2

From the 0th law, Thermal equilibrium is when there is no heat transfer between two objects. So I want to ask is temperature the only "potential"-esque quantity which should be equalized for stop of heat flow? If temperature is the only one then why is it the only one? Could we prove this?

tryst with freedom
  • 7,940

2 Answers2

6

In general, thermal equilibrium means maximizing the entropy. The reason we use temperature is that very often, two systems can do this by exchanging energy. Under an exchange of energy $dE$, $$dS_{\text{tot}} = \frac{dS_1}{dE_1} \, dE + \frac{dS_2}{dE_2} \, (-dE) $$ so the maximum entropy is achieved when this is zero, and the systems have the same $$\frac{dS}{dE} = \frac{1}{T}$$ where this is really a definition of $T$.

In general, you can exchange other things too. For example, if a container is separated in two by a movable piston, then the total volume of the two pieces is conserved, and we can maximize entropy by exchanging volume. Then in thermal equilibrium, they have the same $$\frac{\partial S}{\partial E} \bigg|_V = \frac{1}{T}, \quad \frac{\partial S}{\partial V} \bigg|_{E} = \frac{p}{T}$$ where the second equation serves as the thermodynamic definition of pressure. If the total number of some kind of particle is conserved, and the systems can exchange particles, we equalize $$\frac{\partial S}{\partial E} \bigg|_{V, N} = \frac{1}{T}, \quad \frac{\partial S}{\partial V} \bigg|_{E, N} = \frac{p}{T}, \quad \frac{\partial S}{\partial N} \bigg|_{V, E} = - \frac{\mu}{T}$$ where the third equation defines the chemical potential. If there were $n$ separate types of such particles, we'd have $n$ separate chemical potentials that would be set equal.

There are plenty of more exotic options too. In general, there is a potential for every conserved quantity which is conserved, can be exchanged between the systems, and affects the entropy in the thermodynamic limit. (On the other hand, in an introductory course it's reasonable to focus on systems with only one or two, to avoid too much complication with partial derivatives.)

knzhou
  • 101,976
  • Mind blowing answer!! can you tell why we choose maximizing entropy? Now I understand we can think of any state function as a potential but why does heat transfer stop only when it is in thermal equilibrium?? – tryst with freedom May 09 '20 at 17:39
  • "The reason we use temperature is that very often, two systems can do this by exchanging energy, so the maximum entropy is achieved when they have the same "

    Some maximization is going on but I don't see any first derivative equal to 0 buisness going on here

     – tryst with freedom May 09 '20 at 17:40
  • @DDD4C4U At this level, "maximizing entropy" is just the most fundamental postulate... but if you want to know a bit more about why that is, you can try searching, there are lots of questions that have been asked about this. – knzhou May 09 '20 at 17:40
  • Thank you sir but I don't know what exactly to search. If you could tell me the key terms to search to find my knowledge I would really appreciate it – tryst with freedom May 09 '20 at 17:41
  • @DDD4C4U I edited to clarify the "set the derivative to 0" point. – knzhou May 09 '20 at 17:42
  • 3
    @DDD4C4U You want to look up keywords like "ergodic hypothesis", also see this question. – knzhou May 09 '20 at 17:42
  • could u show some of the exotic ones you mentioned? I am really really interested in knowing – tryst with freedom May 09 '20 at 18:12
  • Let us continue this discussion in chat. – tryst with freedom May 09 '20 at 18:20
  • Being a bit nit-picky here. Technically having equal pressures is "mechanical equilibrium", and having the same chemical potential is "chemical equilibrium". You seem to have lumped it all under "thermal equilibrium" when you introduced the mechanical equilibrium example. Perhaps you meant "thermodynamic equilibrium"? – BioPhysicist Sep 09 '20 at 13:25
1

First, the 0th law is not what you think it is...

From the 0th law, Thermal equilibrium is when there is no heat transfer between two objects.

This is not the 0th law, this is just the definition of thermal equilibrium.

The 0th law is just something needed to make thermal equilibrium a well-defined "equality" between systems at thermal equilibrium. This is taken straight from the mathematics of equivalence relations, where we need a relation to be reflexive, symmetric, and trasitive. By definition of thermal equilibrium, a system is in thermal equilibrium with itself (reflexivity), and if system 1 is in thermal equilibrium with system 2, then it must be that system 2 is in thermal equilibrium with system 1 (symmetry).

However, transitivity is not guaranteed by the definition of thermal equilibrium. Therefore, we need the 0th law, which actually states

If two thermodynamic systems are each in thermal equilibrium with a third one, then they are in thermal equilibrium with each other.

So, contrary to what you have stated that the 0th law is the definition of thermal equilibrium, it is actually something that characterizes thermal equilibrium. The 0th law brings in transitivity so that, mathematically, thermal equilibrium is an equivalence relation.

As for the meat of the question, I think knzhou's answer is great. Anything arising from maximization of entropy can typically be used to describe some sort of equilibrium.

BioPhysicist
  • 56,248
  • hehe the answer came in late, but yes, this answer is more accurate to the actual misconception than knzhou's answer ( which is actually still a high quality one). Now, my understanding is that each law of thermodynamic, states a rule and also concept. The rule is of thermal equilibrium and the concept is of temperature. – tryst with freedom Sep 09 '20 at 14:27
  • @Buraian To be fair knzhou had already addressed the main part of your question. I just wanted to address the 0th law part mostly. – BioPhysicist Sep 09 '20 at 14:36