The Jacobian determinant as the ratio of differential volume elements

Question

In basic courses on analysis, one learns that the Jacobian determinant represents the change of differential volume under a coordinate transformation. For example in 2 dimensions:

If

$$u = f_1(x,y) \\ v = f_2(x,y)$$

our Jacobian matrix would be

$$J = \left( \begin{array}{cc} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{array}\right)$$

and it's determinant

$$|J| = \frac{\partial f_1}{\partial x} \frac{\partial f_2}{\partial y} - \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x}$$ should now be the ratio of the infinitesimal areas $dudv = J dxdy$.

We get the same result if we form the differentials $$du = \frac{\partial f_1}{\partial x} dx + \frac{\partial f_1}{\partial y} dy \\ dv = \frac{\partial f_2}{\partial x} dx + \frac{\partial f_2}{\partial y} dy$$ interpret the $d$-terms as forms and multiply these equations with the wedge product.

$$ du \wedge dv = \frac{\partial f_1}{\partial x}\frac{\partial f_2}{\partial x} dx \wedge dx + \frac{\partial f_1}{\partial x} \frac{\partial f_2}{\partial y} dx \wedge dy + \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x} dy \wedge dx + \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial y} dy \wedge dy $$ Using the properties of the wedge product:

$$ du \wedge dv = \left(\frac{\partial f_1}{\partial x} \frac{\partial f_2}{\partial y} - \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x}\right) dx \wedge dy $$

My question now is: Why does the "naive" approach lead to a wrong result:

$$ dudv = \frac{\partial f_1}{\partial x}\frac{\partial f_2}{\partial x} dx^2 + \left(\frac{\partial f_1}{\partial x} \frac{\partial f_2}{\partial y} + \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x}\right) dxdy + \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial y} dy^2 $$

Why do I need forms (with the concept of orientation) to get the correct result? My math classes have always treated differentials as a quick and "dirty" way to cut short calculations, but are there any texts that provide a consistent and rigorous treatment of differentials, and their interpretation as forms?

Answer 1

In essence, the problem is that the unit volume spanned by $\mathrm du$ and $\mathrm dv$ does not need to be a rectangular parallelogram, in which case its volume is obviously no longer given by the product of the lengths of the sides.

Thus this is an issue with the concept of volume itself, and it will remain even if your transformation $(x,y)\mapsto(u,v)$ is a general linear transformation. The outer product of covectors was built to capture the concept of volume in this context, and the wedge product of one-forms is its transparent generalization to the case of curvilinear coordinates.

Answer 2

Your question had puzzled me too for a while and deserved a more rigorous answer: I think I figured it out.

The total differential of a variable never assumes that any one of the four variables here is constant throughout the equation: $$du = \overbrace{\frac{\partial u}{\partial x} dx}^{\text{change in}\\u\text{ due to }x\\ \;\; dy=0} + \overbrace{\frac{\partial u}{\partial y} dy}^{\text{change in}\\u\text{ due to }y\\ \;\;dx=0},$$ in this case, $v$, though absent in this equation, is not constant because $$v=v(x,y)\cap \{dx\ne 0 \cup dy\ne 0\},$$ whereas, for the purpose of substitution in integration, we are not interested in seeing more than one integration variable change at a time, so this is not applicable, and I don't know how to justify your derivation, which appears to treat the total differentials $du$ & $dv$ as vector sums, prior to cross-multiplying them; why it works may be a fluke.

