Hypothesis about temperature (split from Physical mechanism how matter absorbs radiation.)

[martillo]

On 3/15/2024 at 12:00 PM, swansont said:

There is no “internal EM energy”

Yes there is. It is called thermal radiation. It is the only thing that exist in the interior of a perfect black body and it is the thing it emits.

See: https://en.wikipedia.org/wiki/Black-body_radiation

For instance:

"Black-body radiation is the thermal electromagnetic radiation within, or surrounding, a body in thermodynamic equilibrium with its environment, emitted by a black body (an idealized opaque, non-reflective body). It has a specific, continuous spectrum of wavelengths, inversely related to intensity, that depend only on the body's temperature, which is assumed, for the sake of calculations and theory, to be uniform and constant.^[1][2][3][4]

As the temperature of a black body decreases, the emitted thermal radiation decreases in intensity and its maximum moves to longer wavelengths. Shown for comparison is the classical Rayleigh–Jeans law and its ultraviolet catastrophe.

A perfectly insulated enclosure which is in thermal equilibrium internally contains blackbody radiation, and will emit it through a hole made in its wall, provided the hole is small enough to have a negligible effect upon the equilibrium. The thermal radiation spontaneously emitted by many ordinary objects can be approximated as blackbody radiation."

and:

"Black body[edit]

Main article: Black body

All normal (baryonic) matter emits electromagnetic radiation when it has a temperature above absolute zero. The radiation represents a conversion of a body's internal energy into electromagnetic energy, and is therefore called thermal radiation. It is a spontaneous process of radiative distribution of entropy.

Conversely, all normal matter absorbs electromagnetic radiation to some degree. An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called blackbody radiation."

 

On 3/15/2024 at 12:00 PM, swansont said:

You can’t have motion of atoms, having KE, and have a bunch of EM radiation interior to the system, and have the theory work.

kT is directly related to mv^2 (with a constant related to degrees of freedom)

If there’s energy stored as EM radiation, there’s less KE, but that means a lower temperature.

Yes it works. "kT is directly related to mv^2 (with a constant related to degrees of freedom)" continues to be right. It is just that there's the extra "bunch of EM radiation interior to the system" in the system may be not taken into account.

 

On 3/15/2024 at 2:47 PM, swansont said:

That's the electrostatic interaction, which involves virtual photons. That's typically not referred to as EM radiation, which consists of real photons. The vibrational modes of a solid can be described in terms of phonons (not photons);  there are non-radiative ways of changing those states.

If I can explain the thing with real existent particles (the photons) why would I appeal to the mathematical artifice of "virtual particles" (phonons)?

 

On 3/15/2024 at 4:36 PM, joigus said:

On 3/14/2024 at 4:55 AM, martillo said:

the temperature of an atom is [...]

There is no such thing.

You mean it is not being considered. I can define it as below.

 

On 3/15/2024 at 4:36 PM, joigus said:

Thermodynamics defines temperature based on thermal equilibrium.

I can define the temperature of an atom (as of that any material object) in thermal equilibrium with its surrounding environment as being the same as that of the environment which can be measured by a thermometer.

Edited March 17 by martillo

[KJW]

18 minutes ago, martillo said:

Quote

Thermodynamics defines temperature based on thermal equilibrium.

I can define the temperature of an atom (as of that any material object) in thermal equilibrium with its surrounding environment as being the same as that of the environment which can be measured by a thermometer.

It occurred to me how one might do that. One could define the temperature of a single particle in thermal equilibrium with its surrounding environment in terms of the kinetic energy distribution over time of the particle. Applying the ergodic principle transforms this to an ensemble distribution for which the notion of temperature naturally applies.

 

[martillo]

58 minutes ago, KJW said:

It occurred to me how one might do that. One could define the temperature of a single particle in thermal equilibrium with its surrounding environment in terms of the kinetic energy distribution over time of the particle. Applying the ergodic principle transforms this to an ensemble distribution for which the notion of temperature naturally applies.

