Talk:Temperature/Archive 1

Early text
The actual relationship between temperature and kinetic energy is not as straightforward as is implied by this description. There is no one to one correlation between temperature and average kinetic energy for all materials. While the statements may be true for an ideal monatomic gas, they are not accurate for other materials.

I just removed the reference to "Newton's Law of cooling" because this law refers specifically to convection, while heat transfer may occur via conduction, convection or radiation. (Not to mention that Wikipedia's link to "Newton's Law of cooling incorrectly links to the law of heat conduction). Instead I directed the reader to heat for a more complete description of heat transfer.

I also removed the statement that Statistical mechanics shows that the temperature of a substance is directly related to the average speed of the atoms in that substance. While this statement is reasonably accurate for the case of a monatomic gas as is mentioned further down in the article, the situation is somewhat more complicated for more complex molecules, liquids and solids. It may be worthwhile to include a more detailed analysis of the implications of Statistical mechanics on temperature, but this discussion should be deeper in the article. Ideally, someone would refine and expand the section on "Temperature in gases" to include this discussion and possibly add further discussion for the cases of liquids and solids, although it might be better to make these into a separate page. -- Matt Stoker


 * I think the intuition that temperature has something to do with atoms moving or vibrating is an important one that even people who can't follow the more precise and technical development further down should be given right away. After all, it was a major insight to move from the concept of heat as a "fluid" to the mechanical explanation of temperature. AxelBoldt 19:15 Dec 11, 2002 (UTC)


 * I agree with your reasoning. I only objected to the previous statement because it wasn't completely accurate. I've modified the page accordingly, let me know what you think (or change it if you can do better!).Matt Stoker

WRT temperatures lower than absolute zero, I must admit I'm a little dubious. I'm no physicist, but from the description I read from the link it doesn't seem like ordinary matter can have "negative temperature" as a whole. Could somebody who actually understands the physics please comment? --Robert Merkel

I added the stuff about negative temperatures. I don't really understand much about them, I just remembered hearing once they existed, and when I saw the article saying they didn't, I went looking for some information. Here are some more refs if they are any help:,. Judging by what is said at it appears that negative temperatures can be attained in quantum systems such as lasers operating at very low temperatures. -- SJK

"Temperature is an intensive property of a system." I am likely to have a rough understanding of what "temperature" is long before I understand what one might possibly mean by "intensive property of a system." This is an example of a sort of definition that decidedly should not begin an encyclopedia article--it should be rather far down in the article, once the technical apparatus is on the table, clearly explained, so that the people who need to read this article can actually understand the technical definition. I think the purpose of an encyclopedia is not just to sum up knowledge in technical terms, but involves the more difficult task of making that knowledge as clear as possible to people who don't yet have it. Of course, some scientific concepts simply can't be explained in every encyclopedia article in which they're used, and one shouldn't try to give backround for those concepts--one should just offer pointers to the more basic articles--in many cases, perhaps just links. But in this particular case (temperature), here we have an opportunity to make a relatively simple physical concept clear, both on an intuitive, very basic high school physics level and also on more advanced levels. I'm not trying to tell ya'll to do work you don't want to do--I'm just trying to spell out my vision of what we should expect out of scientific articles about ordinary concepts. --LMS


 * I've tried to address your concern, as well as significantly expand the theoretical basis for temperature, address the concept of negative temperatures, and mention temperature measurement. Let me know what you think. Its probably still a bit on the technical side.  I think that's OK deep in the body, but we probably need to include more elementary information in the introduction. I welcome any suggestions about what could be added to this page.  --Matt Stoker


 * Rather than add to this page, I'd favor deleting from this article (and adding elsewhere, either as new articles or into the thermodynamics article) much of the material that only peripherally concerns temperature per se. Things like the formulas for heat capacity are what I mean. --BRG

There's also a first law basis for temperature. For a monatomic ideal gas: U = (3/2)kT

For more complicated systems (diatomics gases, even solids):

U = .5kT(times the number of quadratic degrees of freedom)

i.e. (5/2)kT for diatoms (not 6/2 because the smallest axis of rotation requires too much energy to access the first h/2&pi; angular momentum)

where the quadratic degrees of freedom are basically the number of ways that particles in the system can store energy (i.e. translational, rotational, vibrational, etc.). What this relies on is the idea of equipartition. Essentially, in order for the system to be in truely random motion, it has to store energy in every accessible resevior equally.

Perhaps it's worth mentioning that absolute zero was originally extrapolated from examining the x-intercept of the pressure vs. temperature line of a gas at constant volume.

It should also be mentioned in the bit on absolue zero that it can not be achieved at all. Heisenberg's uncertainty principle dicates:

&Delta;E&Delta;t > h/2&pi; &Delta;r&middot;&Delta;p > h/2&pi;

In order to drop the energy of microscopic particles to zero, implying that its momentum is zero, too, would require that it be that way for an infinite amount of time and that it takes up an infinite amount of space. The logical contradiction is that a particle at absolute zero would also have a definite position at a definition time, something that the uncertainty principle allows no particles to have. What this means is that the particles will continue to wiggle, no matter what, just enough to satisfy the above.--BlackGriffen


 * From thermodynamics it can be shown that absolute zero cannot be achieved. However, the Heisenberg's uncertainty principle doesn't exclude this possibility.  My understanding is that absolute zero is defined as that point where everything is in the ground state (the lowest energy level).  From the Particle in a box problem, we see that for particles enclosed in a "box" the ground state possesses some amount of energy above zero (the zero point energy).  Thus even at absolute zero, the system contains some energy, so an infinite amount of time and space are not required to achieve absolute zero.  You are correct that for the system to have an energy of zero, it must occupy an infinite space.  However, the system temperature can be reduced to absolute zero without reducing the energy to zero.


 * Note that the zero point energy arises because of the uncertainty principle. Since the "box" constrains the particle within a certain range of positions, the momentum and thus the energy cannot be zero. Also, note that in the ground state the wavefunction occupies the entire box, so in effect the particle size expands to fill the box.--Matt Stoker

I was just playing with the controls on my monitor, and it has a control to adjust "color temperature", taking on values from 5000K up to 9300K. Changing the temperature changes the colour of what is being displayed on the screen. What is color temperature? -- SJK


 * Read this - . It probably merits inclusion in the article, but I'll leave that to someone who properly understands it. - MMGB

That's the "color temperature" you're looking at there--5000K to 9300K--the frequency of the light being emitted affects the color that the objects actually appear. Color temperature is something that people don't often notice except through electrical equipment (one exception--at dusk, when white objects typically appear light purple). You can see experimentation with color temperature a lot in Kubrick films, for instance in _Eyes Wide Shut_ the light coming in from a window was almost always conspicuously blue, whereas the light from lamps on end tables was more orange. Arc-sodium lights give off an orange hue. Video cameras can typically adjust for color temperature by zooming into a white object and setting the white balance (telling the camera "this object is white"); the camera then adjusts the colors to show true white as white. White-balancing is necessary especially indoors under fluorescent lighting. Cinematographers can also white-balance to objects which aren't white, resulting in the color of that object being downplayed in the image. For instance, you can bring more warmth into a picture by white-balancing off something light blue, like faded blue jeans; in this way white-balancing can serve in place of a filter when a filter isn't available. --- A black hole gives of some radiation owing to the splitting of virtual matter and antimatter particles at the event horizon. This gives a black hole a temperature. The entropy of a black hole is directly proportional to its surface area and its energy is directly proportional to its radius. Hence its entropy is proportional to the square of its energy.

This through equation (7) or (8) in the main article, results in the temperature being inversly proportional to the energy, so resulting in a negative heat capacity. The temperature of a black hole formed from a star would be very low (less than the 3 Kelvin of the cosmic background radiation). Karl Palmen --- Note that equations (7) and (8) are essentially identical (one is the reciprocal of the other), since dqrev = dE.

