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Michael Mozina
SFN Regular
1647 Posts |
 Posted - 03/30/2006 : 11:36:54 [Permalink]
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quote:
Originally posted by JohnOAS
quote:
Originally posted by Michael Mozina
http://www.nasa.gov/mpg/124357main_flare_320.mpg
I really liked this movie from the press release. They even show the coronal loop coming up from underneath. How then can the base of these arcs orininate in the corona?
It's a nice animation. I did a quick search and couldn't find the page describing the movie. Without knowing what the different "layers" and features are meant to represent, the intention of the animation and the known limitations, it's little more than a distraction, however pretty.
If you could post the accompanying description (press release?) or, better yet, the actual NASA page the link came from, it might be possible for others to comment in context.
http://www.nasa.gov/vision/universe/solarsystem/clear_weather_feature.html |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/30/2006 : 12:24:05 [Permalink]
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quote:
Originally posted by Michael Mozina
quote:
Originally posted by Dave W.
You're obviously not reading my posts any more, because I asked you pointed questions about the density issue after you brought it up. When are you going to respond to them?
I am reading your posts Dave. I'm also carefully picking and choosing which issues I want to resolve and focus on rather than going through blow by blows with you. I'll be happy to respond to your questions but you need to start responding to a few key issues I've asked you about.
You claimed that I ignored your question, which is simply not true.
I certainly didn't ignore it. Your "pick your poison" options demonstrate very well that you've got some strange ideas about density. I can't answer your question until you tell me why you think a "higher density" would be a "problem," since it is no "problem" for the Earth's "mass separated" atmosphere to have lots of density fluctuations (as I use the term "density").quote:
Specifically the issue I'm interested in at that moment are the temperature/density of the coronal loops vs. the corona itself. That is the issue I keep returning to because it's critical that we resolve this before we try to move forward. This is no "minor" issue, it's a "major" issue.
Fine, then why don't you offer a temperature for the corona, and another one for the loops? Why don't you offer some density measurements for each? And then you can explain how they got that way, using standard laws of physics that apply everywhere. That would be "better" than Lockheed can do, which has been your apparent standard for acceptable criticism.quote:
The other option he mentioned was *density*, which seems to be the one you've latched onto as an "out". This however presents a significant problem for your explanation since you claimed that all this takes place in the corona, and you've provided nothing in the way of explaining this density difference.
If plasmas are "attracted to" an electrical current, won't they be "attracted to" magnetic fields, also? Actually, that's what Kosovichev was talking about when he was discussing magnetic cooling down around sunspots, too.quote:
Now of course the author doesn't mention the fact that it could be more dense *and* a higher temperature, but based on the fact he put the higher temps inside the loop, that seems to be what he's suggesting is happening, even if he's trying to give the other idea some credence.
You're confused about what he was saying. He wasn't ever trying to say that the brightest areas inside a loop (however poorly you've defined them) are also the hottest, he in fact was saying that the dimmer areas inside the loop are the hottest, because the brighter areas inside the loop simply represent a higher density. Bright areas and hot areas are both inside the loops, Michael, they're just not in the same place inside the loops. I asked you before how wide you think the average "arc" is, and you've yet to answer.quote:
The problem of course with the whole bright=dense material claim is that not a single meantion was made of how they "tuned" this technique...
You haven't described how you "tuned" your method, either.quote:
...or even how they tested to see how each wavelength was affected by the atmosphere in terms of absortion/reflection, ect.
You've never offered any such things, yourself.quote:
He also points out that while the method is used, it isn't really agreed by everyone that it even works, and he himself hedged his bets which suggest to me that even he isn't sold on the concept.
Right, it's a very complex science, and you're trying to make it even more oversimplified than you claim Lockheed's method is by saying "brighter equals hotter."quote:
Either way however, you need to commit here in some way so we can move forward. Is the material more dense, hotter, or both? Why do you choose the option(s) you choose?
Which material would you like me to examine? Point to a pixel or two.quote:
quote:
When are you going to respond to my point about the UofM paper, which doesn't even contain the words "current" or "flow" (or other terms related to current flow) at all?
I already did. They use the terms electron density and stuff like that. They are directly suggesting that these emissions are driven by electron excitation and heat.
None of which suggests any net current, and thus isn't evidence for an "electrical discharge."quote:
It suggest *both* aspects are involved. When are you going to answer *my* question about what you think drives electron excitation if *not* the flow of electricity?