Here's the more intuitive approach. It keeps the $x$-$y$ and $u$-$v$ planes separate and looks at mapping each coordinate's displacement vector into a displacement vector in the other plane, one coordinate at a time. Letting $$u_x=\frac{\partial u}{\partial x},\ v_x=\frac{\partial v}{\partial x},\ u_y=\frac{\partial u}{\partial y},\ v_y=\frac{\partial u}{\partial y}, $$ the mapping begins here: $$\begin{align} (x,y) &\leftrightarrow (u,v) \\ (dx,0) &\leftrightarrow (u_x\,dx,v_x\,dx) \\ (0,dy) &\leftrightarrow (u_y\,dy,v_y\,dy) \\ \det\left[\begin{matrix} dx & 0 \\ 0 & dy \end{matrix}\right]&\leftrightarrow \left|\det\left[\begin{matrix} u_x\,dx & v_x\,dx \\ u_y\,dy & v_y\,dy \end{matrix}\right]\right|\quad\text{// differential volumes}\\ dx\,dy &\leftrightarrow |u_x v_y-v_x u_y|\,dx\,dy=du\,dv, \end{align} $$ although one is square and the other a parallelogram, $du\,dv$ refers to a rectangular region having the same area (generally referred to as "volume") as that of the parallelogram mapped by $dx\,dy$.
An integration function at point $(u,v)$, mapped by $(x,y)$, has the same value: $$ \begin{align} f(x,y) &= f(x(u,v),y(u,v))=g(u,v) \\ \therefore f(x,y)\,dx\,dy &= g(u,v)\,dx\,dy=\frac{g(u,v)}{|u_x v_y-v_x u_y|}\,du\,dv. \end{align} $$ Note that there is not enough information to infer the orientation of the axes or their scales from just the functional relationship between these variables or from their Jacobian.

Answer 3

I straggled with the very same question and what I got is that the theorem statement involves integrals and not differentials. In particular the theorem I'm referring to is usually known as the change of variables theorem and it states the following; \begin{equation} \int_{\Omega}{g(\vec{u})d\Omega}=\int_{V}{G(\vec{x})JdV}\qquad (1) \end{equation} and NOT: \begin{equation} d\Omega=JdV\qquad(2) \end{equation} where $\vec{u}(\vec{x}):A\subseteq\mathbb{R}^{n}\longrightarrow\mathbb{R}^{n}\text{ is }C^1[A],J\text{ is the determinant of }D\vec{u}\equiv d\vec{u}/d\vec{x}^T, G\equiv (g\circ\vec{u})(\vec{x})$ and $g:\vec{u}(B)\longrightarrow\mathbb{R}\text{ is integrable with }B\subseteq A$. Furthermore I assumed: $\vec{u}\equiv[u\quad v\quad w]^T, \vec{x}\equiv[x\quad y\quad z]^T\quad\text{s.t.}\quad d\Omega\equiv dudvdw,\quad dV\equiv dxdydz$

This means that the formulation (2) $d\Omega=JdV$ is wrong per se and the correct one is formulation (1). In any physical or scientifically book you usually find (2), but in math books I've not found yet formulation (2). I have to say that usually when (2) is used, it's implied that it's true only if it's replaced inside an integral, but anyway I consider it a misleading formula which should not stand on its own without any further explanation.

What I want to emphasise is that the correct formulation of the statement requires the integral (more precisely a ${measure}$ is required). As a consequence, the wedge product formulation is correct because it implies areas (of a parallelogram for 2D) or volumes (of a parallelepiped for 3D if involved in a triple product). Instead the infinitesimal doesn't imply a measure, at least a priori. In particular: \begin{equation} |d\vec{u}\wedge d\vec{v}|=|d\vec{u}||d\vec{v}|\sin{\gamma}\ne dudv\qquad(3) \end{equation} where $\gamma$ is the angle between the vectors $d\vec{u}$ and $d\vec{v}$. The disequivalence in (3) is due not only for the presence of $\sin{\gamma}$ but also because the module of the vectors $d\vec{u}$ and $d\vec{v}$ is different from the corresponding differentials $du$ and $dv$, namely $du\ne|d\vec{u}|$ and $dv\ne|d\vec{v}|$, in particular: \begin{equation} du\equiv\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy\quad\text{ but }\quad|d\vec{u}|\equiv\sqrt{\Bigl(\frac{\partial u}{\partial x}dx\Bigr)^2+\Bigl(\frac{\partial u}{\partial y}dy\Bigr)^2}\longrightarrow du\ne|d\vec{u}|\\ dv\equiv\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy\quad\text{ but }\quad|d\vec{v}|\equiv\sqrt{\Bigl(\frac{\partial v}{\partial x}dx\Bigr)^2+\Bigl(\frac{\partial v}{\partial y}dy\Bigr)^2}\longrightarrow dv\ne|d\vec{v}| \end{equation}