The Science WikiProject currently focuses on this article
The Temperature article is now the focus of the WikiProject. This project aims to make article understandable by the general public, and interesting to the scientifically inclined. Please visit this project and contribute! Thanks in advance. Pcarbonn 20:53, 16 Jul 2004 (UTC)

Thermodynamics
Looking in here because it's mentioned on Peer Review. Not being much of a peer on questions of thermodynamics, I make only one fairly obvious comment: how should this relate to the Thermodynamics article? There seems to be a good deal of overlap, with perhaps a lot of detailed information here that could fit better in the other article. This one might take a seriously simple-minded approcach (not a pejorative term in a popular work), as it does in places, with an invitation to go to the more technical article for more technical information. Most of the section on the second-law definition would be a candidate for a transfer. Then again, maybe not. But surely the division of labor wants discussion. Dandrake 07:01, Aug 10, 2004 (UTC)


 * Agreed. Much of the material under the second law definition of temperature really belongs in an article about the second law of thermodynamics and/or entropy. It is redundant to argue the second law here and in another article where such an argument naturally belongs. BlackGriffen 10:52, 13 June 2007 (UTC)

Regarding internal and external temperature, black-body radiation, and other concepts
LeBofSportif. You should be more humble before acussing others of inventing concepts just because you've nevery heard of them before. The following article http://physicsweb.org/articles/world/18/3/5/1 is about the quantum behavior of C60 Buckyballs. About half-way down it describes the notion of "internal temperature." It is well-known in physics. Previous versions of this temperature page and similar pages have wrestled with this concept. Before you go back and revert the page—which is childish when done the way you do it —PLEASE read the above-referenced article. I've got twelve patents to my name and many are related to thermodynamics. What are your qualifications? Greg L 22:27, 6 June 2006 (UTC)


 * I already had a section at the bottom for this discussion. Qualifications are irrelevant, understanding of physics is what is important.
 * Firstly, the phrase "external temperature" does not enter into that article you cite. Secondly, the article is about a specific experiment where "internal temperature" is introduced in quotation marks since it is relevant to the experiment in question. "Internal temperature" does not enter main stream discussions about temperature. The vibrational, rotational and translation d.o.f will be at the same temperature in most sensible gases worth talking about in the wikipedia article. Introducing internal temperature is just confusing for readers. LeBofSportif 22:33, 6 June 2006 (UTC)

How would you like to write about how a single, hot molecule emits black-body radiation due to its internal temperature?? Readers should know that not only do molecules move about, but they internally vibrate.
 * Of course readers should know that molecules have internal degrees of freedom. But in thermal equilibrium, each degree of freedom will be at the same temperature. It's only in funny experimental situations, or at very cold extremes where the degrees of freedom will have different temperatures, and these are not worth mentioning in an overview section of an encyclopaedic article. And I really cannot allow you to leave the phrase "This motion is known as its “internal energy” or “internal temperature.”" in the article - it's just plain wrong. energy and temperature are not the same thing, and moreover, internal energy is a totally different concept, which includes translational motion. LeBofSportif 22:48, 6 June 2006 (UTC)

That's exactly the concept that should be conveyed by the article, that internal temperature of molecules is the same the same as its external temperature. Nothing in the article says otherwise. All the energy is shared as equaly as possible. How would you propose to write that the molecules themselves have internal kinetic energy that diminishes when all the other forms of thermal energy diminish? An earler version of the "Temperature" article stated as follows:

"Temperature in respect to matter is a property only of macroscopic amounts and serves to gauge the average intensity of the random actual motions of the individually mobile particulate constituents. Intraparticle motions apparently contribute only to the heat capacity.

This is a way-wrong way of thinking abou the subject. The intraparticle motions contribute to heat capacity becuase heat energy went into them. It is therefore inescapable that the molecules themselves have an intraparticle temperature as a result of “intraparticle motions.” The German researches used a perfectly sensible term (internal temperature) that's been used before elsewhere. The term perfectly describes how heat energy is distributed in a molecular system. If the "intraparticle-motions temperature" is the same as its "actual translational motions-temperature", then a substance comprising molecules is in perfect equilibrium.


 * A single molecule does not have a temperature. Nor does a single atom. Temperature is an emergent property of collections of thermally interacting constituents. It is totally unnecessary to talk about a seperate temperature for the internal degrees of freedom. One can simply observe that they contribute to the heat capacity. LeBofSportif 23:19, 6 June 2006 (UTC)


 * You are presupposing that each degree of freedom is coupled to its own kind significantly stronger than to the other degrees of freedom. This is a bad presupposition.LeBofSportif 23:24, 6 June 2006 (UTC)

Your first sentence two paragraphs up is wrong. That is precisely what Arndt et all were doing; they shot individual hot molecules. The article as I wrote it specifically singled out molecules as a unique class for having internal temperature. Your last sentence seems to be a very weak argument: (let's just observe that something happened but not talk further about why). But WHY do molecules contribute to heat capacity? Because they're vibrating due to all the kinetic energy they've internaly absorbed. How does one talk about and accurately describe this energy so the reader can understand the complete nature of the vibrations in matter?

As regards your next paragraph. I don't think anything I wrote suggests the presupposition that any of the degrees of freedom are unique or special in some way that permits them to have a different temperature. Each vibrational and rotational degree of freedom will have its own frequency and amplitude such that after achieving equilibrium (due to kinetic energy sharing via collisions and thermal photons), all forms of vibration will have the same temperature. That's why I wrote As a substance cools, all forms of heat energy simultaneously decrease in magnitude.

"Temperature" is the average heat energy in a substance. Individual molecules can be agitated in a variety of ways; that's why there is evaporation and sublimation: individual molecules can achieve more than their fair share of kinetic energy and get kicked entirely out of a particular phase. The important point that should be conveyed to the reader is that the intraparticle motions of a molecule—its energy— is shared by the rest of the system. An individual molecule can momentarily be internally agitated, but once it spends any time in contact with other molecules, it will come back into equilibrium.


 * Temperature is NOT the average heat energy in a substance.
 * I repeat. A SINGLE MOLECULE DOES NOT HAVE A TEMPERATURE. Temperature is a parameter describing the distribution of energies across a population. A population of 1 does NOT have a temperature. In a gas at thermal equilibrium, the molecules have different energies, according to maxwell-boltzmann distribution.
 * As for "How does one talk about and accurately describe this energy so the reader can understand the complete nature of the vibrations in matter?" You can talk about the vibrational and rotational energies of molecules all you want in the article. Also you said: "Each vibrational and rotational degree of freedom will have its own frequency and amplitude such that after achieving equilibrium (due to kinetic energy sharing via collisions and thermal photons), all forms of vibration will have the same temperature." EXACTLY. The same temperature. No distinction between external and internal temperatures because there is only one temperature.LeBofSportif 23:47, 6 June 2006 (UTC)

Of course individual molecules have an internal temperature. They can do so because they are comprised of a plurality of atoms that are thermally agitated with respect to one another. An individual molecule is a "population" of atoms. An individual atom can't have an internal temperature. Arndt and his associates (quoting from the article I referenced) Indeed, when we increased the internal temperature of carbon-70 molecules to above 1000 K, the contrast between the interference fringes slowly disappeared. These guys were firing individual molecules through a diffraction gratting. If there is a 1000 K flame, the CO2 molecules in it will have intraparticle motions (due to absorption of kinetic energy) that are most easily described as having an average temperature of 1000 K. Greg L 00:05, 7 June 2006 (UTC)


 * Your confidence in your error is baffling. An individual molecule cannot have a temperature since temperature is a property of an ensemble of thermally interacting constituents. I frankly don't give a damn what Arndt and his associates report but if they said the temperature of the molecules (plural) was 1000K then I've got no problem with that. It is irrelevant to the fact that a single member of an ensemble does not have a temperature. LeBofSportif 00:15, 7 June 2006 (UTC)


 * You have edited your comment since I replied. When I use "molecule" I am not considering giant covalent structures, like graphite, which do have a temperature on account of their macroscopic scale. An oxygen molecule does not. Nor does a nitrogen molecule. Buckyballs presumably come somewhere between. LeBofSportif 00:23, 7 June 2006 (UTC)


 * Please stop editing the main article while this discussion is progressing. We are not in agreement, and I will just revert you. LeBofSportif 00:26, 7 June 2006 (UTC)

Are you suggesting than a diatomic molecule does not have internal vibrations? If you concede that it does, then this is the vibrations of a population. It is why a molecule (a plurality of particles) has a temperature. Greg L 00:32, 7 June 2006 (UTC)

Have you asked anyone you trust what they think of this dispute? Greg L 00:34, 7 June 2006 (UTC)


 * I trust myself and my ability to understand simple thermal physics. A diatomic molecule has a vibrational degree of freedom. It may be excited, or it may not be. This is insufficient to be able to say "the molecule has temperature X". A diatomic molecule does not have a temperature. LeBofSportif 00:38, 7 June 2006 (UTC)


 * I have edited the article to remove the references to internal and external temperatures. Before reverting these changes (if you intend to), please find support for what you are saying and discuss it here. LeBofSportif 17:09, 7 June 2006 (UTC)

So… after researching the issue, you acknowledge that heat energy CAN go into even simple molecules. But even if a molecule is excited, you are unwilling to call that a "temerature,” right? And also, when you say "may be excited, or it may not" you are apparently suggesting that the translational energy (from bouncing around) doesn't necessarily have to  excite the internal degrees of freedom of a molecule and the atoms it is comprised of can remain still with respect to one another even though their translational temperature is substantially above absolute zero and the molecule is constantly being impacted.

