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Michael Mozina
SFN Regular
1647 Posts |
 Posted - 07/26/2006 : 16:28:15 [Permalink]
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You'll note in my model that arcs on the surface cause superheated silicon plasma to rise up from below. That column of superheated plasma "punches through" the relatively cool photosphere. While there may be cool areas seen in sunspot activity, these sunspots are directly related to the increased heat from the arcs below, which is 100% consistent with rising sea temperatures with increased sunspot activity. Now tell me again how you know that the *average* temperature within a sunspot is *on average* 2000 degrees cooler than the rest of the photosphere, and why don't these related to *cooling* trends then here on earth?
I think I just found the topic and the image that I want use in my next blog entry. |
| Edited by - Michael Mozina on 07/26/2006 16:29:25 |
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Dave W.
Info Junkie
USA
26022 Posts |
Posted - 07/26/2006 : 20:12:35 [Permalink]
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quote:
Originally posted by Michael Mozina
quote:
Originally posted by Dave W.
How does scattering explain the presence of the correct number of total neutrinos, Michael?
It doesn't.
Great! Then oscillation is the more parsimonious explanation of all observed phenomena, and Occam's Razor says we should start looking with more parsimonious explanations.quote:
quote:
Being able to explain a small subset of all observations doesn't mean that you can explain all observations.
Oh the irony of that comment considering the fact that you are taking for granted that a couple of experiements from a close distance, in a few isolated conditions can explain *all* observations related to "scattering", even now that we know they have mass.
You haven't explained why neutrinos having mass should make a difference. How many weak nuclear force interactions are mediated by particle mass? Which weak interactions must be viewed differently if one assumes massive neutrinos versus zero-mass neutrinos?quote:
quote:
Except it's based upon the premise that all that needs to be explained is the "missing" neutrinos, when that isn't the case.
Sure it is Dave. Particle physics has nothing to do with solar models or concern about various solar models.
Right, you're just going to ignore all those atmospheric, reactor and accelerator studies with the correct total neutrino flux.quote:
I'm expecting you to note that *wavelengths* have a direct relationship to absortion and scattering.
How have neutrino wavelengths changed with the addition of mass?quote:
Photons are also immune to gravitional forces.
Utterly and completely false, since otherwise Einstein would have been proven wrong in 1919.quote:
Now that you know they have mass, why wouldn't you *insist* we test these scattering predictions through many types of materials over many distances to see it these ideas have merit?
As I've already explained: there have been no changes to our understanding of the weak nuclear force due to neutrinos having mass. If the forces which can affect neutrinos are no different with or without neutrino mass, then why should one expect to seeing "scattering" different from what's already been measured?quote:
Why wouldn't you *insist* that we calibrate our instruments in this way before leaping to any conclusions about what to expect, expecially now that we know they have *mass*.
You can scream about mass all day long, Michael, but until you can show how it will make a difference for neutrino scattering, you're wasting your breath.quote:
http://physicsweb.org/articles/world/13/12/10
Doesn't mention neutrinos at all.quote:
quote:
Again: what do photons have to do with neutrinos? Nothing.
That is a pure statement of faith on your part.
No, it's based on my understanding of quantum physics and the standard particle model.quote:
They are particle waves, just like photons Dave.
Photons are not "particle waves," Michael, and neutrinos aren't, either. Why not at least try to use the same terminology as everyone else?quote:
The difference is that neutrinos have mass and a different spin.
You forgot that photons are carrier particles (they create a field), neutrinos are not.quote:
The are not necessarily identical in behavior, but the scattering/absorption based on wavelength observations we make with photons *may* apply to neutrinos as well. We have no way to know unless we try.
Well, we don't know if neutron bombs cure the common cold, either. Should we try that? Of course not. We've got no reason to suspect that neutron bombs will cure the common cold, just like we've got no reason to think that massive neutrinos will be "scattered" any more (or less) than when we thought they had zero mass.quote:
If you never test it, you'll never have any proof that oscillation is occuring...
