Diferent kind of energy (heat and kinetic)

[Andrew Mason]

The important thing to realize is that, by making the blackbody a *cavity* instead of an object, we have explicity demonstrated that the radiation field *itself* has a temperature, independent of the material that comprises the walls of the cavity.
Matter is not needed for thermal energy to exist. Thermal energy is not mechanical in origin. We can, under some conditions, *model* thermal energy in terms of molecular motion.
Blackbody radiation has nothing to do with the matter that 'created' it.

It has always been my understanding, perhaps wrong, that radiation passing out through the aperture originates from the matter on the interior surface of the cavity. The small aperture compared to the cavity size should mean that virtually all the radiation passing through the hole is absorbed in the interior of the cavity. This means that the interior temperature of the cavity surface will increase until the energy of the radiation emitted back through the hole per unit time is equal to that of the radiation incident on the hole from outside. But that emitted radiation originates from the matter on the interior surface of the blackbody cavity. In measuring the peak of that emitted radiation, you are still measuring, effectively, the temperature of the interior surface of the cavity. That cavity cannot, even in theory, exist without matter.

If I am missing something here, I would appreciate your comments.

AM

 

[Andy Resnick]

It has always been my understanding, perhaps wrong, that radiation passing out through the aperture originates from the matter on the interior surface of the cavity. The small aperture compared to the cavity size should mean that virtually all the radiation passing through the hole is absorbed in the interior of the cavity. This means that the interior temperature of the cavity surface will increase until the energy of the radiation emitted back through the hole per unit time is equal to that of the radiation incident on the hole from outside. But that emitted radiation originates from the matter on the interior surface of the blackbody cavity. In measuring the peak of that emitted radiation, you are still measuring, effectively, the temperature of the interior surface of the cavity. That cavity cannot, even in theory, exist without matter.

If I am missing something here, I would appreciate your comments.

AM

You seem to be hung up on the walls of the cavity, not the radiation field within the cavity. The cavity walls are not the thermal reservior, the outside world is. Or an object within the cavity (see below).

First, the radiation need not originate from the walls of the cavity. The cavity walls can be 100% reflective (i.e. e = a = 0). When the radiation field within the cavity has reached thermal equilibrium with the outside world, the spectrum of radiation emitted *from the hole* will follow Planck's law.

Here's another example: take a cavity, the interior sprayed with gold (gold is highly reflective in the IR), and place a water triple-point cell within the cavity. Let's pretend the triple-point cell can float in the cavity (since we are doing this on the space station). What will be the radiation field within the cavity (to a good approximation)? It will correspond to a blackbody at 273.15 K, even though the cavity walls may be at a different temperature.

The field is primary in blackbody radiation, not matter.

What's interesting is that if that light is then (spectrally) filtered, even by a narrowband notch filter, the radiation field no longer has a temperature.

Now, for the question of the temperature of the walls of the cavity. Blackbody construction is a specialized topic of which I know little; what I do know is that the material should be highly (thermally) conductive. Maintaining a primary radiometric temperature standard is not trivial (the NIST and HGH pages show some of what is involved), but is to some degree routine. NIST's are made out of graphite, and the primary standard has a gold crucible inside.

Does this help?

 

[Andrew Mason]

You seem to be hung up on the walls of the cavity, not the radiation field within the cavity. The cavity walls are not the thermal reservior, the outside world is. Or an object within the cavity (see below).

First, the radiation need not originate from the walls of the cavity. The cavity walls can be 100% reflective (i.e. e = a = 0). When the radiation field within the cavity has reached thermal equilibrium with the outside world, the spectrum of radiation emitted *from the hole* will follow Planck's law.

Thanks. This is very helpful.

I think the difficulty I am having is the difference between the theoretical blackbody and a real one. In a theoretical cavity, perhaps one can imagine perfect reflection and the radiation bouncing around inside indefinitely until it happens to pass back out through the aperture. But in a real blackbody, there cannot be perfect reflection. In a real blackbody, I don't see how reflection can occur without transferring momentum to the particles (atoms, molecules, phonons) in the interior surface. In the real world, the radiation that enters the cavity is absorbed by the interior surface resulting in heating of the interior surface. This heating continues until the rate at which radiation emitted by the interior surface exits the cavity is equal to that of the incident radiation.

