Diferent kind of energy (heat and kinetic)

[AlexB2010]

I have a hypothetical question:
Suppose a use a chemical source of energy to heat a 1Kg Iron ball, or, transform that same chemical energy in kinetic energy to accelerate an equal ball in space.
The heated ball in space will cool down, loosing energy.
The accelerated ball will maintain its velocity and the energy given to then will “stick” to the ball, manteining the energy level.
What is the difference in the energy that makes than stick or run away from the ball?
Thanks in advance.

Alex

 

[Andrew Mason]

There is another way to look at this in which heat "sticks" but the kinetic energy doesn't.

The heating of the ball will increase its temperature. That temperature will be the same as measured by observers in other reference frames (ie. moving with respect to the ball). So, in that sense, the thermal energy sticks to the heated ball.

The kinetic energy of the moving ball will not be the same for other moving observers. For some, the kinetic energy will decrease (ie. the impulse slows it down) and for others it will increase. So kinetic energy of the "moving" ball does not "stick" to it. It depends on who is doing the measuring.

Measurement of kinetic energy is relative to the frame of reference of the observer. Measurement of temperature (molecular kinetic energy) is always related to the frame of reference of the object whose temperature is being measured.

The reason the temperature of the ball drops is: radiation. The molecules of the ball are continually accelerating and decelerating randomly. This results in emission of electromagnetic radiation. The ball will radiate energy at a rate that depends on its temperature. The ball's temperature will drop until the rate of radiation emitted is equal to the rate of radiation absorbed from other sources.

The "moving" ball does not radiate because its motion is uniform.

AM

 

[AlexB2010]

Hi Andrew,
I don’t think the changing of reference can be used here, because since Galileo, to perform a physics experiment you need to determine a reference frame were the measurements are made. On the same frame of reference the accelerated ball will maintain their velocity and the energy given to then will increase the “energy level” of the ball.
On heat, I know that they will be loose in space equilibrating in their environment, irradiating in space, but since you point out relativity: Heat is considered kinetic energy of individual molecules, why they will be the same on all frames of reference? Since they are movement like any other.
My question is, why the difference behavior of a energy that came from the same source? Why kinetic energy not irradiate in space? Why the molecular motion is different from acceleration on the whole ball?
Thanks,
Alex

 

[Andy Resnick]

I have a hypothetical question:
Suppose a use a chemical source of energy to heat a 1Kg Iron ball, or, transform that same chemical energy in kinetic energy to accelerate an equal ball in space.
The heated ball in space will cool down, loosing energy.
The accelerated ball will maintain its velocity and the energy given to then will “stick” to the ball, manteining the energy level.
What is the difference in the energy that makes than stick or run away from the ball?
Thanks in advance.

Alex

Well, you need to be a little careful here- the moving ball will continue to move as long as it doesn't interact with anything else, and the heated ball will radiate only if it is allowed to interact with a cold reservior.

But using common sense, you are basically correct; heat and kinetic energy are two different kinds of energy, and can be interconverted to some degree. Heat can be converted into work, and work can be converted into heat... not with 100% efficiency, but that's entropy for you...

 

[AlexB2010]

Hi Andy, thanks again.
Somehow kinetic energy becomes part of a moving body and heat goes away. Someone know how? What happens in a small particle defined by their momentum if they receive kinetic energy.

Alex

 

[Andrew Mason]

Well, you need to be a little careful here- the moving ball will continue to move as long as it doesn't interact with anything else, and the heated ball will radiate only if it is allowed to interact with a cold reservior.

This requires a bit of clarification. I would not consider space to be a cold reservoir. But the heated ball will radiate heat into space at a rate that is only dependent on its temperature. The ball will lose thermal energy unless and until the rate of absorption of energy from other sources equals or exceeds the rate at which it loses thermal energy. But it always radiates.

But using common sense, you are basically correct; heat and kinetic energy are two different kinds of energy, and can be interconverted to some degree. Heat can be converted into work, and work can be converted into heat... not with 100% efficiency, but that's entropy for you...

One might want to phrase this a little differently. Heat and mechanical energy derive from kinetic energy of matter. Heat is the kinetic energy of random molecular (non-uniform) motion. Mechanical energy is the kinetic energy of uniform, non-random motion of macroscopic pieces of matter.

