point.

from which we can generate estimates...

The trick in science is that when you cannot take a direct observations, you need to find a way to make indirect ones, which are atleast quantifiable.. and repeatable. Thats the hard part.

we can estimate things, based on other values, such as the solar wind outputs of the average star, over time,...
and seeing each star as you descibed as having a high positive potential, then a star is a large capacity(volume, and surface area) with a volatage potantial of billions of volts.. it is then imaginable that we can calculate the capacitive values that exist between the stars.. which of course is effected by the denisty of the ions in space.

all of which suggests that electrons are drawn into solar systems, and that there may exist vast inductive fields from clouds of electrons in vacume moving steadily towards all stars, and as such would act as an electrical circuit.. the question that immediately arise is where do all the electrons come from if they are all falling.?

which then suggest that they cannot be failing but instead may take up stellar orbits as clouds, forming fields with any level of complexity.

Eh.. but what do i know.. i just got this idea from you, and while i and unsure about everything, i think its atleast considerable, but not so much as a powersource, but as mechanism for the expansion of the galaxy.? perhaps.

-MT

[upriver]

I have determined that the interstellar helium flux tubes have power density greater than the output of sun.

Sun power output 4*10^26 watts
Dia 1,390,000 KM
r 695,000 KM
Area of diameter sun in square Kilometers 151,746,791,500 square km
Area of diameter of sun in square meters 151,746,791,500,000
Estimated size of helium flux tube 151,746,791,500,000 * 5 solar diameters= 758,733,957,500,000 square meters based on Ulysses flux density diagram.
Plasma flux tube current density based on a 1,000,000 G magnetic field for an average solar flux tube @ (80amps per sq meter= 1 gauss)

At any one point there is a magnetic field equal to 1,000,000 Gauss which is equal to a current density of 2,500,000 amps per square meter. The number of square meters in the helium flux tube, which in the plasma density diagrams appears to be approx. 5 times larger than the sun, is 758,733,957,500,000 sq/metres. This equals a average current density of any cross section of the flux tube at 1,896,834,893,750,000,000,000 amps assuming a homogenous cross section. If we use the voltage of Jupiter's moon, Io's flux tube as a guess(I would like to use the current values as well but I do not have an area for this value), 400,000v * 1,896,834,893,750,000,000,000A = 7,587,339,575,000,000,000,000,000,000 watts or
7.5 * 10^26 watts per interstellar helium flux tube, of which there are 2 of them crossing the sun. The potential across the sun is the difference of the 2 flux tubes assuming one is positive and one is negative, or one is in and one is out, which is 15 *10^52 watts. Each pole of the sun is responsible for 1/2 of the 4*10^26 watts of the total output of the sun, or 2 * 10^26 watts.
The sun power output is 4*10^26 watts, which is well within the potential of an individual flux tube.

[upriver]

the question that immediately arise is where do all the electrons come from if they are all falling.?

There was plasma before anything else, and plasma is seperated charges, so you have free electrons and ions making currents before you have stars or anything else. Recombination powers everything? If you had a perfect motionless plasma, all you would have to do is move one electron to start the whole process of magnetic field and electric current generation.

[Mosheh Thezion]

Facinating... now, let me ask... are you suggesting that all the energy output of the sun is dirived from this souce?
or is it a component?
and either way... what is the source of the energy initially?

as if not from inside stars then where?

i believe if anything we must imagine both... with the fields and currents being not the source, but the result.. and these fields then play there part in regulating the motions of the stars between each other.?

-MT

[Ricimer]

I have determined that the interstellar helium flux tubes have power density greater than the output of sun.

Sun power output 4*10^26 watts
Dia 1,390,000 KM
r 695,000 KM
Area of diameter sun in square Kilometers 151,746,791,500 square km
Area of diameter of sun in square meters 151,746,791,500,000

by this you mean the cross section right?

Estimated size of helium flux tube 151,746,791,500,000 * 5 solar diameters= 758,733,957,500,000 square meters based on Ulysses flux density diagram.
Plasma flux tube current density based on a 1,000,000 G magnetic field for an average solar flux tube @ (80amps per sq meter= 1 gauss)

where did you get these figures? This is actually a very important detail. If these figures are wrong...there's no reason to continue. Showing work can also help (for instance I don't have to do any calculations from scratch just to check if you're right )

At any one point there is a magnetic field equal to 1,000,000 Gauss which is equal to a current density of 2,500,000 amps per square meter.

Show me. this is where the math comes in.

The number of square meters in the helium flux tube, which in the plasma density diagrams appears to be approx. 5 times larger than the sun, is 758,733,957,500,000 sq/metres. This equals a average current density of any cross section of the flux tube at 1,896,834,893,750,000,000,000 amps assuming a homogenous cross section. If we use the voltage of Jupiter's moon, Io's flux tube as a guess(I would like to use the current values as well but I do not have an area for this value), 400,000v * 1,896,834,893,750,000,000,000A = 7,587,339,575,000,000,000,000,000,000 watts or
7.5 * 10^26 watts per interstellar helium flux tube, of which there are 2 of them crossing the sun. The potential across the sun is the difference of the 2 flux tubes assuming one is positive and one is negative, or one is in and one is out, which is 15 *10^52 watts. Each pole of the sun is responsible for 1/2 of the 4*10^26 watts of the total output of the sun, or 2 * 10^26 watts.
The sun power output is 4*10^26 watts, which is well within the potential of an individual flux tube.

