Inverse of Jacobian matrices

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  • #1
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Homework Statement:

I want to prove that ##\left| \frac{\partial(x,y)}{\partial(u,v)} \right|=\frac{1}{\left|\frac{\partial(u,v)}{\partial(x,y)}\right|}## (If this is true)

Relevant Equations:

##\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix} x_u & x_v\\ y_u&y_v \end {vmatrix}##
##\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix} u_x & u_y\\ v_x&v_y \end {vmatrix}##
I got that ##{x_u}{y_v}-{x_y}{y_u}=####\frac{1}{\frac{1}{{x_u}{y_v}}-\frac{1}{{y_u}{x_v}}}##. But this implies that ##{x_u}{x_v}{y_u}{y_v}=-1## and I don't see how that is true?
 

Answers and Replies

  • #2
anuttarasammyak
Gold Member
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Very simple case of u=x, v=y obviously the statement stands. but
[tex]x_u x_v y_u y_v = 1*0*0*1=0[/tex]
So your result seems wrong.

Find 2X2 matrices A, B by partial differentiation
[tex]
\begin{pmatrix}
du \\
dv \\
\end{pmatrix}

= A

\begin{pmatrix}
dx \\
dy \\
\end{pmatrix}[/tex]

[tex]
\begin{pmatrix}
dx \\
dy \\
\end{pmatrix}

= B

\begin{pmatrix}
du \\
dv \\
\end{pmatrix}[/tex]

So
[tex]AB=BA=E[/tex]
[tex]det\ A\ \ det\ B = 1[/tex]
 
Last edited:
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  • #3
pasmith
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By the chain rule [tex]
1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x}[/tex] and [tex]
0 = \frac{\partial y}{\partial x} = \frac{\partial y}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x}[/tex] and similarly [itex]\frac{\partial y}{\partial y} = 1[/itex] and [itex]\frac{\partial x}{\partial y} = 0[/itex]. Now I suspect that if you expand [tex]
1 = \frac{\partial x}{\partial x}\frac{\partial y}{\partial y} - \frac{\partial y}{\partial x}\frac{\partial x}{\partial y}[/tex] and rearrange it then you will obtain your result.
 
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  • #4
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By the chain rule [tex]
1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x}[/tex] and [tex]
0 = \frac{\partial y}{\partial x} = \frac{\partial y}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x}[/tex] and similarly [itex]\frac{\partial y}{\partial y} = 1[/itex] and [itex]\frac{\partial x}{\partial y} = 0[/itex]. Now I suspect that if you expand [tex]
1 = \frac{\partial x}{\partial x}\frac{\partial y}{\partial y} - \frac{\partial y}{\partial x}\frac{\partial x}{\partial y}[/tex] and rearrange it then you will obtain your result.
I think I understand. But if ##\frac {\partial x}{\partial u}=\frac{1}{\frac {\partial u}{\partial x}}##, then doesn't ##\frac {\partial x}{\partial u}\frac {\partial u}{\partial x}=1##? Then how do you resolve the contradiction that ##1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x} = 2##?
 
  • #5
pasmith
Homework Helper
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I think I understand. But if ##\frac {\partial x}{\partial u}=\frac{1}{\frac {\partial u}{\partial x}}##, then doesn't ##\frac {\partial x}{\partial u}\frac {\partial u}{\partial x}=1##? Then how do you resolve the contradiction that ##1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x} = 2##?
The relation [tex]\frac{\partial x}{\partial u} = \left(\frac{\partial u}{\partial x}\right)^{-1}[/tex] does not hold in general. For example, with plane polar coordinates we have [tex]\frac{\partial r}{\partial x} = \frac xr[/tex] and [tex]\frac{\partial x}{\partial r} = \cos \theta = \frac xr \neq \frac rx.[/tex]

Rather, if that relation does hold then it implies that [itex]x[/itex] is independent of [itex]v[/itex] so that [itex]\frac{\partial x}{\partial v} = 0[/itex].
 
  • #6
anuttarasammyak
Gold Member
333
158
Further to my post #2

[tex]A=
\begin{pmatrix}
u_x & u_y \\
v_x & v_y \\
\end{pmatrix}[/tex]
[tex]B=
\begin{pmatrix}
x_u & x_v \\
y_u & y_v \\
\end{pmatrix}[/tex]
[tex]AB=
\begin{pmatrix}
u_x & u_y \\
v_x & v_y \\
\end{pmatrix}
\begin{pmatrix}
x_u & x_v \\
y_u & y_v \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}[/tex]

It meets with post #3.

[tex]A=B^{-1}[/tex]
[tex]
\begin{pmatrix}
u_x & u_y \\
v_x & v_y \\
\end{pmatrix}
=\frac{1}{x_uy_v-x_vy_u}
\begin{pmatrix}
y_v & -x_v \\
-y_u & x_u \\
\end{pmatrix}[/tex]

For an example (1,1) component says

[tex]u_x=\frac{1}{x_uy_v-x_vy_u}y_v[/tex]

You see in order ##u_x =\frac{1}{x_u}## as you expect, ##x_vy_u=0 ## and ##y_v \neq 0## is required.
 
Last edited:

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