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We seem to be going in circles. Perhaps it would break the cycle if we do this step by step. Damorbel raised an excellent question, What do you think the Maxwell-Boltzmann distribution is? Assuming it was intended rhetorically, it leads naturally into another. Damorbel, what would you say the units are for the average of the Maxwell-Boltzmann distribution? -- Vaughan Pratt ( talk) 22:19, 3 November 2011 (UTC)
For example, consider a 22.4 L box at absolute zero with an interior at absolute zero, into which two 10 gram crystals of solid neon (also at absolute zero, or as close to it as you like) have been fired by cannons in opposite directions, at 1100 m/sec relative to the box. As they head toward impact with each other in the center of the box, it is sealed. Question: what is the temperature of this system, according to you? What is the temperature of each neon atom, as seen in the lab frame? What is the entropy of the system?
Suppose the crystals miss each other and bounce between walls of the box, perfectly elastically. If thermal contact could be make between this box and another box of neon of the same size, filled with gas at STP, which direction would heat flow?
Now, suppose the crystals are allowed to break up and fill the box with neon gas as their kinetic energy is distributed randomly, in a Maxwell-Boltzmann distribution. Has the temperature of the system changed? Has the entropy changed? What is the direction of heat flow now between this box and the box holding the same amount of neon at STP?
Note: I suppose all this amounts to much the same as asking what would happen if Maxwell's demon managed to take the mole of atoms in a 22.4 L box of neon at standard temp and pressure, and get them going all at the same velocity, in 3 orthogonal directions, all while conserving momentum and energy. Clearly the entropy changes. Does the temperature change? That's a question for you. Starting with the crystals, I've merely run that thought experiment in the opposite direction, in the direction of spontaneity. S B H arris 19:16, 4 November 2011 (UTC)
Energy, not temperature. You can't talk about the temperature of an individual photon, either. Energy, yes. Temperature, no. S B H arris 07:28, 7 November 2011 (UTC)
Try going back to your idea that individual photons have a temperature. Certainly a collection of photons at different frequences in a blackbody distribution have a collective temperature, but they are a collection. What about one photon of given energy? What about a monochromatic beam of photons at a single energy?
This is not an academic exercise. Your microwave oven heats things that are already so hot they are emitting in the infrared (at about 0.001 cm). But the photons from the oven have a wavelength of several cm. So why does the energy go from oven to food, and not the other way around? Could it be because you're not talking about two things of different temperature at all? The food has a temperature, but what about the microwave output? Remember, the photons from the Big Bang have a wavelength of about 0.1 cm, so your microwave oven photons are "colder" than that, if you insist that their temperature is an individual thing, and even when there are great numbers, it doesn't require a distribution. So by your theory, they shouldn't be able to heat anything much above 0.2 Kelvin or so. My oatmeal gets considerably hotter. In fact, you can put a neon glow tube into a microwave (with a cup of water next to it for a load) and see the neon plasma ionize at thousands of degrees K. What's going on, pray tell? Something wrong with your theory?
A beam of monochromatic microwaves is a lot like a beam of atoms all moving in one direction at one velocity. You think it has a "temperature" you can calculate as energy/particle. I'm here to tell you that since it has an entropy/energy as low as you like (depending on how monochromatic it is) it can induce temperatures as high as you like. Think about that next time you see plasma sparks at very high temps, from metal points inside your microwave oven. S B H arris 17:29, 7 November 2011 (UTC)
I'm disappointed in you about the microwaves. Yes, photons are particles (see photoelectric effect and Compton effect). You can have one photon. You can't switch to talking about the electric field of an EM wave when it suits you, but talk about individual particles of light (as you do above-- you brought the subject up) when you think that will serve your argument. Nature must do the same thing no matter which way you chose to view it/her.
Yes, you're quite right that the maximum temperature a beam can heat anything to, is the slice out of the Stefan-Bolzmann power law it represents for a black body, but that depends on the narrowness of frequency, and the beam's maldistribution of frequencies away from that of a from black body spectrum (which is the analog of Maxwell-Bolzmann distribution for boson photons). But this does not fit your definition of temperature. First, your definition clearly doesn't work for a single photon. Second, it allows me to adjust the "heating power" of a beam of a given energy, simply by making it more monochromatic, but keeping its power the same. If "temperature" is simply energy per particle in any circumstance (or an average of each particle energy for any particle collection), I shouldn't be able to do that, since the number of particles and the energy are not changed by this, yet the heating power of the beam (what it can do to temperatures of objects) is.
