The Electric sun, energy, and alternative models

Saiph · Mar 28, 2005

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 Ve <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

Saiph · Mar 28, 2005

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. • <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

emperor_of_localgroup · Mar 28, 2005

That was a very good post. More readings to do. I'm not a conspiracy theorist. Existing accepted theories may be correct, but they are not taking us anywhere new, so we have to look at alternative theories also. <div class="Discussion_UserSignature"> Earth is Boring </div>

Saiph · Mar 28, 2005

Actually, there is plenty of work going on with existing theories, and more papers are published now than before, so we're going somewhere. However, it doesn't hurt to look at the other theories, even if wrong they may inspire an idea. Back to the electric universe: 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^-18)^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 this isn't enough charge. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

Saiph · May 9, 2008

bump up and away!This posts relevancy has been restored, feel free to peruse at your leasure (or go blind from math shock).  If you have anything to add to the analysis, such as sustained current flow calculations etc, as apposed to my "instantaneous" approach...   Feel free to post it. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

derekmcd · May 9, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>bump up and away!This posts relevancy has been restored, feel free to peruse at your leasure (or go blind from math shock).  If you have anything to add to the analysis, such as sustained current flow calculations etc, as apposed to my "instantaneous" approach...   Feel free to post it. Posted by Saiph</DIV> I'm not entirely clear on your electrical potential formula.  Where you get the charge 10^-18 from? <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

Saiph · May 9, 2008

seems my posts are oddly cut off when I look a them...anybody else notice this (there should be along calculus derivation of gravitational contraction in the first post.For the time being, a repost in full: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/lectures/clark/LectPH101/starform.htm http://www.astro.uiuc.edu/~kaler/sow/betelgeuse.html http://sdnhm.org/fieldguide/fossils/timeline.html  <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

Saiph · May 9, 2008

repost of the next post, in its entirety.  don't worry, I'll delete or truncate these if the issue gets fixed: 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 10^10 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 ~10^31 coulombs, giving a surface charge density (as the sun should be a conductor) of 10^31 coulombs / 1.5 x 10^18 m2 = ~6 x 10^13 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 10^10 volts (as claimed), Q is ~1 x10^19, and Me is 9 x 10^-31 This gives a velocity, excluding relativistic corrections, of 10^11 meters per second. Clearly exceeding light. Even protons are ejected at relativistic velocities (without relativity the velocity is ~3x10^9 m/s). Observed solar wind velocity is on the order of 106 m/s, fast but hardly relativistic.  <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

Saiph · May 9, 2008

The third post in the series is actually intact in it's entirety..I post it here for continuity though. -----------------ctually, there is plenty of work going on with existing theories, and more papers are published now than before, so we're going somewhere. However, it doesn't hurt to look at the other theories, even if wrong they may inspire an idea. Back to the electric universe: 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^-18)^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 this isn't enough charge.   <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

Saiph · May 9, 2008

Okay, now for my actual response to derekmcd:  10^-19 culomb is the charge of an electron and/or proton.  Right? or did I make a gaff... <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

