Given is a coordinate system $\{x_1,x_2,…,x_n\}$ and another, second coordinate system $\{y_1,y_2,…,y_n\}$, where
$x_1=x_1(y_1,y_2,…,y_n)$
$x_2=x_2(y_1,y_2,…,y_n)$
…
$x_n=x_n(y_1,y_2,…,y_n)$
Then the Jacobian matrix is
$${\mathbf J}=\begin{pmatrix}
\frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} & … & \frac{\partial x_1}{\partial y_n}\\
\frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} & … & \frac{\partial x_2}{\partial y_n}\\
… \\
\frac{\partial x_n}{\partial y_1} & \frac{\partial x_n}{\partial y_2} & … & \frac{\partial x_n}{\partial y_n}\\
\end{pmatrix}$$
Now, consider vector $\vec{u}$. Its coordinates in $\{x_k\}$ are
$\vec{u}=(u_{x1},u_{x2},…,u_{xn})$, while its coordinates in $\{y_k\}$ are
$\vec{u}=(u_{y1},u_{y2},…,u_{yn})$, where
$$\begin{pmatrix}
u_{x1}\\
u_{x2}\\
… \\
u_{xn}\\
\end{pmatrix}=
{\mathbf A} \cdot
\begin{pmatrix}
u_{y1}\\
u_{y2}\\
… \\
u_{yn}\\
\end{pmatrix}$$
What is the difference between the Jacobian matrix ${\mathbf J}$ and the Transformation matrix ${\mathbf A}$? How are they related? Please, write the expression that connects them.
————EXAMPLE———–
Cartesian and cylindrical coordinates are related via
$x=r\cos\theta$
$y=r\sin\theta$
$z=z$
Then the Jacobian is
$${\mathbf J}=\begin{pmatrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z}\\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z}\\
\frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z}\\
\end{pmatrix}=
\begin{pmatrix}
\cos\theta & -r\sin\theta & 0\\
\sin\theta & r\cos\theta & 0\\
0 & 0 & 1\\
\end{pmatrix}$$
and $\det({\mathbf J})=r$. Because ${\mathbf J}$ is orthogonal when $r=1$,
${\mathbf J}^{-1}={\mathbf J}^{T}=\begin{pmatrix}
\cos\theta & \sin\theta & 0\\
-\sin\theta & \cos\theta & 0\\
0 & 0 & 1\\
\end{pmatrix}$
The Cartesian basis vectors are
$\hat{\mathbf i} = \begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}$;
$\hat{\mathbf j} = \begin{pmatrix}
0\\
1\\
0\\
\end{pmatrix}$;
$\hat{\mathbf k} = \begin{pmatrix}
0\\
0\\
1\\
\end{pmatrix}$;
The cylindrical basis vectors are
$\hat{\mathbf r}=
\begin{pmatrix}
\cos\theta\\
\sin\theta\\
0\\
\end{pmatrix}$;
$\hat{\mathbf \theta}=
\begin{pmatrix}
-\sin\theta\\
\cos\theta\\
0\\
\end{pmatrix}$;
$\hat{\mathbf z} = \begin{pmatrix}
0\\
0\\
1\\
\end{pmatrix}$.
It seems that
$\hat{\mathbf r}={\mathbf A}\cdot \hat{\mathbf i}$
$\hat{\mathbf \theta}={\mathbf A}\cdot \hat{\mathbf j}$
$\hat{\mathbf z}={\mathbf A}\cdot \hat{\mathbf k}$
if the transformation matrix ${\mathbf A}$ is given by
${\mathbf A}=\frac{1}{\det({\mathbf J})}{\mathbf J}^{-1}$.
Is this conclusion true? Is this the relationship between ${\mathbf A}$ and ${\mathbf J}$?
Best Answer
The term "Jacobian" traditionally refers to the determinant of the derivative matrix. The derivative matrix can be thought of as a local transformation matrix.
If you want the amount of change ${dx,dy,dz}$ due to a change ${dr,d\theta,dx}$ multiply the derivative matrix by the latter as a column vector. It's just the chain rule.
Think it through, geometrically.