Nobody can prove anything to you that you're dead-set against believing, Michael. Given how you deny the evidence, and twist the evidence you don't deny, it is crystal clear that you're dogmatism prevents you from even trying to listen to what I've been saying.quote:
...and no evidence to demonstrate that oscillation is a "better" explanation than scattering.
Oscillation explains both the "missing" neutrinos and the neutrinos of other flavors "found." Scattering cannot do this. Therefore, oscil |
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Dave W.
Info Junkie
USA
26022 Posts |
Posted - 07/26/2006 : 20:18:52 [Permalink]
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quote:
Originally posted by Michael Mozina
quote:
Originally posted by Dave W.
Astronomers use blackbody calculations to estimate the temperature of distant stars, not to explain it.
But it's not even a valid "estimate" in a mass separated model where the neon layer would release most of the visible light, and every layer would "skew" the coloring scheme.
So what? You made a blatantly false statement about astronomers using blackbody principles to "explain" the temperature of distant stars, and I corrected you on it. Rather than admit that you were wrong, you're now trying to change the subject over to whether or not the estimation technique is "valid." Keep dodging and weaving, Michael, it's good exercise even if it's a transparently poor rhetorical strategy.
quote:
quote:
And I've still yet to see anyone "explain" a sunspot with blackbody principles.
http://cse.ssl.berkeley.edu/bmendez/ay10/2000/notes/dis2.html
Sure Dave.
Hey, now I've seen someone using blackbody principles to explain sunspots. What was the point again? |
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Dave W.
Info Junkie
USA
26022 Posts |
Posted - 07/26/2006 : 20:32:59 [Permalink]
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quote:
Originally posted by Michael Mozina
One really interesting thing I saw on that last link is a graph that shows the relationship between the surface temperature of the earth's oceans and the number of sunspots.
http://cse.ssl.berkeley.edu/bmendez/ay10/2000/notes/1127.jpg
You'll notice that the temperature of the oceans rise as the number of sunspots increase, which would seem to stand in opposition of your belief that sunspots represent areas that are "cooler" on average than the surrounding materials of the photosphere. Some areas of the sunspote may be cooler, but areas of the sunspot may be much hotter as well.
Actually, if you look at the whole Sun, you'll see that areas outside sunspots get hotter when there are more sunspots. The total energy output of the Sun rises with sunspot activity.
Next post:quote:
You'll note in my model that arcs on the surface cause superheated silicon plasma to rise up from below. That column of superheated plasma "punches through" the relatively cool photosphere.
Yeah. Why is superheated silicon darker than hot neon again? Why is it that the neon stays in a nice "layer" even with the violent eight-minute "roiling" of 1,000-km-wide features?quote:
While there may be cool areas seen in sunspot activity, these sunspots are directly related to the increased heat from the arcs below, which is 100% consistent with rising sea temperatures with increased sunspot activity.
Sure it is, as are an infinite number of other explanations. Why is it that we don't see arcs shooting up directly out of sunspots, but have to look in the corona for them (or, as you assert - er, in your opinion - under the sunspots to see them)?quote:
Now tell me again how you know that the *average* temperature within a sunspot is *on average* 2000 degrees cooler than the rest of the photosphere...
Because it makes sense considering the massive magnetic fields in the region.quote:
...and why don't these related to *cooling* trends then here on earth?
Because only someone staring myopically at sunspots would miss the fact that the areas outside of sunspots get hotter? |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 07/27/2006 : 06:31:23 [Permalink]
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I've got a question about your model of the sun Michael.
You have stated many times that the sun is mass seperated down to the isotope. Does that mean that the layers would be as shown below:
Carbon-9
Nitrogen-10
Carbon-10
nitrogen-11
carbon-11
oxygen-12
nitrogen-12
carbon-12
Oxygen-13
nitrogen-13
and so on?
One other question:
When you talk about the Iron surface of the sun, if the sun is mass separated does that mean that the surface should be Mn-Fe-Co with maybe some Cr and Ni?
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If I knew then what I know now then I would know more now than I know. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 07/27/2006 : 08:46:50 [Permalink]
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quote:
Originally posted by furshur
I've got a question about your model of the sun Michael.