DrRocket · May 9, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>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).If this were so and if as the EU people contend the sun is powered by this current then based on the known output of the sun there would be an enormous current flowing into the sun and an enormous magnetic field resulting from it that would, at the surface of the Earth dwarf the observied field.  I posted a really rough estimate of this on another thread. The response was that the mystical plasma flow shields the earth -- no mechanism identified. 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. This would only be true if the electrons existed in some established electric field, as does the pebble in the gravitational field of the Earth.  But an electric field is due to a geometrical arrangement of charges.  If the field is due to the particles in the hypothetical plasma, then it is a plasma flow that does not vary in time.  If the field is externally imposed then must inquire as to the source. • 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 notion of potential is that of a scalar field existing in space, and a potential difference is a concept that relates to a pair of points.  So what this sentence must be implyiing is that the Sun is essentially an equipotential body, the plasma forms an essentially equipotential body, and the difference in potential between any point on one body and any point on the other is about 10 billion volts.  But supposedly the plasma permeates all of space.  So the Sun is at 10 billion volts(let' say postiive for the purposes of discussion) to all of space, which is filled with charged particles.  That would require an enormous and continuous inrush of electrons to the sun and an equally enormoug outrush of electrons from it.  As you calculated later the resulting massive charge buildup would tear the sun apart in short order.Now the EU guys claim that the Sun has current passing through it (see below), while the plasma permeating space is a "near perfect" conductor.  In that case the Sun,  also being made of plasma and also a good conductor with very low resistance, cannot have much, if any voltage drop across it.  So, here we have near perfect conductor. enveloped in a near perfect conductor, which ought to make the entire universe and equipotential body.  So where does the voltage gradient come from to drive all of these currents ? And they claim that this is all governed by "electrical theory" that EEs understand but that leaves astrophysicists mystified.  If that is so then I think I had better hide both of my EE diplomas -- I don't want the association with anyone gullible enough to believe this garbage. • 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. From the perspective of a moderately distant obsever (like us on Earth) this would still look like a huge current tube and generate an enormous magnetic field.  By the way, what determines the direction of current flow ?  Radially in won't work because as noted elsewhere the charge build up would rip things apart in very little time at all.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. •I don't think so.  If someone has every exhibited a plasma discharge over a prolonged period of time (let's say a week given that the claim is the sun has done this for a couple of billioin years) with one-way flow of charge it would have made headlines, among other things.  The necessary conditions to simulate either the real or the imagined solar physics are not amenable to sustained support in a laboratory.While I applaud your attempt to bring rationality to this discussion of EU, I am afraid that the effort is futile.  In order to believe this nonsense. you have to suspend bellief in Maxwell's equations (forget about Kirchoff's  version of circuit theory despite the rants about "electrical theory"), you have to believe in a solid nickle and iron shell just below the photosphere (the heat flux inside a radiating sphere at the temperature of the photosphere would make that rather unlikely), you have to believe that the photosphere is a big neon light (despite the near absence of neon detected by physicists), you have to believe in a mystical current flowing throug the universe with no apparent motive force and no clear direction (or perhaps an invisible extension cord), and you have to believe that this is a rational model that is simply unappreciated by a horde of well-educated astrophysicists who cannot understand it and actively suppress discussion on the topic while it is perfectly obvious to someone who dropped out of electrical engineering.   Posted by Saiph</DIV> <div class="Discussion_UserSignature"> </div>

derekmcd · May 9, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Okay, now for my actual response to derekmcd:  10^-19 culomb is the charge of an electron and/or proton.  Right? or did I make a gaff... Posted by Saiph</DIV>{   The Electrical potential is K*(10^-18)^2 / R giving 10^-40 / R.    } Is what you have.  Also, can you, instead of using 1040 for 10^40 (for stuff you are transferring from word), use 1e40 or 1.+40?  If not, that's ok.  I still understand (as much as my limit allows anyway). <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

Saiph · May 10, 2008

This isn't transfered from word...but from uplink.  Heck the first posts are a direct bump (even if truncated)  I have no idea why the formatting is going to pot. as for the equations you quoted, those are the equations for electrical potential, leaving R as a variable (since it's the same in both cases) that would drop out if we wanted it to.  <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

derekmcd · May 10, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>This isn't transfered from word...but from uplink.  Heck the first posts are a direct bump (even if truncated)  I have no idea why the formatting is going to pot. as for the equations you quoted, those are the equations for electrical potential, leaving R as a variable (since it's the same in both cases) that would drop out if we wanted it to.  Posted by Saiph</DIV>I just don't understand where 10^-18 comes from.  Is it represented in coulombs and, if so, what particle is it?  Or is it a constant I'm obviously not aware of?  <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

DrRocket · May 10, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>... 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....  Posted by Saiph</DIV>We have a small problem here.  Look at the expressioin that you have for U. I think it represents the energy that would be obtained from moving the mass of the sun from infinity to a spherical shell with the radius of the sum.  But that is a spherical shell, and not a ball.  But that is not the real problem.  Note the R in the denominator.  If you shrink the radius to 0, U becomes infinite  -- and that is correct.  Now, when you calculated dR/dt you calculated a formally correct value, but one that is valid at R = 7 x 10^8 m.  The balue of dR/dt depends on R.  So you can't calculate it at one value of R and then apply it as a constant to determine when R becomes 0.  Since what you are doing is relating radiated energy to energy available from gravitational collapse and then applying conservation of energy, it is fairly easily seen that becase U grows without bound as R tends to zero, that in fact R will never actually become zero.I may have an arithmetic error in my calculation, but using the remainder of your figures I calculate that it would take about 30 billion years to shrink to a radius of half of the present value, assuming that total radiation remains constant.It think the approach was actually pretty clever, but with the glitch in the math it just doesn't work out. <div class="Discussion_UserSignature"> </div>

