Jacobian of a coordinate system wrt another system

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Jacobian of the coordinate- system ($u_1, u_2$) with respect to another coordinate- system (x,y ) is given by

J = $\begin{vmatrix} \frac { \partial {u_1 } } {\partial {x } } & \frac { \partial {u_1 } } {\partial {y} } \\ \frac { \partial {u_2 } } {\partial {x } } & \frac { \partial {u_2 } } {\partial {y } }\end{vmatrix}$

Now, $u_1(x,y), u_2(x,y) = ?$

In polar coordinate system,

$x(r, \theta) = \vec r \cdot \hat x$

$y(r, \theta) = \vec r \cdot \hat y$

Applying the same,

$u_1(x,y)= \vec x \cdot \hat u_1 = x$

$u_2(x,y)= \vec x \cdot \hat u_2 = \frac { k}{\sqrt{ (k^2 x^2 + y^2)}}$

Thus, I get J as a function of x.

Is this correct?

 

[Orodruin]

No. The expressions you quote are only valid for sn orthonormal basis.

 

[tnich]

$u_1(x,y)$ and $u_2(x,y)$ are given in the problem statement. Why not just find the partial derivatives directly?

 

[Pushoam]

$u_1(x,y)$ and $u_2(x,y)$ are given in the problem statement. Why not just find the partial derivatives directly?

$u_1(x,y) = x \\ u_2(x,y)= | \vec u_2| = \sqrt{ (k^2 x^2 + y^2)}$

This also gives J as a function of x.

Is this correct?

 

[Dick]

$u_1(x,y) = x \\ u_2(x,y)= | \vec u_2| = \sqrt{ (k^2 x^2 + y^2)}$

This also gives J as a function of x.

Is this correct?

No, you are overthinking this. A point $\vec v$ is represented as a linear combination $x \vec x +y \vec y$, where $x, y$ are your coordinates. You want to map it to a new point $u_1(x,y) \vec x+u_2(x,y) \vec y$ under a mapping where $\vec x$ gets mapped to $\vec u_1=\vec x$ and $\vec y$ is mapped to $\vec u_2=k\vec x+\vec y$. Now can you figure out what $u_1(x,y)$ and $u_2(x,y)$ are?

 

[Pushoam]

You want to map it to a new point $u_1(x,y) \vec x+u_2(x,y) \vec y$ under a mapping where $\vec x$ gets mapped to $\vec u_1=\vec x$ and $\vec y$ is mapped to $\vec u_2=k\vec x+\vec y$.

I couldn't understand this mapping thing. Could you please give a corresponding link for it?

 

[Dick]

I couldn't understand this mapping thing. Could you please give a corresponding link for it?

I really can't think of anything to link to. The point is that vectors don't have a Jacobian. Transformations have a Jacobian. What they are doing here is describing a linear transformation that takes the vector $\vec x$ into the vector $\vec u_1$ and $\vec y$ into $\vec u_2$. That's what they want the Jacobian of. You do know about linear transformations, right?

 

[Pushoam]

What I understood is: A point given by $\vec v_1$ in the Area $A_1$ gets mapped to another point given by $\vec v_2$ in the Area $A_2$.

$\vec v_1 = x ~\hat x + y ~\hat y \\ \vec v_2 = \vec u_1 + \vec u_2 = x (k+1) ~\hat x + y ~\hat y$

Now, $\vec v_2 = \vec u_1 + \vec u_2 = u_1(x,y) ~\hat x + u_2(x,y) ~\hat y$ ...(1)

Using (1), I get $u_1(x,y) = x (k+1)$ ,

$u_2(x,y) = y$

But,is (1) right? Shouldn't it be $\vec v_2 = \vec u_1 + \vec u_2 = u_1(x,y) ~\hat {u_1} + u_2(x,y) ~\hat {u_2}$ ...(2)?

I am not familiar with linear transformations for these vectors. I used to get $u_1, u_2$ as coordinates and then I used to find out the Jacobian. I suddenly got this question to find out the coordinates from the vectors.

 

[Pushoam]

But why should $\vec v_2 = \vec u_1 + \vec u_2$ ?