 

The thermal equilibrium I'm conceiving is something different. Is not related to the kinetic energy of the particle. I consider the particle is capable to absorb and emit EM radiation (photons). All atoms have their characteristic levels of energy in accordance to the levels of its electrons. All atoms have their characteristic spectra. When electrons jump to other levels they absorb or emit photons. In a thermal equilibrium with its environment the particle continuously absorbs and emits photons maintaining an average level of energy. Those photons constitute the EM radiation present in the environment called thermal radiation and it has a temperature associated to it. The temperature is related to the density of the power of the radiation per unity of area according to the Stefan-Boltzmann Law: 

P/A = σT⁴

 

Edited March 17 by martillo

[martillo]

From Enthalpy (https://en.wikipedia.org/wiki/Enthalpy) :

Definition[edit]

The enthalpy H of a thermodynamic system is defined as the sum of its internal energy and the product of its pressure and volume:^[1]

H = U + p V ,

where U is the internal energy, p is pressure, and V is the volume of the system; p V is sometimes referred to as the pressure energy Ɛ_p .^[6]

Physical interpretation[edit]

The U term is the energy of the system, and the p V term can be interpreted as the work that would be required to "make room" for the system if the pressure of the environment remained constant. When a system, for example, n moles of a gas of volume V at pressure p and temperature T, is created or brought to its present state from absolute zero, energy must be supplied equal to its internal energy U plus p V, where p V is the work done in pushing against the ambient (atmospheric) pressure.

 

The internal thermal radiation I'm considering (in its two models, EM waves or photons) has momentum and energy both contributing to the enthalpy. Its momentum contributes in the second pV term exerting a force and work while its energy contributes in the first U term the internal energy of the system.

 

Edited March 17 by martillo

[joigus]

On 3/17/2024 at 5:44 AM, KJW said:

It occurred to me how one might do that. One could define the temperature of a single particle in thermal equilibrium with its surrounding environment in terms of the kinetic energy distribution over time of the particle. Applying the ergodic principle transforms this to an ensemble distribution for which the notion of temperature naturally applies.

 

Only when a system is in thermal equilibrium, and provided it is ergodic, the time average of the kinetic energy coincides with the ensemble average at any one time. So you need the ensemble plus the fact that the system be in thermal equilibrium plus ergodicity. Very very special conditions indeed. And when it works, you've used the whole ensemble to define it... So it's not a property of the particle. It's a property of the ensemble!

So no --I insist--, there is no temperature as a property of a singled-out particle of an ensemble in general. And when there is such a thing, it's only by stretching the concept so that what really is a property of the ensemble is decreed to be a property of any and every one of the members of the ensemble.

I don't see how this definition does anything, really. And believe me, I would like nothing more than to be illuminated about anything physics.

[KJW]

On 3/20/2024 at 8:44 AM, joigus said:

Only when a system is in thermal equilibrium, and provided it is ergodic, the time average of the kinetic energy coincides with the ensemble average at any one time. So you need the ensemble plus the fact that the system be in thermal equilibrium plus ergodicity. Very very special conditions indeed.

Doesn't thermal equilibria imply ergodicity? Statistical thermodynamics is based on this assumption.

 

 

On 3/20/2024 at 8:44 AM, joigus said:

And when it works, you've used the whole ensemble to define it... So it's not a property of the particle. It's a property of the ensemble!

It is often said that one can't assign a temperature to a single particle, and that temperature is a statistical property. But I have shown how temperature can be assigned to a single particle while maintaining the statistical nature of temperature. The ensemble is the particle (over time).

 

 

On 3/20/2024 at 8:44 AM, joigus said:

I don't see how this definition does anything, really.

I'm not the one who started the discussion on assigning temperature to a single particle. Oddly enough, @martillo rejected my suggestion in favour of something that is not going to work.

 

 

[joigus]

1 hour ago, KJW said:

I'm not the one who started the discussion on assigning temperature to a single particle. Oddly enough, @martillo rejected my suggestion in favour of something that is not going to work.

 

Agreed.  @martillo's suggestion is not going to work. Your suggestion is, if I understand correctly, a valiant attempt --let's put it that way-- to try and make sense of their hopeless intention to define temperature as an attribute of one molecule or atom.

