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How to compute a Jacobian using polar coordinates?
How do I convert a vector field in Cartesian coordinates to spherical coordinates?Simple proof of integration in polar coordinates?gradient in polar coordinate by changing gradient in Cartesian coordinateJacobian Determinant of Polar-Coordinate TransformationParameter Transformation with the JacobiandA in polar coordinates using total differentialsElementary JacobianUsing the Jacobian matrix to find surface area without a change of basis.Gradient in polar coordinatesWhat is the Jacobian in this transformation
$begingroup$
Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
$$
F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$
Its Jacobian matrix is
$$tag1
beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$
The following alternative computation is wrong at (!) and (!!), and I cannot see why.
Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$
The result (!!) is off by a factor of $r^-2$ from the correct one, which is given in (1). Equation (!) must also be wrong. Indeed, computing the Jacobian matrix from (2) using the chain rule, and using that
$$
Dphi = beginbmatrix cos theta & sin theta \
-rsin theta & rcos thetaendbmatrix$$
and that
$$tag!!!
D(phi^-1)= beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix,$$
I obtain the result
$$
beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$
which is different from the matrix in (!), and which gives the correct determinant of $-1/r^4$, as it should be.
Can you help me spot the mistake?
SOLUTION (added at a later time). As answerers pointed out, there is a mistake in (!!!). The correct matrix to be used is
$$ D(phi^-1)|_Fcirc phi(r, theta)
= beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix.$$
Had I used this matrix, I would have found the correct result for the Jacobian matrix of $tildeF$, which is the equation marked (!). Thus, (!) is actually correct.
My fundamental misunderstanding was the assumption that, because of (2), the Jacobian determinant should be invariant for coordinate changes. This is not true; what follows from (2) is only that
$$
det DtildeF|_(r, theta)= det Dphi^-1|_Fcircphi(r, theta)det Dphi|_(r, theta) det DF|_phi(r, theta).
$$
The first two factors in the right-hand side need not cancel, as I erroneously thought.
calculus multivariable-calculus differential-geometry
asked Apr 23 at 20:07
Giuseppe NegroGiuseppe Negro
17.7k332128
$endgroup$
add a comment |
$begingroup$
Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
$$
F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$
Its Jacobian matrix is
$$tag1
beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$
The following alternative computation is wrong at (!) and (!!), and I cannot see why.
Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$
The result (!!) is off by a factor of $r^-2$ from the correct one, which is given in (1). Equation (!) must also be wrong. Indeed, computing the Jacobian matrix from (2) using the chain rule, and using that
$$
Dphi = beginbmatrix cos theta & sin theta \
-rsin theta & rcos thetaendbmatrix$$
and that
$$tag!!!
D(phi^-1)= beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix,$$
I obtain the result
$$
beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$
which is different from the matrix in (!), and which gives the correct determinant of $-1/r^4$, as it should be.
Can you help me spot the mistake?
SOLUTION (added at a later time). As answerers pointed out, there is a mistake in (!!!). The correct matrix to be used is
$$ D(phi^-1)|_Fcirc phi(r, theta)
= beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix.$$
Had I used this matrix, I would have found the correct result for the Jacobian matrix of $tildeF$, which is the equation marked (!). Thus, (!) is actually correct.
My fundamental misunderstanding was the assumption that, because of (2), the Jacobian determinant should be invariant for coordinate changes. This is not true; what follows from (2) is only that
$$
det DtildeF|_(r, theta)= det Dphi^-1|_Fcircphi(r, theta)det Dphi|_(r, theta) det DF|_phi(r, theta).
$$
The first two factors in the right-hand side need not cancel, as I erroneously thought.
calculus multivariable-calculus differential-geometry
asked Apr 23 at 20:07
Giuseppe NegroGiuseppe Negro
17.7k332128
$endgroup$
add a comment |
$begingroup$
Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
$$
F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$
Its Jacobian matrix is
$$tag1
beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$
The following alternative computation is wrong at (!) and (!!), and I cannot see why.
Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$
The result (!!) is off by a factor of $r^-2$ from the correct one, which is given in (1). Equation (!) must also be wrong. Indeed, computing the Jacobian matrix from (2) using the chain rule, and using that
$$
Dphi = beginbmatrix cos theta & sin theta \
-rsin theta & rcos thetaendbmatrix$$
and that
$$tag!!!
D(phi^-1)= beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix,$$
I obtain the result
$$
beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$
which is different from the matrix in (!), and which gives the correct determinant of $-1/r^4$, as it should be.
Can you help me spot the mistake?
SOLUTION (added at a later time). As answerers pointed out, there is a mistake in (!!!). The correct matrix to be used is
$$ D(phi^-1)|_Fcirc phi(r, theta)
= beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix.$$
Had I used this matrix, I would have found the correct result for the Jacobian matrix of $tildeF$, which is the equation marked (!). Thus, (!) is actually correct.