I already did - electrons can be driven to extreme velocities by magnetic fields. Since such motion isn't uniform, there will be no net current even with lots of electrons moving very fast.quote:
Since the resolution on that composite image is so limited, and since the brightness on each image is minimal, its likely that everything we see in that image is probably associated with the arc itself rather than any sort of reflection or absortion/emission process.
You've provided no method whereby we can differentiate between them, still. And I see at least four different "footprints" in that image, suggesting at least three different "arcs," |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/30/2006 : 12:28:03 [Permalink]
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quote:
Originally posted by JohnOAS
Perhaps not, but it does illustrate quite clearly that you "see" raw images and calculated images as showing the same types of features and properties. Yet you still claim to be the only person interpreting these images "in detail", regardless of the fact that numerous people have pointed out to you, scientifically, why what you see isn't what you assert. But, I forgot, Michael Mozina doesn't make mistakes in regard to image interpretation.
Huh? That's really unfair of you John. I never claimed I never make mistakes. We *all* make mistakes, including the folks at Lockheed and NASA. No human being is immune. What kind of irks me is that I *immediately* came clean and yet you still condemn me for stepping up to the plate and admitting my mistake? Come on John, that is simply not fair. As I meantioned however, my basic statement was true. We can see these "structure" in raw images. I've shown you such images and you can find such images on my website.
quote:
No one has to use any particular model to point out flaws in your arguments. No matter how many times people say it, you keep assuming that everything has to prove the gas model correct, otherwise, by default, your model is correct.
No, actually that is the logic you and Dave are using. You seem to be of the opinion that I am personally responsible for "proving" every aspect of every question put to me about a Birkeland model or gas model theory is correct. That's your game not mine. I'm suggesting *both* (all) models need to be scrutinized and compared, and no model is right by default.
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My arguments regarding the imaging, and your incorrect interpretations have nothing to do with the gas model. Nothing.
What "incorrect" interpretation do you refer to that I haven't already copped to?
quote:
No one can give an explanation which will satisfy you unless they first agree that what you see in the images represent an actual picture (in the conventional, photographic sense) of solid surfaces and other features.
No, that's not what I'm suggesting. What I'm suggesting is that if you believe that these images can be explained a "better" way, they you are obligated to put such an explantion on the table. It has to be attentive to all the minute details we see in the multimillion dollar images.
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I never tried to explain anything using black body radiation. I simply used it as part of an example illustrating exactly why brighter is not necessarily hotter.
You have not demonstrated your case, and neither has Dave. Brighter is this case *is* hotter. This composite image demonstrates this:
In this case, brighter certainly is hotter. Trace has a temp range of 160K up to 20MK, whereas Yohkoh's temp range is between 2 and 20 million Kelvin. Anything that is between 160K and 20Million Kelvin is glowing in this image. Any area that is dark in this image are simply not that hot.
quote:
Wrong. Wrong again.
Unless you take wavelength into account (and all the associated wavelength dependant properties of the observation system), this is meaningless. Brighter where Michael?
It's brighter *in the arc* John. That Trace/Yohkoh overlay is taking a LOT of wavelengths into account, and specifically the wavelengths that are associated with high temperature plasma. Neither one of these spacecraft sees any of this magic invisible heat Lockheed claims is there. Why not? Where is there any observations evidence to support Lockheeds position?
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The physics I'll grant you, on "simple", I beg to differ.
It is simple in this case IMO. It's Lockheed that is trying to make it complicated. It's not complicated, it's an electrical discharge.
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Logic and common sense can get you into a lot of trouble if you don't understand the context completely.
Yes, I agree. Evidently Lockheed thinks there technique is "logical", but they haven't a clue about the absorbtion rates so they got erroneou results. What they didn't show was a lot of common sense, i.e. going back to the original images and noticing the bright areas of the 195A image. If they had any doubts, that overlay image should have been used to determine that there is no ivisible heat in the corona.
quote:
Trying to apply logic and common sense to situations which are far from "common" to our everyday experience is not even sensible. I'll take the scientific method over common sense every day, and twice on Tuesdays.