Here's another example: take a cavity, the interior sprayed with gold (gold is highly reflective in the IR), and place a water triple-point cell within the cavity. Let's pretend the triple-point cell can float in the cavity (since we are doing this on the space station). What will be the radiation field within the cavity (to a good approximation)? It will correspond to a blackbody at 273.15 K, even though the cavity walls may be at a different temperature.

In this case, if you assume that the radiation that exits the cavity comes from the water (which will necessarily be 273.15 K) you are simply measuring the radiation from the water. Unless the triple point cell itself has an emissivity of 1, the cavity will emit blackbody radiation equivalent to a temperature of something less than 273.15K, would it not?

The field is primary in blackbody radiation, not matter.

What's interesting is that if that light is then (spectrally) filtered, even by a narrowband notch filter, the radiation field no longer has a temperature.

Now, for the question of the temperature of the walls of the cavity. Blackbody construction is a specialized topic of which I know little; what I do know is that the material should be highly (thermally) conductive. Maintaining a primary radiometric temperature standard is not trivial (the NIST and HGH pages show some of what is involved), but is to some degree routine. NIST's are made out of graphite, and the primary standard has a gold crucible inside.

Does this help?

Yes. It is a very interesting area that bridges classical and quantum physics. I am going to do some more reading. Thanks.

AM

 

[AlexB2010]

Updating the thought experiment.
Instead of using an iron ball and give kinetic energy and thermal energy, and leave then in cold space.
What happens if the subject is a single iron atom?
If I give heat to a single iron atom what happens? They can’t have internal molecular movement.

Alex

 

[Andy Resnick]

That's an interesting question, and an active area of research: how does macroscopic behavior arise from microscopic considerations (mesoscopic physics).

http://books.google.com/books?hl=en...=onepage&q=mesoscopic physics thermal&f=false

http://www.nature.com/nature/journal/v427/n6976/abs/nature02276.html

http://www.sciencemag.org/cgi/search?session_query_ref=rbs.queryref_1276787372488&COLLECTIONS=hw1&JC=sci&FULLTEXT=%28mesoscopic+AND+physics+AND+thermal+AND+quantum+AND+dots%29&FULLTEXTFIELD=lemcontent&TITLEABSTRACTFIELD=lemhwcomptitleabs&RESOURCETYPE=HWCIT&ABSTRACTFIELD=lemhwcompabstract&TITLEFIELD=lemhwcomptitle[/URL]

 

[AlexB2010]

That's an interesting question, and an active area of research: how does macroscopic behavior arise from microscopic considerations (mesoscopic physics).

http://books.google.com/books?hl=en...=onepage&q=mesoscopic physics thermal&f=false

http://www.nature.com/nature/journal/v427/n6976/abs/nature02276.html

http://www.sciencemag.org/cgi/search?session_query_ref=rbs.queryref_1276787372488&COLLECTIONS=hw1&JC=sci&FULLTEXT=%28mesoscopic+AND+physics+AND+thermal+AND+quantum+AND+dots%29&FULLTEXTFIELD=lemcontent&TITLEABSTRACTFIELD=lemhwcomptitleabs&RESOURCETYPE=HWCIT&ABSTRACTFIELD=lemhwcompabstract&TITLEFIELD=lemhwcomptitle[/URL][/QUOTE]

If I understand right the links that you post, when you individualize in terms of a single atom, a coherence/decoherence due to clash between quantum and classic mechanics will arise. Some people suppose that some coherent behavior of solids will arise from decoherent interaction between the various molecules.
Don’t having full access to scientific papers that comprise of public knowledge in the internet is paradoxical. In Science fast is better, scientific journal are slow expensive and restrictive. Scientific journals are a barrier to science, people needs to publish their results direct on internet. (just a small off topic, because I can't read the full articles without going to library).
Very Nice to know the field of mesoscopic physics. Looks like a very promising approach to experimental physics.