Now, this end bit is heading in the right direction...but I think your values are a bit off (as it is time for a reality check). What would an energy flux 10^20 times greater than the suns output do to the solar system? It would certainly be detectable...and likely turn jupiter into a blazing star...and likely incinerate everything else.

At the very least, the average temperature of interplanetary gas and dust would be much, much higher.

[Karl]

I have determined that the interstellar helium flux tubes have power density greater than the output of sun.

Sun power output 4*10^26 watts
Dia 1,390,000 KM
r 695,000 KM
Area of diameter sun in square Kilometers 151,746,791,500 square km
Area of diameter of sun in square meters 151,746,791,500,000
Estimated size of helium flux tube 151,746,791,500,000 * 5 solar diameters= 758,733,957,500,000 square meters based on Ulysses flux density diagram.
Plasma flux tube current density based on a 1,000,000 G magnetic field for an average solar flux tube @ (80amps per sq meter= 1 gauss)

At any one point there is a magnetic field equal to 1,000,000 Gauss which is equal to a current density of 2,500,000 amps per square meter.

Ok, good. Now you have made a prediction, now check your predicted magnetic field against the observed values to determine if your hypothesis holds up or not. . .

[taurus26]

As I have said elsewhere physics was certainly not my strong point. I can't understand a thing EXCEPT, was that a cube I saw encasing the Sun?

If it was, it ties up perfectly with my so called "Rubiks Cube" map of the universe!
( located a little down the page!)

I would invite any discussion on that theory, and any comment as to why the logic is not sound!

[Ricimer]

well, that cube was merely an arbitrary arrangement for a computer simulation...I.e. it's no more valid than a child drawing the sun as a square and saying the sun really is square.

[upriver]

where did you get these figures? This is actually a very important detail. If these figures are wrong...there's no reason to continue. Showing work can also help (for instance I don't have to do any calculations from scratch just to check if you're right )

Yes, the cross section. All this stuff is like learning a new language.
This is where I got my magnetic field estimates. For the estimated magnetic density of the interstellar flux tube model, I scaled these numbers up.


"The magnetic energy density of a field of 10^5 G is two orders of magnitude larger than the kinetic energy density of the convective motions in the lower solar convection zone.
This raises serious doubts whether the conventional turbulent dynamo process based upon cyclonic convection can work on the basis of such a strong field. Moreover, it is unclear whether solar differential rotation is capable of generating a toroidal magnetic field of 10^5 G; it is conceivable that thermal processes like an entropy-driven outflow from exploded flux tubes leads to the large field strength required."
http://www.mps.mpg.de/projects/solar...erriz_2002.pdf


I rembered it being 10^6, so remove 1 zero.

What would an energy flux 10^20 times greater than the suns output do to the solar system? It would certainly be detectable...and likely turn jupiter into a blazing star...and likely incinerate everything else.

This is were the observational part comes in. Since I dont know all the components of the circuit I can't do all the math. I already know I'm off, so I will work on it. But it possible that the whole heliosphere is charged and it powers the planets rotation as well, since most of the planets give more heat than can be accounted for?(conjecture). This is where the model needs more development. I'm looking for more Ulysses observations to complete the circuit. The other thing I find interesting is that the tube changes shape from solar max to min.

As I have said elsewhere physics was certainly not my strong point. I can't understand a thing EXCEPT, was that a cube I saw encasing the Sun?

The cube outline is to delineate the area of interest, in this case the measurements made by Ulysses of the neutral helium flux tube.
"Depicts: Galactic dust concentrations in solar system (high concentration: red/yellow, low concentration: blue/green)
Copyright: ESA" It does not say if it is a simulation which I will assume that it is a simulation based on measurements.

[Van Rijn]

"The magnetic energy density of a field of 10^5 G is two orders of magnitude larger than the kinetic energy density of the convective motions in the lower solar convection zone.
This raises serious doubts whether the conventional turbulent dynamo process based upon cyclonic convection can work on the basis of such a strong field. Moreover, it is unclear whether solar differential rotation is capable of generating a toroidal magnetic field of 10^5 G; it is conceivable that thermal processes like an entropy-driven outflow from exploded flux tubes leads to the large field strength required."
http://www.mps.mpg.de/projects/solar...erriz_2002.pdf

Reading this article, it is clear this is talking about the structure of the sun. Solar physics is one thing, but I see nothing about high energy interplanetary (or interstellar) flux tubes. Your evidence for these is ...?

