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I am going through a proof for the existence of temperature (a proof written in C. J. Adkins equilibrium thermodynamics, if anyone is interested) and I am confused about one of the steps. I am hoping someone will be able to clarify it for me.

Before anyone starts up: I would ask in physics but I think the bit I am stuck on is the maths trick being used and I do not want to confuse things even more.

Starting with the condition for equilibrium for 1 and 3.

$$F_1=(p_1,V_1,p_3,V_3)=0$$

Next the condition for equilibrium for 2 and 3.

$$F_2=(p_2,V_2,p_3,V_3)=0$$

Solving both for $p_3$

$$p_3=f_1(p_1,V_1,V_3)=0$$

$$p_3=f_2(p_2,V_2,V_3)=0$$

and equating the two:

$$p_3=f_1(p_1,V_1,V_3)=f_2(p_2,V_2,V_3)$$

Next is the bit I do not get because he solves for $p_1$

$$p_1=g(V_1,p_2,V_2,V_3)$$

What happened between $p_3$ and $p_1$? I have no idea what trick he did to get away with doing that. The bit afterwards is made vague if I just accept it without knowing why so I would be keen to get some ideas here because he then solves for $p_1$ like this

(the physics bit) and by the zeroth law this must be true:

$$F_3=(p_1,V_1,p_2,V_2)=0$$

Ok back to the maths:

$$p_1=f_3(V_1,p_2,V_2)$$

Which apparently means $V_3$ can drop out of

$$p_3=f_1(p_1,V_1,V_3)=f_2(p_2,V_2,V_3)$$

But it comes across like he did it and said "I did it so that proves I can do it". I do not know how he did it and it looks a bit like he is cheating without seeing that step!

$$p_3=f_1(p_1,V_1)=f_2(p_2,V_2)$$

Anyway I get the rest (provided the previous stuff was true) but I may as well finish it for the sake of completeness:

$$\Phi_1(p_1,V_1)=\Phi_2(p_2,V_2)$$

Which proves the existence of temperature.

Adkins has clearly never had a thermostat war or he'd not have needed all that but anyway... I would really appreciate some explanations of the bit I am confused on.

EDIT:

Please observe the bit where I say the next is the bit I do not get because he solves for $p_1$

$$p_1=g(V_1,p_2,V_2,V_3)$$

EDIT:

The amazon look inside shows the proof which is in the first few pages under the zeroth law chapter

Magpie
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  • The notation is unclear. It looks like the first equality sign immediately after $F_1$ and $F_2$ should be removed so that equilibrium conditions are $F_i(a,b,c,d)=0$. Maybe $F_1 = F_2$ if the same physical law is applied to the different quadruples. Is there an online link to the page of the book where this is done? – zyx Mar 24 '13 at 03:02
  • @zyx thanks. I added a link to the question. – Magpie Mar 24 '13 at 03:53
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    Thanks. Adkins does write the equations as $F_i(...) = 0$. He allows for the possibility of different $F_i$ for different types of systems. – zyx Mar 24 '13 at 04:02

3 Answers3

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Well, I truly do not get the point what's wrong? I guess at first he expressed pressure, that is needed to reach equilibrium for the states 1 and 3 then he did the same for the state 2 and 3 and then said that with this pressure ($p_3$) states 1 and 3, and 2 and 3 will be in equilibrium simultaneously, so we can try to connect the parameter in those states. So he equated them:

$f_1(p_1,V_1,V_3)=f_2(p_2,V_2,V_3)$

And equating them he excluded variable $p_3$. I guess this is all about it.

Caran-d'Ache
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By the equilibrium of 1 and 3; and 2 and 3 the first relation of $p_1$ is derived such as $$p_1=g(V_1,p_2,V_2,V_3)$$ As next point transitivity is employed between systems. Because of transivity condition systems 1 and 2 are also in equilibrium and there exists a transitive closure (smallest transitive relation) which is given by below relation $$F_3(p_1,V_1,p_2,V_2)=0$$ which can be solved for $p_1$ $$p_1=g(V_1,p_2,V_2)$$ In summary transivity closure is used to eliminate system 3 from the solution set which also eliminates $V_3$ from the relation.

-------Extension for derivation of $p_1$---------

The equilibrium condition between systems 1 and 3 is given by $$F_1(p_1,V_1,p_3,V_3)=0$$ which can be solved explicitly to find $p_3$ $$p_3=f_1(p_1,V_1,V_3)$$ The equilibrium condition between systems 2 and 3 is given by $$F_2(p_2,V_2,p_3,V_3)=0$$ which can be solved explicitly to find $p_3$ $$p_3=f_2(p_2,V_2,V_3)$$

Since bot definiton of $p_3$ is same and we can eliminate $p_3$ such as $$p_3=f_1(p_1,V_1,V_3)=f_2(p_2,V_2,V_3)\Rightarrow G(p_1,p_2,V_1,V_2,V_3)=f_1(p_1,V_1,V_3)-f_2(p_2,V_2,V_3)=0$$ This equation can be solved explicitly for $p_1$ $$p_1=g(V_1,p_2,V_2,V_3)$$

AnilB
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  • By the equilibrium of 1 and 3; and 2 and 3 the first relation of $p_1$ is derived such as $$p_1=g(V_1,p_2,V_2,V_3)$$? When was it derived? I didn't see it get derived. That is why I am asking about it. – Magpie Mar 23 '13 at 22:44
  • I guess it shall be clear now? – AnilB Mar 24 '13 at 02:54
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The part in question looks to be the implicit function theorem (IFT) to me.

If we define $f = f_1(p_1,V_1,V_3) - f_2(p_2,V_2,V_3)$, then $f$ is a map from $\mathbb{R}^{1+4} \to \mathbb{R}$.

If you can show that $f$ satisfies the conditions of IFT (which I guess is the case...

I don't have the book, so I don't know what conditions of $f_1,f_2$ are), you can just apply the theorem. Let me know what you think.

Magpie
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Euler....IS_ALIVE
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  • You should not need the book the zeroth law isn't the bit I am confused about. I do think it is the IFT but I understood the first application of it (for $p_3$) it was just what he did with $p_1$ that I could not see where it came from. – Magpie Mar 18 '13 at 07:19
  • Then I must be confused. You asked "What happened between $p_3$ and $p_1$? I have no idea what trick he did to get away with doing that." My answer does not answer your question? – Euler....IS_ALIVE Mar 18 '13 at 16:52
  • Nope, sorry it doesn't. because I explained the bit I was stuck on and that's not the bit you clarified. – Magpie Mar 18 '13 at 17:54