# Cyclic rule (thermodynamics)

1. Jan 26, 2014

### Nikitin

1. The problem statement, all variables and given/known data
Derive the cyclic rule in thermodynamics.

$\frac{\partial p}{\partial T} \cdot \frac{\partial T}{\partial V} \cdot \frac{\partial v}{\partial p}=-1$

2. Relevant equations

3. The attempt at a solution

OK, so I write out the total differential of $p$: $dp=\frac{\partial p}{\partial T} dT + \frac{\partial p}{\partial V} dV$. What do I do now? In the assignment a tip says I should set $dp=0$, but why should only dp equal zero, and not also dT or dV? What are the consequences of dp being zero (other than pressure being constant)?

2. Jan 26, 2014

### voko

What you are supposed to understand here is that $p = p(T, V), T = T(p, V), V = V(p, T)$. Now, once you have $dp$ and then you say $dp = 0$, then you can say that $dT = {\partial T \over \partial V} dV$. That gives you some relationship between partial derivatives. Continue from there.

3. Jan 26, 2014

### voko

Just for the record, even if you are supposed to solve the problem in that way, that is quite a awkward way. It works, but then if you ask yourself this question: but isn't this only true when $dp = 0$? you can't easily answer that.

A much more satisfying approach is the following one. Any thermodynamic system has an equation of state: $$G(p, T, V) = 0 = \mathrm {const}$$. Then we have $$dG = G_p dp + G_T dT + G_V dV = 0$$ where subscripts denote partial differentiation. From this differential, one can express a partial derivative of any thermodynamic variable with respect to any other variable. For example: $${\partial p \over \partial T} = -{ G_T \over G_p }$$ Then you can show that the cyclic rule follows directly. Note that if you add more thermodynamic variables (such as entropy), it is obvious that the rule continues to work.

4. Jan 26, 2014

### amattiol

Total differential ->
dP = (∂P/∂T)n,VdT + (∂P/∂V)n,TdV + (∂P/∂n)T,Vdn

What if P is Constant?

0 = (∂P/∂T)VbardT + (∂P/∂Vbar)TdVbar
so...
(∂P/∂T)VbardT = - (∂P/∂Vbar)PdVbar

(∂P/∂T)P = - (∂P/∂T)vbar / (∂P/∂Vbar)T With a constraint of P being constant

Cyclic Rule

Im pretty sure.. I hope that helps?

5. Jan 26, 2014

### Nikitin

Amattiol, thanks, but the problem is you need to assume one of the thermodynamic variables (P,T or V) are constant.

That's exactly my problem. If you assume dp =0, then shouldn't the formula work only when the pressure is constant? And ditto when you derive it by assuming dV or dT are constant.

I really don't understand how (and why) this "cyclic-rule" formula works, when you can derive it with so many different starting conditions..

You lost me here.. How can you do that? thanks for all the help btw, I highly appreciate it.

6. Jan 26, 2014

### voko

Did you understand how I got $\partial p \over \partial T$? Can you derive $\partial T \over \partial V$ and $\partial V \over \partial p$ in the same way?

7. Jan 26, 2014

### amattiol

I think it has to do with 2nd derivatives...

in continuation from where i left of last time:

(∂P/∂Vbar)T = -RT/Vbar2
(∂2P/∂Vbar2)T = 2RT/Vbar3

(∂P/∂T)Vbar = R/Vbar
(∂2P/∂T2)Vbar = O

((∂/∂T)(∂P/∂T)T)Vbar = -R/Vbar2
((∂/∂Vbar)(∂P/∂T)Vbar)T = -R/Vbar2

Since those last 2 mixed partials are the same does that have anything to do with it being cyclic - Im not quite sure, just trying to possibly spark something.

8. Jan 27, 2014

### Nikitin

I suspect you set $dV$ equal to zero? But how is that any better than deriving the cyclic rule by using the total differential of one of the variables?

9. Jan 27, 2014

### voko

If you have variables $x, y, ... z$ and you need a partial derivative of one with respect to another, then by the very definition of partial derivative only $x$ and $y$ are variable during partial differentiation. When you have a relation like $$X dx + Y dy + ... + Z dz = 0$$ it then becomes $$X dx + Y dy = 0$$ so you get $${\partial x \over \partial y} = - {Y \over X}$$

The difference with the previous method is that you start with a differential equation that is always true and then obtain partial derivatives from it using a just their mathematical definition without any hand waving.

This argument can be made even more formal. It is known as the implicit function theorem.