Finally, even if you don't accept the quantum theory of EM radiation, you can replace a microwave beam with a beam of atoms, all moving at a velocity which has a very narrow spectrum (that is, they are all going nearly the same velocity, as closely as physical possible). I did that in my previous examples and you missed the point. The smaller you get such a beam velocity spread, the hotter that beam can heat another collection of gas you aim it at. Do you understand this point? If I shoot a beam of atoms at a bottle of gas, even if the beam atoms have far less kinetic energy than the average energy of the molecules in the gas in the bottle, I will heat the bottle and the gas, so long as the power of the atom beam exceeds the power of that velocity-spread channel in the Maxwell-Bolzmann temp that I aim it at (and assuming that the incoming gas molecules can bounce off the bottle, so they can transfer some energy without having to stay behind and be equilibrated). If atoms in any object are constrained (as in solid objects) a bombardment of atoms of any temperature-- even close to absolute zero, will heat the object, so long as the density of the beam is high enough. That's more or less what you do when you throw a baseball at a wall! The baseball can be cooler than the wall (in terms of what you measure when you stick a themometer in both), and even can be colder when you count its kinetic energy, and yet it will still heat it when it strikes, and you can keep that up till the wall is very hot. The point is that you can arrange this with a baseball so the energy transfer is only in one-direction, because the atoms in the baseball are only moving in one direction.
You could go farther and hit a paddle-wheel with slow baseballs outside a container of gas, and this wheel could turn a crank connected to a fan inside the box that heats the gas by friction. This works no matter what speed of the baseball, since the wheel always turns only one direction, even with a slow cold ball. You could do the same with a gas beam, turing a paddle wheel outside the box connected to a fan inside. You begin to see the role entropy plays in energy transfer, since none of this would work if the gas outside didn't have not only energy/per particle, but a DIRECTED energy per particle (low entropy per energy).
The gas certainly has a temperature, but what about the beam? (or the cold baseball?). If you're going to insist the beam does also, and insist that it is defined as the mean kinetic energy of the beam atoms (even though it is maldistributed) then you must say this is a situation in which heat is transferred from cold to hot, spontaneously. It's exactly the same situation as a microwave beam or laser directed into a hole in a black body radiator. Thus, we either need to give up the second law of thermodynamics (heat goes from hot to cold, those being defined in terms of temperature), or else we must give up your (private) definition of "temperature." Which is it to be? S B H arris 16:14, 8 November 2011 (UTC)
Hello everyone. Just to let you know, I have made a minor edit and replaced the reference "Boltzmann's constant" with "the Boltzmann constant", merely to ensure consistency throughout the article. However, I did not touch the quotation from Planck's Nobel Lecture - which proved to be accurate, nor the occurrence in the title of item #7 in the "References" list - which I cannot verify. -- NicAdi 14:35, 29 December 2011 (UTC)
Personally, I'm learning a lot from this discussion about how to present this difficult subject to readers who don't understand it well. S B H arris 19:26, 7 November 2011 (UTC)
I have removed the text:-
Which is quite wrong. There is no need to 'observe temperature at the bulk level'
Temperature is an intensive quantity which means it is independent of the amount ('number of particles) present.
Surely it takes onl a little imagination to realise that temperature is the measure of the energy of a particle and the average energy of 'N' particles when there are a number (N) particles in the system? (the system must have only one temperature i.e. it must be in equilibrium, of course.
As for observe, what is that supposed to mean? In quantum systems particles with insufficient energy are unable to initiate reactions; check Einstein's Nobel prize winning photo emission paper of 1905. In this paper he showed how only photons with sufficient energy (= high enough temperature) were able to displace electrons. -- Damorbel ( talk) 15:48, 5 December 2012 (UTC)
The argument for the revision above is also applies to all relationships in the article between microscopic (particle level) effects and macroscopic (or bulk) systems.
Macroscopic systems have an undefined number of particles, a Mole (unit) is an example with NA (the Avogadro number) of particles but a number 106 smaller changes nothing.
Having a large number of particles in a system places an additional requirement on the ( Boltzmann constant) relationship between particle energy and temperature. To establish a measurable temperature for such a system the average energy of the particles must be related by the Boltzmann constant.
I have removed or modified a number of texts in the article :=
It is not "a bridge.... etc." The are various requirements for systems with > 1 particle; one is where the particles are able to exchange energy in random way, this is the basis of thermal equilibrium, the Maxwell Boltzmann distribution describes the energy distribtion of such a system. Another multi-particle sytem with a defined temperature is one where all the particles have the same energy; this is unusual because it is not an equilibrium condition but can arise at low densities where the particles seldom exchange energy through collisions.