There are other smart people in the world besides you and I. Many have Ph.D's on thermodynamics but you are unwilling to allow them a chance to edit my article. Have you considered actually leaving my prior version there for more than a few hours? It would be interesting to give others a chance to weigh-in but you are unwilling to permit this to occur—to allow others to weigh-in. Things that are factually incorrect (particularly on a fundamental subject like temperature) would soon be corrected by others. Squatting on an article as if you consider it your own private domain and immediately deleting others’ contributions because you refuse to accept notions like "the internal temperture of molecules," is not what Wikipedia is about. You could have waited a week for my contribution to be considered and weighed by the rest of the world's community but you have taken it upon yourself revert any edit you disagree with because of your fabulously high self-assesment of your ability to so thoroughly understand the issue.

I am going to do exactly as you suggest. I am going to run what I wrote by an authority of some sort, (probably a local University). I don't have time for tit-for-tat reverting. Such behavior is the playground of people who can't let go of things and (an may well have other “issues”). Wikipedia is a place for everyone to contribute and the behavior you've been exhibiting is the sort that just makes people throw up there hands and go elswewhere rather than put up with the frustration of having the "community" nature of Wikipedia subverted by a single individual who acts as an article's policeman.

It may be some time before you see my contribution again here. If you do, first try swallowing a small "humble pill," and then come back here to see why I put it there. And even then, you might give the contribution some time. Just sit back for a while and see how others handle it. You might actually find the process intresting to watch. Greg L 18:33, 7 June 2006 (UTC)


 * Firstly, I only edited the article because you continued to edit without responding to any of the points I made at the bottom of this talk page, which predate my decision to remove some of the content you added. I could only assume you were ignoring the talk page and so I changed back some of the points in your edits to draw attention to it.


 * Secondly, I am not squatting on the article as if it's my private domain. All I am doing is monitoring it, and removing inaccuracies when they occur. Leaving them in as a social experiment to see who else might edit the article does not interest me and that is why I see no point in leaving some of your contributions untouched for others to assess. As you say "Things that are factually incorrect (particularly on a fundamental subject like temperature) would soon be corrected...". Indeed, I correct many.


 * Thirdly, I do not have a "fabulously high self-assesment of [my] ability to so thoroughly understand the issue" but rather a realistically high self-assessment.


 * Fourthly, I do not have "other “issues”".


 * Fifthly, these are some of the reasons I edited the article:
 * Temperature is not a measure of the average kinetic energy, or if it is, it is a different measure for each material. Thus I changed the article back to saying related to average kinetic energy.
 * Use of the phrases subatomic particles and elementary particles is misleading. Quarks have no place in an introductory article about temperature.
 * I do not feel that specifics about metals and metallic thermal conductivity are appropriate in the overview for temperature, especially when the classical theory of metals is totally inadequate and a quantum theory of the harmonic crystal and electron gasses is required for an accurate description. This is too detailed for this section.
 * External temperature is not a common notion or phrase in physics.
 * Internal temperature is not a common phrase in physics.


 * Sixthly, you said:

So… after researching the issue, you acknowledge that heat energy CAN go into even simple molecules. But even if a molecule is excited, you are unwilling to call that a "temerature,” right? And also, when you say "may be excited, or it may not" you are apparently suggesting that the translational energy (from bouncing around) doesn't necessarily have to excite the internal degrees of freedom of a molecule and the atoms it is comprised of can remain still with respect to one another even though their translational temperature is substantially above absolute zero and the molecule is constantly being impacted.
 * I am absolutely saying that! The vibration energies of a diatomic molecule are quantised, and they are unlikely to be excited unless the typical thermal energy is greater than the vibration energy. Give me a hydrogen molecule from a gas at 300K and give me a hydrogen molecule from a gas at 200K and you will most likely find that there is no vibrational excitation. So how do you propose to assign an internal temperature to the molecule, given that a single molecule in equilibrium at 200K and a single molecule in equilibrium at 300K can be the same, i.e. not vibrationally excited? I shall repeat what I have said many times: temperature is only meaningful for a large number of thermally interacting constituents. Otherwise, energy is the only thing worth talking about. LeBofSportif 23:53, 7 June 2006 (UTC)

I'm having the temperature article reviewed by two Ph.D.s. One is an expert in thermodymanics and teaches at a university. The other is a chemist with whom I worked on fuel cells for seven years. Let's see what these two have to say shall we? Greg L 07:30, 8 June 2006 (UTC)

the same as above continued - october
Well, that was in June and now it's October. Over the months, the issue of whether or not molecules have an "internal temperature" has been reviewed by numerous Ph.D.s. Among them has been Dan Cole (who authored three published papers on absolute zero, one of which is Derivation of the classical electromagnetic zero-point radiation spectrum via a classical thermodynamic operation involving van der Waals forces, Daniel C. Cole, Physical Review A, Third Series 42, Number 4, 15 August 1990, Pg. 1847–1862.) In simple terms, the response from all the experts could be characterized as "Well… Duhh!” The vast majority of the correspondence with the experts was usually on a topic of greater importance: absolute zero and the nuances of the phenomenon at that temperature.  The harshest criticism from any Ph.D. was from someone who commented with something along the lines of "The internal temperature of molecules are always, on average, the same as the substance's temperature, so why make the distinction and risk confusing the issue?" So I re-wrote the paragraph to avoid any confusion.

In hindsight, this dispute should never have occurred: “temperature” is always the average kinetic energy of particle motion. If a molecule (as distinct from a monatomic gas) internally absorbs kinetic energy (heat), that kinetic energy is precisely associated with a particular internal temperature. The thermodynamic temperature of any plurality of fundamental particles of nature (atoms in a monatomic gas or 2+ atoms in a molecule) is always a proportional function of the kinetic energy of their movements. Molecular-based substances have greater specific heat than the monatomic gases precisely because the molecules absorb heat energy internally. Nothing can absorb heat energy and not attain a greater temperature. Nothing. Never. In a nutshell: If you heat up a sample of any molecular-based gas to, say, 1000 °C, and then release single molecules across a gap, the molecules will first emerge at 1000 °C and will immediately begin radiative cooling as they emit photons with a black-body spectrum of 1000 °C (on average). If it's monatomic atoms flying across the gap, they just fly and don't radiate; atoms of course, can't have an internal temperature.

I don't have any inclination to argue this point any further because it is so elementary. I did, however, feel compelled to make this "discussion" entry so others might not get thrown off track. If you still feel that molecules can't have an internal temperature, don't bother arguing about how you are right because you "understand thermodynamics" (like it's some sort of X-men power); go find an authoritative source (hopefully a reputable, published Ph.D. in thermodynamics) who supports your argument and cite the individual. And please save the “ARGH!!! MISCONCEPTION ALERT” stuff for some other forum. This behavior comes across as if you are such an awesome expert on a subject, that you just can't contain your shear frustration of how others can be such imbeciles. I just gave up on this "Temperature" article and went elsewhere to make contributions rather than put up with the frustration of all that. Greg L 21:48, 13 October 2006 (UTC)


 * I think a point that is being somewhat ignored is the fact that at equilibrium, at a given temperature, there is a distribution of individual particle energies. If we have a gas consisting of a large number of point particles (no internal degrees of freedom) at equilibrium, and at temperature T, then the average energy of each particle will be E=3kT/2. But there will be some particles with half that energy, some with twice that energy.


 * If the gas is isolated - not connected in any way to an outside system, then the total energy will be conserved, and the temperature will be exactly 2E/3k where E is the total energy divided by the number of particles - the average energy per particle. If we have just one particle, then ok, we could define its temperature to be 2E/3k where E is its (conserved) energy.


 * But this is only good for a single particle in isolation. If that particle is part of a larger system, then the whole idea of "its" temperature breaks down. Its energy will change radically with every collision, and so will its "temperature" defined by the above equation. At one instant it may have a "temperature" of 2T and a nanosecond later it may have a "temperature" of T/2. The formal definition of the "temperature" of a single particle has become useless. You might average its energy over time to get the time-averaged "temperature" of the single particle but this will be equal to the temperature of the whole gas, and so it wont be anything special.


