On Viscount Aero's request, I went back to buff up my model of dusty plasma collapse into a CFDL star, and I finished some calcs I had been working on. The results spell trouble for the standard model of the Sun. All of this is on my website, where I'll be updating it, if/when new info comes along, but here is the current version, for your convenience. (Skip to the end if you just want to see the Conclusions.)
If the Sun's energy originated with the momentum of the particles in the collapsing dusty plasma, we can estimate the kinetic energy using Newtonian laws of motion, if we know the mass and the velocity. The mass of the Sun is a known quantity — 2 × 1030 kg. Unfortunately, we don't know the velocity of particles in a collapsing dusty plasma, but we can guess at it. Comets approaching the Sun achieve a velocity of roughly 480 km/s. Of course, they don't start from as far away as the extents of the dusty plasma, nor do they keep getting accelerated all of the way to the center of gravity — the centrifugal force in their elliptical orbits limits the acceleration. They are also accelerated merely by gravity, while dusty plasmas collapse due to forces that are at least 5× stronger (as acknowledged by CDM proponents). So we'll figure that cometary velocities are probably shy, and perhaps something like 700 km/s would be more like it.
E = ½ · m · v2 = ½ · (2 × 1030 kg) · (700,000 m/s)2 = ½ · (2 × 1030 kg) · (4.9 × 1011 m2/s2) = 4.9 × 1041 Joules
If we figure that the driving force is 5× stronger, and conservatively neglect the integration of the acceleration through time, we can guess that the resultant velocity is 5× faster. This yields 1.22 × 1043 Joules.
Note that this is the amount of energy that the Sun still has left, given its current mass. We have no idea how much energy was originally pumped into the Sun, because we don't know how much mass the Sun has lost to the interstellar winds.
We can also estimate the energy not by starting at the end, with the ultimate velocity achieved nearing the centroid, but rather, at the beginning, with an accounting of the thermal energy that was originally contained in the dusty plasma. On compression, all of that heat will be conserved and concentrated. Knowing the elements and their specific heat capacities, we can then estimate the total energy of the system, and it should match the total estimated by the momentum going into the accretion.
First we have to find the pressure of the primordial dusty plasma. The Sun's mass is 1.99 × 1030 kg, and it condensed from a dusty plasma with a volume of something like 7.48 × 1046 m3, for a density of 2.66 × 10−17 kg/m3. The temperature would have been roughly 10 K. For a first pass, we can run the numbers assuming that it is all diatomic hydrogen, with a molar mass of 2.02 × 10−3 kg/mol.
Knowing the temperature, we can double-check the joules by calculating the specific heats of the various elements within the Sun.
The Sun has a total mass of 1.99 × 1030 kg, and the standard model asserts that it is 75% hydrogen and 25% helium, with traces of heavier elements. So the average heat capacity would be 12026.25 J·kg·K.
But the Elements section demonstrated that there are a wide variety of elements in the Sun, and each element has a different specific heat capacity. So it is more accurate to read the elements table to find the specific heats, and to find the masses of each element from the properly weighted abundances (using this code). Those masses can be multiplied by the compacted temperature of 7.76 × 108 K, and by their respective heat capacities, to get the Joules per element, which can be summed to get the total. The result returned by this code is 1.92 × 1042 Joules, which is within an order of magnitude of the 4.9 × 1041 Joules based on momentum. And recall that if the implosion velocity was 5× faster, the resultant energy came out to 1.22 × 1043 Joules. So the specific heat estimate sits right in the middle of the kinetic energy estimates, and 1042 Joules is a good working average.
Conclusions
Thus two fundamentally different methods of calculating the energy of the Sun both return numbers within an order of magnitude out of 42, and we can confidently draw conclusions. The significance for solar modeling is that this rules out the standard model, where an average temperature of 7.76 × 108 K, and an average pressure of 4.52 × 1015 Pa, simply are not possible. Given only gravity pulling in, and hydrostatic pressure pushing back out, the standard model finds an equilibrium with a maximum of 1.5 × 107 K and 1017 Pa in the core, dropping down to virtually nothing at the surface. But in so doing, that model rejects almost all of the thermal energy that went into the Sun's accretion. It then has to assert the existence of an internal heat source (i.e., the "fusion furnace") to get a steady power output. But then it's only solving a problem that it created. And while the present thermodynamic calculations are based on measurable conditions in dusty plasmas, the standard model has to assume the presence of a fusion furnace. Since the expectations of that model are not met in many other ways (e.g., how do gamma rays from nuclear fusion in the core get converted to 6000 K black-body radiation?), we can only conclude that the standard model is fundamentally flawed.