Saiph · May 10, 2008

derekmcd:  It's a typo, should be 10^-19 Culombs...and it's supposed to be the charge of an electron (or proton, as they are equivelant). Dr Rocket...Saiph is sleepy now...I'll peruse my math soon.  This was a back of the envelope calculation from...3 years ago.  Pretty standard derivation when I was actually taking a stellar modeling course...not so much now.With some digging I'm sure we could find a third party derivation...but I'll try and check my work first.I do, however, recognize your critique.  Basically, for those who don't speak math, he's saying that I've determined the rate the sun needs to shrink to produce the light we see...at it's current radius.  As the sun gets smaller, the rate the sun has to shrink changes as well.  Now, how this balances with the decreased surface area in the luminosity, vs liberated energy due to the collapse... I'm to frazzled to say just now. Anyway, one of the major points I was trying to make with this post, as the context of 3 years ago is sorta lost, was to demonstrate the sort of work, analysis, and research I would need from anybody espousing an alternate model for the sun.  Usually all we see are assumptions and hand waving, no numbers, no formal logic, nothing.  Just, "Well, it's gotta be some billion volt potential!" which is sorta vague to begin with.  <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

aulfat_hussain · May 10, 2008

• 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. The mentioned line    "the sun is powered"  this point has no such kind of possiblity and prove because sun itself is very massive source of energy and it would not effected by other source of energy easily and also the distance between sun and other stars are so far. If we think about radiated energy from other source of energy then there are alot of matters in unverse which obsorbe radiated energy. May be very little energy would repel and obsorb by sorrounding of sun which has no effective effect. On other point we know that universe is in status of expanding so universe has alots of matters which are thirsty of radiated energy. It is obvious that our galaxy is in status of bombardment by different amount of energy and different scattered particles but their ratio is very less as we compare them with matter of our galaxy like black body, which they are obsorbed by them in very thirsty mode.      <div class="Discussion_UserSignature"> </div>

DrRocket · May 10, 2008

[QUOTE]derekmcd:  It's a typo, should be 10^-19 Culombs...and it's supposed to be the charge of an electron (or proton, as they are equivelant). Dr Rocket...Saiph is sleepy now...I'll peruse my math soon.  This was a back of the envelope calculation from...3 years ago.  Pretty standard derivation when I was actually taking a stellar modeling course...not so much now.With some digging I'm sure we could find a third party derivation...but I'll try and check my work first.I do, however, recognize your critique.  Basically, for those who don't speak math, he's saying that I've determined the rate the sun needs to shrink to produce the light we see...at it's current radius.  As the sun gets smaller, the rate the sun has to shrink changes as well.  Now, how this balances with the decreased surface area in the luminosity, vs liberated energy due to the collapse... I'm to frazzled to say just now. Anyway, one of the major points I was trying to make with this post, as the context of 3 years ago is sorta lost, was to demonstrate the sort of work, analysis, and research I would need from anybody espousing an alternate model for the sun.  Usually all we see are assumptions and hand waving, no numbers, no formal logic, nothing.  Just, "Well, it's gotta be some billion volt potential!" which is sorta vague to begin with.  Posted by Saiph[/QUOTE]  I understand the sleepy problem -- I was in that condition also, but the wife wanted a cable TV movie recorded,so there I was.When you get around to trying to check the calculation you might want to consider some of the following.  It might help to use slightly more accurate figures for electron charge, sun mass, etc.  You do have the correct order of magnitude and that is OK for your main point, but in the final analysis is as easy to use more accurate figures, if you use a calculator or a computer to do the arithmetic.  The two main issues are to look at the derivation and satisfy yourself that the gravitational potential formula models the geometry of the sun accurtely enough for your purposes.  I think it represents a thin spherical shell of radius R rather than a solid ball of radius R, but I could be mistaken.  I need to think about that issue a bit more.  The other, and easier to handle, thing is that if you pursue your derivation what you get is an expressioin for dR/dt that involves R, a non-linear first-order ordinary differential equation.  It is separable and integrable.  You can solve it fairly easily and get and expression for R as a function of t.  You can also find t as a function of R and it is relatively easy from that to compute a "half-life" for R.Trying to actually show the math in this format is something that I have not found a reasonable approach for at this point.  I have not been able to apply anything from my scanner so far, as apparently either the files are too big or I don't quite know the trick to get the job done.  <div class="Discussion_UserSignature"> </div>