 * If the particles have internal degrees of freedom, like vibrational modes, then each will have an additional energy of kT/2. (I'm assuming that temperature is high enough so that there are no quantum effects). With regard to the "internal temperature" of molecules, the same ideas mentioned above hold. The internal vibrational energy of an individual molecule will be jumping around with each collision just as drastically as its kinetic energy. The concept of the internal temperature of a single molecule only becomes useful if you average its internal energy over time, or average it over a large number of such molecules, both of which will give you the same answer, assuming equilibrium. PAR 03:10, 14 October 2006 (UTC)

subconversation 1

 * PAR: You are exactly right. It's nice to have support on the issue that molecules have an internal temperature.  But no, the concept that individual molecules will have a distribution of energies and velocities hasn't been lost in the argument (at least mine). What you're describing is the Maxwell–Boltzmann distribution of speeds. That's why I wrote “…emit photons with a black-body spectrum of 1000 °C (on average). ”   Greg L 06:08, 14 October 2006 (UTC)


 * Well, my support is not total - There is the "no quantum effects" condition on the above discussion. That means the internal degrees of freedom will be strongly linked to the kinetic degrees of freedom so that the "internal temperature" will be the same as the "external" temperature (in equilibrium). But out of equilibrium, if the rate of energy flow between internal and kinetic degrees of freedom were different, the temperatures could be different. Also, I think it is a valid statement to say that a single particle has no temperature. It seems valid to say that each degree of freedom can have its own temperature, which is 2E/k where E is the average energy in that degree of freedom. Average over all particles, that is. An individual degree of freedom in an individual particle has no temperature. The statement that "temperature is not proportional to the average energy" means that its not proportional to the average kinetic energy, because different particles have different internal degrees of freedom availiable to carry more or less energy. PAR 14:12, 14 October 2006 (UTC)


 * PAR: All sorts of special conditions like "out of equilibrium" can be considered. The issue is strictly whether or not molecules have an internal temperature. One simply can't prevail with any argument predicated on the logic that “just one molecule can't have an internal temperature so none of them do.”  Asserting that “[a]n individual degree of freedom in an individual particle [molecule] has no temperature ” is like asserting that 50+ helium atoms can have a "temperature" (average kinetic energy) but two helium atoms can't have an average kinetic energy. Note that a single "degree of freedom" in a molecule requires at least two atoms (a plurality of fundamental particles).  If I told you a single helium atom had a speed of 4780 meters per second, you could conclude that the bulk sample from which that atom originated had a most probable temperature of 5500 K. It's only a likelihood; not a guarantee.  If ten such atoms were isolated within a container with a wall temperature of 5500 K (thus the system was in equilibrium), and if one waited for several million collisions (less than a second of observation time), one could properly draw a statistically valid conclusion about the temperature of the sample based on their average speed (and thus, their average kinetic energy). Such a sample obeys the Maxwell-Boltzman distribution as regards the likelihood of any particular atom having a particular velocity at any given instant. One might correctly argue that one can't have an average kinetic energy for a single helium atom which is in equilibrium with its environment (although I think even that statement isn't true), but it's certainly not true to say that two atoms (a plurality) can't have an average kinetic energy (temperature). You are correct when you state that “different particles [atoms] have different internal degrees of freedom [within a molecule]” . It's true that different degrees of freedom within molecules radiate black-body radiation, and when summed, this spectrum doesn't take the form of the classic black-body curve. It would be a mistake to conclude that this means that individual molecules can't have an internal temperature.  Let's take the example of a molecule that has only one degree of internal freedom.  Clearly, there is only one kinetic energy to measure so even by your argument, it would have an internal temperature.  This concept doesn't break down just because there are two or three degrees of freedom.  If one measured the effective black-body temperature of the spectra for each individual molecule that had 2+ degrees of freedom, one would find that their internal temperatures (average kinetic energy within a single molecule) were, averaged across many molecules, the exact same temperature as their translational temperatures. There is no magical minimum quantity of the fundamental particles of nature at which a temperature can no longer be assigned; one can always rely on the probabilities.


 * We're getting bogged down in minutia intended to show off how damn smart we are. The only two points I've been trying to make (and what previously got deleted ad nauseam) are these two statements: 1) Heat energy is stored in molecules’ internal motions which gives them an internal temperature. 2) The internal temperature of the molecules in any bulk quantity of a substance are, on average, equal to the temperature of their translational motions. This is well established and elementary.  Greg L 23:40, 14 October 2006 (UTC)

subconversation 2

 * Greg L, I disagree with you again. For brevity, I shall point out only one of your errors - most of your others are repeats and you can find their refutations above somewhere. Error no 1: you said "Nothing can absorb heat energy and not attain a greater temperature. Nothing. Never." I say "Take a glass of water with some ice in it and let it absorb some heat energy. Observe whether it attains a greater temperature." If you concede this obvious point, then we can move onto some less elementary thermodynamics. LeBofSportif 14:16, 14 October 2006 (UTC)


 * Who do you think you're trying to kid? You're skirting the point and you know it. Yes, you correctly pointed out a hole in my written argument. I should have said "with the exception of phase changes and ionization etc. etc". This concept isn't lost on me and I doubt it's lost on most contributors who might try to make improvements to this article.  You know full well what the point of this whole argument had been about: Whether or not molecules have an internal temperature or not.  At first you fed me a bunch of your “ARGH” stuff, squatted on the Temperature article, and kept reverting anyone who added words to that effect.  After reading some supporting documents I posted, you (sort of) conceded that *maybe* big complex molecules might have an internal temperature (as if something the laws of physics are different for 30 atoms and two atoms). I have zero interest in the no-doubt impossible effort of trying to get you to concede for the record that molecules have internal temperatures. The whole point of lowering myself into this muck was to get it into the talk page that it is a simple fact (I can't say “indisputable” fact because you will dispute it) that molecules have internal temperatures.  Your argument (paragraph #7 from the top in this section) was “A single molecule does not have a temperature.” Unless the molecule is at T=0, this is simply flat wrong.  All who come to this discussion page have a right to know this so this article and others on the Web can be improved without misinformation being propagated.


 * These two statements are true: 1) Heat energy is stored in molecules’ internal motions which gives them an internal temperature. 2) The internal temperature of the molecules in any bulk quantity of a substance are, on average, equal to the temperature of their translational motions.


 * It's a simple concept: Whether it's a gas sample of a molecule with 60 atoms or a simple diatomic gas like oxygen, if the sample has a temperature of, for instance, 1000 kelvin, the individual molecules have, on average, internal temperatures of 1000 kelvin. Also, as I stated above, releasing these molecules individually for inspection will reveal that they radiate black-body radiation (as shown in illustration b here). Averaging a statistically significant quantity will reveal that when first released from the 1000 kelvin sample, their black-body spectrum will be declining on a curve indicating that when they were first released from the sample, they had an internal temperature of 1000 K.  This is an important concept and is at the heart of molecular interferometry (as written about in Physics World: Probing the limits of the quantum world Markus Arndt et al.). It's also an important underpinning in understanding where heat energy goes. Greg L 20:01, 14 October 2006 (UTC)


 * The simplest approach for me is to attack your two assertions. Since 2) depends on the existence of internal temperature (which I dispute), it depends on 1), so I shall only attack 1). My attack on 1) is thus:


 * Given an isolated hydrogen molecule which is in its rotational first-excited state and its vibrational groundstate, what is its internal temperature? And what is the general method whereby we determine the internal temperature of similar molecules? These both boil down to the question: what, precisely, is the definition of your 'internal temperature'? LeBofSportif 01:23, 15 October 2006 (UTC)

Regarding some of Greg's quotes made above, I think this would also give my answer to LeBofSportif's questions as well. In short, the internal temperature of molecules does exist, but only for large numbers of molecules, just as temperature may be assigned to the "external" degrees of freedom for a large number of molecules.

"'There is no magical minimum quantity of the fundamental particles of nature at which a temperature can no longer be assigned; one can always rely on the probabilities.'"

But you have to somehow measure these probabilities. And you do this with a thermometer, which averages over a large number of particles. Even if you wish to assign a temperature to a single particle as the temperature of the population it came from, that temperature must be measured by averaging over the population. If you wish to assign a temperature to a single particle as the temperature of the population it came from, go ahead, but its never done in any literature I have ever seen. We agree its temperature cannot be related to its individual energy right? Because that would be jumping all over the place.

"'Heat energy is stored in molecules’ internal motions which gives them an internal temperature.'"

Yes, as long as we average their internal energies over a large number of molecules which are in equilibrium, just like we do with the three translational degrees of freedom.

"'The internal temperature of the molecules in any bulk quantity of a substance are, on average, equal to the temperature of their translational motions. This is well established and elementary.'"

Yes, at equilibrium. Otherwise maybe not.