This statement:-
is wildly untrue!
In particle physics it is the molecule, not the atom that is the relevant unit. Polyatomic molecules such as carbon dioxide have a much higher heat capacity than monatomics such as helium, diatomics (H2, O2 ) are in between. To understand the reasons why, the article Degrees of freedom (physics and chemistry) helps explain.
Please, gentlemen, argue and discuss before inserting such statements!
Therefore the section "Bridge from macroscopic to microscopic physics" contributes nothing to the Boltzmann constant article and will be deleted. -- Damorbel ( talk) 07:38, 6 December 2012 (UTC)
The Boltzmann constant will shortly replace the Kelvin as one of the seven base units in the International System of Units (SI) because it can be determined more accurately than the triple point (see http://iopscience.iop.org/0026-1394/48/3/008/pdf/0026-1394_48_3_008.pdf) more accurately
Currently the Boltzmann constant can be determined to 6 places (kB = 1.380 651(17) × 10−23 JK−1) by measuring johnson noise. There is no way the Boltzmann constant can be regarded as:-
There are a number of notable features here, the constant is measured using electron, not gas, temperature, so the article, when limiting itself to atoms of gas is far too restrictive.
The Boltzmann constant is applicable to all particles, even grains of pollen, as observed by Einstein in his 1905 paper on Brownian motion entitled :-
The section Role in the equipartition of energy has no content about the equipartition of energy. I have inserted a link. -- Damorbel ( talk) 08:58, 6 December 2012 (UTC)
The law of equipartition breaks down when the thermal energy kBT is significantly smaller than the spacing between energy levels. Equipartition no longer holds because it is a poor approximation to assume that the energy levels form a smooth continuum, which is required in the derivations of the equipartition theorem above.
Jheald, commenting upon my assertion that:-
And then you write:-
What you refer to as these rotational and vibrational energies are not quantum states of the particles and I certainly do not exclude them from the particle energy contributing to heat capacity etc., etc.; of course they do! All I am saying is that the energy locked in particle quantum states is not part of the kinetic energy = 1/2v2that is exchanged during thermal collisions.
Again you write:-
Who ever said they did? The Maxwell Boltzmann distribution only applies in equlibrium conditions i.e. when the particles exchange energy freely and with equal probability.
You write :-
Yes, because at low temperatures the vibration energies are too small to overcome the diatomic binding forces, thus there is little or no energy in the (elastic) binding at low temperatures.
Further, you write:-
Not true. When energy is locked in a quantum state it only has quantum degrees of freedom which will not be activated by quantum events with lower energy, that is why Einstein got the 1922 Nobel prize. The energy in the inner quantum states of atoms is relatively high, for the inner electrons of metals it is extremely high e.g. X-Rays. The quantum states of molecules are in the range of thermal energy, 1.24 meV - 1.7 eV, which corresponds to the kinetic energy of molecules at intermediate tempertures so there is a great deal of overlap, effectively leading to a continuum, as can be seen from the Planck energy spectrum which never goes to zero.
You write:-
Ultimately, at high enough temperatures, all atoms will be stripped of their electrons i.e. 100% ionised and all the electrons and protons will be thermal and none of them will be in particle aggregates such as atoms and molecules. Such conditions are said to exist at the centre of stars.
The reason for my original deletion, which you 'undid' was to eliminate the very real confusion between the Bolztmann constant and the application of it. The Boltzmann constant is a very simple but important ratio between temperature and particle energy, it has the same dimension as entropy, joules/K but entropy is about a system of particles whereas the Boltzmann constant is about a single degree of freedom. In view of all this I would like you to undo the deletion of my contribution.-- Damorbel ( talk) 10:18, 8 December 2012 (UTC)
The article is about the Boltzmann constant, but Jheald writes about :-
The article is about the Boltzmann constant which is the energy of a single particle per Kelvin, but Jheald writes about:-
The article is about the Boltzmann constant, but Jheald writes about :-
Oh dear! Electrons? Photons? The kinetic rotation energy of the molecules? Yes Jheald, I know what your talking about and it isn't the Boltzmann constant!
Oh dear! Where is the Boltzmann constant in this?
Now I know that you have nothing to contribute to the article on the Boltzmann constant, what a shame!
