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04-10-2016 10:48
Surface Detail
★★★★☆
(1673)
Into the Night wrote:
jwoodward48 wrote:
If the argument is reduced to "no, YOU have the burden of proof", then it's pretty much over.


Incorrect. The burden of proof is on the one trying to change what already is in place, such as what Surface is attempting to do.

I'm not trying to change anything. Every physics paper and textbook in existence that refers to the subject describes the emission of radiation from gases in terms of line spectra resulting from changes in atomic or molecular energy levels. It is you and IBdaMann who claim to have invented some new theory of domains to explain this radiation. It is you two who are challenging (I use the word loosely!) established quantum theory.

Atomic hydrogen, for example, emits strongly at a wavelength of 656.281 nm. Conventional quantum theory says that this is due to electrons transitioning from the third to the second principal energy levels of hydrogen. How does your theory of domains explain this emission?
04-10-2016 13:20
IBdaMann
★★★★★
(4578)
Surface Detail wrote: No gas radiates in accordance with Planck's law.

That's a strange way of wording "I am aware of no example of any gas whose E differs from what Planck's predicts at a given temperature and wavelength in the gas' domain." Your wording appears to be a combination of poor English and inadequate math acumen (maybe you weren't paying attention in the 2nd grade when your British professors were teaching you how to redesign the space station)



This would be the fifth time I have spelled it out for you. You OBVIOUSLY will never admit when you are mistaken on matters of your WACKY religion, which explains your compulsive dishonesty.

Just one example that can be verified. That's all you need. Why haven't you supplied it already?


.


Global Warming: The preferred religion of the scientifically illiterate.

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
04-10-2016 13:42
Surface Detail
★★★★☆
(1673)
IBdaMann wrote:
Surface Detail wrote: No gas radiates in accordance with Planck's law.

That's a strange way of wording "I am aware of no example of any gas whose E differs from what Planck's predicts at a given temperature and wavelength in the gas' domain." Your wording appears to be a combination of poor English and inadequate math acumen (maybe you weren't paying attention in the 2nd grade when your British professors were teaching you how to redesign the space station)



This would be the fifth time I have spelled it out for you. You OBVIOUSLY will never admit when you are mistaken on matters of your WACKY religion, which explains your compulsive dishonesty.

Just one example that can be verified. That's all you need. Why haven't you supplied it already?


As anybody with any knowledge of physics is aware, Planck's law applies to black bodies, not gases. Why would anyone try to verify that gas behaves like a black body when it is clearly not one? A line spectrum is very obviously not a black body emission spectrum.

You're the one who is trying to rewrite the textbooks. So the onus is on you to show us one example of a gas that does emit in accordance with Planck's law. We're waiting.
04-10-2016 14:25
jwoodward48
★★★★☆
(1537)
Or how about explaining how DOMAINS can make a peak out of thin air? After all, the spectrum I posted appeared to have the same domain as the Planck spectrum, but maybe it had a different DOMAIN.
04-10-2016 14:59
jwoodward48
★★★★☆
(1537)
Even if you could get gas spectra from the Planck spectra using the MAGIK POWER O' DOMAINS, you couldn't use S-B anymore. It also wouldn't inviolate AGW.
04-10-2016 19:02
IBdaMann
★★★★★
(4578)
Surface Detail wrote: As anybody with any knowledge of physics is aware, Planck's law applies to black bodies, not gases.

As anyone with even an American education can see is that you still haven't provided so much as a single example. If Planck's flatly does not apply across the board then just grab the first verifiable example of a gas whose measured E at a given temperature and wavelength of its domain differs from what Planck's states (sixth time, not that I'm counting).

Just the first example you find. Any example will do. Just one will be fine.

[@ jwoodward48, you're free to find one as well. Don't get the impression that the example has to come from Surface Detail ]


... wait a minute, when your were in nursery school and your British professors were explaining the nuances of Chaucer's style, were you perhaps ill the day they covered what the words "example" and "instance" mean?

Am I going to quickly for you?