Also, individual molecules do not emit black body radiation. They emit single photons with a specific single energy. If the emitting molecules are in thermal equilibrium and the photons are being constantly emitted and reabsorbed, then these photons will have a distribution of energies, just like the kinetic energy of the molecules have a distribution of energies. Like the molecules, this photon distribution will be an equilibrium distribution. An equilibrium distribution of photon energies is called black body radiation, while an equilibrium distribution of molecular kinetic energies is called a Maxwell-Boltzmann distribution. PAR 02:05, 15 October 2006 (UTC)


 * Par: You are wrong in both sentences where you state “Also, individual molecules do not emit black body radiation. They emit single photons with a specific single energy.” Read again Physics World: Probing the limits of the quantum world Markus Arndt et al.:


 * Again, this is paper deals with molecular interferometry where they fire individual, hot molecules through a slit. The individual molecules are emitting photons. Beyond a certain temperature, the wavelength of the photons becomes less than twice the distance between the separated atomic wavefronts and this carries sufficient "which path" information to destroy the interference pattern in the inteferometer.  Anytime photons are emitted due to the collisions of atomic orbitals due to temperature-induced vibrational motion, it is called black-body radiation.  The only exceptions I know of are ZPE-related photons (mostly a T=0 issue).  True, any single photon will be at a single wavelength.  However, a heated molecule doesn't cool all the way to T=0 with that one emission.  If you capture the spectrum of the emmissions from an individual molecule, it takes the form of a cooling black-body. And even if you capture and observe only one black-body photon and can't establish a curve, that photon is still called a “black-body photon”.  You should read the article. I found it very interesting.   Greg L 02:34, 15 October 2006 (UTC)


 * Looking at that article, I searched on "black" and found "just as a little block of hot solid material glows, emits black-body radiation and cools through evaporation." I couldn't find anything that said an individual molecule emits black body radiation. In the example you gave, I could see how multiple emissions would approximate a black body spectrum, but the error would be roughly 1 over the square root of the number of emissions. I have never heard of a "black body photon", but I think we agree that if you specify the energy and spin of a photon, you have specified everything that can be known about that photon by measuring that photon alone. If I tell you the energy and spin of a photon, you cannot tell me if it is a black body photon. In order to be a black body photon, you must know that it came from a population of other photons in equilibrium. It's "black body" nature is not a piece of information that it carries with it. PAR 03:04, 15 October 2006 (UTC)


 * PAR: I think you're getting too hung up with quantum stuff. I think it's not so complicated.  Everything that's hotter than absolute zero always emits black-body radiation. Even when the emitting object is not a perfect black body (as happens with anything that's not in equilibrium and even many things that are such as tungsten filaments), the radiation being emitted is still black-body radiation. Any given photon needn't be classified as a "black-body photon" by analyzing a particular property of a given photon; black-body photons can be classified as such simply by having knowledge of why they were emitted. Fluorescent emissions aren't black-body photons and neither are lasers. But pretty much anything a soldier every sees in a thermal imaging sight is black-body radiation.  The cosmic background radiation is black-body radiation.  Of course, if you receive unknown photons and a spectrum analysis shows the radiation has a black-body distribution, that tells you what you need to know about the source without directly confirming the fact.  Black-body radiation is simply the product of collisions of atomic orbitals surrounding atoms as a result of thermal agitation.  That's why the black-body spectrum and the Maxwell distribution appear so similar.  I spent a lot of time corresponding with Ph.D. experts on black-body and ZPE photon subjects and still don't pretend to be an expert on the subject; particularly since most of my correspondence was in making sure what was going on at absolute zero was down pat.  So maybe I'm about to learn something new.  Why is the spin property of a photon a necessary bit of information to have when it comes to black-body radiation? No speculation now… quote real hard facts.   Greg L 05:44, 15 October 2006 (UTC)

Sorry to interrupt...Greg, could you address my question above about the internal temperature of an isolated hydrogen atom? The way you answer this will likely explain your opinions on what counts as black body radiation and that sort of thing, so we may kill 2 birds with one stone, so to speak. Also, you say to PAR that "you're getting too hung up on quantum stuff". The world is a quantum world - a proper understanding of statistical mechanics and how it applies to the real world requires the "quantum stuff". Some of the points I make only make sense if you accept the "quantum stuff". Absolute zero, the quenching of heat capacities, the third law of thermodynamics - these have their proper context when viewed with the "quantum stuff". LeBofSportif 11:07, 15 October 2006 (UTC)


 * LeBofSportif : Quantum stuff is important. Yes. But there's no point arguing such details when one starts off with statements like “Temperature is not average kinetic energy. This is a misconception, and a dangerous one.” What I'm saying is that all atomic and molecular motion (whether internal degrees of freedom or translational motion), at temperatures greater than T=0, has a kinetic energy associated with it as fixed by the Boltzmann constant (Kb = 1.380 6505(24) × 10–23 J/K). I'm also saying that for a system in equilibrium, the internal temperature of molecules (2+ atoms) are always—on average for a statistically significant quantity of molecules—equal to the temperature of their translational motions. This is the notion that you and I first collided on because you violently objected (“ARGH!!!”) to the notion that molecules can have an internal temperature. Or course they do!  And this internal temperature is always the same, on average, as the substance's temperature. I'm also saying that for systems in equilibrium, all internal and external thermal motion of molecules and all external thermal motion of atoms, while this motion has a mean temperature (and a mean kinetic energy associated with it), actually occurs across a range of velocities as described by the Maxwell distribution. I'm also saying that all atom-based matter (monatomic atoms and molecular-based substances) radiate thermal photons whenever electron orbitals are perturbed by thermal agitation and the relationship of temperature to the peak emission wavelength of the resulting black-body radiation is given by Wien's displacement law constant b (2.897 7685(51) × 10–3 m K). I'm also saying that individual molecules with temperatures greater than T=0 radiate thermal photons.  Greg L 17:05, 16 October 2006 (UTC)

And for the record, I'm with PAR on this one. Call them thermal photons if you want, but not black-body photons, since black-body is a term which can only be meaningfully applied to describe a distribution of photons. Any one-photon can never be a black-body photon. It is like saying "I measured my classmates' heights and discovered that each person's height is normally distributed." When read carefully, what I just said in quotes is nonsense. If you want to express the "reason" it was emitted, call it a thermal photon. And since you are making claims about molecules, your claims should be valid for smaller molecules like hydrogren and chlorine, not just whatever beasts they were using in that experiment you keep referencing, so perhaps the examples you give could be framed in terms of these molecules. LeBofSportif 11:18, 15 October 2006 (UTC)


 * I don't have a problem with that. However, the experts I've been corresponding with have used "black body" more broadly than that.  Of course, that was in the context of drawing a distinction to ZPE photons (which is a T=0 phenomenon).  At temperatures greater than T=0, they've called them "black-body photons."  Maybe they shouldn't and are just being a little loose when engaging in less rigorous e-mail correspondence. I'll read some of the papers I've received to see what terminology is used for individual photons.   Greg L 17:05, 16 October 2006 (UTC)

I know this discussion is a bit old, but I can't help but respond to Greg L.'s continued appeal to his “experts”. I am a PhD Physical Chemist – my graduate work was with statistical thermodynamics. I'm intimately acquainted with this topic, and LeBofSportif is quite right that a single molecule (or a handful, for that matter) does not have “temperature”. Temperature is a description of an ensemble quantity. Calling excited states (rotational, translational, vibrational, whatever) temperature is akin to calling an individual's lifespan “life expectancy”. It may seem like a silly play in semantics, but folks in my line of work spend a great deal of time worrying about correctly reproducing ensemble averages, like temperature, from properties of individual atoms and molecules. The difference is profound - I really can't understand why your “experts” were so cavalier as to conflate the two. But, the energy of an isolated molecule is no more “temperature” than my birth-date is “a generation”, or my car is “traffic” - period.134.253.26.9 (talk) —Preceding comment was added at 22:27, 28 February 2008 (UTC)

The natural degrees of temperature
Maybe we should include a section (or at least mention) on the temperature scale based on natural units (the gravitational constant, planck distance, speed of light). Since this degree (absolute zero at one end, infinant temperature at the other, considered the original temperature of the big bang) is too large, bringing it down by a factor of 32, we get degrees roughly 3 degrees farenheit apart. Natural units are thought to be more universal, and less arbitrary than conventional units (ie, 100 units between the point at which the vapor pressure of H2O reaches 101.3 kPa, and where it condenses, again only at 101.3 kPa), as they are based on physical constants. With this system, the temperature of freezing water comes to around 193 degrees, and room temperature to 208, and the boiling point to 264. If we fix a parallel system to the freezing point as in Kelvin and Celsius, we get room temperature to be 15 degrees, and boiling to be 71. These units could be called centigrade (discarded by the metric system), or degrees Planck.

On another note; this may be misunderstanding on my part, but it seems to me that temperature is just energy/volume, as it is the average kinetic energy. This seems to make sense, but no where, on the internet or in chemistry, have I ever heard this even contemplated. GWC 10:50 EST 25 Oct 2004


 * It is the Heat that is an Job and hence an Energy. Changing the temperature of a mass (or of a volume) of a gas (or of an object) give a difference of the energy.
 * Note that what can be measure is only difference of energy not energy. This difference of energy is proportional to the mass (or volume) to the difference of temperatura. The proportional factor (that depends on the material) is called specific heat. That temperature and kinetic energy are related (and notably in an direct way) is the result of the [kinetic theory of gas]], but this is true only for a gas (indeed only for an ideal gas). Temperature and work (or energy) and the corresponding units of measure, kelvin and joule, are related by the Boltzmann constant.
 * Note that what can be measure is only difference of energy not energy. This difference of energy is proportional to the mass (or volume) to the difference of temperatura. The proportional factor (that depends on the material) is called specific heat. That temperature and kinetic energy are related (and notably in an direct way) is the result of the [kinetic theory of gas]], but this is true only for a gas (indeed only for an ideal gas). Temperature and work (or energy) and the corresponding units of measure, kelvin and joule, are related by the Boltzmann constant.