Ultimately, the cyclic rule holds because at least three variables are related with one another via at least one equation. It is similar to the simpler rule $${\partial x \over \partial y} = \left[ {\partial y \over \partial x} \right]^{-1}$$ All this was known long before thermodynamics came into existence and is really a subject of calculus, pure and simple.

10. Jan 27, 2014

### Nikitin

OK, I might have to just work with the equation for a few months and then look at this thread again. I'll probs understand much more.

But there's one thing:

Could you explain further, and if possible without using the implicit function theorem? You see I'm familiar with implicit functions, but I can't remember having learned the "implicit function theorem" in multivariable calculus.

11. Jan 27, 2014

### voko

When you have $$G(x, y, z) = 0$$ that means you have implicit functions $x = f(y, z)$, $y = g(x, z)$ and $z = h(x,y)$ such that if, for example, you take $f(y, z)$ then $$G(f(y, z), y, z) = K(y, z) = 0$$ Now we can apply the chain rule $${\partial K (y, z) \over \partial y} = \left[ {\partial G (x, y, z)\over \partial x} \right]_{x = f(y, z)} {\partial f (y, z) \over \partial y} + \left[ {\partial G (x, y, z)\over \partial y} \right]_{x = f(y, z)}$$ That is usually written in a simplified form $${\partial K \over \partial y} = {\partial G \over \partial x} {\partial f \over \partial y} + {\partial G \over \partial y}$$ assuming we know what arguments apply to what. Which, admittedly, may be confusing. The "bracketed" notation used in many texts on thermodynamics originally tried to address this problem, by mentioning all the other arguments outside the brackets, but this is rarely explained properly and is thus confusing by and in itself. Then we take another liberty: we identify $x$ with $f(y, z)$ (because $x = f(y, z)$) and that becomes $${\partial K \over \partial y} = {\partial G \over \partial x} {\partial x \over \partial y} + {\partial G \over \partial y}$$ Recalling that $$K(y, z) = 0 = \mathrm{const}$$ we have $${\partial K \over \partial y} = {\partial G \over \partial x} {\partial x \over \partial y} + {\partial G \over \partial y} = 0$$ which gives $${\partial x \over \partial y} = -{ {\partial G \over \partial y} \over {\partial G \over \partial x} }$$ That is, essentially, the implicit function theorem (in its simplest form). It is surprising that it was not taught in your course of calculus.

12. Jan 27, 2014

### Nikitin

OK, thanks for your time! I think I understand the main parts now.

As for the Implicit Function theorem: hmm, well it is very possible that I've just forgotten about it. I'll take a look in my calculus book tonight.

Last edited: Jan 27, 2014
13. Jan 27, 2014

### Nikitin

Well, it seems our book (University Calculus by Hass, Weir and Thomas) only mentioned a simple version of it for two variables. At any rate, I get all of what you're saying except here:

I stared at the screen for a long time, but I still don't get why you're just ignoring the z variable. I know you say it's a function of (x,y), but how does that make dz/dy=0?

I mean, shouldn't it go like this:
$${\partial K \over \partial y} = {\partial G \over \partial x} {\partial x \over \partial y} + {\partial G \over \partial y} + {\partial G \over \partial z}{\partial z \over \partial y}$$?

14. Jan 28, 2014

### voko

$K(y, z)$ is a function of $y$ and $z$ by definition. So its partial derivative with regard to $y$ cannot take $z$ into account - by definition again.

15. Jan 28, 2014

### Nikitin

Oh I see. You define one of the variables (x, in the case above) as dependant on the two other independent variables (y and z), and thus when you carry out the partial differentiation of G(x,y,z) with regards to one of the independent variables, the other independent variable dies (as the two independent variables aren't functions of each other). All of this being completely legal since G(x,y,z)=0 => you can order the function so that x=x(y,z), y=y(x,z) and z=z(x,y).

Then, to derive the triple product rule you calculate:

1) $G_y$ with $x=x(y,z)$
2) $G_z$ with $y=y(x,z)$
3) $G_x$ with $z=z(y,z)$

order the expressions into a suitable form (I ordered them into dx/dy =.., dy/dz=.. and dz/dx=...) and multiply them together to end up with the triple product rule.

Okay I got it. Kind of annoying I had to spend so much time just because I didn't understand the hand-waving used in the assignment (set dp to equal zero), but oh well. At least now I know more math because of you, voko.

Thanks allot!! :)

Last edited: Jan 28, 2014