Surface Detail wrote:You're the one who is trying to rewrite the textbooks


Excuse me, I posted this:


Did you miss it? What exactly am I "rewriting"? Wait! Could you point me to the example you posted of a gas' radiation that verifiably violates this at a given temperature and wavelength of the domain? I must have missed it.

Surface Detail wrote: So the onus is on you to show us one example of a gas that does emit in accordance with Planck's law. We're waiting.

That was the lamest attempt to shift your burden of proof that you could have possibly made. You even used as your basis the classic "everyone knows." Even while citing the science I posted.

You are certifiably a moron. Where were you educated, if you don't mind me asking?

Too funny.


.


Global Warming: The preferred religion of the scientifically illiterate.

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
04-10-2016 19:34
IBdaMann
★★★★★
(4578)
jwoodward48 wrote: It also wouldn't inviolate AGW.

1. "inviolate" is not the correct word. I think you meant "falsify."

2. AGW is unfalsifiable. It is a religion. It cannot be shown to be false any more than Christianity can be shown to be false.


.


Global Warming: The preferred religion of the scientifically illiterate.

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
04-10-2016 19:47
jwoodward48
★★★★☆
(1537)
I meant violate.

You keep using that word. I do not think it means what you think it means.
04-10-2016 19:58
Surface Detail
★★★★☆
(1673)
IBdaMann, I don't know how to put this any more simply: Planck's law does not apply to gases. Planck derived the law specifically for black bodies, not gases. It describes the spectral density of electromagnetic radiation emitted by a black body, not a gas. Don't take my word for it; read a physics book.

I can't answer your question because it is a stupid question. The intensity of light emitted by a gas doesn't just depend on temperature and wavelength; it also depends on the pressure and density of the gas. There is no straightforward relationship between temperature and radiation intensity as a function of wavelength for a gas.

You might as well ask for examples of solids that don't obey the ideal gas law. You won't find any because the ideal gas law doesn't apply to solids, just as Planck's law doesn't apply to gases.
Edited on 04-10-2016 20:17
04-10-2016 20:50
jwoodward48
★★★★☆
(1537)
That is SO PERFECT. That is the analogy that I wanted to make.

Also, the identity of the gas would matter too, right? (Identity as in chemical identity.)

Edit: Apparently that isn't a thing. I meant... something sort of similar to element. Which chemical the gas is.
04-10-2016 20:56
IBdaMann
★★★★★
(4578)
Surface Detail wrote: IBdaMann, I don't know how to put this any more simply:

Correction, you don't know how to pout this any more simply. You whine like a fukcing baby. Go ahead and give us a big "Waaaaaaaaa!"

Surface Detail wrote: Planck's law does not apply to gases.

So we have Planck's law which you wish would just go away. Unfortunately for you and your WACKY religion it isn't going anywhere.

We also have your insistence, ... your repeated, pouting, whining insistence that gases radiate in some manner other than according to Planck's law.

When asked for just one example, you rapid-fire lame excuses, and I mean lame.

Until something changes, I think we've effectively concluded this topic. You know what you need to produce, otherwise you can't use the "Planck's law doesn't apply" angle to preach your "greenhouse effect" crap.

Surface Detail wrote:I can't answer your question because it is a stupid question.

It's Planck's Law. It's science. I know, that makes it stupid.

Surface Detail wrote: The intensity of light emitted by a gas doesn't just depend on temperature and wavelength;

...and that isn't what Planck's law says.

Surface Detail wrote: ... it also depends on the pressure and density of the gas. There is no straightforward relationship between temperature and radiation intensity as a function of wavelength for a gas.

I know I have posted Planck's for you many times but have you ever seen it? Your posts indicate that you are talking about something else.


Do they have science over there in the UK?


.


Global Warming: The preferred religion of the scientifically illiterate.

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist

Edited on 04-10-2016 21:05
04-10-2016 22:00
jwoodward48
★★★★☆
(1537)
Cargo cult scientist. There's no talking with somebody like you. "It's science! IT'S SCIENCE!" Yeah, a key part of science is that you should never be so sure about yourself that you dismiss everybody else with a "well you're just not science, I am".