Alex

 

[Andrew Mason]

Updating the thought experiment.
Instead of using an iron ball and give kinetic energy and thermal energy, and leave then in cold space.
What happens if the subject is a single iron atom?
If I give heat to a single iron atom what happens? They can’t have internal molecular movement.

Alex

In classical physics, temperature is a macroscopic property. It requires a large number of particles whose distribution of kinetic energies follow a Maxwell-Boltzmann curve. You do not have that if you have only one atom. So, in classical physics, the temperature of an atom has no meaning.

AM

 

[|<ings]

this has probably been said already:

Heat flows from objects of higher temperature to objects of lower temperature until both (or all if there are more than two objects) are at the same temperature (i.e. thermal equilibrium). If the space around the ball is colder (at a lower temperature) than the ball, then heat will flow or radiate out of the ball and into that space. Note that the heat gained by the space is equal to the heat lost from the ball, so in terms of the ball-space system (if it is isolated), no heat will be lost.

 

[Count Iblis]

There is no problem in defining a temperature, but there is an issue with getting the usual laws of thermodynamics. For a system with only a few degrees of freedom, you would need to consider the so-called canonical ensemble. Here one imagines that the system is in contact with a heat bath at some temperature T. Considering the system to be isolated (i.e. describing it according to the microcanonical ensemble) won't yield the usual thermodynamical equations. For systems with a large number of degrees of freedom, the two ensembles are equivalent.


You can see this issue very clearly in this http://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Proof_of_for_reversible_processes"

You see that in the microcanonical case, you only get the equation dS = dQ/T if you ignore a term that scales as the inverse system size. So, the relation only holds in the thermodynamic limit.

In the derivation for the canonical ensemble, no such terms are generated, so the relation holds in general.

 

[Andy Resnick]

If I understand right the links that you post, when you individualize in terms of a single atom, a coherence/decoherence due to clash between quantum and classic mechanics will arise. Some people suppose that some coherent behavior of solids will arise from decoherent interaction between the various molecules.

That's how I understand it. There's too many cool experiments to keep track of! :)

 

[Andy Resnick]

In classical physics, temperature is a macroscopic property. It requires a large number of particles whose distribution of kinetic energies follow a Maxwell-Boltzmann curve. You do not have that if you have only one atom. So, in classical physics, the temperature of an atom has no meaning.

AM

That's true, but ultimately limiting. For example, we can put a single atom in a cavity and let it interact with the radiation field. What are the thermodynamics of the system? There's experiments using Rydberg atoms and resonant microcavities being done, and the best I can understand is that "strange things happen".

http://hal.archives-ouvertes.fr/docs/00/22/18/32/PDF/ajp-jphyscol198243C221.pdf

Ugh... this was 1982? I am so far behind...

 

[AlexB2010]

That's true, but ultimately limiting. For example, we can put a single atom in a cavity and let it interact with the radiation field. What are the thermodynamics of the system? There's experiments using Rydberg atoms and resonant microcavities being done, and the best I can understand is that "strange things happen".

http://hal.archives-ouvertes.fr/docs/00/22/18/32/PDF/ajp-jphyscol198243C221.pdf

Ugh... this was 1982? I am so far behind...

On 82’s Fabre paper they use a group of 100 molecules. I find some recent data showing that using a single organic molecule and given heat to then you will generate electrical current. That’s can show thermal energy given to a single molecule will excite the electron field.
http://www.lbl.gov/msd/assets/docs/highlights/07_5Segalman_thermoelectrics.pdf

What appears in absence of contact of other molecules only the electrical field will increase energy and molecular motion will not happen?
If you think in macro phenomena as a result of non deterministic quantum interactions, looks like each single interaction will be contributing to a mean result and the quantum indeterminism will be seen only on the precision limit of measures.