I think you're right to say that the ergodic theorem is essential to define thermodynamic equilibrium. If most typical physical systems we deal with were not ergodic, I don't think statistical ensembles would work at all within the context of variables such as the partition function, temperature, Helmholtz's free energy, entropy, and such. It would be a disaster. The least I can say is those variables would be as good as useless.

So why bother trying to make sense of something that just doesn't? Just to spite me? Temperature is an ensemble-related parameter. There is no operational definition that would allow us to measure the temperature of a molecule either. There is no theoretical framework that allows us to define it in such a way except by way of the ensemble.

Temperature is an ensemble property. Even more so than entropy is. At least the microscopic entropy of a molecule can be defined as the volume of phase space for that molecule, which is always the same. Not so for temperature.

 

[martillo]

6 hours ago, KJW said:

I'm not the one who started the discussion on assigning temperature to a single particle. Oddly enough, @martillo rejected my suggestion in favour of something that is not going to work.

 

3 hours ago, joigus said:

Agreed.  @martillo's suggestion is not going to work. Your suggestion is, if I understand correctly, a valiant attempt --let's put it that way-- to try and make sense of their hopeless intention to define temperature as an attribute of one molecule or atom.

I think you're right to say that the ergodic theorem is essential to define thermodynamic equilibrium. If most typical physical systems we deal with were not ergodic, I don't think statistical ensembles would work at all within the context of variables such as the partition function, temperature, Helmholtz's free energy, entropy, and such. It would be a disaster. The least I can say is those variables would be as good as useless.

So why bother trying to make sense of something that just doesn't? Just to spite me? Temperature is an ensemble-related parameter. There is no operational definition that would allow us to measure the temperature of a molecule either. There is no theoretical framework that allows us to define it in such a way except by way of the ensemble.

Temperature is an ensemble property. Even more so than entropy is. At least the microscopic entropy of a molecule can be defined as the volume of phase space for that molecule, which is always the same. Not so for temperature.

 

Fine, if you can't manage my definition of the temperature for a single atom just forget it. My consideration that in the Kinetic Theory of Gases the internal thermal radiation is not taken into account still holds.

As I said I'm considering https://en.wikipedia.org/wiki/Black-body_radiation particularly the following:

On 3/17/2024 at 12:57 AM, martillo said:

"Black body[edit]

Main article: Black body

All normal (baryonic) matter emits electromagnetic radiation when it has a temperature above absolute zero. The radiation represents a conversion of a body's internal energy into electromagnetic energy, and is therefore called thermal radiation. It is a spontaneous process of radiative distribution of entropy.

Conversely, all normal matter absorbs electromagnetic radiation to some degree. An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called blackbody radiation."

The thermal radiation is described mathematically by the Stefan-Boltzmann Law: Pot/A = σT⁴ where Pot is the Potential of the radiated energy by unity of time.

If the volume of gas cannot be considered as a black body it is just to make the appropriated correction in the law: εσT⁴ .

The thermal radiation is an EM radiation (photons) and has momentum and energy.

The main claim I'm pointing out is that in the volume of gas in a thermal equilibrium not only the atoms/molecules of the gas are present. There are photons in the volume with momentum and energy and they both contribute to the respective momentum and energy of the total system of gas plus photons in the volume everything counting for the total enthalpy of the system: H = U + PV

If the system is externally heated in some amount there is an increment ΔH = ΔU + Δ(PV).

The problem is in how the increment of this ΔH "heat" is distributed to the ΔU and Δ(PV) in the system. I tried a first mathematical account for everything together but I found it rather complicated for me and I'm not trying anymore for now. If I find something appropriated at some time I would return here.

 

On 3/15/2024 at 12:00 PM, swansont said:

You can’t have motion of atoms, having KE, and have a bunch of EM radiation interior to the system, and have the theory work.

kT is directly related to mv^2 (with a constant related to degrees of freedom)

If there’s energy stored as EM radiation, there’s less KE, but that means a lower temperature.

There would be less kinetic energy in the atoms/molecules of gas (with less average velocity) but this would not imply less temperature. It is compensated by an amount of temperature given by the photons according to the Stefan-Boltzmann Law.