You have stated many times that the sun is mass seperated down to the isotope. Does that mean that the layers would be as shown below:
and so on?
No, it means that the layers will separate by tbe element, calcium, silicon, neon, helium, and hydrogen. The flow of electricity will tend to arrange each of these layers by the isotope.
quote:
One other question:
When you talk about the Iron surface of the sun, if the sun is mass separated does that mean that the surface should be Mn-Fe-Co with maybe some Cr and Ni?
When we reach the solid surface, there is mass separation taking place in the solids. It operates much like the crust of the earth, only it's got a higher iron content. All the elements can be contained in the solid crust, but the plasma layers in the solar atmosphere are arranged by weight, and electrically arranged by the isotope within each layer. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 07/27/2006 : 09:01:16 [Permalink]
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quote:
Originally posted by Dave W.
So what?
So what? How can you even ask that question?
So what if it's totally wrong to be trying to apply this method to distant stars to attempt to "estimate" the temperature Dave? What if the color of the star has very little if anything to do with the temperature, but more to do with the thicknesses and layouts of the various elemental layers?
quote:
You made a blatantly false statement about astronomers using blackbody principles to "explain" the temperature of distant stars, and I corrected you on it. Rather than admit that you were wrong, you're now trying to change the subject over to whether or not the estimation technique is "valid." Keep dodging and weaving, Michael, it's good exercise even if it's a transparently poor rhetorical strategy.
This is a blatently false statement on your part Dave. The issue here is whether or not the *method* is even valid for *estimating* or "explaining" (God do you like to nitpick over words) the temperature of a distant star. If the sun's layers are mass separated by the element, then this "method" may not work at all to "estimate", or "explain" the surface temperature of a distant star! There's no dodge here Dave. It's simply a question of whether such a *method* can even be reasonably applied to a distant star. You seem to assume it can be used to "estimate" the surface temperatures of distant stars. The bulk of the white light from our own star however comes mainly from the neon photosphere. The "color" is dictated by the elements present, and their relative thicknesses, not necessarily anything directly related to it's temperature.
quote:
Hey, now I've seen someone using blackbody principles to explain sunspots. What was the point again?
The point is that you are being intentionally evasive and you're dragging your feet, and you are making this process a *lot* more difficult than it needs to be. Blackbody principles are used a *lot* in astronomy Dave. Denying this fact is not helping to move this conversation along. It's certainly not adding to your credibility in this conversation. |
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Dave W.
Info Junkie
USA
26022 Posts |
Posted - 07/27/2006 : 09:37:11 [Permalink]
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quote:
Originally posted by Michael Mozina
This is a blatently false statement on your part Dave. The issue here is whether or not the *method* is even valid for *estimating* or "explaining" (God do you like to nitpick over words) the temperature of a distant star.
No, Michael, the main issue is that your own words continue to impeach your alleged ability to have a scientific discussion. When you make blatantly false statements like "Photons are also immune to gravitional forces," or "Astronomers use blackbody principles to explain everything from the temperature of distance suns, to sunspots," you're simply telling everyone that you don't know what you're talking about. Science is a realm of precise language, Michael, and we cannot talk science if you continue to use sloppy language while attempting to make a point.quote:
If the sun's layers are mass separated by the element, then this "method" may not work at all to "estimate", or "explain" the surface temperature of a distant star!
And if the Sun's photosphere were made of fondue, we'd also be hard-pressed to "explain" the temperature. So what? Show that the blackbody estimates must be completely wrong, and scientists will stop using them. You won't be able to show that based upon your "thin layer" argument, since that's just an argument from incredulity which is demolished by your own use of Thompson scattering.quote:
There's no dodge here Dave. It's simply a question of whether such a *method* can even be reasonably applied to a distant star. You seem to assume it can be used to "estimate" the surface temperatures of distant stars.
I don't assume any such thing. The idea that star temperatures can be estimated from their spectrum is based upon many decades of experimental results.quote:
The bulk of the white light from our own star however comes mainly from the neon photosphere. The "color" is dictated by the elements present, and their relative thicknesses, not necessarily anything directly related to it's temperature.