AM

 

[Andy Resnick]

This requires a bit of clarification. I would not consider space to be a cold reservoir. But the heated ball will radiate heat into space at a rate that is only dependent on its temperature. The ball will lose thermal energy unless and until the rate of absorption of energy from other sources equals or exceeds the rate at which it loses thermal energy. But it always radiates.

One might want to phrase this a little differently. Heat and mechanical energy derive from kinetic energy of matter. Heat is the kinetic energy of random molecular (non-uniform) motion. Mechanical energy is the kinetic energy of uniform, non-random motion of macroscopic pieces of matter.

AM

The heated ball will not lose heat energy if it is at the same temperature as it's surroundings.

Heat does not have a mechanical origin.

 

[Andrew Mason]

The heated ball will not lose heat energy if it is at the same temperature as it's surroundings.

Can the surroundings not be empty space? The temperature of a vacuum is undefined, since there are no molecules in a vacuum. But a heated ball in space can still lose heat energy.

Heat does not have a mechanical origin.

I would have to disagree. Heat can originate from mechanical motion. The flow of heat is essentially a mechanical flow - molecules transferring kinetic energy to other molecules through physical collisions.

AM

 

[Andy Resnick]

.I would have to disagree. Heat can originate from mechanical motion. The flow of heat is essentially a mechanical flow - molecules transferring kinetic energy to other molecules through physical collisions.

AM

This discussion has been had numerous times on other threads; I have no desire to cover that ground again unless you have something original to say. I'll simply mention the following:

1) You would agree (hopefully) that electromagnetic energy does not have a mechanical origin; why insist that thermal energy have a mechanical origin?

2) Your statement only has meaning in the context of equilibrium/partition functions/statistical considerations. Dissipative process (e.g. friction, viscous losses) are exempt from these considerations. There is no (AFAIK) mechanical microscopic model for friction. Do you know of one?

As to your first comment, blackbody radiation has a well-defined temperature and requires no atoms. Put the ball in a cavity held at the same temperature, and it cannot lose energy to the surroundings.

 

[AlexB2010]

Coming to the basic question again:
Thermal energy given or taken to a body will equilibrate on their surroundings. Kinetic energy that accelerates the body will stay glued to de body.
Why this happens? Why differences, since both energy are movement?
Answers exist? or is this an unsolved question to physics?

Alex

 

[Andy Resnick]

AlexB2010,

As I mentioned earleir, kinetic energy does not 'stay glued' to the body, if the body can interact with another body (e.g. friction). Heat will not 'stay glued' to the body either, if it can interact with other bodies (e.g. thermal equilibration).

 

[AlexB2010]

AlexB2010,

As I mentioned earleir, kinetic energy does not 'stay glued' to the body, if the body can interact with another body (e.g. friction). Heat will not 'stay glued' to the body either, if it can interact with other bodies (e.g. thermal equilibration).

Hi Andy,
I do a little thinking based in your explanation; correct me if I am wrong.
In my think experiment, in releasing the chemical bonds I liberate thermal energy that are used to heat a ball or run some kind of machinery that will convert thermal energy to motion that will be applied to the ball given acceleration and increasing the speed.
The ball traveling in open space in uniform motion will not interact with other solids and then will go forever.
The heated ball have the internal molecular motion increased, the molecules will be given kinetic energy. The internal collisions will transform the kinetic energy in thermal radiation that is capable of moving in free space, making the ball loose energy (In colder space).
Since the accelerated ball don’t have increased molecular motion, the kinetic energy given to then will be all in one direction, molecular interaction don’t occur and the kinetic energy will stay in the ball as long they are in uniform motion.
Thermal energy will be defined as wave of amplitude and frequency defined; can I define kinetic energy the same manner?
What happens if I give kinetic energy to a photon?

Alex

 

[Andrew Mason]

1) You would agree (hopefully) that electromagnetic energy does not have a mechanical origin; why insist that thermal energy have a mechanical origin?

I am not insisting that thermal energy must have a mechanical origin. I am just disagreeing with your general statement that thermal energy (heat) does not have a mechanical origin.

2) Your statement only has meaning in the context of equilibrium/partition functions/statistical considerations. Dissipative process (e.g. friction, viscous losses) are exempt from these considerations. There is no (AFAIK) mechanical microscopic model for friction. Do you know of one?

I agree that friction is necessarily a macroscopic phenomenon. All collisions at the molecular level are elastic.

As to your first comment, blackbody radiation has a well-defined temperature and requires no atoms. Put the ball in a cavity held at the same temperature, and it cannot lose energy to the surroundings.

