Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

OKay my book just starts out talking about the vague definition we have for temperature and we ended up with the Zeroth law of Thermodynamics which states:

Two systems are in thermal equilibrium if and only if they have the same temperature.

So does that mean, the (brief) meaning of the property temperature is a measure of whether one system is in thermal equilibrium with another?

share|improve this question
3  
Yes, that's the definition. –  Alexander Sep 8 '12 at 1:27
    
But this yes refers only to the Zeroth law, not to the question appended to it, for which the answer is no. –  Arnold Neumaier Sep 9 '12 at 13:44
    
@ArnoldNeumaier, could you elaborate? –  Hawk Sep 9 '12 at 18:33
    
I had explained this in my answer. –  Arnold Neumaier Sep 9 '12 at 19:59
    
Woops, for some reason I didn't see your name looks different in another answer... –  Hawk Sep 10 '12 at 0:44

3 Answers 3

The temperature is not a measure whether a system is in equilibrium. Instead, it is one of the parameters characterizing a system _that_is_ in equilibrium.

It is that particular parameter that tells how much volume the quicksilver in a conventional thermometer occupies at a fixed pressure. Thus one can read the temperature of a thermometer from a scale attached to the quicksilver container. (From a microscopic point of view, it is a measure of the speed of the random movement of molecules in a substance. The faster the motion the more volume is needed to accomodate the motion.)

The role of the zeroth law is somewhat different. If a system is in equilibrium with a thermometer then the combined system has a well-defined temperature, so you are entitled to take what you read off the thermometer as the temperature of the system measured.

On the other hand, if the system is in equilibrium and the thermometer is in equilibrium, but the combined system is not (namely for instance when you just brought the two in contact), they will have different temperatures, and it takes a while till a joint equilibrium is reached. (The while may be a lot, depending on how well the system is insulated.)

Thus the zeroth law explains why (and under which conditions) one can use thermometers to measure temperature. You must bring the thermometer into contact with the system and then wait until equilibrium is established. Only then the measurement at the thermometer gives the temperature of the system.

Edit: If a system is not in equilibrium, there is no unique temperature that can be assigne to the system. Thus the notion of temperature makes sense only in equilibrium. Out of equilibrium you have instead a temperature field. Differences in temperature at neighboring places introduce a thermal force (entropy production) that steers the system to equilibrium. Therefore, a correct version of the statement in your final question is: ''Equality of temperature is a necessary condition for equilibrium.'' It doesn't suffice though, as also pressure and chemical potentials must be equal. Equality of temperature, pressure and chemical potentials indeed implies (for a chemical system) that the system is in equilibrium.

See also Chapter 7 ''Phenomenological thermodynamics'' of my book http://lanl.arxiv.org/abs/0810.1019 (The chapter is readable independent of the remainder of the book, and needs little background only.)

share|improve this answer
    
Instead, it is one of the parameters characterizing a system _that_is_ in equilibrium., isn't that just changing the wording though or looking at it in a different view? –  Hawk Sep 10 '12 at 0:44
    
It means something quite different. See the edit at the end. –  Arnold Neumaier Sep 10 '12 at 9:48
    
I totally agree with Arnold's answer and I would like to add a comment on the size of the system. Thermodynamics is defined for systems where N, the number of particles, and V the volume of the system are such that: N->infinity, V-> infinity and N/V is finite. –  Shaktyai Sep 10 '12 at 10:32
    
To further complete Arnold's answer: temperature is a macroscopic parameter with no equivalent at the microscopic level that characterize macroscopic systems at equilibrium. It can be measured by letting the system reach an equilibrium (until no heat is exchanged), with an other system called a thermometer. –  Shaktyai Sep 10 '12 at 10:40

That would be definition of temperature in thermodynamic framework. However, as Ron has remarked, it can be understood better in framework of statistical mechanics which in some sense is a more fundamental science than thermodynamics.