[upriver]

Reading this article, it is clear this is talking about the structure of the sun. Solar physics is one thing, but I see nothing about high energy interplanetary (or interstellar) flux tubes. Your evidence for these is ...

Ulysses observations.......
Again;

[Van Rijn]

Ulysses observations.......
Again;

Whew! That is a large image. Would you mind explaining where in those diagrams I can find the answer to my question? With numbers? I'm not asking if there are relative concentrations of hydrogen gas in space, I am asking where there is evidence of interplanetary or interstellar flux tubes with extreme magnetic fields and high energy density.

Images are irrelevant. Information on the sun's photosphere is irrelevant. I'm asking about quantitative evidence about the interplanetary and interstellar medium.

[VanderL]

Upriver,

I'm interested in the calculations you are trying to make, but I'm afraid these calculations will convince nobody. Ultimately the energy source of an electrically powered Sun is provided by Birkeland currents from our Milky Way, I don't think anything more can be said at this moment than "the power input equals the power output", more or less the same as in a fusion driven situation. I don't think the conditions inside the Sun's core have been measured and subsequently used to show what the power output will be. It is always the other way around; the Sun's core "must have" densities and temperatures that allow fusion to happen etc, etc. (And remember that the diagnostic for fusion, the neutrino numbers, needed "new physics" to match observations, a situation that would be used to condemn any alternatives that are proposed.) The assumption that the Sun is powered electrically can be asserted through other means, like for instance showing that comets are an electrical phenomenon, and have compositions comparable to asteroids.

Btw:

Upriver:
Area of diameter sun in square Kilometers 151,746,791,500 square km
Area of diameter of sun in square meters 151,746,791,500,000

I think you forgot 3 zeroes. Good luck with the OOM calculations.

Cheers.

[upriver]

I'm researching what I can right now.
Give me a few days.
Thanks.

[VanderL]

I'm researching what I can right now.
Give me a few days.
Thanks.

Sure, if solving the neutrino problem took 30-odd years, you're entitled a few days!

Cheers.

[Ricimer]

Upriver,

I'm interested in the calculations you are trying to make, but I'm afraid these calculations will convince nobody. Ultimately the energy source of an electrically powered Sun is provided by Birkeland currents from our Milky Way, I don't think anything more can be said at this moment than "the power input equals the power output", more or less the same as in a fusion driven situation. I don't think the conditions inside the Sun's core have been measured and subsequently used to show what the power output will be. It is always the other way around; the Sun's core "must have" densities and temperatures that allow fusion to happen etc, etc. (And remember that the diagnostic for fusion, the neutrino numbers, needed "new physics" to match observations, a situation that would be used to condemn any alternatives that are proposed.) The assumption that the Sun is powered electrically can be asserted through other means, like for instance showing that comets are an electrical phenomenon, and have compositions comparable to asteroids.

Btw:



I think you forgot 3 zeroes. Good luck with the OOM calculations.

Cheers.

Actually, central pressures and temperatures can be obtained without looking at the requirements for adequate fusion.

We know the energy out, we know the flux, the area, the volume, and the mass. It's quite easy then to work backwards using gas laws, hydrostatic equilibrium, conservation of energy, gravitational potential and flux (every "shell" can be said to have the same energy in as goes out, due to hydrostatic equilibrium, except the outer edge). These produce values for the central temperature and pressure that are adequate for the fusion process to supply the right amount of energy.

So there is no need to start with the fusion, and build the sun on top of that. You can take the sun, as is, and work down to the pressures and temperatures...and realize fusion is a viable process in those conditions.

[Nereid]

Actually, central pressures and temperatures can be obtained without looking at the requirements for adequate fusion.

We know the energy out, we know the flux, the area, the volume, and the mass. It's quite easy then to work backwards using gas laws, hydrostatic equilibrium, conservation of energy, gravitational potential and flux (every "shell" can be said to have the same energy in as goes out, due to hydrostatic equilibrium, except the outer edge). These produce values for the central temperature and pressure that are adequate for the fusion process to supply the right amount of energy.

So there is no need to start with the fusion, and build the sun on top of that. You can take the sun, as is, and work down to the pressures and temperatures...and realize fusion is a viable process in those conditions.

That it is easy to do these calculations, and discover many (internal and external) inconsistencies with the ES idea has, no doubt, lead some ES proponents to introduce a solid surface to the Sun!

That way, at least, the gas laws (relating pressure, temperature, etc) aren't applicable.

I'm not sure if this is relevant to what upriver is trying to achieve here ...

[Ricimer]

but the solid surface hypothesis directly contradicts Heliosiesmology. And since heliosiesmology is able to accurately and correctly predict the formation of sunspots, even those on the opposite side of the sun (tracked and confirmed when they rotate into view)...the "solid" surface doesn't cut it.

Also, the visible portion of the sun definetly displays gas like behaviors in the spectra lines and such, all the way down to as deep as we can see. So any pictures of a "solid" surface are of this layer, which is decidely a gas due to spectral features.