Edited March 22 by martillo

[swansont]

16 minutes ago, martillo said:

There would be less kinetic energy in the atoms/molecules of gas (may be less average velocity) but this would not imply less temperature. It is compensated by an amount of temperature given by the photons according to the Stefan-Boltzmann Law.

This discussion was originally about a solid, and my posts were in that context, but go ahead and calculate the amount of EM energy. The S-B law gives you the radiated power from a surface. What’s the surface area of the interior of a gas? How do you ger an energy content from it?

If the inside of a solid material the system is at the same temperature, there is no net radiation. Any photon emitted is absorbed, and since c is a big number, the time between these events will be small

 

[joigus]

On 3/15/2024 at 5:08 PM, sethoflagos said:

And yet when we measure temperature gradients between systems at different temperatures, in non-extreme conditions they are generally linear in agreement with the dominant mechanism for transfer of heat being by momentum exchange.

If the dominant mechanism were EMR as you suggest, then the measured gradients would be highly non-linear (cubic in delta T I think).

I understand your POV but I think it's a misleading one. Particle collisions in gases that support rotational and vibrational modes only follow conservation of linear momentum on average. In the general case, some momentum is transferred via the other modes.

Absolutely. Sorry I missed this very good argument for so long. It's only because of what you say that different molecules have different specific heats as a function of temperature. The internal degrees of freedom are totally relevant. This is exactly the reason why different molecular components have different specific heats. What other reason could there be for different gases to display different specific heats if only the CoM DOF were relevant?

Quite a different matter is how quantum mechanics introduces a cutoff for short-length degrees of freedom (independently of how poly-atomic a gas is), and how this played a crucial role in the dawning of quantum mechanics itself. (Birth of the old quantum theory as a mechanism to freeze the short-wavelength DOF.)

[martillo]

26 minutes ago, swansont said:

If the inside of a solid material the system is at the same temperature, there is no net radiation. Any photon emitted is absorbed, and since c is a big number, the time between these events will be small

For any material system, solid, liquid or gas an average temperature is maintained by the atoms/molecules which stay interchanging only some quantity of photons with the environment. Others are absorbed by them. Remember that after all the there's a lot of space between them.

26 minutes ago, swansont said:

This discussion was originally about a solid, and my posts were in that context, but go ahead and calculate the amount of EM energy. The S-B law gives you the radiated power from a surface. What’s the surface area of the interior of a gas? How do you ger an energy content from it?

The surface to be considered is just a representative little "differential" area inside the system through which a "differential" quantity of photons are passing in two directions.

23 minutes ago, joigus said:

Absolutely. Sorry I missed this very good argument for so long. It's only because of what you say that different molecules have different specific heats as a function of temperature. The internal degrees of freedom are totally relevant. This is exactly the reason why different molecular components have different specific heats. What other reason could there be for different gases to display different specific heats if only the CoM DOF were relevant?

Quite a different matter is how quantum mechanics introduces a cutoff for short-length degrees of freedom (independently of how poly-atomic a gas is), and how this played a crucial role in the dawning of quantum mechanics itself. (Birth of the old quantum theory as a mechanism to freeze the short-wavelength DOF.)

I'm sorry but I cannot follow you in this. I don't understand the vocabulary properly and may be I'm not capable to put the subject in your terms.

Edited March 22 by martillo

[joigus]

11 minutes ago, martillo said:

For any material system, solid, liquid or gas an average temperature is maintained by the atoms/molecules which stay interchanging photons with the environment. Remember that after all the there's a lot of space between them.

The surface to be considered is just a representative little "differential" area inside the system through which a "differential" quantity of photons are passing in two directions.

I'm sorry but I cannot follow you in this. I don't understand the vocabulary properly and may be I'm not capable to put the subject in your terms.

Sorry. Here's a lowdown of the vocabulary I've used. Tell me, please, where the problem is:

Degree of freedom

Temperature

Specific heat

internal (rotational/vibrational) vs external (CoM)

cutoff (making some energy --or wavelength-- domain irrelevant; see next)

freezing (as in freezing degrees of freedom by making them very unlikely to store energy under thermal-equilibrium conditions).