Where are your experimental results showing this to be true, Michael? Oh, that's right! I'm not supposed to assume that you're stating is a "fact," here. I apologize profusely for asking you to provide evidence to support your opinion.quote:
The point is that you are being intentionally evasive and you're dragging your feet, and you are making this process a *lot* more difficult than it needs to be. Blackbody principles are used a *lot* in astronomy Dave. Denying this fact is not helping to move this conversation along. It's certainly not adding to your credibility in this conversation.
There you go again, Michael, putting words in my mouth. I never once said that blackbody principles weren't used in astronomy. What I have been arguing against is your assertion that blackbody principles are some sort of "gospel" upon which the standard solar model rests. You've evasively moved the goalposts several times on that subject already, not wanting to admit that you don't know enough about any standard solar model to be able to support your assertion, and so instead you switch to talk about "distant stars" and "sunspots," and now finally to the general claim that astronomers "use" blackbody principles.
If you'd said that in the first place, I would have agreed with you, Michael. But you didn't say that, you said something entriely different, and now you're backpedalling so fast it's a wonder your tires don't catch fire. |
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Evidently, I rock!
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 07/27/2006 : 10:46:36 [Permalink]
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quote:
Originally posted by Dave W.
No, Michael, the main issue is that your own words continue to impeach your alleged ability to have a scientific discussion. When you make blatantly false statements like "Photons are also immune to gravitional forces,"
Ok, that statement was in fact "sloppy" on my part.
quote:
or "Astronomers use blackbody principles to explain everything from the temperature of distance suns, to sunspots," you're simply telling everyone that you don't know what you're talking about.
Boloney Dave. Evidently you've never seen anyone explain sunspots uning blackbody principles, but you yourself said temperature was the issue too! Talk about not having credibility Dave.
quote:
Science is a realm of precise language, Michael, and we cannot talk science if you continue to use sloppy language while attempting to make a point.
Ya, you'll focus on the trivia, and ignore the main issues like those *brightly lit coronal loops*. Nevermind the fact you can't explain Lockheeds math, you're sure the loops aren't a greater temperature than the dark areas of the corona. Go figure.
quote:
And if the Sun's photosphere were made of fondue, we'd also be hard-pressed to "explain" the temperature. So what?
This is why I'm about to bail out of this conversation. Instead of dealing with real issues, you're being flippent, difficult, and just plain evasive on all the core issue. How do you expect to make any progress if you refuse to put up math to support Lockheed's interpretation, and you refuse to consider the math, images and logic I've presented related to the temperature of the coronal loops?
quote:
Show that the blackbody estimates must be completely wrong, and scientists will stop using them.
They don't work on thin plasma Dave! Less than aerogel thin plasma is not going to act like perfect absorber or a perfect emitter. It's not even certain what that plasma is made of! To then claim it's "Ok", to suggest it's "opaque" enough to be considered a "blackbody" is simply absurd.
quote:
You won't be able to show that based upon your "thin layer" argument, since that's just an argument from incredulity which is demolished by your own use of Thompson scattering.
It is not Dave. You've not "demolished" anything. You're just winging it here when you claim it's "opaque". You have no idea if it's "opaque". You don't even know for sure what it's even made of, let alone know that it's "opaque". You certainly have no proof that this material acts like a "blackbody".
quote:
I don't assume any such thing. The idea that star temperatures can be estimated from their spectrum is based upon many decades of experimental results.
Experimental results? Like what? We've never been to another star to know if any of these concepts work anywhere else, so how could you have "experimental results" to support you yet?
quote:
Where are your experimental results showing this to be true, Michael? Oh, that's right! I'm not supposed to assume that you're stating is a "fact," here.
Where is yours Dave? Oh that's right, I'm not suppose do assume anything you're state is a "fact" either. Care to demonstrate your point using alternative methods for measuring the temperatures of stars?
quote:
I apologize profusely for asking you to provide evidence to support your opinion.
Likewise Dave, I'm asking you for the same thing.
quote:
There you go again, Michael, putting words in my mouth. I never once said that blackbody principles weren't used in astronomy. What I have been arguing against is your assertion that blackbody principles are some sort of "gospel" upon which the standard solar model rests.