 * Of course you can change the definition of the unit of measure to have the Boltzmann constant equal to one.
 * This is related to the above discussion on natural unit (Since the natural unit sistem is not easy for those who have not had theoretical physics studies, I suggest not to include this. It is an Encyclopedia, not a book of advanced physics.) The problem of the connection (and the problem of terminology) between temperature and energy is similar to the one between mass and energy.
 * If you do not understand the difference between temperature and heat (and heat is energy as shown by Joule's experiments) consider the diference between electrical voltage and electrical current (this parallelism is not easy to see, indeed). AnyFile 13:50, 12 Nov 2004 (UTC)
 * If you do not understand the difference between temperature and heat (and heat is energy as shown by Joule's experiments) consider the diference between electrical voltage and electrical current (this parallelism is not easy to see, indeed). AnyFile 13:50, 12 Nov 2004 (UTC)
 * If you do not understand the difference between temperature and heat (and heat is energy as shown by Joule's experiments) consider the diference between electrical voltage and electrical current (this parallelism is not easy to see, indeed). AnyFile 13:50, 12 Nov 2004 (UTC)

Explain Counterintuitiveness
In part of the dicussion this is said: "This average energy is independent of particle mass, which seems counterintuitive to many people. Although the temperature is related to the average kinetic energy of the particles in a gas, each particle has its own energy which may or may not correspond to the average." Perhaps an explanation of why this relationship exists would be useful? I'd be happy to write it, and so I'll do it here (in the discussion, so that if you like it you can just copy/paste). (I'm afraid I'm not too familiar with the Wiki Markup for math, so you'll probably have to clean that up a bit but otherwise it should be ok.) Right after the sentence about counter-intuitiveness, insert this:


 * However, after an examination of some basic physics equations it makes perfect sense. The second law of thermodynamics states that two systems when interacting with each other will reach the same average energy. Temperature is a measure of the average kinetic energy of a system. The formula for the kinetic energy of an object (in this case a molecule) is:
 * $$\ K_E=\frac{1}{2} m v^2$$
 * So a particle of greater mass (say a neon atom relative to a hydrogen molecule) will move slower than a lighter counterpart, but will have the same average energy. This average energy is independent of the mass because of the nature of a gas, all particles are in random motion with collisions with other gas molecules, solid objects that may be in the area and the container itself (if there is one). A visual illustration of this from Oklahoma State University makes the point more clear. Not all the particles in the container have the velocity, regardless of whether there are particles of more than one mass in the container, but the average kinetic energy is the same because of the ideal gas law. EagleFalconn 19:13, 30 Nov 2004 (UTC)

I cleaned up the formula, but I disapprove that "Temperature is a measure of the average kinetic energy of a system". Temperature is the increment of energy per unit of increment of entropy.
 * $$\ T=\frac{dE}{dS}$$

If the entropy is proportionate to the number of particles in the system, then the temperature is proportionate to the average energy per particle, not the average energy of the system. Light molecules move fast, heavy molecules move less fast, and a cannon ball hanging in a wire from the ceiling moves too slowly for it to be noticed (when the transient oscillatory motion has faded away and the system is in a state of thermodynamic equilibrium), even if all these particles have the same average kinetic energy. Bo Jacoby 09:24, 14 September 2005 (UTC)

link requires subscription
The third link down in External Links requires a subscription to view the site.

Is there a policy against such links? SOP?


 * I don't know about any policy (but I've been away from WP for a while). Anyway I could not reach the site either and I got a spyware warning. So I'll delete it and keep it here just in case someone disagrees. The link was:Nanotubes may have no 'temperature'. Also, something with "no temperature" sounds weird...--Nabla 23:29, 2005 Apr 28 (UTC)
 * Here's a non-subscription mirror. (And it is weird...) Rd232 17:13, 31 July 2005 (UTC)

Negative temperature
This section read to me as non-sensical, until I realized it was talking about spin and other quantum weirdness. If it's accurate, it needs references, or else it's likely to be removed as inaccurate. It could also be written to be clearer. What does it mean for something to have "hotter than infinite temperature"? Most people would assume that this means that it's very, very hot, as in hot to the touch, but in a non-sensical way. We think of temperature as being proportional to kinetic energy; does spin state really count toward kinetic energy? And if the total kinetic energy of a system is finite, it might be sensible to talk about a negative "spin temperature" but not a negative (or greater-than-infinite) overall temperature, right? -- Beland 09:17, 10 July 2005 (UTC)
 * About you question the answer is yes. Sometimes various degrees of freedom of a system have a different temperature, if their energy coupling is weak. So that here the heat would flow away from the spins, not only toward the outside of the system, but also towards other degrees of freedom of the system like the kinetic energies. And yes, negative temperature does exist. Here's a link with pretty clear explanations:

http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/neg_temperature.htmlThorinMuglindir 23:18, 4 November 2005 (UTC)

Oh, and I don't think we need to be that cautious with the question of neg temp and the definitions of temperature. The defintion of temperature from heat and entropy works, and the Maxwell Boltzman weights, too. The 0th law definition, I don't know, but anyway this one is just the one you use when you can't have statistical mechanics and for some reason you have to stick with just thermodynamics. Which means it's a bit outdated.ThorinMuglindir 23:28, 4 November 2005 (UTC)


 * The zeroth law isn't applicable; if a system initially at negative temperature is brought into thermal equilibrium with a gas, the resulting equilibrium temperature will always be positive. So you can't use a gas to measure negative temperatures. Melchoir 03:08, 11 February 2006 (UTC)

I run NMR (nuclear magnetic resonance) everyday. So, if I apply to the sample a 90 degree proton pulse which makes the populations of the two spin states equal, I have achieved a temperature of infinity? And if I apply a 180 degree pulse, I have achieved a negative absolute temperature? But when I look at the temperature reading, it still reads 25 oC, exactly what I have set. Furthermore, many people have taken MRI (magnetic resonance imaging) scans of their brain. And they still behave normal. I mean their brain is not fried. Both NMR and MRI are known as non-destructive analytical techniques, meaning samples are not destroyed by these analytical methods. This cannot be true if the samples have experienced a temperature of infinity. MRI is called MRI and not NMRI (nuclear magnetic resonance imaging) is because the perceived public fear of the word "nuclear" makes the word dropped. Perhaps, it's time to clarify the "negative temperature" because people start to question about MRI. 24.74.135.9 22:52, 18 April 2007 (UTC)

Correct usage
Anybody have a "definitive" on the correct usage of the ° sign? I notice that most WP pages have it as X °C but others are X° C. I'm looking at Enviroment Canada's manuals and it uses the second version in older manuals but both versions on different pages in some of the newer manuals.CambridgeBayWeather 19:46, 31 July 2005 (UTC)


 * It has been discussed some on Wikipedia talk:Manual of Style (dates and numbers). NIST Special Publication 811 calls for the space (there is no space, however, in 30° of arc).  See the part of the talk page where NPL and IUPAC and BIPM and ISO are discussed, all supporting the space, with the degree sign butting up to the C.  There may be some style guides still supporting omitting the space (the one you haven't mentioned, x°C).  I doubt that any modern guide supports using the space but butting the degree sign to the number rather than to the letter identifying the temperature scale. Gene Nygaard 20:28:19, 2005-07-31 (UTC)


 * Thanks. I fixed the one page where I'd put it in the wrong way.CambridgeBayWeather 20:42, 31 July 2005 (UTC)

I, too prefer “x °C” to “x° C”, but I am not pedantic about it. My pet peeve in this regard is the need to distinguish “°C” from “C°”. That is, “10 °C” is a temperature, but “10 C°” is a temperature difference. Everyone knows this, but almost no one observes it. Georgeisomorphism (talk) 22:46, 17 February 2009 (UTC)

Personally, I tend use just "C" (and "K"): "°C" effectively amount to "degrees centigrade", which can only be justifed as a disambiguation from e.g. "degrees Fahrenheit", where "degree" is the original unit. The rational thing to do is to declare "C" a unit in its own right. Similarly, we do not use phrases like "scale-marks kilogram", but rather just "kilogram". 88.77.129.121 (talk) 04:41, 24 December 2009 (UTC)

vacuum
can a vacuum have an associated temperature? i had a friend who argued the space would feel "freezing", whereas i said that it would be both extremely cold (due to evapouration from bare skin to vaccum) and extremely hot (from ambiant radiation and solar currents - instant sunburn), and oddly neither at the same time, as there isnt enough matter for a temperature to be possible. was i close? mastodon 16:29, 13 December 2005 (UTC)


 * I'd say temperature is where an object reaches thermal equilibrium (which, in space, would be 2.7 Kelvins due to thermal radiation far away from the sun or in a shadow, or much higher while exposed to sunlight). Astronomers, however, might equate it to the average kinetic energy of the few random particles zipping about (which can be up to 10^7 Kelvins... though it wouldn't even be felt). Personally, I'd use the former; having temperature disassociated with "hot" and "cold" as it does for gasses in deep space seems counter-intuitive to me.  Issues of evaporation and energy loss due to boiling are important, of course, but they can't really be considered part of the temperature itself.  --Obsidian-fox 08:06, 11 February 2006 (UTC)

I think the section in the article on temperature of a vacuum is muddled, and inaccurate. I will change it unless anyone justifies what it says here. LeBofSportif 09:55, 13 May 2006 (UTC)

Why can't temperature be expressed in Hz?
Keep in mind that I'm only a grade 11 student, but it seems to me that if heat energy is caused by the vibration of particles, then it could theoeretically be measured in terms of cycles per second, yes?

