Given is a coordinate system $\{x_1,x_2,...,x_n\}$ and another, second coordinate system $\{y_1,y_2,...,y_n\}$, where





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.


Cartesian and cylindrical coordinates are related via




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 ${\mathbf J}$?

  • 1
    $\begingroup$ Only when the change of variable is linear you get the coordinates changing by multiplication by a matrix. $\endgroup$
    – OR.
    Jun 22, 2017 at 0:13
  • $\begingroup$ The transformation matrix is a Jacobian matrix limited to linear transformations. Also the example Jacobian matrix you provided is not orthogonal as the magnitude of the elements of the 2nd column != 1, rather = r. $\endgroup$
    – Sentient
    Jun 22, 2017 at 0:15
  • $\begingroup$ The example you used is misleading because it is not interpreting the introduction correctly. if the coordinates of a vector in Cartesian coordinates are $(x,y,z)$, then its coordinates in the cylindrical system should be $(r,z,\theta)$ for the corresponding values of $r$, $z$ and $\theta$. $\endgroup$
    – OR.
    Jun 22, 2017 at 0:17
  • 1
    $\begingroup$ Yes, it is confusing. But forget about vectors. Vectors in $R^n$ look too much like their own coordinates and the multiple dimensions don't add anything to the understanding. Just look in $R$. Take a coordinate change like $x = y^3+y$. Say that the vector $1$ has coordinate $y=1$. Then in the $x$-coordinates it has coordinate $x=1^3+1=2$. If the change of coordinates were given by a multiplication by a matrix the matrix would have to be $(2)$ and in fact $2=(2)\cdot1$. $\endgroup$
    – OR.
    Jun 22, 2017 at 2:11
  • 1
    $\begingroup$ But the same matrix would not work for other vectors. For the vector $3$ with $y$-coordinate $y=3$, we get $x=3^3+3=30$ and $30\neq (2)\cdot 3$. $\endgroup$
    – OR.
    Jun 22, 2017 at 2:11

1 Answer 1


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.

  • $\begingroup$ wikipedia $\endgroup$
    – OR.
    Jun 22, 2017 at 2:13
  • $\begingroup$ I defer to the authoritative sources. Jacobi, as I understand it, did not invent the Jacobian determinat, per se. The determinant of the vector gradient (also known as the derivative matrix) was nominated "the Jacobian" by Arthur Cayley to honor Jacobi's contribution to the study of determinants, in general. $\endgroup$ Jun 22, 2017 at 2:57
  • 4
    $\begingroup$ Naming doesn't follow authority, but usage. $\endgroup$
    – OR.
    Jun 22, 2017 at 3:01
  • $\begingroup$ @StevenHatton, +1 for your answer and your time. I was trying to understand vectors better. How do they transform? Is there a difference in the transformation law when transforming between two orthogonal coordinate systems and when transforming between orthogonal and non-orthogonal one? $\endgroup$
    – user142523
    Jun 22, 2017 at 15:41
  • $\begingroup$ That is a different question, and should probably posted as such. The short answer is yes. It should also be noted that transformation between locally orthogonal coordinate systems can be fairly complicated because, for example $d\phi$ may be measured in radians, and $dx$ in units of length. en.wikipedia.org/wiki/… $\endgroup$ Jun 22, 2017 at 15:49

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