My fundamental misunderstanding was the assumption that, because of (2), the Jacobian determinant should be invariant for coordinate changes. This is not true; what follows from (2) is only that
$$
det DtildeF|_(r, theta)= det Dphi^-1|_Fcircphi(r, theta)det Dphi|_(r, theta) det DF|_phi(r, theta).
$$
The first two factors in the right-hand side need not cancel, as I erroneously thought.
calculus multivariable-calculus differential-geometry
asked Apr 23 at 20:07
Giuseppe NegroGiuseppe Negro
17.7k332128
$endgroup$
Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
$$
F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$
Its Jacobian matrix is
$$tag1
beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$
The following alternative computation is wrong at (!) and (!!), and I cannot see why.
Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$
The result (!!) is off by a factor of $r^-2$ from the correct one, which is given in (1). Equation (!) must also be wrong. Indeed, computing the Jacobian matrix from (2) using the chain rule, and using that
$$
Dphi = beginbmatrix cos theta & sin theta \
-rsin theta & rcos thetaendbmatrix$$
and that
$$tag!!!
D(phi^-1)= beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix,$$
I obtain the result
$$
beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$
which is different from the matrix in (!), and which gives the correct determinant of $-1/r^4$, as it should be.
Can you help me spot the mistake?
SOLUTION (added at a later time). As answerers pointed out, there is a mistake in (!!!). The correct matrix to be used is
$$ D(phi^-1)|_Fcirc phi(r, theta)
= beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix.$$
Had I used this matrix, I would have found the correct result for the Jacobian matrix of $tildeF$, which is the equation marked (!). Thus, (!) is actually correct.
My fundamental misunderstanding was the assumption that, because of (2), the Jacobian determinant should be invariant for coordinate changes. This is not true; what follows from (2) is only that
$$
det DtildeF|_(r, theta)= det Dphi^-1|_Fcircphi(r, theta)det Dphi|_(r, theta) det DF|_phi(r, theta).
$$
The first two factors in the right-hand side need not cancel, as I erroneously thought.
calculus multivariable-calculus differential-geometry
calculus multivariable-calculus differential-geometry
asked Apr 23 at 20:07
Giuseppe NegroGiuseppe Negro
17.7k332128
asked Apr 23 at 20:07
Giuseppe NegroGiuseppe Negro
17.7k332128
edited Apr 24 at 10:34
Giuseppe Negro
asked Apr 23 at 20:07
Giuseppe NegroGiuseppe Negro
17.7k332128
asked Apr 23 at 20:07
Giuseppe NegroGiuseppe Negro
17.7k332128
asked Apr 23 at 20:07
Giuseppe NegroGiuseppe Negro
17.7k332128
17.7k332128
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't think there is any contradiction here.
Consider the volume form
$$ omega_rm Cart = dx wedge dy.$$
Your first calculation shows that the pullback $F^star(omega_rm Cart)$ is given by
$$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$
Now consider the volume form
$$ omega_rm Polar = dr wedge dtheta.$$
Your second calculation shows that
$$ F^star(omega_rm Polar)=-frac 1 r^2 omega_rm Polar. $$
We can use this to recompute $F^star(omega_rm Cart)$. In view of the fact that
$$ omega_rm Cart = r omega_rm Polar,$$
we have:
beginalign
F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
endalign
which is consistent with the first calculation!
As for the application of the chain rule, we have:
$$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$
The key point is that you must evaluate $D(phi^-1)$ at the point $left(frac x (x^2 +y^2), frac y (x^2 + y^2)right)$, not at the point $(x, y)$.
This is equal to
$$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
which is not the inverse of $(Dphi)|_(r, theta)$.
answered Apr 23 at 20:54
Kenny WongKenny Wong
20.1k21442
$endgroup$
$begingroup$
I couldn't expect a better answer. You nailed it completely. Also, thank you for suggesting the viewpoint of volume forms; that makes for much slicker computations.
$endgroup$
– Giuseppe Negro
Apr 23 at 22:10
add a comment |
$begingroup$
The Jacobians of the two functions aren't equal by the chain rule.
In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$
answered Apr 23 at 21:04
George DewhirstGeorge Dewhirst
1,86025
$endgroup$
$begingroup$
I accepted the other answer, but this one is also correct and useful, as I remarked in the "Solution" section, at the end of the question. Thank you.
$endgroup$
– Giuseppe Negro
Apr 24 at 11:27
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I don't think there is any contradiction here.