[martillo]

22 minutes ago, joigus said:

Sorry. Here's a lowdown of the vocabulary I've used. Tell me, please, where the problem is:

Degree of freedom

Temperature

Specific heat

internal (rotational/vibrational) vs external (CoM)

cutoff (making some energy --or wavelength-- domain irrelevant; see next)

freezing (as in freezing degrees of freedom by making them very unlikely to store energy under thermal-equilibrium conditions).

Thanks. More clear now. You are talking about the vibrations approach.

[joigus]

20 minutes ago, martillo said:

Thanks. More clear now. You are talking about the vibrations approach.

Molecules can vibrate, rotate, twist, and scissor, etc. In fact, there are DOF that are not obviously rotational/vibrational, etc. but some complicated so-called normal modes like,

Quote

The normal modes of vibration are: asymmetric, symmetric, wagging, twisting, scissoring, and rocking for polyatomic molecules. Figure 1: Six types of Vibrational Modes.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Vibrational_Spectroscopy/Vibrational_Modes/Number_of_Vibrational_Modes_in_a_Molecule

Recently, a key to why CO₂ (kinda mysterious, as it's just a boring non-polar linear molecule) is such an important agent in global warming has been found to have a root in resonances of such non-obvious normal modes due to Fermi resonances:

https://arxiv.org/pdf/2401.15177.pdf

[swansont]

2 hours ago, martillo said:

For any material system, solid, liquid or gas an average temperature is maintained by the atoms/molecules which stay interchanging only some quantity of photons with the environment. Others are absorbed by them. Remember that after all the there's a lot of space between them.

Not all that much. In copper it’s about 0.25 nm. Light would take ~10^-18 sec to traverse the distance

If you had a block of copper, 0.01m on a side, it’s going to radiate 2.75 W at 300K

At 4 atoms per nm, there will be 4 x 10^7 atoms along one dimension. 6 x 1.6 x^10^15 atoms on the surface, so perhaps 10^17 atoms near the surface can radiate outwards. at 300K, the photon energy peak is about 0.1 eV. 2.75 W needs about 2 x 10^20 of these photons per second. So each atom is responsible for about 200 photons/sec, or one every 5 milliseconds. But we know these photons can only live for a time that’s around 10^14 times shorter.

These are estimations, so there might be a factor of 2 here and there that could be off. But you’re ~14 orders of magnitude short 

1 cm^3 of copper is about 9g, so the block is somewhere around 10^23 atoms, or 2 x 10^25 photons/s for 10^-18 sec. 2 x 10^7 photons let’s call it 10^8, just to be safe, at 0.1 eV. Which is around 10^-12 J. Compare with the heat capacity of 0.385 J/gK, and we’re at 300K, so we have around a kilojoule of thermal energy in our 9g block.

No, it’s not photons.

[martillo]

1 hour ago, swansont said:

Not all that much. In copper it’s about 0.25 nm. Light would take ~10^-18 sec to traverse the distance

If you had a block of copper, 0.01m on a side, it’s going to radiate 2.75 W at 300K

At 4 atoms per nm, there will be 4 x 10^7 atoms along one dimension. 6 x 1.6 x^10^15 atoms on the surface, so perhaps 10^17 atoms near the surface can radiate outwards. at 300K, the photon energy peak is about 0.1 eV. 2.75 W needs about 2 x 10^20 of these photons per second. So each atom is responsible for about 200 photons/sec, or one every 5 milliseconds. But we know these photons can only live for a time that’s around 10^14 times shorter.

These are estimations, so there might be a factor of 2 here and there that could be off. But you’re ~14 orders of magnitude short 

1 cm^3 of copper is about 9g, so the block is somewhere around 10^23 atoms, or 2 x 10^25 photons/s for 10^-18 sec. 2 x 10^7 photons let’s call it 10^8, just to be safe, at 0.1 eV. Which is around 10^-12 J. Compare with the heat capacity of 0.385 J/gK, and we’re at 300K, so we have around a kilojoule of thermal energy in our 9g block.

No, it’s not photons.