You still haven't shown how MAGIK DOMAINS can create peaks out of non-peaks.
04-10-2016 22:03
Surface Detail
★★★★☆
(1673)
jwoodward48 wrote:
That is SO PERFECT. That is the analogy that I wanted to make.

Also, the identity of the gas would matter too, right? (Identity as in chemical identity.)

Edit: Apparently that isn't a thing. I meant... something sort of similar to element. Which chemical the gas is.

Yes, the line spectrum emitted by a gas depends on the composition of the gas. Each atom or ion has its own set of electronic energy levels, and each type of molecule has its own vibrational and rotational energy levels. The resulting line spectrum is unique to each type of gas; this forms the basis of spectroscopy.

We can also observe absorption spectra, where a cool gas that is illuminated from behind absorbs and re-emits radiation at its characteristic wavelengths. The classic example is the Fraunhofer lines in the solar spectrum, which originate from absorption by the (relatively!) cooler gas above the surface. These allow us to determine the chemical constituents of the sun and, indeed, led to the discovery of helium in the sun before it was discovered on Earth!
04-10-2016 22:14
Surface Detail
★★★★☆
(1673)
IBdaMann, consider this little thought experiment.

Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.
04-10-2016 22:15
jwoodward48
★★★★☆
(1537)
I'm not sure what you are trying to say, IB.

1. The atmosphere cannot affect in any way the light radiated by Earth.
2. The atmosphere cannot affect the energy radiated by Earth.

Which of these is your claim?
04-10-2016 23:39
Into the Night
★★★★★
(9164)
Surface Detail wrote:
IBdaMann, consider this little thought experiment.

Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.


It applies to gases. Making it less or more mass and noting no change in the spectra is no different than any other material; solid, liquid, or gas.

A small bit of steel will still glow red at the same temperature as a large piece of steel.


The Parrot Killer
04-10-2016 23:42
Into the Night
★★★★★
(9164)
jwoodward48 wrote:
Cargo cult scientist. There's no talking with somebody like you. "It's science! IT'S SCIENCE!" Yeah, a key part of science is that you should never be so sure about yourself that you dismiss everybody else with a "well you're just not science, I am".

You still haven't shown how MAGIK DOMAINS can create peaks out of non-peaks.


What is magik? Are you trying to use the word 'magick' or 'magic" (they are two different things)?

In any case, I suggest you look up how domains combine between two different functions,


The Parrot Killer
04-10-2016 23:44
Surface Detail
★★★★☆
(1673)
Into the Night wrote:
Surface Detail wrote:
IBdaMann, consider this little thought experiment.

Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.


It applies to gases. Making it less or more mass and noting no change in the spectra is no different than any other material; solid, liquid, or gas.

A small bit of steel will still glow red at the same temperature as a large piece of steel.

Steel, yes. Gas, no. Each square inch of the surface of the steel will glow with the same colour and brightness, no matter how big or small it is. Each square inch of the gas in the vessel with glow more or less brightly, depending on its density. Ergo, gases cannot follow Planck's law. Which we all know, apart from you and IB apparently.

Edit: The gas in the vessel will also glow with a colour that depends on its identity. Neon, for example, glows with a reddish colour, while argon gives a purplish hue.
Edited on 04-10-2016 23:48
05-10-2016 00:16
IBdaMann
★★★★★
(4578)
Surface Detail wrote:Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.

No, that's not what it tells me ... and I still see no example from you of a gas that radiates in a manner that does not adhere to Planck's.

Suppose we have a body A in space that has a fixed energy source S. Body A reaches equilibrium at temperature T and absorbs and radiates per its surface area.

Now we replace body A with smaller body B that also reaches equilibrium at temperature T.

Which one radiates with greater "intensity"? A or B?

In your "hot gas in a glass vessel" example you started with the equivalent of a bigger body A at temperature T, but then you took some gas away and effectively left yourself smaller body B at the same temperature. Of course the smaller body B has less radiance at any given wavelength.

Notice that Planck's is not necessary here. Stefan-Boltzmann answers this one.


.


Global Warming: The preferred religion of the scientifically illiterate.