The current-free double-layer (CFDL) model takes a different approach. The thermal energy of the primordial dusty plasma is incontrovertible, so it has to be taken as a given. This begs the question of where all of that thermal energy went. The answer is that it got converted to electrostatic potential. Under the enormous pressure of implosion, and with the force of gravity thereafter, electron degeneracy pressure (EDP) separated charges into CFDLs. The resulting electric fields removed the degrees of freedom, thereby reducing the temperature, down to the 6000 K we observe at the Sun's surface. The heat is recoverable if the electric charges are allowed to recombine, where arc discharges regenerate thermal energy from electrostatic potential. Thus the conservation of energy is maintained throughout the life cycle of the system.
Sparky
Re: CFDL solar model passes another test
I can't contribute anything of importance, but I wanted to thank you.. You continue to exhibit a beautiful mind.
CharlesChandler
Re: CFDL solar model passes another test
Thanks, Sparky.
BTW, on closer scrutiny, I realized that both of these methods for estimating energy have to be added together to get the final result. I guess that this got past me because these calcs are rarely used in conjunction like this.
Normally, if you're compressing a gas, the speed of the compression is small compared to the particle velocity itself, so it isn't taken into account. In other words, in an internal combustion engine, when the piston is compressing the air, the speed of the piston (i.e., 20 m/s) is slight compared to the speed of the air molecules at room temperature (i.e., 500 m/s), and the piston velocity doesn't add much heat. So you just calculate the consolidation of the existing heat in the air, to figure the increase in temperature. But if you're talking about a dusty plasma that is collapsing at 700 km/s, that's actually a very high speed compared to the particle velocity at 10 K, so yes, the thermalization of that kinetic energy has to be calculated.
Likewise, if you're calculating the Newtonian force, to see how much energy will be released when two objects collide, you wouldn't ordinarily care how hot the objects were in the first place, because that energy doesn't come into play. So if you swing a hammer to drive a nail, you don't bother measuring the temperature of the hammer first, because that doesn't matter.
But in a collapsing dusty plasma, there is a very significant compression ratio (i.e., 1:1019), so that will jack up the temperature all by itself, and extreme velocities will be achieved by the time everything gets to the centroid, and all of the energy stored in momentum needs to be added to the total. In engineering terms, this is an unusual case, where significant quantities of both of these forms of energy are present in the same system, but there it is.
So, going with the conservative numbers, 1.92 × 1042 Joules of adiabatic compression, plus 4.90 × 1041 Joules of momentum, equals 2.41 × 1042 Joules total.
These are still just order-of-magnitude numbers, and this doesn't really change the conclusions. It's just that the standard model is in even deeper hot water. But for the sake of clear thinking, it should be acknowledged that these are two different forms of energy, and they get added together.
As an aside, it never ceases to amaze me how fundamental the problems with the standard model are. Surely scientists ran these numbers, since the laws of thermodynamics are universal, and are regularly used to double-check the energy budget of a system. You'd think that this kind of overview of the energy in our solar system would be covered in the opening paragraphs on the Wikipedia page. For other types of systems, this is common. But I didn't find these numbers anywhere, so I had to learn how to do them myself. And what I found is that the entire premise of the standard model is false. The Sun's gravity can contain the standard model's average pressure of 8.98 × 1013 Pa, but the actual pressure after adiabatic compression should be 4.52 × 1015 Pa. That isn't a star — it's a hydrostatic bomb. So the standard model blows up, and what do scientists do about it? They neglect to mention the simplest of energy calcs, and hope that everybody will take their word that the standard model is still OK.
The bottom line here is that we can't assume that scientists have already done the simplest of sanity checks on their models. To spot errors like these, you have to learn to think like an engineer, and not like a theoretical scientist. Imagine that you're trying to build one, and you have to make a shopping list of all of the things that you'll need in order to get it running. Then go through the whole thing in your mind, checking to see if each piece and each process will combine to produce the desired results. Just remember that if it doesn't work, you have just a tad more than a broken model — you just set yourself up to make a scientific discovery, but you just found where there is a missing piece. Then you just figure out what it would take to plug the whole, and bam! you get something named after you.
CharlesChandler
Re: CFDL solar model passes another test
The CFDL model also explains why the Birkeland currents in the aurora do not project all of the way to the ground. See this for more info:
This shows cme's , solar wind impactions, weather, and volcano activity. Each video is short and relative current.