And now I feel free to restore my contribution after your deletion. -- Damorbel ( talk) 18:00, 8 December 2012 (UTC)
This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 |
This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 |
We seem to be going in circles. Perhaps it would break the cycle if we do this step by step. Damorbel raised an excellent question, What do you think the Maxwell-Boltzmann distribution is? Assuming it was intended rhetorically, it leads naturally into another. Damorbel, what would you say the units are for the average of the Maxwell-Boltzmann distribution? -- Vaughan Pratt ( talk) 22:19, 3 November 2011 (UTC)
For example, consider a 22.4 L box at absolute zero with an interior at absolute zero, into which two 10 gram crystals of solid neon (also at absolute zero, or as close to it as you like) have been fired by cannons in opposite directions, at 1100 m/sec relative to the box. As they head toward impact with each other in the center of the box, it is sealed. Question: what is the temperature of this system, according to you? What is the temperature of each neon atom, as seen in the lab frame? What is the entropy of the system?
Suppose the crystals miss each other and bounce between walls of the box, perfectly elastically. If thermal contact could be make between this box and another box of neon of the same size, filled with gas at STP, which direction would heat flow?
Now, suppose the crystals are allowed to break up and fill the box with neon gas as their kinetic energy is distributed randomly, in a Maxwell-Boltzmann distribution. Has the temperature of the system changed? Has the entropy changed? What is the direction of heat flow now between this box and the box holding the same amount of neon at STP?
Note: I suppose all this amounts to much the same as asking what would happen if Maxwell's demon managed to take the mole of atoms in a 22.4 L box of neon at standard temp and pressure, and get them going all at the same velocity, in 3 orthogonal directions, all while conserving momentum and energy. Clearly the entropy changes. Does the temperature change? That's a question for you. Starting with the crystals, I've merely run that thought experiment in the opposite direction, in the direction of spontaneity. S B H arris 19:16, 4 November 2011 (UTC)
Energy, not temperature. You can't talk about the temperature of an individual photon, either. Energy, yes. Temperature, no. S B H arris 07:28, 7 November 2011 (UTC)
Try going back to your idea that individual photons have a temperature. Certainly a collection of photons at different frequences in a blackbody distribution have a collective temperature, but they are a collection. What about one photon of given energy? What about a monochromatic beam of photons at a single energy?
This is not an academic exercise. Your microwave oven heats things that are already so hot they are emitting in the infrared (at about 0.001 cm). But the photons from the oven have a wavelength of several cm. So why does the energy go from oven to food, and not the other way around? Could it be because you're not talking about two things of different temperature at all? The food has a temperature, but what about the microwave output? Remember, the photons from the Big Bang have a wavelength of about 0.1 cm, so your microwave oven photons are "colder" than that, if you insist that their temperature is an individual thing, and even when there are great numbers, it doesn't require a distribution. So by your theory, they shouldn't be able to heat anything much above 0.2 Kelvin or so. My oatmeal gets considerably hotter. In fact, you can put a neon glow tube into a microwave (with a cup of water next to it for a load) and see the neon plasma ionize at thousands of degrees K. What's going on, pray tell? Something wrong with your theory?
A beam of monochromatic microwaves is a lot like a beam of atoms all moving in one direction at one velocity. You think it has a "temperature" you can calculate as energy/particle. I'm here to tell you that since it has an entropy/energy as low as you like (depending on how monochromatic it is) it can induce temperatures as high as you like. Think about that next time you see plasma sparks at very high temps, from metal points inside your microwave oven. S B H arris 17:29, 7 November 2011 (UTC)
I'm disappointed in you about the microwaves. Yes, photons are particles (see photoelectric effect and Compton effect). You can have one photon. You can't switch to talking about the electric field of an EM wave when it suits you, but talk about individual particles of light (as you do above-- you brought the subject up) when you think that will serve your argument. Nature must do the same thing no matter which way you chose to view it/her.
Yes, you're quite right that the maximum temperature a beam can heat anything to, is the slice out of the Stefan-Bolzmann power law it represents for a black body, but that depends on the narrowness of frequency, and the beam's maldistribution of frequencies away from that of a from black body spectrum (which is the analog of Maxwell-Bolzmann distribution for boson photons). But this does not fit your definition of temperature. First, your definition clearly doesn't work for a single photon. Second, it allows me to adjust the "heating power" of a beam of a given energy, simply by making it more monochromatic, but keeping its power the same. If "temperature" is simply energy per particle in any circumstance (or an average of each particle energy for any particle collection), I shouldn't be able to do that, since the number of particles and the energy are not changed by this, yet the heating power of the beam (what it can do to temperatures of objects) is.
