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Multivariate analogue of triple product rule

Physics Asked by Nanite on March 28, 2021

I was doing some thermodynamic calculations and I found the need for a multivariable analogue of the triple product rule. Basically I had a set of $m$ functions $z_i(y, x_1, x_2, ldots, x_m)$, and I needed to calculate partial derivatives of changes in $x$ resulting from a change in $y$ where all $z_i$ values are simultaneously held constant, i.e., I want to calculate:

$$ left(frac{partial x_1}{partial y}right)_{z_1, z_2, ldots, z_m}, ldots, left(frac{partial x_m}{partial y}right)_{z_1, z_2, ldots, z_m} . $$

As a reminder, the regular triple product rule would tell us the following, which unfortunately is not helpful in the multivariable case:

$$ left(frac{partial x}{partial y}right)_{z} = – frac{left(frac{partial z}{partial y}right)_{x}}{left(frac{partial z}{partial x}right)_{y}} qquad qquad text{when $m=1$}. $$

To compactify, let’s use linear algebra notation and write our problem as us knowing $mathbf z (y, mathbf x)$, and wanting to find:

$$ left(frac{partial mathbf x}{partial y}right)_{mathbf z} = ~ ??? $$

I believe I have the answer to this (which I will post as an Answer), but I am unsure as I cannot find it in standard places. This raises a follow-up question for me, which is whether this has a name?


Example

I came across the need for the above rule when trying to work out heat capacity in the grand canonical ensemble, in the case of a multi-component system (multiple chemical potentials $mu_1, ldots, mu_m$ and multiple particle counts $N_1, ldots, N_m$). In this ensemble our natural variables are temperature $T$ and $boldsymbol mu$, but heat capacity is a constant-$mathbf N$ derivative. We have:

begin{align}
C_V & = T left(frac{partial S}{partial T}right)_{V, mathbf N}

& = T left(frac{partial S}{partial T}right)_{V, boldsymbol mu} + T { left(frac{partial S}{partial boldsymbol mu}right)_{V, T} }^T left(frac{partial boldsymbol mu}{partial T}right)_{V, mathbf N} qquad text {(using regular multidimensional chain rule)}
end{align}

So the question is then how to get $ left(frac{partial boldsymbol mu}{partial T}right)_{V, mathbf N} $, i.e., here $mathbf x = boldsymbol mu$ and $mathbf z = mathbf N $. Besides grand canonical ensemble, similar examples would appear if we are considering ensembles involving vectorial quantities, such as magnetization or angular momentum.

One Answer

Essentially we are asking, when $y$ changes, what movement in $mathbf x$ is needed to ensure that $mathbf z$ stays constant. We have this exact differential for changes in $mathbf z$:

$$mathrm dmathbf z = mathbf J , mathrm dmathbf x+ left(frac{partial mathbf z}{partial y}right)_{mathbf x} , mathrm dy $$

where $mathbf J$ is a Jacobian-like matrix, with elements given by:

$$ J_{ij} = left(frac{partial z_i}{partial x_j}right)_{y,x_1,ldots,x_n text{ excluding } x_i}. $$

Now just set $mathrm d mathbf z = 0$, and isolate $mathrm dmathbf x/mathrm d y$ to get the answer:

$$ left(frac{partial mathbf x}{partial y}right)_{mathbf z} = - mathbf J ^{-1} left(frac{partial mathbf z}{partial y}right)_{mathbf x}. $$

Alternatively, we might write this as:

$$ left(frac{partial mathbf x}{partial y}right)_{mathbf z} = - [(boldsymbol nabla_{mathbf x} mathbf z)_y] ^{-1} left(frac{partial mathbf z}{partial y}right)_{mathbf x}. $$


So for my heat capacity example, we get:

$$ C_V = T left(frac{partial S}{partial T}right)_{V, boldsymbol mu} + T { left(frac{partial S}{partial boldsymbol mu}right)_{V, T} }^T left[ (nabla_{boldsymbol mu}mathbf N)_{V, T} right]^{-1} left(frac{partial mathbf N}{partial T}right)_{V, boldsymbol mu} . $$

What's interesting is that the matrix $(nabla_{boldsymbol mu}mathbf N)_{V, T} $ is actually the covariance matrix of particle numbers, multiplied by $kT$. The same matrix also shows up when we try to calculate isothermal compressibility $beta_T$, where $1/beta_T = V (partial p / partial V)_{T, boldsymbol N}$ -- here it is slightly less of a surprise since it's know that particle number fluctuations are exactly related to compressibility, but it's interesting that the inverse of the covariance matrix (i.e., the precision matrix) is what appears in these calculations.

Answered by Nanite on March 28, 2021

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