Not in this case. In this case we're looking at an electrical discharge. It's hot. It's very hot. It's much hotter than the photosphere and chromosphere. We're talking hot hot hot. You wouldn't take two images of a lightening bolt, not know anything about the absortion rates of the wavelengths and then try to do a math manipulation that suggests the air around the lightning bolt was hotter than the plasma inside the arc. That is however what Lockheed it doing with this technique.
quote:
No, but unless I understood everything that goes on during this brief, electrostatic discharge, I also wouldn't claim the converse.
And if I had not cut my teeth on Yohkoh solar images, I wouldn't be claiming the converse either. Since however I can see with my own eyes from Rhessi, Geos and especially my favorite spacecraft (Yohkoh) that there is no missing heat, I have no trouble saying loud and clear that Lockheed blew it. They don't have the first clue about heat signatures in the corona. From the article that Dave posted, even they aren't sure the technique works properly and the author himself put the high temperature plasma in the loop as I do. About all I can say at this point is that there is no observational support for their position and without knowing the absortion r |
| Edited by - Michael Mozina on 03/30/2006 12:32:14 |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/30/2006 : 12:30:30 [Permalink]
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quote:
Originally posted by Michael Mozina
This isn't a strawman Dave, this is a reality check and this is what science is all about.
Nonsense. You've claimed that if Lockheed is right, then there should be images of a bright corona with dark loops in it, which is not what they're saying (nor what I've been saying), which is why it's a strawman.quote:
You are essentially proposing some sort of magic, invisible heat...
No, that's your strawman again.quote:
Contrary to your statements you did *not* "disprove" that brighter areas are hotter areas, you "aledged" this based on the statements of a guy that put the high temperature plasma *inside* the arc as I do. Hoy Vey!
You're the only person here making the claim that brighter equals hotter, so you've got the burden of proof, Michael. Nitta offered an expert opinion which doesn't agree with your assessment. That's a high bar you've got to overcome to make your case. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/30/2006 : 12:44:22 [Permalink]
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quote:
Originally posted by Michael Mozina
They use the terms electron density and stuff like that. They are directly suggesting that these emissions are driven by electron excitation and heat. It suggest *both* aspects are involved. When are you going to answer *my* question about what you think drives electron excitation if *not* the flow of electricity?
Sorry to butt in Dave, but that first line is just so funny I had to comment.
An electron absorbs a photon which results in electron exitation.
That does not appear to be due to a flow of electricity.
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/30/2006 : 12:59:13 [Permalink]
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quote:
Originally posted by Michael Mozina
Anything that is between 160K and 20Million Kelvin is glowing in this image. Any area that is dark in this image are simply not that hot.
Well, that contradicts your earlier claim that anything which is glowing is either that hot or reflecting the light or abosrbing and re-emiting those wavelengths. Make up your mind.quote:
It is simple in this case IMO. It's Lockheed that is trying to make it complicated.
This contradicts your earlier contention that Lockheed is oversimplifying the problem.quote:
The problem here has nothing to do with math or a lack of math. It's related to people and the resistence people have to new ideas.
Which is still why the actual standard solar model and what it may or may not explain about the images is irrelevant. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/30/2006 : 13:02:04 [Permalink]
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quote:
Originally posted by furshur
Sorry to butt in Dave...
Please, butt in all you like. Especially with good stuff about photon absorbtion that I neglected. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/30/2006 : 15:37:34 [Permalink]
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quote:
Originally posted by Dave W.
Nonsense. You've claimed that if Lockheed is right, then there should be images of a bright corona with dark loops in it, which is not what they're saying (nor what I've been saying), which is why it's a strawman.
I've spoken with you online for many months now Dave, and I know you to be an intelligent person and a science oriented person as well. I guess that's why I'm so frustrated with you right now.
You've told me in the past that you wanted math. I'll do the math for you on this issue Dave, and I will spell it out for you in black and white. I took the two images in question and I added the two images together to see exactly where *all* the photons were concentrated and to see where they are coming from. The math is really quite simple. A+B=C on a pixel by pixel basis.
