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FriendlyHamiltonian

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The thermodynamic identity is usually expressed in the following differential form

$$

dU = TdS - PdV + \mu dN,

$$

where [itex] U [/itex] , [itex] T [/itex], [itex] S [/itex], [itex] P [/itex], [itex] V [/itex], [itex] \mu [/itex] and [itex] N [/itex] are the internal energy, temperature, entropy, pressure, volume, chemical potential and number of particles of the system respectively. If I am not mistaken, I can act with a vector, say [itex]\frac{\partial}{\partial N}[/itex], to yield

$$

\frac{\partial U}{\partial N} = T \frac{\partial S}{\partial N} - P \frac{\partial V}{\partial N} + \mu \implies \mu = \frac{\partial U}{\partial N} - T \frac{\partial S}{\partial N} + P \frac{\partial V}{\partial N}.

$$

Consider the following question:

*Consider a monoatomic ideal gas that lives at height [itex]z[/itex] above sea level, so each molecule has potential energy [itex]mgz[/itex] in addition to its kinetic energy. Show that the chemical potential [itex]\mu[/itex] is the same as if the gas were at sea level, plus am additional term [itex]mgz[/itex]:*

$$

\mu(z) = -k_b T \text{ln}\left[\frac{V}{N}\left(\frac{2\pi m k_bT}{h^2}\right)^{3/2}\right] + mgz.

$$My attempt was knowing that:

The

*"ideal monoatomic gas"*implies [itex]U = \frac{3}{2}k_bT[/itex] (by equipartition theorem) and the validity of Sackur-Tetrode equation:

$$

S=k_bN\ln \left[{\frac {V}{N}}\left({\frac {4\pi m}{3h^{2}}}{\frac {U}{N}}\right)^{3/2}\right]+{\frac {5}{2}},

$$

together with the assumption that [itex]V \neq V(N)[/itex]. If one uses the above formula for [itex]\mu[/itex] and takes the partial derivatives I yield

$$

\mu(z) = -k_b T \text{ln}\left[\frac{V}{N}\left(\frac{2\pi m k_bT}{h^2}\right)^{3/2}-\frac{3}{2}\right] + mgz,

$$

which is almost correct except for that [itex]-\frac{3}{2}[/itex], although it still exhibits the problems described below.

**I came to the conclusion that I don't know how to manipulate these equations in differential form, am I allowed to do the above "act with [itex]\frac{\partial}{\partial N}[/itex]" business?**The solution provided by the book is to say, hey hold [itex]U[/itex] and [itex]V[/itex] fixed so that the thermodynamic identity now reads

$$

0 = TdS - 0 + \mu dN \implies \mu = T \left(\frac{\partial S}{\partial N}\right)_{V,U \text{ fixed}}

$$

but [itex]U = U(N)[/itex], in particular [itex]U = \frac{3}{2} k_b N T[/itex] I could litterally make all the [itex]N[/itex]s in [itex]S[/itex] disappear by substituting [itex]N = \frac{2 U}{3 k_b T}[/itex] and claim that

$$

\mu = T \left(\frac{\partial S}{\partial N}\right)_{V,U \text{ fixed}} = 0,

$$

which is ridiculous. I'm really lost with the mathematics behind this type of calculations... Which would be the correct way to proceed?