# Calculation of Jacobian

#### mathmari

##### Well-known member
MHB Site Helper
Hello!!!

Having the transformation t=τ+p, I want to calculate the jacobian $\frac{J(t,τ)}{J(τ,p)}$.
Isn't it $$\frac{J(t,τ)}{J(τ,p)}=\begin{vmatrix} \frac{ \vartheta t}{\vartheta τ}& \frac{\vartheta t}{\vartheta p} \\ \frac{\vartheta τ}{\vartheta τ} & \frac{\varthetaτ }{\vartheta p} \end{vmatrix}=\begin{vmatrix} 1& 1 \\ 1 & -1 \end{vmatrix}=-1-1=-2$$? But the absolute value of the Jacobian for the convolution is $1$..What did I wrong?

#### Fantini

MHB Math Helper
The transformation you actually have is this:

$$\begin{cases} t = \tau + p, \\ \tau = \tau. \end{cases}$$

Seems tautological? Yes, but it's the same way you do cylindrical coordinates:

$$\begin{cases} x = r \cos \theta, \\ y = r \sin \theta, \\ z = z. \end{cases}$$

We see that $\tau$ is independent of $p$. Therefore your Jacobian is

$$\frac{\partial (t, \tau)}{\partial (\tau, p)} = \begin{vmatrix} \frac{\partial t}{\partial \tau} & \frac{\partial t}{\partial p} \\ \frac{\partial \tau}{\partial \tau} & \frac{\partial \tau}{\partial p} \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = -1.$$

#### mathmari

##### Well-known member
MHB Site Helper
The transformation you actually have is this:

$$\begin{cases} t = \tau + p, \\ \tau = \tau. \end{cases}$$

Seems tautological? Yes, but it's the same way you do cylindrical coordinates:

$$\begin{cases} x = r \cos \theta, \\ y = r \sin \theta, \\ z = z. \end{cases}$$

We see that $\tau$ is independent of $p$. Therefore your Jacobian is

$$\frac{\partial (t, \tau)}{\partial (\tau, p)} = \begin{vmatrix} \frac{\partial t}{\partial \tau} & \frac{\partial t}{\partial p} \\ \frac{\partial \tau}{\partial \tau} & \frac{\partial \tau}{\partial p} \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = -1.$$