I think something is not right in your calculation. You have a piece of copper radiating 2.75 W. If it is in thermal equilibrium the internal thermal radiation is the same as that external radiation of 2.75 W. That is 2.75 Joules per second. It is not about a kilojoule of thermal energy as you say.

[sethoflagos]

4 hours ago, joigus said:

Absolutely. Sorry I missed this very good argument for so long. It's only because of what you say that different molecules have different specific heats as a function of temperature. The internal degrees of freedom are totally relevant. This is exactly the reason why different molecular components have different specific heats. What other reason could there be for different gases to display different specific heats if only the CoM DOF were relevant?

Quite a different matter is how quantum mechanics introduces a cutoff for short-length degrees of freedom (independently of how poly-atomic a gas is), and how this played a crucial role in the dawning of quantum mechanics itself. (Birth of the old quantum theory as a mechanism to freeze the short-wavelength DOF.)

Glad someone was listening 🙂

2 hours ago, joigus said:

Recently, a key to why CO₂ (kinda mysterious, as it's just a boring non-polar linear molecule) is such an important agent in global warming has been found to have a root in resonances of such non-obvious normal modes due to Fermi resonances:

Worth a +1 just for the mention. Hope someone else is listening.

[swansont]

1 hour ago, martillo said:

I think something is not right in your calculation. You have a piece of copper radiating 2.75 W. If it is in thermal equilibrium the internal thermal radiation is the same as that external radiation of 2.75 W.

Why should it be? The 2.75 W is dependent on the surface area. Show your work if you disagree.

Thermal equilibrium means the surroundings are at 300K, and 2.75 W is also being absorbed. But that’s at the surface. 

 

 

1 hour ago, martillo said:

That is 2.75 Joules per second. It is not about a kilojoule of thermal energy as you say.

The kilojoule is the thermal energy content (mass* specific heat capacity * temperature), not the radiated power. They are two different things. The radiated power will tend to reduce the thermal energy by reducing the temperature. 

[sethoflagos]

6 hours ago, martillo said:

Fine, if you can't manage my definition of the temperature for a single atom just forget it. My consideration that in the Kinetic Theory of Gases the internal thermal radiation is not taken into account still holds.

The internal energy of thermal radiation within a space occupied by a gas is accounted for by the internal energy of the photon gas that co-occupies that same space, Look at the wikipedia page on 'photon gas' for an explanation.

On 3/17/2024 at 6:08 AM, martillo said:

The thermal equilibrium I'm conceiving is something different. Is not related to the kinetic energy of the particle. I consider the particle is capable to absorb and emit EM radiation (photons). All atoms have their characteristic levels of energy in accordance to the levels of its electrons. All atoms have their characteristic spectra. When electrons jump to other levels they absorb or emit photons. In a thermal equilibrium with its environment the particle continuously absorbs and emits photons maintaining an average level of energy.

At everyday temperatures, black body photons capable of inducing electron orbital jumps are to all intents and purposes non-existent. The dominant process for generating black body radiation is via the acceleration of charged particles, and as @swansont has pointed out, this can be many orders of magnitude below the internal energy of the matter phase.

There is interchange between the matter phase and proton gas phase, and this can have effects during dynamic changes in thermal equilibrium. Off the top of my head, it's one reason we can never quite achieve the theoretical adiabatic combustion temperature in fuel burning processes. But if you only start getting a glimpse of a phenomenon at >1500 K, there really are no grounds whatsoever for claiming it to be a dominant process at normal, lower temperatures. 

[martillo]

3 hours ago, swansont said:

Why should it be? The 2.75 W is dependent on the surface area. Show your work if you disagree.

Thermal equilibrium means the surroundings are at 300K, and 2.75 W is also being absorbed. But that’s at the surface. 

When a system is in thermal equilibrium it has the same temperature in any place so the radiated power anywhere is also the same. The Power is the Energy emitted by the area of any radiating volume you can consider inside the system. Particularly a enough small one to identify it as a "place" in the system. The Power divided by the area gives the same value everywhere and it is the temperature in that place.

2 hours ago, sethoflagos said:

The internal energy of thermal radiation within a space occupied by a gas is accounted for by the internal energy of the photon gas that co-occupies that same space, Look at the wikipedia page on 'photon gas' for an explanation.