Finally, even if you don't accept the quantum theory of EM radiation, you can replace a microwave beam with a beam of atoms, all moving at a velocity which has a very narrow spectrum (that is, they are all going nearly the same velocity, as closely as physical possible). I did that in my previous examples and you missed the point. The smaller you get such a beam velocity spread, the hotter that beam can heat another collection of gas you aim it at. Do you understand this point? If I shoot a beam of atoms at a bottle of gas, even if the beam atoms have far less kinetic energy than the average energy of the molecules in the gas in the bottle, I will heat the bottle and the gas, so long as the power of the atom beam exceeds the power of that velocity-spread channel in the Maxwell-Bolzmann temp that I aim it at (and assuming that the incoming gas molecules can bounce off the bottle, so they can transfer some energy without having to stay behind and be equilibrated). If atoms in any object are constrained (as in solid objects) a bombardment of atoms of any temperature-- even close to absolute zero, will heat the object, so long as the density of the beam is high enough. That's more or less what you do when you throw a baseball at a wall! The baseball can be cooler than the wall (in terms of what you measure when you stick a themometer in both), and even can be colder when you count its kinetic energy, and yet it will still heat it when it strikes, and you can keep that up till the wall is very hot. The point is that you can arrange this with a baseball so the energy transfer is only in one-direction, because the atoms in the baseball are only moving in one direction.
You could go farther and hit a paddle-wheel with slow baseballs outside a container of gas, and this wheel could turn a crank connected to a fan inside the box that heats the gas by friction. This works no matter what speed of the baseball, since the wheel always turns only one direction, even with a slow cold ball. You could do the same with a gas beam, turing a paddle wheel outside the box connected to a fan inside. You begin to see the role entropy plays in energy transfer, since none of this would work if the gas outside didn't have not only energy/per particle, but a DIRECTED energy per particle (low entropy per energy).
The gas certainly has a temperature, but what about the beam? (or the cold baseball?). If you're going to insist the beam does also, and insist that it is defined as the mean kinetic energy of the beam atoms (even though it is maldistributed) then you must say this is a situation in which heat is transferred from cold to hot, spontaneously. It's exactly the same situation as a microwave beam or laser directed into a hole in a black body radiator. Thus, we either need to give up the second law of thermodynamics (heat goes from hot to cold, those being defined in terms of temperature), or else we must give up your (private) definition of "temperature." Which is it to be? S B H arris 16:14, 8 November 2011 (UTC)
Hello everyone. Just to let you know, I have made a minor edit and replaced the reference "Boltzmann's constant" with "the Boltzmann constant", merely to ensure consistency throughout the article. However, I did not touch the quotation from Planck's Nobel Lecture - which proved to be accurate, nor the occurrence in the title of item #7 in the "References" list - which I cannot verify. -- NicAdi 14:35, 29 December 2011 (UTC)
Personally, I'm learning a lot from this discussion about how to present this difficult subject to readers who don't understand it well. S B H arris 19:26, 7 November 2011 (UTC)
I have removed the text:-
Which is quite wrong. There is no need to 'observe temperature at the bulk level'
Temperature is an intensive quantity which means it is independent of the amount ('number of particles) present.
Surely it takes onl a little imagination to realise that temperature is the measure of the energy of a particle and the average energy of 'N' particles when there are a number (N) particles in the system? (the system must have only one temperature i.e. it must be in equilibrium, of course.
As for observe, what is that supposed to mean? In quantum systems particles with insufficient energy are unable to initiate reactions; check Einstein's Nobel prize winning photo emission paper of 1905. In this paper he showed how only photons with sufficient energy (= high enough temperature) were able to displace electrons. -- Damorbel ( talk) 15:48, 5 December 2012 (UTC)
The argument for the revision above is also applies to all relationships in the article between microscopic (particle level) effects and macroscopic (or bulk) systems.
Macroscopic systems have an undefined number of particles, a Mole (unit) is an example with NA (the Avogadro number) of particles but a number 106 smaller changes nothing.
Having a large number of particles in a system places an additional requirement on the ( Boltzmann constant) relationship between particle energy and temperature. To establish a measurable temperature for such a system the average energy of the particles must be related by the Boltzmann constant.
I have removed or modified a number of texts in the article :=
It is not "a bridge.... etc." The are various requirements for systems with > 1 particle; one is where the particles are able to exchange energy in random way, this is the basis of thermal equilibrium, the Maxwell Boltzmann distribution describes the energy distribtion of such a system. Another multi-particle sytem with a defined temperature is one where all the particles have the same energy; this is unusual because it is not an equilibrium condition but can arise at low densities where the particles seldom exchange energy through collisions.
