- #1
AxiomOfChoice
- 533
- 1
Suppose I am changing variables from [itex](x,y)[/itex] to [itex](s,t)[/itex], where
[tex]
\begin{align*}
s & = \frac 12 (x+y),\\
t & = y - x
\end{align*}
[/tex]
According to Wikipedia, if I want to see how the measure [itex]dx dy[/itex] changes, I need to compute the Jacobian matrix [itex]J[/itex] associated with this variable transformation and take its determinant. It will then follow that [itex]dx dy = \det J ds dt[/itex]. The Jacobian matrix takes the form
[tex]
\begin{bmatrix}
\partial x / \partial s & \partial x / \partial t \\
\partial y / \partial s & \partial y / \partial t
\end{bmatrix}
=
\begin{bmatrix}
1 & -1/2 \\
1 & 1/2
\end{bmatrix}
[/tex]
Is it just coincidence that the matrix [itex]J[/itex] is identical to the matrix of the transformation; i.e., the matrix that shows up in the identity
[tex]
\begin{bmatrix}
x \\ y
\end{bmatrix} =
\begin{bmatrix}
1 & -1/2 \\
1 & 1/2
\end{bmatrix}
\begin{bmatrix}
s \\ t
\end{bmatrix}
[/tex]
[tex]
\begin{align*}
s & = \frac 12 (x+y),\\
t & = y - x
\end{align*}
[/tex]
According to Wikipedia, if I want to see how the measure [itex]dx dy[/itex] changes, I need to compute the Jacobian matrix [itex]J[/itex] associated with this variable transformation and take its determinant. It will then follow that [itex]dx dy = \det J ds dt[/itex]. The Jacobian matrix takes the form
[tex]
\begin{bmatrix}
\partial x / \partial s & \partial x / \partial t \\
\partial y / \partial s & \partial y / \partial t
\end{bmatrix}
=
\begin{bmatrix}
1 & -1/2 \\
1 & 1/2
\end{bmatrix}
[/tex]
Is it just coincidence that the matrix [itex]J[/itex] is identical to the matrix of the transformation; i.e., the matrix that shows up in the identity
[tex]
\begin{bmatrix}
x \\ y
\end{bmatrix} =
\begin{bmatrix}
1 & -1/2 \\
1 & 1/2
\end{bmatrix}
\begin{bmatrix}
s \\ t
\end{bmatrix}
[/tex]