The black-body consists of matter in some form (plasma, atoms) and has a temperature. But the radiation does not really have a temperature. We refer to radiation temperature but what we are really referring to is the temperature of the blackbody matter that produces the observed electro-magnetic spectrum.

AM

 

[Andrew Mason]

What happens if I give kinetic energy to a photon?

Kinetic energy is defined only for bodies of matter. You can't do work on a photon. You can't change its speed so you can't give it kinetic energy. It does have momentum, though.

AM

 

[AlexB2010]

Kinetic energy is defined only for bodies of matter. You can't do work on a photon. You can't change its speed so you can't give it kinetic energy. It does have momentum, though.

AM

The ability of receive kinetic energy, is it exclusive to fermions ?

 

[Andy Resnick]

But the radiation does not really have a temperature. We refer to radiation temperature but what we are really referring to is the temperature of the blackbody matter that produces the observed electro-magnetic spectrum.

AM

Well.. this *is* an original statement. It's also at variance with Planck's law. No, blackbody radiation is the spectral distribution of a free field (with infinite degrees of freedom) at thermal equilibrium with a reservior. It is material independent, especially since there's no such thing as 'black body matter'.

 

[Andrew Mason]

Well.. this *is* an original statement. It's also at variance with Planck's law. No, blackbody radiation is the spectral distribution of a free field (with infinite degrees of freedom) at thermal equilibrium with a reservior.

You are misinterpreting what I said. My point was that only matter can really have a temperature, at least in the classical sense of temperature. Radiation temperature is assigned to a spectrum of radiation that is emitted by a blackbody. The radiation temperature is the surface temperature of the blackbody that would produce that radiation distribution. That is all I am saying. If I am wrong on that, please correct me.

It is material independent, especially since there's no such thing as 'black body matter'.

My reference to "the blackbody matter" is to the matter that makes up the blackbody. A blackbody has to be made up of matter. It doesn't matter what the matter is. Radiation can only be emitted from matter. If I am wrong on that, please correct me.

We seem to be getting a little far from the original question here.

AM

 

[Andy Resnick]

You are misinterpreting what I said. My point was that only matter can really have a temperature, at least in the classical sense of temperature. Radiation temperature is assigned to a spectrum of radiation that is emitted by a blackbody. The radiation temperature is the surface temperature of the blackbody that would produce that radiation distribution. That is all I am saying. If I am wrong on that, please correct me.
My reference to "the blackbody matter" is to the matter that makes up the blackbody. A blackbody has to be made up of matter. It doesn't matter what the matter is. Radiation can only be emitted from matter. If I am wrong on that, please correct me.

We seem to be getting a little far from the original question here.

AM

I'm just following where you are leading.

Edit: Let me try a different track, and describe blackbody radiation.

It is true that blackbody radiation can be defined as "the electromagnetic radiation emitted by a blackbody object at temperature 'T'". But since a blackbody is defined as a material for which the emissivity 'e'= 1 for all wavelengths, and since no real matter has that property, it's not a particular useful way to define blackbody radiation.

A better way to construct a blackbody is to instead realize that e = 1 also means the absorptivity 'a' = 1. That's Kirchoff's law. We *can* construct an object that absorbs all wavelengths equally (and perfectly): a cavity containing a tiny hole. In fact, making the hole not tiny doesn't really affect things too much:

http://physics.nist.gov/Divisions/Div844/facilities/pyro/pyro.html
http://www.hgh.fr/corps-noir-infrarouge-infrared-source-blackbody-en.php
http://www.npl.co.uk/engineering-measurements/thermal/temperature/products-and-services/npl-fixed-point-blackbody-sources

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.

Blackbody radiation has nothing to do with the matter that 'created' it.

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.

 

[AlexB2010]

I'm just following where you are leading.

Blackbody radiation has nothing to do with the matter that 'created' it.

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.

Thermal energy can interact with matter generating kinetic molecular motion. How these interactions happen?

Alex

 

[Count Iblis]

Perhaps it can help to think of this as follows. Picture the ball as consisting of atoms. Indeed, everything consists of elementary particles. These elementary particles interact with each other according to certain laws of physics. In principle this allows you to describe how the ball will move, cool down etc. etc.

But this is not a very practical way of doing computations. You don't want to account for what every atom in the ball is doing when you want to compute how fast it is moving. So, wat you do is you split the motion of the ball in two parts. One part is the center of mass motion of the ball, the other part is the motion of the atoms relative to the center of mass motion. And then you describe the latter in a statistical way.