$\vec u_1 , \vec u_2$ are not mutually orthogonal.

 

[tnich]

This is a little tricky. The system of equations you want to solve is:
$$u_1 \vec u_1 + u_2 \vec u_2 = x \vec x + y \vec y$$
It helps me to write this in matrix format like this:
$$\begin{bmatrix}
\vec u_1 & \vec u_2
\end{bmatrix}
\begin{bmatrix}
\ u_1 \\ u_2
\end{bmatrix}
=
\begin{bmatrix}
\vec x & \vec y
\end{bmatrix}
\begin{bmatrix}
\ x \\ y
\end{bmatrix}
$$
You already know that $\vec u_1 = \vec x$ and $\vec u_2 = k \vec x + \vec y$, so you can substitute for $\vec u_1$ and $\vec u_2$.
$$\begin{bmatrix}
\vec x & k \vec x + \vec y
\end{bmatrix}
\begin{bmatrix}
\ u_1 \\ u_2
\end{bmatrix}
=
\begin{bmatrix}
\vec x & \vec y
\end{bmatrix}
\begin{bmatrix}
\ x \\ y
\end{bmatrix}
$$
$$\begin{bmatrix}
\vec x & \vec y
\end{bmatrix}
\begin{bmatrix}
1 & k\\
0 &1
\end{bmatrix}
\begin{bmatrix}
\ u_1 \\ u_2
\end{bmatrix}
=
\begin{bmatrix}
\vec x & \vec y
\end{bmatrix}
\begin{bmatrix}
\ x \\ y
\end{bmatrix}
$$
As long as $\begin{bmatrix}\vec x & \vec y\end{bmatrix}$ is invertible (and it is because $\vec x$ and $\vec y$ form a basis), this implies that
$$
\begin{bmatrix}
1 & k\\
0 &1
\end{bmatrix}
\begin{bmatrix}
\ u_1 \\ u_2
\end{bmatrix}
=
\begin{bmatrix}
\ x \\ y
\end{bmatrix}
$$
which you can solve for $u_1(x, y), u_2(x, y)$.

 

[vela]

Are you familiar with how the Jacobian relates the two areas, $A_1$ and $A_2$?

 

[LCKurtz]

Are you familiar with how the Jacobian relates the two areas, $A_1$ and $A_2$?

I've been wondering how long it would be before someone mentioned that, especially since the OP appears to have posted a "quickie" test question.

 

[Pushoam]

Are you familiar with how the Jacobian relates the two areas, $A_1$ and $A_2$?

No.

But, I think the following:

Area $A_1$ consists of all of the vectors which are a linear combinations of $\vec x , \vec y$ ( basis vectors of this vector space) and $A_2$ consists of all of the vectors which are a linear combinations of $\vec u_1, \vec u_2$ .

Now, a vector $\vec r$ is same in both basis system.

So, $\vec r = x \vec x + y \vec y = u_1 \vec u_1 + u_2 \vec u_2$ .

 

[Dick]

No.

But, I think the following:

Area $A_1$ consists of all of the vectors which are a linear combinations of $\vec x , \vec y$ ( basis vectors of this vector space) and $A_2$ consists of all of the vectors which are a linear combinations of $\vec u_1, \vec u_2$ .

Now, a vector $\vec r$ is same in both basis system.

So, $\vec r = x \vec x + y \vec y = u_1 \vec u_1 + u_2 \vec u_2$ .

You seem to be thinking of this as some kind of change of basis problem. I don't think it is. I think they want you to find the Jacobian of the linear transformation $f$ defined by $f(\vec x)=\vec u_1$ and $f(\vec y)=\vec u_2$.

 

[tnich]

You seem to be thinking of this as some kind of change of basis problem. I don't think it is. I think they want you to find the Jacobian of the linear transformation $f$ defined by $f(\vec x)=\vec u_1$ and $f(\vec y)=\vec u_2$.

I thought the same initially, until you pointed out that vectors don't have a Jacobian, transformations do. The change of basis interpretation is also consistent with the observation by @vela about the relationship between the two areas $A_1$ and $A_2$ and the vectors $\vec x, \vec y, \vec u_1$ and $\vec u_2$.