CharlesChandler
Re: CFDL solar model passes another test
As a follow-up to my OP, I completed work on my write-up of the solar Energy Budget, and I got another unexpected revelation out of it.
First, I calculated how much time the Sun has left, given its mass, and the mass loss rate. (For the purposes of these calcs, it doesn't matter that the Sun is recycling mass, as I'll explain later.)
ΔTime = solar mass / mass loss rate = 1.99 × 1030 kg / 5.75 × 109 kg/s = 3.46 × 1020 seconds = 1.10 × 1013 years
Knowing the rate at which matter is expelled by CMEs, and the current rate of power production (1026 watts) from that expulsion, and knowing the total mass, we can easily find the total amount of energy that was pumped into the system in the first place. In other words, if every time the Sun lost 1 kg of mass, it produced 1 watt of power, and if it had 100 kg left, it has 100 seconds left, and it will produce 100 watts of power in those 100 seconds. Since J = W × s, we can then say that the system has a total of 100 joules. And it doesn't matter if the system is recycling mass, because the energy budget works out the same way — it just means that with time, CMEs expelling the same amount of matter will release less energy per kilogram, but in the end, it will release the same total amount of energy.
Eelec = watts × seconds = 1.93 × 1026 W × 3.46 × 1020 s = 6.67 × 1046 J
All of that energy could have only come from two sources: the heat that was already contained in the dusty plasma before it imploded, and heat that was generated by the implosion itself (i.e., the thermalization of all of that momentum). I did a bunch of reading, and came up with much more accurate numbers for the thermal energy in the original dusty plasma, but it only came out to 1.11 × 1043 J, which is less than 1/1000 of the total amount of energy. The only other source is the momentum from the implosion. Knowing the amount of energy, I could then calculate the velocity of the implosion, using Ek = ½·m·v2.
This came out to 2.59 × 108 m/s. That's 86.38% of the speed of light! This, of course, is certainly possible, since the force causing the implosion is 3 orders of magnitude greater than gravity, so I got a resultant velocity that was 3 orders of magnitude greater than the speed of comets that are accelerated toward the Sun just by gravity.
But there is a hidden significance here — approaching the speed of light, it takes more and more energy to continue the acceleration. Thus we can consider the speed of light to be a limiting factor in how much velocity can be developed, pretty much regardless of the force acting on the matter. And of course the energy stored in momentum is a function of the velocity. So there is a limit to the amount of energy that can be developed in a collapsing dusty plasma due to velocity. The implication is that there is only one other remaining factor in the equation for kinetic energy that can vary the total: mass. So dusty plasmas implode with a speed limit, and the energy so produced varies just with the mass of the dusty plasma. Thus heavy dusty plasmas will build up a lot of energy, and light dusty plasmas will only build up a little bit of energy. And where do we see that same relationship, between mass and energy release? In the Hertzsprung-Russell diagram. Heavy stars burn really brightly, while small stars are much dimmer. If dusty plasmas could implode at varying rates, depending on how much electric force was present, you'd get a random distribution of sizes and luminosities, such as small bright stars that imploded really fast, and large dim stars that had a lot of mass, but imploded slowly, so they didn't develop much energy. Yet the main sequence in the HR diagram is very regular, where luminosity varies directly with mass. So the speed of the implosion had to be regulated, and I found the regulator.
Chuh-ching (again).
Lloyd
Re: CFDL solar model passes another test
Say that more clearly. Okay?
You're saying the regulator of star formation is the speed limit of light?
Or the regulator is the mass of the dusty plasma?
Regulate means what exactly?
And how does light speed, or mass, or something, regulate star formation?
It limits the velocity of implosion of dusty plasma in space and that causes big stars to shine brightly and small stars to shine dimly how?
Sparky
Re: CFDL solar model passes another test
Yet the main sequence in the HR diagram is very regular, where luminosity varies directly with mass.
CharlesChandler
Re: CFDL solar model passes another test
Lloyd wrote: Say that more clearly. Okay?
I'm saying that dusty plasmas all implode at roughly the same velocity, because they hit a speed limit. So in the following equation...
Ek = ½·m·v2
..."v" is always the same. Therefore, the only thing that varies is the mass. Thus there is a direct relationship between the mass of a star, and the amount of energy it has. And that relationship shows up quite clearly as the main sequence in the Hertzsprung-Russell diagram.
The "speed limit" is the speed of light. Approaching c, it takes more and more force to accelerate objects, until it gets to the point that it just doesn't matter how much more force there is — the velocity just isn't going to increase. This leaves mass as the only variable.