Now, we can describe the motion of the ball using conservation of energy and momentum. But because momentum is linear in the velocity, the total momentum of the system equals the momentum of the center of mass of the ball. In case of energy this is not the case. First the atoms can have potential energy, but even the sum of the kinetic energies of the atoms is not the kinetic energy of the center of mass.

This means that the statistical description you get when you average out what the atoms are doing will have to take into account the energy of the atoms. Changes in this part of the energy is what we call heat.

 

[AlexB2010]

Perhaps it can help to think of this as follows. Picture the ball as consisting of atoms. Indeed, everything consists of elementary particles. These elementary particles interact with each other according to certain laws of physics. In principle this allows you to describe how the ball will move, cool down etc. etc.

But this is not a very practical way of doing computations. You don't want to account for what every atom in the ball is doing when you want to compute how fast it is moving. So, wat you do is you split the motion of the ball in two parts. One part is the center of mass motion of the ball, the other part is the motion of the atoms relative to the center of mass motion. And then you describe the latter in a statistical way.

Now, we can describe the motion of the ball using conservation of energy and momentum. But because momentum is linear in the velocity, the total momentum of the system equals the momentum of the center of mass of the ball. In case of energy this is not the case. First the atoms can have potential energy, but even the sum of the kinetic energies of the atoms is not the kinetic energy of the center of mass.

This means that the statistical description you get when you average out what the atoms are doing will have to take into account the energy of the atoms. Changes in this part of the energy is what we call heat.

Very nice explanation. How internal motion will became thermal radiation and equilibrates in their surroundings?

 

[Andy Resnick]

Thermal energy can interact with matter generating kinetic molecular motion. How these interactions happen?

Alex

An excellent question, except that many more phenomena are possible than simple changes to motion. Consider the flow of heat during a phase change: evaporation, or sublimation, or more general phase changes like the glass transition. Thermal energy can change viscosity, density, or optical properties of matter as well.

A useful way to model these interactions is by phenomenological constitutive relations; for example the heat capacity and the latent heat, which you may be familiar with, but things like thermo optic coefficients as well:

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TXW-4K3D36B-8&_user=10&_coverDate=06%2F28%2F2006&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1368350345&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=6fc247327a61c5bd41edf757b0ab3a9a

By going over to a mechanical-only theory (statistical mechanics is a *mechanical* theory), some information is gained: the modeling of heat as motion. However, much more information is lost: the origin of dissipation, for example. Statistical mechanics is totally inappropriate (and unable) to discuss this:

http://www.google.com/url?sa=t&sour...q3iaYL&usg=AFQjCNEYgNrfANj8Mip5H7RgeNgk-NbI4A

 

[Count Iblis]

Statistical mechanics is totally inappropriate (and unable) to discuss this

That's nonsense. Statistical mechanics is the fundamental basis of all of thermal physics. The very concept of dissipation can only be properly formulated within a statistical theory. Very general theorems like the fluctuation-dissipation theorem, Onsager relations, Green-Kubo relations etc. etc. can be derived using statistical mechanics methods.

 

[Count Iblis]

Very nice explanation. How internal motion will became thermal radiation and equilibrates in their surroundings?

This is a manifestation of the second law of thermodynamics. It is a consequence of the fact that all accessible states of a system are a priori equally likely. This is just like throwing dice, the numbers 1 to 6 are all eqally likely. It is the same with the huge number of microstates of a system.

Now when you throw two dice and you add up te numbers you will see that the total of 7 is ore likely, because there are mnore states that corresond to a sum of 7 than there are for, say, 2.

Similarly, energy is dissipated as a consequence of the number of possible states increasing in a certain direction. i.e. when the energy ids competely dissipated, there are far more states the whole system (the ball plus environment in this case) can be in than in the original state. The logarithm of te number of states is called entropy; the second law says that the total entropy can only increase.

 

[AlexB2010]

An excellent question, except that many more phenomena are possible than simple changes to motion. Consider the flow of heat during a phase change: evaporation, or sublimation, or more general phase changes like the glass transition. Thermal energy can change viscosity, density, or optical properties of matter as well.