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
05-10-2016 00:18
jwoodward48
★★★★☆
(1537)
Ah, so identity does mean something here! Yet another way IB and Into are wrong.
05-10-2016 00:19
IBdaMann
★★★★★
(4578)
Into the Night wrote:A small bit of steel will still glow red at the same temperature as a large piece of steel.

Nail hit squarely on the head.


.


Global Warming: The preferred religion of the scientifically illiterate.

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
05-10-2016 00:20
Into the Night
★★★★★
(9164)
Surface Detail wrote:
Into the Night wrote:
Surface Detail wrote:
IBdaMann, consider this little thought experiment.

Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.


It applies to gases. Making it less or more mass and noting no change in the spectra is no different than any other material; solid, liquid, or gas.

A small bit of steel will still glow red at the same temperature as a large piece of steel.

Steel, yes. Gas, no. Each square inch of the surface of the steel will glow with the same colour and brightness, no matter how big or small it is. Each square inch of the gas in the vessel with glow more or less brightly, depending on its density. Ergo, gases cannot follow Planck's law. Which we all know, apart from you and IB apparently.

I've never measured volume with square inches. I don't measure gases by volume either. Like cereal, gases are measured and sold by weight, not by volume (there may be some settling of the contents in shipping).

Surface Detail wrote:
Edit: The gas in the vessel will also glow with a colour that depends on its identity. Neon, for example, glows with a reddish colour, while argon gives a purplish hue.

So? How does this violate Planck's law?


The Parrot Killer
05-10-2016 00:23
Surface Detail
★★★★☆
(1673)
IBdaMann wrote:
Surface Detail wrote:Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.

No, that's not what it tells me ... and I still see no example from you of a gas that radiates in a manner that does not adhere to Planck's.

Suppose we have a body A in space that has a fixed energy source S. Body A reaches equilibrium at temperature T and absorbs and radiates per its surface area.

Now we replace body A with smaller body B that also reaches equilibrium at temperature T.

Which one radiates with greater "intensity"? A or B?

In your "hot gas in a glass vessel" example you started with the equivalent of a bigger body A at temperature T, but then you took some gas away and effectively left yourself smaller body B at the same temperature. Of course the smaller body B has less radiance at any given wavelength.

Notice that Planck's is not necessary here. Stefan-Boltzmann answers this one.

Nope. Both Planck's law and the Stefan-Boltzmann law refer to emission per unit area. The smaller body B has the same radiance as the bigger body A because each square inch of it radiates at the same rate. Each square inch of gas of the same temperature and different densities does not radiate at the same rate. Hence gases cannot obey Planck's law.
05-10-2016 00:25
Surface Detail
★★★★☆
(1673)
Into the Night wrote:
I've never measured volume with square inches. I don't measure gases by volume either. Like cereal, gases are measured and sold by weight, not by volume (there may be some settling of the contents in shipping).

05-10-2016 00:27
jwoodward48
★★★★☆
(1537)
Into the Night wrote:
What is magik? Are you trying to use the word 'magick' or 'magic" (they are two different things)?


Its an intenshonal mispeling to sho hou litl i thinc of ur claems.

In any case, I suggest you look up how domains combine between two different functions,


No clue what this means. Either you've put it weirdly, and you actually mean something, or you're just throwing words at me. Please let it be the first.


"Heads on a science
Apart" - Coldplay, The Scientist

IBdaMann wrote:
No, science doesn't insist that, ergo I don't insist that.

I am the Ninja Scientist! Beware!
05-10-2016 00:30
jwoodward48
★★★★☆
(1537)
IBdaMann wrote:
Surface Detail wrote:Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.

No, that's not what it tells me ... and I still see no example from you of a gas that radiates in a manner that does not adhere to Planck's.

Suppose we have a body A in space that has a fixed energy source S. Body A reaches equilibrium at temperature T and absorbs and radiates per its surface area.

Now we replace body A with smaller body B that also reaches equilibrium at temperature T.

Which one radiates with greater "intensity"? A or B?