Saiph · May 10, 2008

drRocket:  Your scanned files are likely to big. I was going to host my images using Google's Picasa program, though I suppose Flickr is another option.  Then just link to them from here.  Also keeps load times down for those who don't wish to see it, or just wish to see a text reply. One reason the figures are so general is the original reason I did those figures, was for a "rapid estimation" paper in one of my advanced lab courses in college.  Granted, I did so to tie into the topic on SDC at the time, but that's why the analysis is so loose.  <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

origin · May 10, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'> Trying to actually show the math in this format is something that I have not found a reasonable approach for at this point.  I have not been able to apply anything from my scanner so far, as apparently either the files are too big or I don't quite know the trick to get the job done.  Posted by DrRocket</DIV>The easiest way to do this (I think) is to do the work on Powerpoint and save it as a .jpg.  You can then upload it on one of the many sites that hold pictures for free and subsquetly download it to the forum.  Be sure and compress the picture - it is just a calculation so it won't really lose anything.   <div class="Discussion_UserSignature"> </div>

DrRocket · May 10, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>The easiest way to do this (I think) is to do the work on Powerpoint and save it as a .jpg.  You can then upload it on one of the many sites that hold pictures for free and subsquetly download it to the forum.  Be sure and compress the picture - it is just a calculation so it won't really lose anything.  Posted by origin</DIV>I'm not sure that trying to do mathematics on PowerPoint would be successful at anything more than driving me nuttier than I already am.  Trying to use real symblols and show calculation using a computer has been really frustrating.  I know people who set things up for exposition and publication using LaTex and they have become very adept at its use.  It is rather standard with university mathematicians these days.  Lots of lecture notes are prepared that way.  But, as far as I know they do research calculations the old-fashioned way -- pencil and paper.I'm not sure that I want to take the time to learn LaTex.  The software is available for free, and I think I have it stored on disk somewhere.  But I am not sufficiently committed to spend the time to actually learn to use it. I can do everything that I normally want to do more efficiently on paper. <div class="Discussion_UserSignature"> </div>

drwayne · May 10, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I'm not sure that trying to do mathematics on PowerPoint would be successful at anything more than driving me nuttier than I already am.  Trying to use real symblols and show calculation using a computer has been really frustrating.  I know people who set things up for exposition and publication using LaTex and they have become very adept at its use.  It is rather standard with university mathematicians these days.  Lots of lecture notes are prepared that way.  But, as far as I know they do research calculations the old-fashioned way -- pencil and paper.I'm not sure that I want to take the time to learn LaTex.  The software is available for free, and I think I have it stored on disk somewhere.  But I am not sufficiently committed to spend the time to actually learn to use it. I can do everything that I normally want to do more efficiently on paper. Posted by DrRocket</DIV>Wow! It's been 25 years since I touched TeX (I never really did LaTeX - though the girl that did the typesetting on my dissertation did).  My rememberance of it was it was addictive, more like programming a document than writing one.Wayne <div class="Discussion_UserSignature"> "1) Give no quarter; 2) Take no prisoners; 3) Sink everything." Admiral Jackie Fisher </div>

DrRocket · May 11, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Wow! It's been 25 years since I touched TeX (I never really did LaTeX - though the girl that did the typesetting on my dissertation did).  My rememberance of it was it was addictive, more like programming a document than writing one.Wayne Posted by drwayne</DIV>I didn't use either one.  I used a symbol ball on an IBM Selectric, and LOTS of whiteout.  Typing services were sufficiently expensive that it was actually cheaper to buy that typewriter (they were'nt cheap at that time) and do it myself rather than to hire it done.  Fortunately it was only 31 pages.  Unfortunately that was pretty much 31 pages of gory calculation.  But other guys wrote dissertations that went into several hundred pages, usually with lots of explanatory stuff.  I felt fortunate with the shorter document.Mine was closer to a paper.  In fact I basically wrote the paper for publication and then just added a couple of pages to it to fit the requirements for a dissertation.  Didn't have to type the thing twice that way.  The sequence is actually a bit more unusual than just that.  Before I wrote up the paper I had it solicited and essentially accepted for publication by the editor of the journal to which I eventually sent it.  I gave a talk at a meeting about the result.  The editor told my advisor that he thought that was plenty good enough for a dissertation, and to write up the paper and send it to him.  My advisor agreed and off I went.  The guy was pretty well known generally and particularly well known to my advisor so there was no hesitancy on his part -- I think he was glad for the advice as I was his first student. <div class="Discussion_UserSignature"> </div>

