# Notation in thermodynamics

1. Mar 16, 2012

### VantagePoint72

I'm having trouble understanding some of the notation used in thermodynamics. The textbook will write things like:

$$\frac{1}{T} = \left(\frac{\partial S}{\partial E}\right)_{N,V}$$

such as you might use in the thermodynamic temperature in the microcanonical ensemble, and it says that the subscripts denote constant volume and number of particles. But, when you take a partial derivative you always hold all the non-differentiating variables constant. That's what a partial is. So what information do the subscripts add?

2. Mar 17, 2012

### Philip Wood

Because you know this came from
$$dE = TdS â€“ PdV + \mu dN$$
it's obvious what variables need holding constant.
But if you didn't know where it came from, you might think that $$\frac{1}{T} = (\frac{dS}{dE})_{N, P}$$
was perfectly reasonable.

BTW: How do you do brackets which expand properly?

Last edited: Mar 17, 2012
3. Mar 17, 2012

### VantagePoint72

But my point is that it's a partial derivative, so in both cases you hold all the variables constant except for E. Sorry, but I still don't understand.

For the brackets, type \left( and \right) wherever you use them in a formula. LaTeX will automatically make them the right size based on context.

4. Mar 17, 2012

### VantagePoint72

(you can also right-click on any LaTeX and choose Show Math As > TeX Commands and see the source code)

5. Mar 17, 2012

### vanhees71

In thermodynamics it's very important to take the partial derivatives with holding the right other variables fixed. Thus, one gives as subscripts these variables that are kept constant. E.g., the specific heat of a gas depends on whether you keep to pressure or volume fixed (of course in addition also the total particle number is kept fixed), i.e., one has

$$c_p=\left (\frac{\partial U}{\partial T} \right )_{p,N}, \quad c_V=\left (\frac{\partial U}{\partial T} \right )_{V,N}.$$

Note that $c_p \neq c_V$ although in both cases you take the derivative of the internal energy with respect to temperature. The difference is in which other variables are hold constant when taking the derivative.

6. Mar 17, 2012

### Philip Wood

If we're dealing with a one-constituent fluid and an open system (N not constant), we need 3 independent variables. We might use (N, S, V,), or (p, S, T), or (N,S, p) or ((N, V, T) and so on. Any function of state can be expressed in terms of any of these trios, though sometimes not all three will be needed, and sometimes the expression will be awkward. When you take a partial derivative with respect to one variable you need to hold constant the other 2 in your chosen trio. In your example you can't hold EVERYTHING except E constant, because E is related to other variables!

Here's a simple example, for a closed system, for which there are only two independent variables. We can choose (P, V), (V,T), (P,T)... We have the two heat capacities:
$$c_{p} = T \left(\frac{dS}{dT}\right)_{p}$$ and $$c_{v} = T \left(\frac{dS}{dT}\right)_{V}$$.

Thanks very much for the latex instruction.

7. Mar 17, 2012

### DrDu

The problem is that thermodynamics uses a somewhat different convention than mathematics. In mathematics,what physicists call e.g. U(S,V) and U(p,T) are considered different functions and would not be given both the same name U. Hence it is always clear on which variables a function depends and they have not to be specified in taking partial derivatives. In thermodynamics these two functions are considered as merely two different parametriizations of the same physical quantity whence the same symbol is used for both. The price you have to pay is that you have to specify all the variables when taking partial derivatives.

8. Mar 18, 2012

### Philip Wood

DrDu. I'm very pleased that you have made this point explicitly. There seems to be a conspiracy of silence about it. Although my discipline is Physics, I very much dislike what I consider to be thermal physicists' misuse of function notation. For me, U(S, V) should be the name of a recipe to get a value of U from two values of S and V. Thus U(P,T) should have the same value as U(S, V) if P and T have the same values as S and V respectively. [I know they can't have, because of units, but that makes the problem worse rather than better.]

What would mitigate the problem is for physicists to use a different style of brackets, for example U[S, V], U[P, T] when they don't mean that U is the same function of S and V as it is of P and T, but I don't suppose many people care enough...

9. Mar 18, 2012

### vanhees71

I have to correct the 2nd formula. By definition the heat capacity at constant pressure is

$$c_p=T \left (\frac{\partial S}{\partial T} \right )_{N,p}.$$

The correct energy-like potential is the enthalpy rather than internal energy. By definition it's the Legendre transform of the internal energy wrt. the pair (p,V):

$$H=U+p V \; \Rightarrow \; \mathrm{d} H=\mathrm{d} U + p \mathrm{d} V + V \mathrm{d} p = T \mathrm{d}S + V \mathrm{d} p - \mu \mathrm{d} N.$$

From this we get

$$c_p=\left (\frac{\partial H}{\partial T} \right )_{N,p}.$$

For $c_V$ my previous definition is correct since

$$c_V=T \left (\frac{\partial S}{\partial T} \right )_{N,V}=\left (\frac{\partial U}{\partial T} \right )_{N,V}.$$

10. Mar 18, 2012

### DrDu

I don't think there is a conspiracy. I remember having learned this in my first year in "Mathematics for chemists". It is rather common for physicists to use different conventions than mathematicians.