It is treated as applicable to solar physics Dave, and it's used to explain everything from the temperature of a remote star, to the cause of a sunspot on our own star. No one can demonstrate that sunspots have anything whatsoever to do with blackbody principles. This is a "myth", and gas model theory is propogating myths related to how a sunspots form in the photosphere.
quote:
You've evasively moved the goalposts several times on that subject already, not wanting to admit that you don't know enough about any standard solar model to be able to support your assertion, and so instead you switch to talk about "distant stars" and "sunspots," and now finally to the general claim that astronomers "use" blackbody principles.
Oh boloney! You are simply nitpicking on trivial wording, ignoring the importance of what I'm talking about, and getting hot and bothered over nothing.
quote:
If you'd said that in the first place, I would have agreed with you, Michael. But you didn't say that, you said something entriely different, and now you're backpedalling so fast it's a wonder your tires don't catch fire.
What I said was true Dave. Blackbody concepts are used in astronomy to explain (estimate) everyting from the temp of a star to the cause of a sunspot. It's part of the "dogma" of gas model theory. I even showed you links to how that dogma is used in the classroom. Since you seem to have a problem with the term "gospel", maybe we'll just call it a bunch of false "dogma". Does that make you happier? |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 07/27/2006 : 11:29:26 [Permalink]
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quote:
No, it means that the layers will separate by tbe element, calcium, silicon, neon, helium, and hydrogen. The flow of electricity will tend to arrange each of these layers by the isotope.
I don't get what you mean. How will electicity arrange the isotopes? For instance Carbon 11, 12, 13 and 14; they all have the same charge.
quote:
When we reach the solid surface, there is mass separation taking place in the solids. It operates much like the crust of the earth, only it's got a higher iron content.
Are you saying that the surface of the sun is a solid that is layered with Cr, Mn, Fe, Co, and Ni? Are you saying also that the crust of the earth is mass separated?
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If I knew then what I know now then I would know more now than I know. |
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Dave W.
Info Junkie
USA
26022 Posts |
Posted - 07/27/2006 : 12:18:05 [Permalink]
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quote:
Originally posted by Michael Mozina
When you get around to explaining how much pressure sits on the surface of the photosphere, then maybe we'll continue that conversation.
From this data set, for the r/R values over one, I used an online function fitter to find that the densities of the solar "atmosphere" from the photosphere up to above 500 km above it approximate the function d = 25,677.88/r2 when r is measured in kilometers from the center of the Sun, and d is measured in grams per cubic centimeter.
From Wikipedia, I found that the density of the interplanetary medium near Earth is approximately five particles per cubic centimeter, and that the density of the interplanetary medium drops off with the square of the distance from the Sun. Using those figures - and assuming all five particles are neutrons, to be as heavy as possible - I was able to create several data points to feed into the function fitter, and it came up with d = 0.0000001874/r2 (with d and r measuring the same things as previously). These two functions are off by twelve orders of magnitude for all values of r, and so there is no obvious "break" between them (where the solar atmosphere suddenly gives way to the interplanetary medium).
So, for the purposes of calculating the pressure on the photosphere, I will use the more-massive function (by a factor of 1,000,000,000,000), the one generated by the standard solar model's 500 km of data.
Now, Newton showed that F = Gm1m2/r2. m1 will be the mass of the Sun, 1.9891×1030kg. m2 will be the mass of a cubic meter of "atmosphere" at distance r from the Sun's center. Since we know the mass of a single cubic centimeter in grams, we need only multiply the prior equation by 1,000,000 (to get from cm3 to m3) and then divide by 1,000 (to go from grams to kilograms), so m2 = 25,677,884.69/r2. Oh, wait... r has to be in meters for Newton's formula to pump out Newtons, so m2 = 25,677,884,687,978.0/r2.
So, plug those three values in, and we find the gravitational force on any particular cubic meter of solar "atmosphere" to be F = 3.409×1033/r4, and when r is in meters from the center of the Sun, F will be in Newtons. (r, of course, must be at least 696,000,000.0 for this function to have meaning.)