A useful way to model these interactions is by phenomenological constitutive relations; for example the heat capacity and the latent heat, which you may be familiar with, but things like thermo optic coefficients as well:

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TXW-4K3D36B-8&_user=10&_coverDate=06%2F28%2F2006&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1368350345&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=6fc247327a61c5bd41edf757b0ab3a9a

By going over to a mechanical-only theory (statistical mechanics is a *mechanical* theory), some information is gained: the modeling of heat as motion. However, much more information is lost: the origin of dissipation, for example. Statistical mechanics is totally inappropriate (and unable) to discuss this:

http://www.google.com/url?sa=t&sour...q3iaYL&usg=AFQjCNEYgNrfANj8Mip5H7RgeNgk-NbI4A

In my simple thinking, matter has some properties that make molecules liking to be together. The likeness to be together will depend on atomic properties of molecule, on the think experiment I just use pure Iron because is a simple chemical element and all molecules will share the same properties. Increasing heat will make molecules move and being able to counteract that's properties, passing some point and the solid will be liquid and another point vapor.
In complex materials like the ones that you show, will be an increased complexity of states since there are more diverse chemical bonds changing the basic behavior of molecules.
But my question is very simple. How thermal radiation increases kinetic motion of molecules? How this waves interact to generate molecular motion?

 

[AlexB2010]

This is a manifestation of the second law of thermodynamics. It is a consequence of the fact that all accessible states of a system are a priori equally likely. This is just like throwing dice, the numbers 1 to 6 are all eqally likely. It is the same with the huge number of microstates of a system.

Now when you throw two dice and you add up te numbers you will see that the total of 7 is ore likely, because there are mnore states that corresond to a sum of 7 than there are for, say, 2.

Similarly, energy is dissipated as a consequence of the number of possible states increasing in a certain direction. i.e. when the energy ids competely dissipated, there are far more states the whole system (the ball plus environment in this case) can be in than in the original state. The logarithm of te number of states is called entropy; the second law says that the total entropy can only increase.

In giving energy to a molecule I know that the electronic energy levels will increase, making some electrons jump from their orbit, did is this electronic jumps that are responsible to increased molecular motion? The heat wave will increase energy level of electrons making then jump from one state to another.

 

[Andy Resnick]

In my simple thinking, matter has some properties that make molecules liking to be together. The likeness to be together will depend on atomic properties of molecule, on the think experiment I just use pure Iron because is a simple chemical element and all molecules will share the same properties. Increasing heat will make molecules move and being able to counteract that's properties, passing some point and the solid will be liquid and another point vapor.
In complex materials like the ones that you show, will be an increased complexity of states since there are more diverse chemical bonds changing the basic behavior of molecules.
But my question is very simple. How thermal radiation increases kinetic motion of molecules? How this waves interact to generate molecular motion?

My problem is that you have thrown all all the relevant issues in order to create an overly simplified model. Even so- your too-simple model will predict the wrong behavior for liquid water between 0 and 4 deg. C. How can you modify your simple model to account for the anomalous thermal expansion of water?

 

[AlexB2010]

My problem is that you have thrown all all the relevant issues in order to create an overly simplified model. Even so- your too-simple model will predict the wrong behavior for liquid water between 0 and 4 deg. C. How can you modify your simple model to account for the anomalous thermal expansion of water?

I am just trying to figure out some basic principles and avoiding complicating when is not really necessary. If I am guessing right this anomalous water expansion will be due to crystalline arrangement of water molecules.

 

[Andy Resnick]

I am just trying to figure out some basic principles and avoiding complicating when is not really necessary. If I am guessing right this anomalous water expansion will be due to crystalline arrangement of water molecules.

Guessing is not science. What is your evidence?

 

[AlexB2010]

Guessing is not science. What is your evidence?

You are right guessing is only part of science, first you need to see a phenomena, after you make educated guesses (hypothesis), after you experimentally test your guesses and if everything goes well make a theory or go back to phase 2, guessing.

Water isn’t a simple molecule because they are made of two different substances, and do not share the properties with neither, oxygen or hydrogen. The two hydrogen atoms are attached to oxygen in an approximate angle of 105 degrees. When water is over 4 degrees the molecular motion weakens the hydrogen bonds and the oriented intermolecular interaction don’t occur. Below 4 degrees water takes their oriented bonds and their volume increase due to the geometrical conformation of these bonds.

My intention with the think experiment is to clarify for me the relation between kinetic and thermal energy, and how one can be converted into another. So, chemical quantum electron dynamics is a little out of my focus now. If complicating the model is essential to understand the kinetic/ thermal phenomena fine, if not I am just staying in the basics.

 

[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]