michaelmozina · May 13, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'> If this were so and if as the EU people contend the sun is powered by this current then based on the known output of the sun there would be an enormous current flowing into the sun and an enormous magnetic field resulting from it that would, at the surface of the Earth dwarf the observied field.  I posted a really rough estimate of this on another thread. The response was that the mystical plasma flow shields the earth -- no mechanism identified. Posted by DrRocket</DIV>The magnetosphere of planet Earth is "emprical science" Doctor.  It has been studied now for some time, and it has been studied with empirical in-situ satellite measurements since the 70's when Birkeland's aurora theories were first confirmed from space.  To suggest now that the magnetosphere of Earth is "mystical plasma" is bewildering to say the least.  It is simply "normal plasma" doing what normal plasma does, namely shelding us from the elecctriclally charged particles blowing off the sun.  I have also shown you evidence that the solar wind process selectively favors He+2 ion to He+1 ions by 20 to 1 demonstrating that charge attraction is the driving force behind solar wind. There is nothing "mystical" about electron flow through plasma, charge attraction, or particle physics.  Alfven mathematically explained these empirical physical processes back in the 70's, using a lot of material from Birkeland and what he learned about plasma while developing MHD theory.  Coincidently, Alfven did predict that the sun acts as unipolar inductor, and indeed we see evidence of ring currents around Saturn that do appear to be induction driven based on the rotation rate of Saturn.  Every single part of Alfven's EU theories are based known laws of physics, electrical theory, and MHD theory.  Every single bit of it is based upon empirical laboratory experimentation.  There is nothing "mystical" involved in EU theory.   Sun's and galaxies are simply large Farraday disks, or unipolar inductors.  There is no "missing momentum" in solar system formation, that moment is converted to electrical energy via the process of induction like we see in Saturn's ring currents.   Alfven pretty much wrote the book on Plasma Cosmology theory.  It's called Cosmic Plasma.   All of these issues and ideas are carefully explained in that book in a step by step manner.   It's heavy duty reading, but it's solid emprical science, not mysticism.  It's expensive however.  I think I spent $120 on that book, but it was the best investment I ever made as it relates to astronomy and physics.  I highly recommend it.  I promise you if you read it, nothing about EU theory will seem the least bit "mystical'.   It is built entirely upon emprical science, emprical testing and emprical science.  The only thing really left to really be careful about  is the scaling factor.  In other words, all of the physics involved in EU theory comes straight from plasma physics and it is simply scaled up to the size of a universe.  If there is any valid criticism of EU theory, it is the fact the we cannot be absolutely certain that plasma physics principles will scale properly to that size and to that EM field strength.  Thet's really it's only "unproven" aspect, and it's only "Achilles heal".  If we use Birkeland's emprical approach to science, we can only really "confirm" these scaling factors by getting out there into space and taking in situ measurements. What you need to appreciate here however is that every important aspect of EU theory/plasma cosmology theory comes straight from plasma physics and MDH theory.  It is simply scaled up to size.  There is nothing here that is "mystical".  At worst case there are some "unknowns" related to scaling factors, but many aspects of EU theory have already been "confirmed" by in-situ measurements, including the high wind speeds in the polar regions as Alfven predicted would occur in a unipolar inductor.  That same induction model is then just scaled up to a galactic size, and the large, rapidly rotating magnetic field at the core of a galaxy sends induction currents through the spiral arms of the Milky Way.  The induction currents generated by all of the stars in the spiral arms are adding current to the system.  Essentially the momentum of the spinning objects are being converted into electrical current.  The total amount of current that flows though an individual sun however is not simply the "extra" current it converts to electrical energy, but all the energy flowing through that region of the Spiral arms, which is a much larger current.  A galaxy might eventually "spin down", but the galactic current flows that go through our sun may also add momentum back into single suns in the form of spin momentum, meaning that a star may last for as long as the spin energy of the whole galactic systems lasts, and galaxies be "powered" externally for all we know.  The most important point to remeber here , is it that while some aspects of EU theory may be "undemonstrated" by empirical in-situ measurements in space, nothing about EU theory is based upon mysticism.  That is in fact why I am drawn to the theory in the first place.   <div class="Discussion_UserSignature"> It seems to be a natural consequence of our points of view to assume that the whole of space is filled with electrons and flying electric ions of all kinds. - Kristian Birkeland </div>

DrRocket · May 13, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'> nothing about EU theory is based upon mysticism.  That is in fact why I am drawn to the theory in the first place.   Posted by michaelmozina</DIV>The effect of the magnetosphere on charged particles is well known.  But what does that have to do with shielding Earth from the enormous magnetic field that would result from an equally enormous current powering the sun ?EVERYTHING about EU relies on a mystical current flow through the universe, that has not been detected and that would more obvious than an elephant in the strawberry patch if it did. <div class="Discussion_UserSignature"> </div>

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