Now here is where my calculus gets rusty, and I'll have to do things "by hand" that should otherwise just be "find the area under the curve" easy. But either way, we envision a square-meter column of atmosphere rising from the surface of the photosphere. The lowest meter of it is putting 3.409×1033/696,000,000.04 Newtons of pressure on the photosphere, or 0.01452747186 N. The next cubic meter on top of that adds another 0.01452747177 N. Keep going, and sum up however many cubic meters of atmosphere we want.
Why not out to one Astronomical Unit? Sure the last cubic meter only adds another 0.00000000000681 N, but every little bit counts, right?
Since the result actually comes out in Newtons per square meter, the total comes to 3,370,380.0 Pascals of pressure. This is clearly far too high (just like I said above), since the standard model only predicts a pressure of 7,742.3 Pascals just below the surface of the photosphere. However, even if I only sum out to a billion meters, the pressure (using the fitted density) comes out to over two million Pascals. Obviously, my equation for density is massively incorrect.
Interestingly enough, Wiki says that the corona is less dense than the photosphere by 12 orders of magnitude. Where have I heard that before? So if we use the solar-model fitted data up to 10,000 km from the photosphere (706,000,000.0 m), and then the interplanetary-medium fitted equation after that, what do we get for total pressure? We still calculate 141,255 Pa, too high by a factor of 18. There's gotta be a steeper drop in density across the chromosphere than what I've got, but I don't have enough detail to properly model the Sun across those layers.
So, Michael, what does your solar model say the pressure on the photosphere will be? Oh, that's right: it doesn't say anything at all on that issue because the mass of the Sun is an unknown value. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 07/27/2006 : 12:27:02 [Permalink]
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quote:
Originally posted by furshur
I don't get what you mean. How will electicity arrange the isotopes? For instance Carbon 11, 12, 13 and 14; they all have the same charge.
You are right, that was very poorly worded. Gravity and the magnetic fields created by the flow of electrity are the forces that separate the heavier isotopes. These are the same forces that separate the plasmas themselves into well defined layers.
quote:
Are you saying that the surface of the sun is a solid that is layered with Cr, Mn, Fe, Co, and Ni? Are you saying also that the crust of the earth is mass separated?
I inadvertantly left out the word "no" in my sentence.
It should have read: When we reach the solid surface, there is *no* mass separation taking place in the solids. It operates much like the crust of the earth, only it's got a higher iron content.
Geez, between those two comments, it's no wonder that you were confused by my statements. *That* was truly a sloppy response on my part. I'm sorry about that. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 07/27/2006 : 12:31:20 [Permalink]
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Michael, since we know the distance from the sun and we know the orbital period of the earth why not use the following formula to find the mass of the sun? It has nothing to do with any model of the sun.
M = 4*pi(squared)*r(cubed)/G*T
M = mass of the sun
r = distance from the sun
G = gravitation constant 6.67 * 10^-11 N*m^2/kg^2
T = Period (seconds)
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If I knew then what I know now then I would know more now than I know. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 07/27/2006 : 13:20:49 [Permalink]
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quote:
Originally posted by Dave W.
Great! Then oscillation is the more parsimonious explanation of all observed phenomena, and Occam's Razor says we should start looking with more parsimonious explanations.
That *might* be true *if* you already knew that gas model theory was a "law" Dave. Otherwise you're tyring to tie two "theories" together that are not necessarily even related in any way.
Occum's razor arguments also would suggest we should be ruling out the more mundain options like scattering before we start creating frankenparticles that violate lepton conservation principles. This is a particle physics question and the answer can be demonstrated *entirely* through particle physics *without* regard to solar models. Since we do not know if the current solar model is accurate or inaccurate, there is nothing gained by trying to tie the two together, and there is the potential of being mislead by that "coincidence" or to "write off" the solar theory because it didn't predict the right type of neutrinos. We don't know if these are related issues, and since it's a particle phyiscs question that can be entirely answered by particle physics, there is no need to worry about it.