Completely agree with this. 

2 hours ago, sethoflagos said:

At everyday temperatures, black body photons capable of inducing electron orbital jumps are to all intents and purposes non-existent. The dominant process for generating black body radiation is via the acceleration of charged particles, and as @swansont has pointed out, this can be many orders of magnitude below the internal energy of the matter phase.

There is interchange between the matter phase and proton gas phase, and this can have effects during dynamic changes in thermal equilibrium. Off the top of my head, it's one reason we can never quite achieve the theoretical adiabatic combustion temperature in fuel burning processes. But if you only start getting a glimpse of a phenomenon at >1500 K, there really are no grounds whatsoever for claiming it to be a dominant process at normal, lower temperatures. 

Well, I'm currently developing (trying) a different explanation through a different cause for the phenomenon. I'm thinking in the photons continuously exchanged by the particles of a system with their surrounding environment. For instance, in a perfect black body of an empty reflective cavity the particles would be the molecules of the wall of the cavity where it can be considered the reflections the same as an instantaneous absorption and emission of photons.

Edited March 23 by martillo

[swansont]

7 hours ago, martillo said:

When a system is in thermal equilibrium it has the same temperature in any place so the radiated power anywhere is also the same. The Power is the Energy emitted by the area of any radiating volume you can consider inside the system. Particularly a enough small one to identify it as a "place" in the system. The Power divided by the area gives the same value everywhere and it is the temperature in that place.

But the issue is the energy content. You’ve provided no analysis to show that the thermal energy of the object is from the photons. Just assertion.

[martillo]

1 hour ago, swansont said:

But the issue is the energy content. You’ve provided no analysis to show that the thermal energy of the object is from the photons. Just assertion.

My reasoning is that the energy that is supplied to the systems we are analyzing comes in the form of electromagnetic radiation commonly called "heat" which is, in my understanding, entirely composed by photons.

To begin I think in the "empty" cavity with totally reflective walls of a perfect black body is full of photons. It is under this assumption that Planck and Einstein derived their famous "Planck Law" of a black body radiation and "Planck-Einstein formula" E = hf. I then conclude that "heat" is actually composed by photons.

The "heat" transferring between two systems is always through the interchange of photons which can pass through the walls separating them. Photons can pass through atoms and molecules (with scattering dispersion sometimes) carrying energy.

Finally I consider that these photons come into the heated system and cannot just disappear. I can identify that part of them are converted into other forms of energy like the energy levels of electrons in the atoms and also other part is involved in the bonds between the atoms of the molecules. This energies compose the internal energy U in the equation H = U + PV.

My claim is that the energy U is completely ignored in the called Kinetic Theory. In the Kinetic Theory we have U = 0 and H = PV = Kinetic Energy of the particles (translational motion in the ideal gases and vibrational motion in solids). I think that this is not what really happens in reality. As I said the incoming photons do not disappear and part of them is converted to structural forms of energies of the molecules, other part to their kinetic energy and other part just remains as photons in the internal thermal energy.

 

As I said at some time I'm currently trying to take a mathematical account for how those are energies are distributed in the enthalpy H = U +PV equation. I think I advanced a bit in this direction:

A system (solid, liquid or gas) is heated with an energy ΔH.

Its internal energy U is incremented in ΔU = ΔStr + ΔRad where ΔStr is about the structural energy and ΔRad is about the thermal radiation.

In agreement with Einstein I assume a possible increment Δm in the mass of the particles due to the absorption of energy being Δm = E₀/c² = ΔH/c² as negligible and so the increment is in PV is Δ(PV) = ΔKE (the increment in the kinetic energy of the particles) .

Then I have: ΔH = ΔU + Δ(PV)= ΔStr + ΔRad + ΔKE.

I think now the Equipartition Principle applies here and so ΔStr = ΔRad = ΔKE = KT and I can have ΔH = 3KT in all cases, solids, liquids and gases.

I just must mention here that for ideal gases we  have ΔH = 3KT and not 3KT/2 in the generalized formulation of the Equipartition Theorem applied to ideal gases as it can be seen at Wikipedia page of the theorem: https://en.wikipedia.org/wiki/Equipartition_theorem .