Here's the result of the math:
By adding the original 171 and 195 images together, we see where all the light from high temperature plasma is concentrated. While the 171A image *may* peak in the 1 million K range, that is simply a minimum temperature for the loops. The 195A tends to peak in the 1.5 million range and as we can see the coronal loops is brightly lit in that image as well, meaning that the loops now have to be *at least* 1.5 million Kelvin (using Lockheed's logic). No dark areas in the 195A image can be considered "hotter" than the bright regions of the 195A image *without* further corroboration of some sort. These dark regions regions are dark in BOTH images and the third image as well. All three images A, B and C are all in agreement that the light is concentrated *in* the loop. If there is heat in the corona that exceeds the range of these two filters, it has to be in the 20Mk range+. If the plasma of the corona is not emitting photons, we cannot assume it is millions of degrees.
http://trace.lmsal.com/Science/ScientificResults/Publications/phillips_tr_resp_apj.pdf
According to Lockheed the light that we see in the 171A image are "generally" indicative of a temperature ranges of between 160,000 Kelvin up to about 1 Million , with the peak being in the 1 million Kelvin range (typically). Of course we also know that this is actually a gross oversimplification since we know that the full heat spectrum also includes several Calcium ion photons and FeXX ion photons which puts us into the 20 MK range according to all your experts.
The images tell us that the light is concentrated *inside* the coronal loops and the show us that the highest temperatures that we can see are found inside the coronal loops. We have no idea if the dark region around the loops is hot. In fact we have reason to believe it's not hot since this plasma isn't emitting much in the way of high energy photons, certainly nothing on the order of what we see in those coronal loops. The only high energy photons we see, comes from the loops themselves and the light from high temperature plasma is concentrated in the loops. While the loops may be upwards of 20Mk for all we know based on these two images, that is not true of the dark plasma around the loops. There is no observational evidence to suggest that the dark regions are anywhere near the temperature range of the coronal loops. Period.
The fact you expect me to "prove" what Lockeed is utlimately "assuming" in their method simply shows how irrational this conversation has now become. Worse yet, you expect me to believe that this heat is invisible not only to both the 171A and 195A image, but to Yohkoh's view as well. That is simply unacceptable from a scientific perspective. If you expect me to believe these dark regions are hot, you have to confirm this observationally! You can't take the method we are debating and attempt to use it to prove that it works! You need some sort of corroborating evidence that this is true. You are essentially trying to claim there is an invisible heat unicorn hidind in the bottom right corner of this image. Where's the heat Dave? Why can't we see it?
I don't see any signs of your invisible heat unicorn in lower right corner of this image Dave. I see nothing there but black. I see nothing that leads me to believe that anything that is dark in this image is anywhere near a million degrees in fact. Where is the heat and why can't we see it, and why are these atoms not emitting photons that either satellite can image?
quote:
Nitta offered an expert opinion which doesn't agree with your assessment. That's a high bar you've got to overcome to make your case.
No, actually he did *not* disagree with my assessment in the final analysis since he himself put the high temperature plasmas *inside* the loop at the end of that same article! He simply explained how the method (in question) worked, and he even pointed out that experts don't even all agree that this method works as advertized. He then gave us his "expert opinion" about where the 20Mk temps were located and placed these high temperature plasmas exactly where I did, specifically *inside* the loop! Grrrrr! |
| Edited by - Michael Mozina on 03/30/2006 16:38:20 |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/30/2006 : 16:10:56 [Permalink]
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quote:
Originally posted by Dave W.
Nonsense. You've claimed that if Lockheed is right, then there should be images of a bright corona with dark loops in it, which is not what they're saying (nor what I've been saying), which is why it's a strawman.
Where in any of the three images (four if you include the Yohkoh/Trace overlay) is there *any* observational evidence to support Lockheed's methods actually works as advertized? How will you provide *any* evidence at all to demonstrate that this method Lockheed is using is actually capable of showing heat signatures instead of just demonstrating differing photon absortion rates of two different wavelengths? |
| Edited by - Michael Mozina on 03/30/2006 16:11:18 |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/30/2006 : 16:26:01 [Permalink]
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quote:
Originally posted by Dave W.
Well, that contradicts your earlier claim that anything which is glowing is either that hot or reflecting the light or abosrbing and re-emiting those wavelengths. Make up your mind.
I have made up my mind. For purposes of this discussion, and based on the resolution and settings we're looking at in that composite image, I'd say the most if not all the light we can actually see in that particular image is directly related to an emission from an electrical arc. In most images, particularly large resolution images, we'd have to consider the other issues I mentioned. In this particular case, it really doesn't affect much since most of the light is a direct result of the light released in those arcs.
quote:
This contradicts your earlier contention that Lockheed is oversimplifying the problem.