In your "hot gas in a glass vessel" example you started with the equivalent of a bigger body A at temperature T, but then you took some gas away and effectively left yourself smaller body B at the same temperature. Of course the smaller body B has less radiance at any given wavelength.

Notice that Planck's is not necessary here. Stefan-Boltzmann answers this one.


.


Another thing to add to the pile of "IB and Into making themselves look stupid and contradicting themselves" quotes. You're making this too easy for me.

Of course Planck's and S-B don't predict that the smaller body will have less radiance! Tell me, where in the formula is "surface area," "volume", or "mass" included? It isn't.


"Heads on a science
Apart" - Coldplay, The Scientist

IBdaMann wrote:
No, science doesn't insist that, ergo I don't insist that.

I am the Ninja Scientist! Beware!
05-10-2016 00:33
IBdaMann
★★★★★
(4578)
jwoodward48 wrote:No clue what this means.

I appreciate the honesty.

Do you know what a domain is?


.


Global Warming: The preferred religion of the scientifically illiterate.

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
05-10-2016 00:36
jwoodward48
★★★★☆
(1537)
Yes. It is the set of all values that can be the input for a given function. The function maps from the domain to the codomain.

What is "combining domains between functions"? In all my years of college Calc, etc., I've never heard of it. Maybe I know of the same thing by another name. Care to describe it a bit?
05-10-2016 00:39
Into the Night
★★★★★
(9164)
Surface Detail wrote:
IBdaMann wrote:
Surface Detail wrote:Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.

No, that's not what it tells me ... and I still see no example from you of a gas that radiates in a manner that does not adhere to Planck's.

Suppose we have a body A in space that has a fixed energy source S. Body A reaches equilibrium at temperature T and absorbs and radiates per its surface area.
Now we replace body A with smaller body B that also reaches equilibrium at temperature T.

Which one radiates with greater "intensity"? A or B?

In your "hot gas in a glass vessel" example you started with the equivalent of a bigger body A at temperature T, but then you took some gas away and effectively left yourself smaller body B at the same temperature. Of course the smaller body B has less radiance at any given wavelength.

Notice that Planck's is not necessary here. Stefan-Boltzmann answers this one.

Nope. Both Planck's law and the Stefan-Boltzmann law refer to emission per unit area. The smaller body B has the same radiance as the bigger body A because each square inch of it radiates at the same rate. Each square inch of gas of the same temperature and different densities does not radiate at the same rate. Hence gases cannot obey Planck's law.



Where in Planck's law is square inches (or any other measurement of length, area, or volume even used?


The Parrot Killer
05-10-2016 00:45
jwoodward48
★★★★☆
(1537)
It isn't. You don't understand what the result is - the number you get out is radiance. Radiance is emission per unit area.
05-10-2016 00:47
Into the Night
★★★★★
(9164)
jwoodward48 wrote:
Into the Night wrote:
What is magik? Are you trying to use the word 'magick' or 'magic" (they are two different things)?


Its an intenshonal mispeling to sho hou litl i thinc of ur claems.

Done. You yourself have decided to bring the conversation down to this level again, and again without cause. I have given you three very clear warnings.

It will be very difficult to go back now, ****. You've built your steaming pile, now you can go stew in it.

jwoodward48 wrote:
In any case, I suggest you look up how domains combine between two different functions,


No clue what this means. Either you've put it weirdly, and you actually mean something, or you're just throwing words at me. Please let it be the first.

I do not just throw words at you. I was trying to let you discover this on your own. Instead, you decided to lower the conversation to a shit show.


The Parrot Killer
05-10-2016 00:49
IBdaMann
★★★★★
(4578)
jwoodward48 wrote:Of course Planck's and S-B don't predict that the smaller body will have less radiance! Tell me, where in the formula is "surface area," "volume", or "mass" included? It isn't.

I should not have used the word "radiance." I meant overall total energy radiated.


Global Warming: The preferred religion of the scientifically illiterate.

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
05-10-2016 00:50
Surface Detail
★★★★☆
(1673)
Into the Night wrote:
Surface Detail wrote:
IBdaMann wrote:
Surface Detail wrote:Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.