BTW, there is also a lower limit to the speed at which a dusty plasma can implode, and still form into a star — I just don't know how to quantify it yet. All other factors being the same, an imploding dusty plasma should just bounce off of itself. With any momentum at all, the imploding matter will overshoot the hydrostatic equilibrium. But when it does, it "should" guarantee that there will be a rebound, just like a superball bouncing on a sidewalk, where gravitational potential gets converted to kinetic energy, but on impact with the sidewalk, that kinetic energy gets converted to elastic potential, which then is the source of the energy that motivates the rebound. If all of the conversions are non-lossy, the matter should rebound back out to precisely its starting point. And compressing/expanding gases/plasmas are supposed to be non-lossy, since any energy that is "lost" is converted to heat, and since heat is just another factor in hydrostatic potential. So why don't dusty plasmas just bounce off of themselves? The reason can only be that some other set of factors kicks in. I'm contending that under sufficient pressure, charges get separated, producing charged double-layers, and thereafter, the layers will cling tightly to each other, due to the electric force, though the pressure prevents charge recombination. Thus they are current-free double-layers (CFDLs). One of the implications to this is that there has to be a minimum amount of pressure developed by the implosion, to initiate the charge separation process. I just don't know how to calculate that minimum pressure yet. But we can look at other cases where gases are getting compressed, and they rebound instead of collapsing. For example, basketballs wouldn't work properly if the gas inside didn't rebound — you'd try to bounce the ball on the floor, and it would just go "thud" and all of your friends would just stand there and laugh at you. So we know that gases that had overshot the hydrostatic equilibrium will rebound. But if the basketball has a radius of 1010 m, and you throw it down at 86% the speed of light, it collapses into a star. Then all of your friends start cheering. So what is the minimum threshold? This is what I don't know. Anyway, with both a lower and an upper speed limit, you get a very regular main sequence in the Hertzsprung-Russell diagram. IMO, the stars in the Asymptotic Giant Branch (i.e., the Red Giants) are not dying — they're in the process of being born. Perhaps some of them don't make it. The ones that do follow the AGB onto the main sequence. Perhaps somewhere in those data are the telltale signs of a minimum amount of energy necessary to form a star.
Lloyd
Re: CFDL solar model passes another test
CC said: Anyway, with both a lower and an upper speed limit, you get a very regular main sequence in the Hertzsprung-Russell diagram. IMO, the stars in the Asymptotic Giant Branch (i.e., the Red Giants) are not dying — they're in the process of being born. Perhaps some of them don't make it. The ones that do follow the AGB onto the main sequence. Perhaps somewhere in those data are the telltale signs of a minimum amount of energy necessary to form a star.
Brant said something like that I guess about 3 years ago, probably when I was discussing his Aether Battery Iron Sun model with him, just before the 4 of us started our Electric Sun Discussion. I mean I think he said maybe the stars toward the upper right in the HR diagram belong in the lower right. I forget what his reasoning was.
Wouldn't the smallest molecular cloud implosion be the kind that produces the smallest rocky body that has CFDLs?
It seems to me that there could be a mega gas cloud made up of different sized smaller gas clouds and all of them could implode together or separately into several CFDL bodies of various sizes. The moons that have geysers etc may be close to the minimum size CFDL body, I suppose. I think the smallest body that becomes a gravitational sphere is about 200 miles in diameter. Any body bigger than that is unable to be non-spherical. Right? But I don't have a clue yet about whether such a sized body would have to have CFDLs. Do you?
If astronauts pulled a few big asteroids together, I think they would collapse gravitationally into a sphere. Do you agree? Would there be enough pressure inside them to form CFDLs? Or do CFDLs need a certain amount of heat too?
CharlesChandler
Re: CFDL solar model passes another test
Lloyd wrote: If astronauts pulled a few big asteroids together, I think they would collapse gravitationally into a sphere. Do you agree? Would there be enough pressure inside them to form CFDLs? Or do CFDLs need a certain amount of heat too?
I don't know if gravity alone could do it. The scenarios that I'm running through involve a great deal of pressure from momentum, way above what gravity alone would provide. Still, enough gravity would invoke electron degeneracy pressure (EDP), and ionize the core. Thereafter, the core will flow like a liquid, even if it's cold enough to be a solid. This is because the ions lack the covalent bonds necessary for crystal lattices. Likewise, the Earth's mantle flows instead of fracturing, though the mainstream doesn't have such a great explanation for this — rocks get more brittle with pressure, until they get pulverized. But if EDP kicks in, the rock can flow. So I guess you could make a spherical planet out of so much asteroidal rubble. But perhaps that's a different threshold. If the orb was formed by an implosion, perhaps that's one way, and then there's the "cold gravity" way, which would be different. (?)