I don't know...Maybe the frequency doesn't actually change as particles vibrate more, because even though they vibrate faster, the distance required to be travelled to make one full cycle also increases, and so the time would as well--in other words, it balances out, and the frequency always stays the same. I honestly have no idea. It's just a thought, I guess. Remember, I'm a grade 11 student, so please don't chastise me for these seemingly unintelligent thoughts.


 * First of all, heat is not about vibration of particles. It is about the kinetic energy of the particles. HOWEVER, in a solid, where particles are constrained to a lattice, the kinetic energy is mainly vibrational - hence your misconception. In a gas, where there is no such constraint, the particles don't vibrate, they can just move fast.


 * Secondly, you are correct that in most cases, vibrational frequency doesn't change as the energy changes. Think of a pendulum - the frequency is the same even when you vary the amplitude. LeBofSportif 15:40, 17 May 2006 (UTC)

In fact, temperature is often expressed in Hz, but not for the reasons suggested (vibrational frequency of particles). Temperature is rigorously defined using quantum mechanical theory, where microscopic objects (electrons, atoms, photons, etc.) are described as existing in discrete states of different energies. For an ensemble of these objects, temperature is a measure of how these objects are distributed over all their possible energy states. When the states are widely separated in energy, high temperatures are needed to "jostle" the objects from one quanta to another. In almost all cases, a transition from one state to another is accompanied by the emission (or absorption) of a photon (light particle) of energy. That's why things glow when they are hot, and the hotter something is, the higher the frequency of the emitted light. For example, blue flames are hotter than red flames, just as blue light (~6x10^14 Hz) is at a higher frequency than red light (~5x10^14 Hz). And even at room temperature, objects "glow" in the lower-frequency infrared spectrum (thus night-vision goggles). Even outer space glows, but it is so cold that that it only contains microwaves (1x10^10 Hz). This relationship between temperature and frequency can be quantified, and it turns out to be: T = hf/k, where h is Planck's constant, k is the Boltzmann constant, and f is frequency. But remember that this frequency is not how fast a particle physically vibrates; rather, it's the frequency of the light emitted when the particle makes a thermal transition between two quantized states.

Average Kinetic Energy
Temperature is not average kinetic energy. This is a misconception, and a dangerous one. It appears on the talk page in a number of places, and now someone has changed the article to read "temperature is a measure of the average kinetic energy..." If this was true, then all gases would have the same heat capacity. They don't, therefore it is not true. Temperature is related to the average kinetic energy, in that, for a given body, an increase in temperature will be proportional to the increase in energy. The proportionality is different for different bodies. When two things are brought into contact, equilibrium will be when they have the same temperature, not the same K.E.LeBofSportif 10:17, 4 June 2006 (UTC)


 * Hey, what d’ya know, you're right! Temperature isn't “average kinetic energy ”; it's as written in the article: temperature is a function of the average kinetic energy of the particles’ translational motions. Please don't change it! In fact LeBofSportif, all the monatomic gases have the exact same molar heat capacity: 20.7862 J mol-1 K-1 (see table). Molecular-based gases always have greater specific heat capacities. And they typically vary from each other because they have differing abilities to absorb heat energy into their internal degrees of freedom.  Also, hydrogen bonding (as with ethanol, water and ammonia) strongly influences molar heat capacity. At 25 °C, the mean kinetic energy of the translational motion of each molecule (it doesn't matter what kind) is 6.174<FONT SIZE="-1"> </FONT>61 × 10–21 Joule. Differences in molecular weight just make for different speeds; the kinetic energy of translational motion stays the same.  For instance, at 4200 K, helium atoms will be moving at a mean speed (not vector-isolated velocity) of 5,176.55 m/s whereas uranium gas atoms will be moving much slower at a mean speed 671.2701 m/s, but both will have exactly the same kinetic energy bound in their translational speeds: 8.905<FONT SIZE="-1"> </FONT>20 × 10–20 J if they’re both at 4200 K. I just don't know how anyone who's made Wikipedia contributions to technical articles could utter a statement like “<font color="#721F01">If this was true, then all gases would have the same heat capacity. They don't, therefore it is not true. ” The only thing that's "dangerous" is confusing heat capacity with temperature and kinetic energy.   Greg L 00:48, 15 October 2006 (UTC)


 * "Temperature is a function of the average kinetic energy" - fine, of course this is true. Except, from a pedagogical point of view it is problematic since you would need a different function for each substance you consider. On the 4th of june, when I wrote that comment, the article contained the line "temperature is a measure of the average kinetic energy". In my world, this implies that there exists some way of using temperature to work out average kinetic energy of a general substance. A 'measure' of height is pretty useless if we can't infer the height it measures, after all. Thus the statement that "Tmp is a measure of avg k.e" implies the existence of some unique function T(E), which takes energy and gives us temperature. Since specific heat is basically dT/dE, if there was a unique T(E), then all substances would have identical specific heat behaviour at any temperature. This is manifestly not the case, and hence the original supposition is false. Hence why I complained about the choice of wording "Tmp is a measure of avg k.e", since this leads, as I have just shown, to a contradiction. While not everyone necessarily will fail to realise that there are many different functions which describe the energy dependence of temperature, I felt it was necessary to make this explicit, since inferred misconception is as bad as implied when we consider the pedagogical context which many Wikipedia articles are viewed in. And for the record, I am not confusing heat capacity with temperature and kinetic energy. LeBofSportif 00:32, 15 October 2006 (UTC)


 * It is exceedingly bad form for you to keep on changing what you have written after I have replied to it. Can I suggest you decide what to write, then post it and then leave it alone. Any after thoughts beyond simple spelling/grammar corrections should really come under a separate post. LeBofSportif 00:56, 15 October 2006 (UTC)


 * Sorry, I agree. Edit conflict and I went ahead anyway. Shouldn't do it. My bad.   Greg L 00:59, 15 October 2006 (UTC)


 * LeBofSportif: Getting back to your statement that “<font color="#721F01">Temperature is not average kinetic energy. This is a misconception, and a dangerous one.” as well as “<font color="#721F01">…from a pedagogical point of view it is problematic since you would need a different function for each substance you consider. ”, neither argument is at all true. It's really, really simple LeBofSportif.  The kinetic energy of the temperature-induced translational motion of particles relates to temperature via Boltzmann’s constant. Thus…


 * $$k = \ $$ 1.380 6505(24) joule/kelvin


 * If you’re still not convinced, see also Wikipedia’s Electronvolt article which relates electronvolts to joules as follows: 1 eV = 1.602 176 53 (14) J and then see the subsection of that article, Electronvolts and temperature, which again relates energy to temperature by stating that one electronvolt is equal to 11,604.5 kelvins. Note that no term is necessary in either formula to account for different “<font color="#721F01">substance[s] ” as  you alleged above. If you have a problem with the first-paragraph definition of the Temperature article, which—as of this writing—makes the following (very) simple statement:


 * “[Temperature] is a function of the average kinetic energy of a certain kind of vibrational motion of matter’s constituent particles called translational motions”


 * …then you need to go revise the Boltzmann constant and Electronvolt articles and all the physics books of the world on the same subjects. And again, your statement that “<font color="#721F01">If this [Temperature / kinetic energy relationship] was true, then all gases would have the same heat capacity. They don't, therefore it is not true., well… that too is astonishingly incorrect. First, as I stated above, different molecular substances have varying molar heat capacities because they have varying abilities to absorb heat energy into their molecules’ internal degrees of freedom. Moreover, your statement is further incorrect because one actually can relate molar heat capacity (joules/mole) to absolute temperature for the monatomic gases such as helium, neon, and argon because they have no internal degrees of freedom; they have only translational motions into which they can absorb heat energy (see table).


