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Jacobian of a coordinate system wrt another system

  • #1
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Homework Statement


upload_2018-2-9_23-40-41.png


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?
 

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Answers and Replies

  • #2
Orodruin
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No. The expressions you quote are only valid for sn orthonormal basis.
 
  • #3
tnich
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## u_1(x,y)## and ##u_2(x,y)## are given in the problem statement. Why not just find the partial derivatives directly?
 
  • #4
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## 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?
 
  • #5
Dick
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## 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?
 
  • #6
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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?
 
  • #7
Dick
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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?
 
  • #8
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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.
 
  • #9
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But why should ## \vec v_2 = \vec u_1 + \vec u_2 ## ?

## \vec u_1 , \vec u_2 ## are not mutually orthogonal.
 
  • #10
tnich
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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)##.
 
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  • #11
vela
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Are you familiar with how the Jacobian relates the two areas, ##A_1## and ##A_2##?
 
  • #12
LCKurtz
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Are you familiar with how the Jacobian relates the two areas, ##A_1## and ##A_2##?
:oldsmile: 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.
 
  • #13
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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 ## .
 
  • #14
Dick
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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##.
 
  • #15
tnich
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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##.
 

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