I'm yet still thinking in how the Equipartition Theorem is applied in this approach. I mean thinking in the physical considerations on the ΔStr and ΔRad energies for the Equipartition Theorem to apply.

 

 

 

Edited March 23 by martillo

[Mordred]

I would suggest you examine the Bose Einstein and Fermi Dirac statistics and how it applies to blackbody temperatures in particular interest is that all effective degrees of freedom inherent in any particle species do contribute to that temperature  and further allow those statistic to predict the number density of particle states at any given blackbody temperature. They are direct applications of the ideal gas laws and can work at any temperature extreme (at least until you hit a singularity state)

[swansont]

3 hours ago, martillo said:

My reasoning is that the energy that is supplied to the systems we are analyzing comes in the form of electromagnetic radiation commonly called "heat" which is, in my understanding, entirely composed by photons.

Radiant heat. There is also conduction and convection.

3 hours ago, martillo said:

To begin I think in the "empty" cavity with totally reflective walls of a perfect black body is full of photons. It is under this assumption that Planck and Einstein derived their famous "Planck Law" of a black body radiation and "Planck-Einstein formula" E = hf. I then conclude that "heat" is actually composed by photons.

Planck was modeling the radiation spectrum, not the behavior of atoms in a solid.

 

3 hours ago, martillo said:

The "heat" transferring between two systems is always through the interchange of photons which can pass through the walls separating them. Photons can pass through atoms and molecules (with scattering dispersion sometimes) carrying energy.

Copper is not transparent

3 hours ago, martillo said:

Finally I consider that these photons come into the heated system and cannot just disappear.

Photons disappear all the time. Ever turn off a light at night and notice how a room immediately gets dark?

3 hours ago, martillo said:

I can identify that part of them are converted into other forms of energy like the energy levels of electrons in the atoms and also other part is involved in the bonds between the atoms of the molecules. This energies compose the internal energy U in the equation H = U + PV.

Both sethoflagos and I have pointed out that these photons don’t have enough energy to cause atomic excitations. You can’t just wish these reactions into existence 

3 hours ago, martillo said:

My claim is that the energy U is completely ignored in the called Kinetic Theory. In the Kinetic Theory we have U = 0 and H = PV = Kinetic Energy of the particles (translational motion in the ideal gases and vibrational motion in solids). I think that this is not what really happens in reality. As I said the incoming photons do not disappear and part of them is converted to structural forms of energies of the molecules, other part to their kinetic energy and other part just remains as photons in the internal thermal energy.

What you think isn’t nearly as important as what you can show, both theoretically and experimentally. You talk of photons passing through materials like copper to transfer heat, when empirically we know copper is opaque. 

All you’ve done here is regurgitate some stuff from wikipedia. There’s no actual analysis to see if the claims are reasonable, and match up with what we observe.

The thermal energy content of the copper in my example is around 1 kjoule. The blackbody photons average around 0.1 eV, meaning there need to be around 6 x 10^22 photons in existence at all times in the material. (this is not an emission rate, this is a population)

But the block is only 1 cm on a side. A photon can travel, at most, 1 cm before it either is absorbed, or leaves the block. It takes less than 0.1 ns to travel that far. An atom has to be emitting 10^10 photons a second to have just one be existing at any given time. That’s an emission rate of 6 x 10^32 photons a sec for there to be a kilojoule of photon energy in the material

Atoms near the surface are the ones that can have photons leave the material. If that’s only the monolayer at the surface. 10^16 atoms. Half will emit in a direction that leaves the material. That’s 10^38 photons a second. 10^18 watts. Slightly higher than what the Stefan-Boltzmann law predicts.

Your idea doesn’t hold up to scrutiny. It’s not even close.

[martillo]

1 hour ago, swansont said:

Your idea doesn’t hold up to scrutiny. It’s not even close.

It's your opinion and your decision. Seems other ones' too. I will continue further with my approach although not here in the forum. Thanks a lot for your comments, they have let me advance a bit while trying to give support to it. 

Best regards.

Edited March 23 by martillo