This statement:-
is wildly untrue!
In particle physics it is the molecule, not the atom that is the relevant unit. Polyatomic molecules such as carbon dioxide have a much higher heat capacity than monatomics such as helium, diatomics (H2, O2 ) are in between. To understand the reasons why, the article Degrees of freedom (physics and chemistry) helps explain.
Please, gentlemen, argue and discuss before inserting such statements!
Therefore the section "Bridge from macroscopic to microscopic physics" contributes nothing to the Boltzmann constant article and will be deleted. -- Damorbel ( talk) 07:38, 6 December 2012 (UTC)
The Boltzmann constant will shortly replace the Kelvin as one of the seven base units in the International System of Units (SI) because it can be determined more accurately than the triple point (see http://iopscience.iop.org/0026-1394/48/3/008/pdf/0026-1394_48_3_008.pdf) more accurately
Currently the Boltzmann constant can be determined to 6 places (kB = 1.380 651(17) × 10−23 JK−1) by measuring johnson noise. There is no way the Boltzmann constant can be regarded as:-
There are a number of notable features here, the constant is measured using electron, not gas, temperature, so the article, when limiting itself to atoms of gas is far too restrictive.
The Boltzmann constant is applicable to all particles, even grains of pollen, as observed by Einstein in his 1905 paper on Brownian motion entitled :-
The section Role in the equipartition of energy has no content about the equipartition of energy. I have inserted a link. -- Damorbel ( talk) 08:58, 6 December 2012 (UTC)
The law of equipartition breaks down when the thermal energy kBT is significantly smaller than the spacing between energy levels. Equipartition no longer holds because it is a poor approximation to assume that the energy levels form a smooth continuum, which is required in the derivations of the equipartition theorem above.
Jheald, commenting upon my assertion that:-
And then you write:-
What you refer to as these rotational and vibrational energies are not quantum states of the particles and I certainly do not exclude them from the particle energy contributing to heat capacity etc., etc.; of course they do! All I am saying is that the energy locked in particle quantum states is not part of the kinetic energy = 1/2v2that is exchanged during thermal collisions.
Again you write:-
Who ever said they did? The Maxwell Boltzmann distribution only applies in equlibrium conditions i.e. when the particles exchange energy freely and with equal probability.
You write :-
Yes, because at low temperatures the vibration energies are too small to overcome the diatomic binding forces, thus there is little or no energy in the (elastic) binding at low temperatures.
Further, you write:-
Not true. When energy is locked in a quantum state it only has quantum degrees of freedom which will not be activated by quantum events with lower energy, that is why Einstein got the 1922 Nobel prize. The energy in the inner quantum states of atoms is relatively high, for the inner electrons of metals it is extremely high e.g. X-Rays. The quantum states of molecules are in the range of thermal energy, 1.24 meV - 1.7 eV, which corresponds to the kinetic energy of molecules at intermediate tempertures so there is a great deal of overlap, effectively leading to a continuum, as can be seen from the Planck energy spectrum which never goes to zero.
You write:-
Ultimately, at high enough temperatures, all atoms will be stripped of their electrons i.e. 100% ionised and all the electrons and protons will be thermal and none of them will be in particle aggregates such as atoms and molecules. Such conditions are said to exist at the centre of stars.
The reason for my original deletion, which you 'undid' was to eliminate the very real confusion between the Bolztmann constant and the application of it. The Boltzmann constant is a very simple but important ratio between temperature and particle energy, it has the same dimension as entropy, joules/K but entropy is about a system of particles whereas the Boltzmann constant is about a single degree of freedom. In view of all this I would like you to undo the deletion of my contribution.-- Damorbel ( talk) 10:18, 8 December 2012 (UTC)
The article is about the Boltzmann constant, but Jheald writes about :-
The article is about the Boltzmann constant which is the energy of a single particle per Kelvin, but Jheald writes about:-
The article is about the Boltzmann constant, but Jheald writes about :-
Oh dear! Electrons? Photons? The kinetic rotation energy of the molecules? Yes Jheald, I know what your talking about and it isn't the Boltzmann constant!
Oh dear! Where is the Boltzmann constant in this?
Now I know that you have nothing to contribute to the article on the Boltzmann constant, what a shame!
And now I feel free to restore my contribution after your deletion. -- Damorbel ( talk) 18:00, 8 December 2012 (UTC)
This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 |