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My physics textbook presented me with the following relationship:

$$\left( \dfrac{\partial x}{\partial t} \right)_\varphi = \dfrac{-(\partial \varphi / \partial t)_x}{(\partial \varphi / \partial x)_t}$$

In researching this relationship, I encountered this Wikipedia article on the triple product rule. The article says the following:

The triple product rule for such interrelated variables $x$, $y$, and $z$ comes from using a reciprocity relation on the result of the implicit function theorem and is given by

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

Note: In each factor the variable in the numerator is considered to be an implicit function of the other two. In each factor the subscripted variable is being held constant.

Here the subscripts indicate which variables are held constant when the partial derivative is taken. That is, to explicitly compute the partial derivative of $x$ with respect to $y$ with $z$ held constant, one would write $x$ as a function of $y$ and $z$ and take the partial derivative of this function with respect to $y$ only.

The advantage of the triple product rule is that by rearranging terms, one can derive a number of substitution identities which allow one to replace partial derivatives which are difficult to analytically evaluate, experimentally measure, or integrate with quotients of partial derivatives which are easier to work with. For example,

$$\left({\frac {\partial x}{\partial y}}\right)_{z}=-{\frac {\left({\frac {\partial z}{\partial y}}\right)_{x}}{\left({\frac {\partial z}{\partial x}}\right)_{y}}}$$

However, I'm unsure of how $\left({\frac {\partial x}{\partial y}}\right)_{z}=-{\frac {\left({\frac {\partial z}{\partial y}}\right)_{x}}{\left({\frac {\partial z}{\partial x}}\right)_{y}}}$ was derived from $\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}\left({\frac {\partial z}{\partial x}}\right)_{y}=-1$, unless there was some, what seems to me to be, abuse of notation? If we proceed as is typical, we get

$$\begin{align} \left({\frac {\partial x}{\partial y}}\right)_{z} \left({\frac {\partial y}{\partial z}}\right)_{x} = \dfrac{-1}{\left({\frac {\partial z}{\partial x}}\right)_{y}} \\ \Rightarrow \left({\frac {\partial x}{\partial y}}\right)_{z} &= \dfrac{ \left( \dfrac{-1}{\left({\frac {\partial z}{\partial x}}\right)_{y}} \right) }{\left({\frac{\partial y}{\partial z}}\right)_{x}} \\ &= \dfrac{-1}{\left({\frac{\partial z}{\partial x}}\right)_{y} \left({\frac{\partial y}{\partial z}}\right)_{x}} \end{align}$$

I would greatly appreciate it if people could please take the time to clarify this.

Community
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The Pointer
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1 Answers1

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I think it's the other way around. That is, the triple product formula follows from a more basic analysis. Namely let $(x,y,z)\in \mathbb R^3$ and suppose $F:\mathbb R^3\to \mathbb R$ with $F(x,y,z)=0$. Then, if none of the partials vanish at $(x,y,z)$, the implicit function theorem provides functions $f,g,h$ such that $x=f(y,z),\ y=g(x,z),\ z=h(x,y),$ which are differentiable, at least in small neighborhoods contained in their domains.

Then, the matrix representation of the (total) derivative of the composition $F(f(y,z),y,z)$ is given by

$$\begin{pmatrix} F_x &F_y &F_z \end{pmatrix}\begin{pmatrix} f_y & f_z\\ 1 & 0 \\ 0&1 \end{pmatrix}=\begin{pmatrix} 0& 0 \end{pmatrix}$$

and then $F_x\cdot f_y+F_y\cdot 1=0$ and $F_x\cdot f_z+F_z\cdot 1=0$ from which we obtain the familiar formulas

$$\frac{\partial x}{\partial y}=f_y=-\frac{F_y}{F_x}\quad \text{and}\quad \frac{\partial x}{\partial z}=f_z=-\frac{F_z}{F_x}.$$

The other well-known formulas are obtained in the same way using $g$ and $h$ and then the triple product follows immediately by multiplying the appropriate formulas. For example

$$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=\left(-\frac{F_y}{F_x}\right)\left(-\frac{F_z}{F_y}\right)\left(-\frac{F_x}{F_z}\right)=-1.$$

Matematleta
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  • Thanks for the answer. If I'm interpreting this correctly, you've derived the triple product formula from more foundational principles, but I don't see how this addresses my question of deriving $\left({\frac {\partial x}{\partial y}}\right){z}=-{\frac {\left({\frac {\partial z}{\partial y}}\right){x}}{\left({\frac {\partial z}{\partial x}}\right)_{y}}}$? – The Pointer Nov 28 '19 at 02:28
  • It's a direct application of the formulas I derived. If you follow my answer, you will get other similar formulas from which the identity you want pops out immediately. – Matematleta Nov 28 '19 at 02:37
  • Namely, $x_y\cdot z_x=\left(-\frac{F_y}{F_x}\right)\left(-\frac{F_x}{F_z}\right)=\left(\frac{F_y}{F_z}\right)=-z_y$ which is your formula – Matematleta Nov 28 '19 at 02:41
  • Hmm, how did you get $= \left( -\frac{F_y}{F_z} \right)$ in $\left(-\frac{F_y}{F_x}\right)\left(-\frac{F_x}{F_z}\right)=\left(-\frac{F_y}{F_z}\right)$? – The Pointer Nov 28 '19 at 02:51
  • it's $F_y/F_z$ and it comes from canceling the $F_x$ – Matematleta Nov 28 '19 at 03:03
  • But is this abuse of notation valid? That was one of my concerns in my post. – The Pointer Nov 28 '19 at 03:03
  • No this is not an abuse of notation, Did you read my answer from start to finish? – Matematleta Nov 28 '19 at 03:03
  • Oh, yes, you're right, since the $F$ are individual terms and not differentials. My apologies, I confused myself. – The Pointer Nov 28 '19 at 03:04
  • There are a lot of symbols running around, but it is really just an application of the implicit function theorem. Pugh has a very good treatment of the theorem. In case you haven't seen the proof, I suggest you go through it, to see what is really going on here. – Matematleta Nov 28 '19 at 03:07
  • I am going to take the time to go through the steps and try to derive $\left({\frac {\partial x}{\partial y}}\right){z}=-{\frac {\left({\frac {\partial z}{\partial y}}\right){x}}{\left({\frac {\partial z}{\partial x}}\right)_{y}}}$, as you said. – The Pointer Nov 28 '19 at 03:09
  • You basically compute the Jacobian matrices of the composed functions and multiply them and set to $0$. (btw I confuse myself all the time, so you are not alone.) Math is hard – Matematleta Nov 28 '19 at 03:10