If neutrinos can oscillate, the tests should demonstrate that fact without making a bunch of *unnecessary assumptions* about the validity of gas model solar theory. Occum's razor arguements also require that we keep things as simple as possible. There is no need to drag solar theory into this issue in any way. All of these ideas should be able to be demonstated though particle physics and the belong to the realm of particle phyiscs. Solar models are irrelevant.
quote:
You haven't explained why neutrinos having mass should make a difference.
The fact they have mass suggests they they can interact with Higgs Bosons inside the atom. A proximity to such a particle may have significant consequences for a particle with an unknown mass. We won't know until we check it out using *direct observation* to answer that question.
quote:
How many weak nuclear force interactions are mediated by particle mass?
That isn't the issue to begin with!
http://en.wikipedia.org/wiki/Gravitation
As it relates to scattering, this is the issue we need to be concerned about Dave, expecially now that we know they have mass.
quote:
Right, you're just going to ignore all those atmospheric, reactor and accelerator studies with the correct total neutrino flux.
Absolutely not Dave. Those studies showed a high level of scattering on the far side on the earth. I think that's important data just like you.
quote:
How have neutrino wavelengths changed with the addition of mass?
I have no idea Dave. I don't know how much mass each neutrino contains. Is there one amount of mass and three wavelenghts, or three different amounts of mass and three unique wavelengths? I really don't know yet.
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Utterly and completely false, since otherwise Einstein would have been proven wrong in 1919.
Ya, Ok, I'll give you that one.
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As I've already explained: there have been no changes to our understanding of the weak nuclear force due to neutrinos having mass.
And as I've already explained, the issue here relates to gravity, gravitional force, gravitons and the differences between GR and QM.
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If the forces which can affect neutrinos are no different with or without neutrino mass, then why should one expect to seeing "scattering" different from what's already been measured?
Because the neutrino has mass, the "forces" that can effect neutrinos are no longer limited to *only* the weak nuclear force. You keep overlooking that Dave. Higgs Bosons, or whatever gives rise to "mass" are going to have to be factored into these discussions and considered in these calculations.
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You can scream about mass all day long, Michael, but until you can show how it will make a difference for neutrino scattering, you're wasting your breath.
In QM theory, we're talking about the possible exchange of gravitons.
http://en.wikipedia.org/wiki/Graviton
What other explanation do you need as a reason that we need to "calibrate" the equipment for these possibilities that are directly dependent on whether or not the particles have mass?
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No, it's based on my understanding of quantum physics and the standard particle model.
QM suggests that gravitons will pass between particles with mass. Are you going to embrace *that* aspect of QM anytime soon?
I've got a call to take. I'll look at the rest of this post later to see if there is anything new that needs to be addressed.
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| Edited by - Michael Mozina on 07/27/2006 14:35:29 |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 07/27/2006 : 14:48:08 [Permalink]
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It's really funny how none of our measurements changed when someone said "neutrinos have mass!"
Why would that be "funny"? Who would expect the actual *measurements* to change only by changing our theories? What's "funny" is that the scattering *predictions* never changed as a result of us changing one of the key aspects of our *theory*. That's the part that is "funny" here. No one expects the observations to change. We observe there is a *lot* of scattering over the course of an entire planet. These "observations" have never changed, but our expectations about neutrino scattering never changed based on our discovery of mass, and that is precisely why the estimates don't jive with direct *observation". It's just that simple.
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That would have been a real treat, actually, and would have caused every particle theory to have been rechecked. But no, neutrinos kept on behaving as they'd always behaved, except we knew about some new things to look for, and when we looked for them, we found them.
And now I've shown you eactly where to look to explain that observation of excess scattering that was never factored into early estimates of neutrino scattering interactions. These predictions were based on a theory of a *massless* particle. There's no mystery here why "predictions" never jived with real life observation. Early predictions were based on belief that these are massless particles. We now know that original premise was false and they actually do have mass. There was a big change made to the core theory and absolutely *no* change was made to the scattering prediction. Therefore the "prediction" doesn't match "observation". What a big surprise. |
| Edited by - Michael Mozina on 07/27/2006 15:21:39 |
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