I am currently working on this. More specifically my question is about Problem 2.5 b). In the solution they get from $$ Nd\mu=-SdT+VdP $$ to $$ N\Big(\frac{\partial\mu}{\partial N}\Big)_{T,V}=V\Big(\frac{\partial P}{\partial N}\Big)_{T,V} $$ which I dont fully understand. I get that the $dT$ term is gone because $T$ is constant and so $dT=$. And because $V$ is constant there is no $\frac{\partial V}{\partial N}$. But I not yet came up with a mathmatical proof. My Problem is that the only way I know to start is to take the partial derivative over $N$ on both sides and let $T,V$ be fixed. Then on the right side I get the right thing but on the left side I get $$ \frac{\partial}{\partial N}\Big(Nd\mu\Big)_{T,V}=\Big(\frac{\partial N}{\partial N}d\mu\Big)_{T,V}+\Big(N\frac{\partial\mu}{\partial N}\Big)_{T,V}=\Big(d\mu\Big)_{T,V}+N\Big(\frac{\partial\mu}{\partial N}\Big)_{T,V} $$ The $d\mu$ should not be there. Can someone help me?
2 Answers
If you know a little differential geometry (vector fields and $1$-forms) and linear algebra (dual bases) then this question becomes almost a triviality/definition. At the outset, let me remark that the result is obtained, not by “differentiating” the equation given (because differentiating an equation of forms is only achieved by exterior derivatives), but rather by “evaluating” the given equation along a specific direction. Just in case this sounds mysterious, I’ll provide a rapid review.
Things like $d\mu,dT,dP$ etc are called $1$-forms, meaning they are objects which eat vector fields and produce a function. So, if $X$ is a vector field, and $f$ is a function, then $(df)(X)$ means the application of the $1$-form $df$ on the vector field $X$. What it measures is the directional derivative of $f$ along $X$. Now, consider the special case where you have a coordinate system $(x^1,\dots, x^n)$. This gives rise to $n$ different vector fields $\xi_{(x),1},\dots, \xi_{(x),n}$, where the $\xi_{(x),i}$ is the vector field which is tangent to the $x^i$ coordinate line (obtained by keeping the coordinates $x^1,\dots, x^{i-1},x^{i+1},\dots, x^n$ fixed). Pictorially, draw a ‘curvy grid’ if you wish, then along the coordinate liens, draw the tangent arrows. If you evaluate $df$ on this particular coordinate-induced vector field $\xi_{(x),i}$, then $df(\xi_{(x),i})$ gives the directional derivative of $f$ in the direction of the $x^i$ coordinate curve when all other coordinates are held fixed… I hope this sounds familiar; it is exactly the definition of the partial derivative $\frac{\partial f}{\partial x^i}$. One thing which I will emphasize (as hopefully you’ve already been told before), the $i^{th}$ vector field $\xi_{(x),i}$ depends on not just $x^i$, but it depends on the remaining coordinates $x^1,\dots, x^{i-1},x^{i+1},\dots, x^n$ as well, since we by definition have to keep them constant. This is why I’ve labored to write $\xi_{(x),i}$ rather than simply $\xi_i$, to emphasize that the definition of this object depends on the entire coordinate system $x=(x^1,\dots, x^n)$.
Next, if we still fix a coordinate system $x=(x^1,\dots, x^n)$ then taking $f$ to be one of the coordinate functions $f=x^j$, we can consider its exterior derivative $dx^j$. This can of course act on the various coordinate vector fields $\xi_{(x),i}$. The result is then $dx^j(\xi_{(x),i})=\delta^j_i$.
Ok, so for your question specifically, we have a $3$-dimensional system, and we consider a coordinate system $x=(x^1,x^2,x^3)=(N,T,V)$. We thus have three coordinate-induced vector fields $\xi_{(x),1},\xi_{(x),2},\xi_{(x),3}$. Now, we consider the equality $N\,d\mu=-S\,dT+V\,dP$, and we evaluate both sides on the vector field $\xi_{(x),1}$. This gives \begin{align} N\,(d\mu)(\xi_{(x),1})&=-S(dT)(\xi_{(x),1})+V(dP)(\xi_{(x),1}).\\ &=0+V(dP)(\xi_{(x),1}), \end{align} where the $0$ is because $(dT)(\xi_{(x),1})=(dx^2)(\xi_{(x),1})=0$, since by definition $x^2$ is kept constant along the vector field $\xi_{(x),1}$. So, this equality can now be written (by definition/notation!) as \begin{align} N\frac{\partial \mu}{\partial x^1}&=V\frac{\partial P}{\partial x^1}. \end{align} Or, what means the same thing, but in more traditional notation, \begin{align} N\left(\frac{\partial \mu}{\partial N}\right)_{T,V}&=V\left(\frac{\partial P}{\partial N}\right)_{T,V}. \end{align}
Notational Remark.