Lloyd
Re: CFDL solar model passes another test
CC said: So what is the minimum threshold [speed limit]? This is what I don't know. Anyway, with both a lower and an upper speed limit, you get a very regular main sequence in the Hertzsprung-Russell diagram.
Can you show at what implosion velocity range and what mass range each type of star formed from? And what's your formula? And does the same formula work for the exotics? A similar list of velocities and masses for the exotics would be helpful, if it ain't too much trouble. I'm aware that I may be asking too many questions to answer in a short time. By the way, is lightning or a solar arc discharge a similar implosion process? If so, that fact may attract EU fans.
I'm aware that I may be asking too many questions to answer in a short time.
The way of the Lloyd. Too many questions, too many links. too Irrelevant or not, to make noise, should we all wear a pink tutus too?
Lloyd
Re: CFDL solar model passes another test
By the way, CC, David just posted at http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=15364&p=101123#p101110 showing that batteries apparently don't short circuit in vacuum. So it looks like there needs to be re-analysis of the rocket tests of upper atmosphere conductance as well as of why batteries don't short out in vacuum.
CharlesChandler
Re: CFDL solar model passes another test
The Sun formed in just 54 minutes!
The cool thing about physics is that if you know two things, you can derive a third thing from them. And the fun has only just begun! Now you know three things, and you might be able to derive a fourth. In the end, you might just be amazed at how much information you can derive.
So in previous posts, I described how I estimated the total amount of energy in the Sun, knowing the mass, and knowing the temperature of the dusty plasma from which it condensed, and knowing the rate at which it condensed (adding the energy of momentum). I correlated that with the rate at which energy is released, and that allowed me to find-tune the estimate of the total amount of energy, and especially, the component of it that came from momentum. This proved that the force causing the collapse of the dusty plasma had to be three orders of magnitude more powerful than gravity. Elsewhere I calculated the forces in dusty plasmas, and sure enough, the electric force between Debye cells (after getting shocked by a supernova) is three orders of magnitude more powerful than gravity. So all of that checked out.
But wait — there's more!
Knowing the velocity of the implosion at the end (2.59 × 108 m/s), and knowing the pressure developed by the adiabatic compression of the dusty plasma (2.60 × 1016 Pa), we can now calculate the amount of time that it took the implosion to come to a full stop. We know that gravity inside the Sun can contain an average presssure of 8.98 × 1013 Pa, so that's the hydrostatic equilibrium. Any pressure above that is a hydrostatic overshoot from momentum (which is stable in the present model, with the additional electric force between CFDLs offsetting the pressure). Since the hydrostatic equilibrium is 3 orders of magnitude less, it doesn't even show up when rounded to 2 significant digits, but the calculations used the full precision.
Povershoot = 2.60 × 1016 Pa − 8.98 × 1013 Pa = 2.60 × 1016 Pa
To find the total amount of force of this pressure, we just multiple it by the surface area of the Sun.
Fovershoot = 2.60 × 1016 Pa × 6.08 × 1018 m2 = 1.58 × 1035 N
This is the force that decelerated the imploding dusty plasma. So we divide that by the mass, to find the (negative) acceleration.
Acceleration = Force / mass = 1.58 × 1035 N / 1.99 × 1030 kg = 7.93 × 104 m/s2
Finally, we apply the acceleration to the initial velocity, to find how long it will take to come to a full stop.
And is there any reason to believe that there are star-formation events within that timeframe?
Gamma-ray bursts (GRBs) can last from ten milliseconds to several minutes, and where the initial burst is usually followed by a longer-lived "afterglow" emitted at longer wavelengths (X-ray, ultraviolet, optical, infrared, microwave and radio). (See Vedrenne, G.; Atteia, J., 2009: Gamma-Ray Bursts: The brightest explosions in the Universe. Springer Science & Business Media)
We found that the temperatures and pressures in the implosion are sufficient for nuclear fusion, which will release gamma-rays, until the continued implosion supplies sufficient matter to absorb them. So it makes sense that gamma-rays are emitted for something less than the total time to stop for the implosion. And it makes sense that thereafter, there is an "afterglow" — that's the star that formed.
So yet again I find a new way of double-checking the CFDL model, and instead of getting impossible numbers, I get numbers that fall right in line with what we already know about other stuff. I'm thinking that I'm on the right track here.