Is this "take my comments out of context day" or what? They are oversimplying the process by not using *all* the images and adding them up. They are also overcomplicating an otherwise simple matter of addition and trying to apply a complicated set of mathematics (which may not even apply) to what should be simple matter off adding up photons. Worse yet, they oversimplified their overly complicated math presentation by not even bothering to consider the different absortion rates of the different filters. What their math *really* seems to show us is that different wavelengths are absorbed at different rates. Instead, they misrepresented the results of a overly complex math formula by oversimplifying the other aspects of this issue that must also be considered.
All they had to do to isolate the heat and light signatures was to add *all* the images together. By trying to compare ratios with complicated mathematical formulas, without regard to absortion rates of wavelengths, they turned what should have been a simple task into a complex excersize in futility that turns physics on its head and gives us invisible heat unicorns hiding in the shadows. What a bunch of boloney. This method doesn't work as advertized and even the author you cited hedged his bets and picked the loops to hold the highest temperature plasma anyway. Even he doesn't believe it works as advertized! |
| Edited by - Michael Mozina on 03/30/2006 16:39:43 |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/30/2006 : 17:14:36 [Permalink]
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quote:
Originally posted by Dave W.
I already did - electrons can be driven to extreme velocities by magnetic fields. Since such motion isn't uniform, there will be no net current even with lots of electrons moving very fast.
This comment literally blew my mind Dave. If you have electrons being driven to extreme velocities by *anything* and it flows *anywhere* including along magnetic field lines that is the very *definition* of current flow. When electrons flow and move through material that *is* current flow. Holy Cow is that a major rationalization on your part. You have fast moving electrons following magnetic field lines but you refuse to see that as current flow. Hoy!
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JohnOAS
SFN Regular
Australia
800 Posts |
Posted - 03/30/2006 : 17:34:21 [Permalink]
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quote:
Originally posted by Michael Mozina
quote:
Originally posted by Dave W.
I already did - electrons can be driven to extreme velocities by magnetic fields. Since such motion isn't uniform, there will be no net current even with lots of electrons moving very fast.
This comment literally blew my mind Dave. If you have electrons being driven to extreme velocities by *anything* and it flows *anywhere* including along magnetic field lines that is the very *definition* of current flow. When electrons flow and move through material that *is* current flow. Holy Cow is that a major rationalization on your part. You have fast moving electrons following magnetic field lines but you refuse to see that as current flow. Hoy!
I don't generally like to answer questions to other people, and I'm confident Dave will respond in kind anyway, but...
What part of "net" don't you understand?
Here, take these five twenty dollar bills. Now pass me twenty five dollar bills, really quickly. What was the net cash flow? But, but, but, giving me money is the very definition of cash flow. Talk about mind blowing. Hoy indeed. |
John's just this guy, you know. |
| Edited by - JohnOAS on 03/30/2006 17:35:14 |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/30/2006 : 17:49:46 [Permalink]
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quote:
Originally posted by JohnOAS
What part of "net" don't you understand?
Here, take these five twenty dollar bills. Now pass me twenty five dollar bills, really quickly. What was the net cash flow? But, but, but, giving me money is the very definition of cash flow. Talk about mind blowing. Hoy indeed.
But John, I could argue that from a surface perspective, that arc might transfers a "net" of "zero" electrons into space, and all electrons were conserved (for the most part), the arc simply moved the electrons from one place on the surface to another area of the sun. The electrons left the surface, and they came back to the surface, so the "net" is zero.
The movement of the electrons through the atmosphere is what's driving these brightly lit and very hot arcs. That's the magic of magnetic fields. The tend to direct and channel the flow of electrons. The flow of electrons will create it's own magnetic fields as well. It's the current *flow* that is critical, not the "net". |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/30/2006 : 21:16:04 [Permalink]
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quote:
Originally posted by Michael Mozina
I've spoken with you online for many months now Dave, and I know you to be an intelligent person and a science oriented person as well. I guess that's why I'm so frustrated with you right now.
I'll assume you're frustrated because I'm not willing to bow down to your unscientific methodologies and agree that they are meaningful.quote:
You've told me in the past that you wanted math. I'll do the math for you on this issue Dave, and I will spell it out for you in black and white. I took the two images in question and I added the two images together to see exactly where *all* the photons were concentrated and to see where they are coming from. The math is really quite simple. A+B=C on a pixel by pixel basis.