For a non-isolated system (i.e. a system which is allowed to exchange energy with its surroundings) temperature is a parameter which tells how energies are distributed in the system$^{**}$. More precisely consider a system which can exist in various states (configurations) $|1>, |2>, |3>,...$ of energies $E_1,\:E_2,\:E_3\:,..$ respectively. Then saying that this system has temperature $T$ means that probability for this system to be found in state $|i>$ of energy $E_i$ is ~$exp(-E_i/kT)$.

Now consider two systems $A$, and $B$.

Suppose $A$ can exist in states $|A1>, |A2>, |A3>,...$ of energies $E^A_1,\:E^A_2,\:E^A_3\:,..$ respectively; and $B$ can exist in states $|B1>, |B2>, |B3>,...$ of energies $E^B_1,\:E^B_2,\:E^B_3\:,..$ respectively. Now suppose we allow these systems to exchange energy with each other. We say that systems $A$, and $B$ have attained equilibrium when the energy distribution in the combined system $A+B$ as well as as in its subsystems $A$, and $B$ is no more changing with time, i.e. when

$1.$ The combined system $A+B$ is at a definite temperature $T$ (i.e. energy distribution in $A+B$ is given according to parameter $T$ )

$2.$ Systems $A$ and $B$ themselves are at some definite temperatures $T_A$ and $T_B$.

With these definitions of temperature and of equilibrium one can show that at equilibrium we should have $T_A=T_B=T$. (You can try to prove it yourself for the simple case when both the systems $A$ and $B$ have finitely many energy states of distinct energy).


** for an isolated system at energy $E$ temperature is defined in a different way.

share|improve this answer

The answer is yes, as Alexander comments, but one should say that it is a little surprising--- it is saying that two systems, no matter their internal dynamics, will exchange heat according to a single real valued parameter. So that if you adjust this one parameter to be equal, then the systems will not exchange energy when in contact.

This can be stated axiomatically, if A doesn't exchange heat with B, and B doesn't exchange heat with C, then A doesn't exchange heat with C, also A doesn't exchange heat with a copy of A. These axiomatically define equivalence classes of systems "at the same temperature".

Further, if A is hotter than B (so energy flows from A to B on average) and B is hotter than C, then A is hotter than C (and this is a relation on equivalence classes). For any two systems A,B with A hotter than B, there is a C which is hotter than B and less hot than A. This tells you that temperature is linearly ordered.

All these axiomatic properties are a-priori surprising, since I didn't say anything about the Hamiltonian of the systems involved. They become obvious once you consider that systems maximize their phase-space volume (entropy). Then the temperature is defined statistically as the reciprocal of the rate of change of entropy with energy, so you always increase entropy by flowing heat from hot to cold. This explains why temperature is a linearly ordered real-valued equivalence-class type thing from simple first principles.

share|improve this answer
    
My book said something about For systems A and B to be in thermal equilibrium, all the information that is needed is that both A and B are in thermal equilibrium with C. This is not true. I don't understand what that means. It seems to say the opposite of what the Zeroth Law say –  Hawk Sep 8 '12 at 2:45
    
@jak: It is true--- if A is at the same temperature as C, and if B is too, then A is at the same temperature as B. This is a little surprising a-priori, but not really surprising to human beings, because we come with little temperature sensors on our skin, so we find it intuitive. –  Ron Maimon Sep 8 '12 at 3:15
    
I am a beginner in Thermodynamics, and we use an ancient book called "heat and thermodynamics by Zemansky and Dittman". It's so confusing! You know a better book? I grabbed a first year physics book, but that wasn't sufficient. –  Hawk Sep 8 '12 at 3:23
    
@jak: The ancient books are usually no worse than the modern ones. Research people generally stopped writing about thermodynamics in the 20th century, since it was subsumed into statistical mechanics. I think the only books one can recommend are little volumes by Planck and Fermi, and they're older than your book I bet. –  Ron Maimon Sep 8 '12 at 3:58
    
Regarding your last sentence, I don't think it says anything about the linearity of the temperature. Couldn't any one-to-one transformation of the temperature scale still fulfill the requirement? I always found this point confusing. $T$ reflects a molecular kinetic energy metric, but why not the square root of the kinetic energy? –  Alan Rominger Sep 9 '12 at 15:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.