No, that's not what it tells me ... and I still see no example from you of a gas that radiates in a manner that does not adhere to Planck's.

Suppose we have a body A in space that has a fixed energy source S. Body A reaches equilibrium at temperature T and absorbs and radiates per its surface area.
Now we replace body A with smaller body B that also reaches equilibrium at temperature T.

Which one radiates with greater "intensity"? A or B?

In your "hot gas in a glass vessel" example you started with the equivalent of a bigger body A at temperature T, but then you took some gas away and effectively left yourself smaller body B at the same temperature. Of course the smaller body B has less radiance at any given wavelength.

Notice that Planck's is not necessary here. Stefan-Boltzmann answers this one.

Nope. Both Planck's law and the Stefan-Boltzmann law refer to emission per unit area. The smaller body B has the same radiance as the bigger body A because each square inch of it radiates at the same rate. Each square inch of gas of the same temperature and different densities does not radiate at the same rate. Hence gases cannot obey Planck's law.



Where in Planck's law is square inches (or any other measurement of length, area, or volume even used?

The most commonly used unit of spectral radiance (which is what Planck's law tells you) in wavelength is watts per steradian per square metre per nanometre (W·sr−1·m−2·nm−1).
05-10-2016 00:53
Into the Night
★★★★★
(9164)
jwoodward48 wrote:
It isn't. You don't understand what the result is - the number you get out is radiance. Radiance is emission per unit area.


No such unit in Planck's law. Planck's law does not calculate radiance.


The Parrot Killer
05-10-2016 00:58
Surface Detail
★★★★☆
(1673)
Into the Night wrote:
jwoodward48 wrote:
It isn't. You don't understand what the result is - the number you get out is radiance. Radiance is emission per unit area.


No such unit in Planck's law. Planck's law does not calculate radiance.

You guys certainly know how to take denial to the next level!
05-10-2016 00:59
Into the Night
★★★★★
(9164)
Surface Detail wrote:
Into the Night wrote:
Surface Detail wrote:
IBdaMann wrote:
Surface Detail wrote:Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.

No, that's not what it tells me ... and I still see no example from you of a gas that radiates in a manner that does not adhere to Planck's.

Suppose we have a body A in space that has a fixed energy source S. Body A reaches equilibrium at temperature T and absorbs and radiates per its surface area.
Now we replace body A with smaller body B that also reaches equilibrium at temperature T.

Which one radiates with greater "intensity"? A or B?

In your "hot gas in a glass vessel" example you started with the equivalent of a bigger body A at temperature T, but then you took some gas away and effectively left yourself smaller body B at the same temperature. Of course the smaller body B has less radiance at any given wavelength.

Notice that Planck's is not necessary here. Stefan-Boltzmann answers this one.

Nope. Both Planck's law and the Stefan-Boltzmann law refer to emission per unit area. The smaller body B has the same radiance as the bigger body A because each square inch of it radiates at the same rate. Each square inch of gas of the same temperature and different densities does not radiate at the same rate. Hence gases cannot obey Planck's law.



Where in Planck's law is square inches (or any other measurement of length, area, or volume even used?

The most commonly used unit of spectral radiance (which is what Planck's law tells you) in wavelength is watts per steradian per square metre per nanometre (W·sr−1·m−2·nm−1).


Planck's law does not calculate radiance. It calculates energy in joules at a given wavelength.


The Parrot Killer
05-10-2016 01:01
jwoodward48
★★★★☆
(1537)
Yep, he's gone off the deep end.
05-10-2016 01:05
Surface Detail
★★★★☆
(1673)
Into the Night wrote:
Surface Detail wrote:
Into the Night wrote:
Surface Detail wrote:
IBdaMann wrote:
Surface Detail wrote:Imagine a glass vessel containing hot gas at a temperature T that is emitting radiation with an intensity I at a wavelength w. Now let us remove some of the gas from the vessel, while keeping the temperature of the gas remaining in the vessel constant. Do you think the intensity of the radiation of wavelength w will remain unchanged? What if we remove almost all of the gas? Still no change?