What I called $\xi_{(x),i}$ is more traditionally written simply as $\frac{\partial}{\partial x^i}$, or $\left(\frac{\partial}{\partial x^i}\right)_{(x^1,\dots, x^n)}$ if we wish to emphasize the entire coordinate system (so we know which other coordinates to keep fixed!), and we have the equation (a definition really) $\frac{\partial f}{\partial x^i}:=(df)\left(\frac{\partial}{\partial x^i}\right)$. So I could have simply said that “the equality you asked for follows immediately by evaluating both sides of $N\,d\mu=-S\,dT+V\,dP$ on the coordinate-induced vector field $\left(\frac{\partial}{\partial N}\right)_{(N,T,V)}$.”
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I will try to provide an answer without using differential geometry. In thermodynamics, the state of the system can be fully determined by knowing some thermodynamical variables. In most cases, we need three (This depends on how complex the system is). The most common ones are usually the total energy $E$, the total volume $V$ and the total number of particles $N$. Since there is an equation of state for the system, where the different quantities are related to each other, one can always switch them around and describe the system with another set of variables, for example $N$, $T$ and $V$. The important thing is that we need to consider three of the thermodynamical variables as independent and all other variables as functions of them.
It is also important to notice that $d\mu$ is not on the same footing as $\mu$. $\mu$ is a thermodynamical function of $N$, $T$ and $V$ while $d\mu$ represents a small variation on $\mu$ when you do a small variation on $N$, $T$ and $V$.
Ok, so now let's move on to answer the question. On one hand, we have a relation between small variations of thermodynamical variables
\begin{equation} N d\mu = -S dT+VdP \end{equation}
This equations tells you how $d\mu$ changes if you change temperature or pressure. On the other hand, the variation of any quantity can be written in terms of variations of $N$, $T$ and $V$, since those are our thermodynamical variables. We get
\begin{equation} d\mu = \big(\frac{\partial \mu}{\partial N}\big)_{T,V} dN + \big(\frac{\partial \mu}{\partial T}\big)_{N,V}dT+ \big(\frac{\partial \mu}{\partial V}\big)_{N,T}dV \end{equation}
and
\begin{equation} dP = \big(\frac{\partial P}{\partial N}\big)_{T,V} dN + \big(\frac{\partial P}{\partial T}\big)_{N,V}dT+ \big(\frac{\partial P}{\partial V}\big)_{N,T}dV \end{equation}
Notice that the identity of the factor in frond of any small variation is always the rate of change of $\mu$ respect to the variation of that variable while keeping everything else fixed. Now, we can plug these equations into the first one to get
\begin{equation} N\Big\{\big(\frac{\partial \mu}{\partial N}\big)_{T,V} dN + \big(\frac{\partial \mu}{\partial T}\big)_{N,V}dT+ \big(\frac{\partial \mu}{\partial V}\big)_{N,T}dV \Big\}=-SdT+V\Big\{\big(\frac{\partial P}{\partial N}\big)_{T,V} dN + \big(\frac{\partial P}{\partial T}\big)_{N,V}dT+ \big(\frac{\partial P}{\partial V}\big)_{N,T}dV\Big\} \end{equation}
Now, this long equation is a bit like a vector equation where you have three orthonormal vectors for your basis. We said in the beginning that we would consider $N$, $T$ and $V$ to be our independent variables. That means that a small variation on the number of particles $N$ can't really affect or be caused by a small variation on the temperature $T$. That means that we can group each of those terms and actually get three equations from this lengthy mess. The first one is
\begin{equation} N\big(\frac{\partial \mu}{\partial N}\big)_{T,V}=V\big(\frac{\partial P}{\partial N}\big)_{T,V} \end{equation}
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