For instance, in the 2D case, the edges $d\vec{x}$ and $d\vec{y}$ (respectively along $\hat{x}$ axis and $\hat{y}$ axis) identify a square of area $dxdy$. They are mapped into two infinitesimal vectors $d\vec{\alpha}$ and $d\vec{\beta}$, which identify a parallelogram: \begin{equation} d\vec{\alpha}=D\vec{u}\,d\vec{x}=D\vec{u}[dx\quad 0]^T=\frac{\partial\vec{u}}{\partial x}dx=\Bigl[\frac{\partial u}{\partial x}dx\quad\frac{\partial v}{\partial x}dx\Bigr]^T \end{equation} \begin{equation} d\vec{\beta}=D\vec{u}\,d\vec{y}=D\vec{u}[0\quad dy]^T=\frac{\partial\vec{u}}{\partial y}dy=\Bigl[\frac{\partial u}{\partial y}dy\quad\frac{\partial v}{\partial y}dy\Bigr]^T \end{equation}

The couple of vectors $d\vec{\alpha}$ and $d\vec{\beta}$ is different, in general, from the couple of vectors $d\vec{u}$ and $d\vec{v}$. In particular these two couples represent the edges of two different parallelograms but with the same area. Indeed the following relations hold recalling that the area of a parallelogram can be calculated as the determinant of the matrix which rows (or columns) are the edges of the parallelogram itself: \begin{equation}[d\vec{\alpha}\quad d\vec{\beta}]=\Bigl[\begin{smallmatrix}d\vec{u}^T\\d\vec{v}^T\end{smallmatrix}\Bigr]\text{ since } \begin{cases} d\vec{u}\equiv\bigl[\frac{\partial u}{\partial x}dx\quad\frac{\partial u}{\partial y}dy\bigr]^T\\ d\vec{v}\equiv\bigl[\frac{\partial v}{\partial x}dx\quad\frac{\partial v}{\partial y}dy\bigr]^T \end{cases} \end{equation}

\begin{equation} |d\vec{\alpha}\wedge d\vec{\beta}|=|d\vec{u}\wedge d\vec{v}|=det\Bigl(\Bigl[ \begin{smallmatrix} \frac{\partial u}{\partial x}dx & \frac{\partial v}{\partial x}dx \\ \frac{\partial u}{\partial y}dy & \frac{\partial v}{\partial y}dy \\ \end{smallmatrix}\Bigr]\Bigr)=det\Bigl(\Bigl[ \begin{smallmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \\ \end{smallmatrix}\Bigr]dxdy\Bigr)=det\Bigl(\Bigl[ \begin{smallmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \\ \end{smallmatrix}\Bigr]\Bigr)dxdy\equiv Jdxdy \end{equation}

Such that $|d\vec{u}\wedge d\vec{v}|=Jdxdy$. But since in general $|d\vec{u}\wedge d\vec{v}|\ne dudv$, it's not possible to state a priori that $dudv=Jdxdy$. For instance, let's consider the 2d polar coordinates: \begin{equation} \begin{cases}u=x\cos{y}\\v=x\sin{y}\end{cases}\longrightarrow \begin{cases} d\vec{u}=\bigl[\frac{\partial u}{\partial x}dx\quad\frac{\partial u}{\partial y}dy\bigr]^T=\bigl[\begin{smallmatrix}\cos{y}dx\\-x\sin{y}dy\end{smallmatrix}\bigr]\\ d\vec{v}=\bigl[\frac{\partial v}{\partial x}dx\quad\frac{\partial v}{\partial y}dy\bigr]^T=\bigl[\begin{smallmatrix}\sin{y}dx\\x\cos{y}dy\end{smallmatrix}\bigr] \end{cases} \end{equation} It's not possible to write $dudv=Jdxdy$, but only $|d\vec{u}\wedge d\vec{v}|=Jdxdy$, indeed: \begin{equation} |d\vec{u}\wedge d\vec{v}|=x\cos{y}dx\cos{y}dy-\sin{y}dx(-x\sin{y}dy)= x((\cos{y})^2+(\sin{y})^2)dxdy=xdxdy\implies J=x \end{equation} \begin{equation} dudv\equiv(\cos{y}dx−x\sin{y}dy)(\sin{y}dx+x\cos{y}dy)=\cos{y}\sin{y}dx^2-x^2\cos{y}\sin{y}dy^2+x(\cos{y})^2dxdy-x(\sin{y})^2dxdy\ne xdxdy \end{equation}