 * By the way, one uses Boltzmann’s constant to relate temperature to the mean, or average energy of the translational motion of particles in a system (a statistically significant quantity of particles), as follows:


 * Emean = 3/2KbT


 * where…
 * Emean = Joules
 * Kb = 1.380<FONT SIZE="-1"> </FONT>6505 × 10–23
 * T = temperature in kelvins


 * This well-recognized formula shouldn’t have to be defended against obfuscations such as “doesn’t apply for systems that aren’t in equilibrium” and similar circumstances, like “the system is being hit by a train.” It's the law.  Again, I'm not necessarily trying to convince you about this; it's much more important that all others who are contributing to the Temperature article should have confidence in the fundamental underpinnings of what temperature is all about: temperature is a function of the kinetic energy of a particlular kind of particle motion.  Extra specificity is still required in the Temperature article with regard to how temperature and the energy of translational motion are related.  The Boltzmann constant and related articles on thermodynamics can be useful resources for contributors.  Also different sections could flow together more fluidly (a common problem with collaborative writing)


 * I have no intention of making too many contributions to this article since, as is evidenced by these discussions, contributing to this article is time-consuming, unrewarding and is so not fun. Maybe others will find it easier to contribute here. Greg L 21:54, 15 October 2006 (UTC)

Temperature is a measure of the average kinetic energy per degree of freedom. At equilibrium, the average energy is kT/2 per degree of freedom. Translational motion in 3-dimensional space has three degrees of freedom, therefore an average energy of three times kT/2 or 3kT/2. Internal degrees of freedom will increase the total energy of a particle (atom or molecule). If there is one internal degree of freedom, then the total energy of the particle at equilibrium will be 2kT. If there are two internal degrees of freedom, then the total energy per particle will be 5kT/2. Different particles, different relationships between energy and temperature, therefore different heat capacities, but the same relationship between energy per degree of freedom and temperature. PAR 22:38, 15 October 2006 (UTC)


 * PAR: Whether it's “3kT/2” or “3/2KbT”, it sounds good to me. Do you think there's anything you wrote that is at variance with what I wrote(?) or are you just expanding?  Greg L 22:47, 15 October 2006 (UTC)

Well, going thru what you have said in this section, I would disagree that temperature is a function of the average energy of only translational motion. Temperature is not tied to the translational motion energy alone, it is tied to particular degrees of freedom. If all those degrees of freedom have the same temperature (i.e. they are in equilibrium with each other), then fine, but out of equilibrium, they may have different temperatures, or they may have such large statistical variations in their temperature that the argument that they have no temperature at all begins to rear its ugly head.

Also, in the classical case, at equilibrium, its not that molecules have varying abilities to absorb heat into their internal degrees of freedom, its simply that they have those degrees of freedom. The rate of absorption is only a factor for non equilibrium situations. The specific heat of a classical ideal gas is Dk/2 where D is the number of degrees of freedom per molecule (internal plus external). The variation of specific heat with temperature is a quantum effect. PAR 23:31, 15 October 2006 (UTC)


 * PAR: I think you are correctly quoting formulas but are drawing incorrect conclusions as regards their causes and effects. For instance, I think you are in error when you state “<font color="#721F01">…its not that molecules have varying abilities to absorb heat into their internal degrees of freedom, its simply that they have those degrees of freedom. ” In fact, different molecules have different molar heat capacities—even if they posses all three internal degrees of freedom—because different molecules have different constructions (that’s why they’re “different” molecules) and therefore exhibit different quantum properties. I’m excluding hydrogen bonding and crystal latice issues for the moment. The enormously complex ways that only three internal degrees of freedom can be expressed is perfectly well demonstrated by this animation. I do however, agree with your later statement that “<font color="#721F01">[t]he variation of specific heat with temperature is a quantum effect. ” I'm also puzzled at your quick tendency to cite out-of-equilibrium exceptions to various rules. In equilibrium, the temperature of the internal degrees of freedom of molecules are, on average, always equal to the temperature of their translational motions. That much is pretty much fixed by definition whenever one constrains the issue by specifying "in equilibrium” isn't it?  Greg L 01:25, 16 October 2006 (UTC)


 * Greg L, you say that "The enormously complex ways that only three internal degrees of freedom can be expressed is perfectly well demonstrated by this animation". It is very clear to me that looking at the animation, you have misunderstood what a "degree of freedom" is in statistical mechanics. For the purposes of the equipartition theorem, a degree of freedom is any quadratic term in the expression for total energy. That animated molecule has many more than three degrees of freedom. When you say "In fact, different molecules have different molar heat capacities—even if they posses all three internal degrees of freedom—because different molecules have different constructions (that’s why they’re “different” molecules) and therefore exhibit different quantum properties." this also betrays your lack of understanding of what a degree of freedom really is. What you are trying to do, is define temperature backwards via the equipartition theorem, choosing only translational modes and using the 1/2 kT result. This is certainly problematic since not all systems which have temperature have translational modes, and moreover, the equipartition theorem as used in this context is only valid in the classical regime, and only valid for quadratic terms in the Energy. LeBofSportif 10:14, 16 October 2006 (UTC)


 * Yes, Greg, I was being too "classical". Classically, for an ideal gas, the "dimensionsless specific heat capacity at constant volume" is a fixed constant D/2 where D is the number of degrees of freedom per particle. (My error writing Dk/2 above). Deviations from this are quantum effects or non-ideality (i.e. long range forces between particles), but an argument can be made that long range forces introduce additional degrees of freedom. Anyway, If you have two different molecules with the same internal degrees of freedom, and they have different heat capacities, then that will be due to a quantum effect, as you stated.


 * Regarding the equality of translational and internal temperature, yes, they are equal at equilibrium. The reason I keep qualifying things with "in equilibrium" because it is the non-equilibrium situations which give a lot of insight into the dynamics of equilibrium. For example, you can have a plasma of neutral atoms, ions, and electrons in an electric field. The atoms and the ions will have one temperature, and the electrons will have a higher temperature because they pick up energy so much more quickly than do the ions from the electric field. The atoms then pick up their energy from collisions with the ions and electrons. All three particle types have a Maxwellian distribution because they collide enough to form one, but their temperatures are different, with the temperature of the atoms being practically the same as that of the ions. When you have non-equilibrium situations, then the average rates of energy transfer between degrees of freedom become important. At equilibrium, the average rates are equal and the average energies are the same.


 * This is all classical. The point I was trying to make is that I don't think that rates of energy transfer from translational to internal degrees of freedom is a factor in modifying the internal temperature of a bunch of molecules, which is what I took your statement about molecules having "differing abilities to absorb heat energy into their internal degrees of freedom". Maybe I misunderstood your meaning with that statement.


 * Also, yes, there are what look like hundreds of internal degrees of freedom in that animation you mentioned. A simple particle (i.e. an atom of a monatomic gas) has three degrees of freedom. That means you need 3 numbers to specify its configuration, and those three numbers are x, y, and z - its positional coordinates. The energy associated with these three degrees of freedom are its external, or translational energy. If you have a molecule composed of two such particles tied together, then you need to specify the distance between them to fully specify their configuration. (their orientation doesn't affect their energy, so that degree of freedom doesn't apply). For energy purposes, there are now four degrees of freedom, three external, one internal. The energy of the internal degree of freedom is carried by the vibration of the molecule in which the distance between them varies sinusoidally. If this can be viewed as a classical vibration, then the average energy of such a molecule will be four times kT/2 or 2kT. But if the vibration is quantized, and its first excited level is at an energy which is huge compared to kT, then that excited level will practically never be excited, and that internal degree of freedom is non-operational, or "frozen out" or whatever you want to call it, and the average energy of these molecules will be back down to 3kT/2. PAR 14:44, 16 October 2006 (UTC)


 * Geez guys. Take all this energy and make some good and correct contributions to this article; it needs a lot of improvements. I took the time to create graphics and animations for articles. Making one single graphic that displays nice and clear takes a boat-load of time.  The “translational motion” animation I've placed in this article was a collaboration between me and a friend in New York.  I've been working with a University on making a better one.  I've taken the time to compress 6 MB animations that someone else made into an (actually useful) 280 kB version so all the articles that use it load in a reasonable time.  And I've cited my work where necessary.  I encourage you to do the same.  Greg L 17:20, 16 October 2006 (UTC)

Ok, I will do. LeBofSportif 19:17, 16 October 2006 (UTC)


 * Minor correction, PAR. The degrees of freedom only contribute if they appear in the energy, and contribute kT/2 only when they appear quadratically. Thus, x, y, and z are not the degrees of freedom - vx, vy, and vz are. More commonly it is written using px, py, and pz, the three components of momenta. For instance, if all of the particles were hooked to springs so that the Hamiltonian was: H = p^2/2/m + m*omega^2 * x^2/2 then the contribution would be 6 kT/2 per particle.

Assessment comment
Substituted at 20:50, 4 May 2016 (UTC)