Here's the result of the math:
[image snipped]
By adding the original 171 and 195 images together, we see where all the light from high temperature plasma is concentrated. While the 171A image *may* peak in the 1 million K range, that is simply a minimum temperature for the loops. The 195A tends to peak in the 1.5 million range and as we can see the coronal loops is brightly lit in that image as well, meaning that the loops now have to be *at least* 1.5 million Kelvin (using Lockheed's logic).
No, that's clearly not Lockheed's "logic," otherwise they wouldn't be using a ratio to determine temperature. Unless you can show me where they've added images together (a meaningless exercise) and claimed that the material must be at least as hot as the highest temperature peak in a wavelength passband.quote:
No dark areas in the 195A image can be considered "hotter" than the bright regions of the 195A image *without* further corroboration of some sort.
Okay, fine. All the temperature measurements are thus invalid for the purposes of this discussion, because neither one of us has "further corroboration" of any of the temperature readings done by anyone to support our respective points.quote:
These dark regions regions are dark in BOTH images and the third image as well.
The intrinsic brightness of a pixel doesn't directly relate to the temperature of the material found there. You know this because you know that you can't possible say that a pixel with a value of 73 (out of 255) in a TRACE image is 1 Mk or 20 Mk. It's impossible to determine from a single image because such a huge range of temperatures is captured by each pixel.quote:
All three images A, B and C are all in agreement that the light is concentrated *in* the loop. If there is heat in the corona that exceeds the range of these two filters, it has to be in the 20Mk range+. If the plasma of the corona is not emitting photons, we cannot assume it is millions of degrees.
Yeah, just go ahead and ignore everything that quantum physics tells us about photon emission.quote:
http://trace.lmsal.com/Science/ScientificResults/Publications/phillips_tr_resp_apj.pdf
According to Lockheed the light that we see in the 171A image are "generally" indicative of a temperature ranges of between 160,000 Kelvin up to about 1 Million , with the peak being in the 1 million Kelvin range (typically). Of course we also know that this is actually a gross oversimplification since we know that the full heat spectrum also includes several Calcium ion photons and FeXX ion photons which puts us into the 20 MK range according to all your experts.
They're your experts, Michael. Except for Nitta, you are the one who has presented all these papers as evidence of your claims. And then, after some discussion, you claim that your own experts are wrong about this or that, but you still expect us to take the bits and pieces left over as "correct" only because those are findings which you don't dispute.quote:
The images tell us that the light is concentrated *inside* the coronal loops and the show us that the highest temperatures that we can see are found inside the coronal loops.
Just go ahead and ignore the density issue, why don't you?quote:
We have no idea if the dark region around the loops is hot.
It's not emitting zero photons, Michael.quote:
In fact we have reason to believe it's not hot since this plasma isn't emitting much in the way of high energy photons, certainly nothing on the order of what we see in those coronal loops.
Since brightness doesn't directly relate to temperature, I don't know how you can assert the truth of that statement.quote:
The only high energy photons we see, comes from the loops themselves and the light from high temperature plasma is concentrated in the loops.
The lowest energy photons we can see are also "concentrated" in the loops, and the light from the lowest-temperature plasma is also "concentrated" in the loops.quote:
While the loops may be upwards of 20Mk for all we know based on these two images, that is not true of the dark plasma around the loops.
How the hell can you say that? The 20 darkest pixel levels in the 171A image are all between 5 and 51 (out of 255). None of the pixels in that image collected zero photons. The 20 darkest pixel levels in the 195A image are all between 27 and 63 (again, out of 255). Any or all of those pixels (and they cover over 26% of both images) may be reading 20 Mk plasma, but we can't tell because brighter doesn't equal hotter (nor does dimmer equal cooler).quote:
There is no observational evidence to suggest that the dark regions are anywhere near the temperature range of the coronal loops. Period.
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/30/2006 : 21:19:28 [Permalink]
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quote:
Originally posted by Michael Mozina
Where in any of the three images (four if you include the Yohkoh/Trace overlay) is there *any* observational evidence to support Lockheed's methods actually works as advertized? How will you provide *any* evidence at all to demonstrate that this method Lockheed is using is actually capable of showing heat signatures instead of just demonstrating differing photon absortion rates of two different wavelengths?
Nice try at shifting the burden of proof, but I'm only interested in your claim that "brighter equals hotter," and showing it to be the crap that it is. You'll note that I am not claiming that Lockheed's method is valid. Neither you nor I actually know what Lockheed's method is, so neither of us can verify or disprove that it "works as advertised." |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
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