Answer: Of course not. The intensity of the radiated light also depends on the density of the gas. That alone should tell you that Planck's law cannot apply to gases.

No, that's not what it tells me ... and I still see no example from you of a gas that radiates in a manner that does not adhere to Planck's.

Suppose we have a body A in space that has a fixed energy source S. Body A reaches equilibrium at temperature T and absorbs and radiates per its surface area.
Now we replace body A with smaller body B that also reaches equilibrium at temperature T.

Which one radiates with greater "intensity"? A or B?

In your "hot gas in a glass vessel" example you started with the equivalent of a bigger body A at temperature T, but then you took some gas away and effectively left yourself smaller body B at the same temperature. Of course the smaller body B has less radiance at any given wavelength.

Notice that Planck's is not necessary here. Stefan-Boltzmann answers this one.

Nope. Both Planck's law and the Stefan-Boltzmann law refer to emission per unit area. The smaller body B has the same radiance as the bigger body A because each square inch of it radiates at the same rate. Each square inch of gas of the same temperature and different densities does not radiate at the same rate. Hence gases cannot obey Planck's law.



Where in Planck's law is square inches (or any other measurement of length, area, or volume even used?

The most commonly used unit of spectral radiance (which is what Planck's law tells you) in wavelength is watts per steradian per square metre per nanometre (W·sr−1·m−2·nm−1).


Planck's law does not calculate radiance. It calculates energy in joules at a given wavelength.

Most precisely: Planck's law tells you the amount of energy (in joules) radiated per second per unit solid angle per square meter per unit of wavelength. This is called the spectral radiance.
05-10-2016 01:42
jwoodward48
★★★★☆
(1537)
Into the Night wrote:
jwoodward48 wrote:
It isn't. You don't understand what the result is - the number you get out is radiance. Radiance is emission per unit area.


No such unit in Planck's law. Planck's law does not calculate radiance.


Planck's Law: I(v, T) = 2hv^3 / c^2 * 1/(e^(hv/kT)-1)

The right part has some stuff we don't care about for dimensional analysis, so we'll ignore it. We now have

h v^3 / c^2

h is Planck's constant, in units of J/Hz.
v is frequency, in units of Hz.
c is the speed of light, in units of m*Hz.

So we have J/Hz * Hz^3 / (m^2 Hz^2), and cancelling out, we get:

J / m^2

We can, of course, divide by 1, or s/s, or s*Hz.

J / (s * m^2 * Hz)

Since square angle is unitless, we can put that in, too.

J / (s * mm^2 * sr * Hz)

Hey! This looks like the unit of spectral radiance! Energy per unit time per unit surface area per unit solid angle per unit frequency!

That's because it is. QED.


"Heads on a science
Apart" - Coldplay, The Scientist

IBdaMann wrote:
No, science doesn't insist that, ergo I don't insist that.

I am the Ninja Scientist! Beware!
05-10-2016 10:22
Surface Detail
★★★★☆
(1673)
jwoodward48 wrote:
Into the Night wrote:
jwoodward48 wrote:
It isn't. You don't understand what the result is - the number you get out is radiance. Radiance is emission per unit area.


No such unit in Planck's law. Planck's law does not calculate radiance.


Planck's Law: I(v, T) = 2hv^3 / c^2 * 1/(e^(hv/kT)-1)

The right part has some stuff we don't care about for dimensional analysis, so we'll ignore it. We now have

h v^3 / c^2

h is Planck's constant, in units of J/Hz.
v is frequency, in units of Hz.
c is the speed of light, in units of m*Hz.

So we have J/Hz * Hz^3 / (m^2 Hz^2), and cancelling out, we get:

J / m^2

We can, of course, divide by 1, or s/s, or s*Hz.

J / (s * m^2 * Hz)

Since square angle is unitless, we can put that in, too.

J / (s * mm^2 * sr * Hz)

Hey! This looks like the unit of spectral radiance! Energy per unit time per unit surface area per unit solid angle per unit frequency!

That's because it is. QED.

I have a feeling that your dimensional analysis may go over a few heads here
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