What is heat?

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.

Surface Detail wrote: No gas radiates in accordance with Planck's law.

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?

Surface Detail wrote: As anybody with any knowledge of physics is aware, Planck's law applies to black bodies, not gases.

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

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.

jwoodward48 wrote: It also wouldn't inviolate AGW.

Surface Detail wrote: IBdaMann, I don't know how to put this any more simply:

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

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

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

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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).

Into the Night wrote:
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,

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.


.

jwoodward48 wrote:No clue what this means.

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.

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.

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.

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.

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?

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.

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.

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).

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.

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.

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.