Consider the volume form
$$ omega_rm Cart = dx wedge dy.$$
Your first calculation shows that the pullback $F^star(omega_rm Cart)$ is given by
$$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$
Now consider the volume form
$$ omega_rm Polar = dr wedge dtheta.$$
Your second calculation shows that
$$ F^star(omega_rm Polar)=-frac 1 r^2 omega_rm Polar. $$
We can use this to recompute $F^star(omega_rm Cart)$. In view of the fact that
$$ omega_rm Cart = r omega_rm Polar,$$
we have:
beginalign
F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
endalign
which is consistent with the first calculation!
As for the application of the chain rule, we have:
$$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$
The key point is that you must evaluate $D(phi^-1)$ at the point $left(frac x (x^2 +y^2), frac y (x^2 + y^2)right)$, not at the point $(x, y)$.
This is equal to
$$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
which is not the inverse of $(Dphi)|_(r, theta)$.
answered Apr 23 at 20:54
Kenny WongKenny Wong
20.1k21442
$endgroup$
$begingroup$
I couldn't expect a better answer. You nailed it completely. Also, thank you for suggesting the viewpoint of volume forms; that makes for much slicker computations.
$endgroup$
– Giuseppe Negro
Apr 23 at 22:10
add a comment |
$begingroup$
I don't think there is any contradiction here.
Consider the volume form
$$ omega_rm Cart = dx wedge dy.$$
Your first calculation shows that the pullback $F^star(omega_rm Cart)$ is given by
$$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$
Now consider the volume form
$$ omega_rm Polar = dr wedge dtheta.$$
Your second calculation shows that
$$ F^star(omega_rm Polar)=-frac 1 r^2 omega_rm Polar. $$
We can use this to recompute $F^star(omega_rm Cart)$. In view of the fact that
$$ omega_rm Cart = r omega_rm Polar,$$
we have:
beginalign
F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
endalign
which is consistent with the first calculation!
As for the application of the chain rule, we have:
$$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$
The key point is that you must evaluate $D(phi^-1)$ at the point $left(frac x (x^2 +y^2), frac y (x^2 + y^2)right)$, not at the point $(x, y)$.
This is equal to
$$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
which is not the inverse of $(Dphi)|_(r, theta)$.
answered Apr 23 at 20:54
Kenny WongKenny Wong
20.1k21442
$endgroup$
$begingroup$
I couldn't expect a better answer. You nailed it completely. Also, thank you for suggesting the viewpoint of volume forms; that makes for much slicker computations.
$endgroup$
– Giuseppe Negro
Apr 23 at 22:10
add a comment |
$begingroup$
I don't think there is any contradiction here.
Consider the volume form
$$ omega_rm Cart = dx wedge dy.$$
Your first calculation shows that the pullback $F^star(omega_rm Cart)$ is given by
$$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$
Now consider the volume form
$$ omega_rm Polar = dr wedge dtheta.$$
Your second calculation shows that
$$ F^star(omega_rm Polar)=-frac 1 r^2 omega_rm Polar. $$
We can use this to recompute $F^star(omega_rm Cart)$. In view of the fact that
$$ omega_rm Cart = r omega_rm Polar,$$
we have:
beginalign
F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
endalign
which is consistent with the first calculation!
As for the application of the chain rule, we have:
$$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$
The key point is that you must evaluate $D(phi^-1)$ at the point $left(frac x (x^2 +y^2), frac y (x^2 + y^2)right)$, not at the point $(x, y)$.
This is equal to
$$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
which is not the inverse of $(Dphi)|_(r, theta)$.
answered Apr 23 at 20:54
Kenny WongKenny Wong
20.1k21442
$endgroup$
I don't think there is any contradiction here.
Consider the volume form
$$ omega_rm Cart = dx wedge dy.$$
Your first calculation shows that the pullback $F^star(omega_rm Cart)$ is given by
$$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$
Now consider the volume form
$$ omega_rm Polar = dr wedge dtheta.$$
Your second calculation shows that
$$ F^star(omega_rm Polar)=-frac 1 r^2 omega_rm Polar. $$
We can use this to recompute $F^star(omega_rm Cart)$. In view of the fact that
$$ omega_rm Cart = r omega_rm Polar,$$
we have:
beginalign
F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
endalign
which is consistent with the first calculation!
As for the application of the chain rule, we have:
$$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$
The key point is that you must evaluate $D(phi^-1)$ at the point $left(frac x (x^2 +y^2), frac y (x^2 + y^2)right)$, not at the point $(x, y)$.
This is equal to
$$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
which is not the inverse of $(Dphi)|_(r, theta)$.
answered Apr 23 at 20:54
Kenny WongKenny Wong
20.1k21442
edited Apr 23 at 21:23
answered Apr 23 at 20:54
Kenny WongKenny Wong
20.1k21442
answered Apr 23 at 20:54
Kenny WongKenny Wong
20.1k21442
answered Apr 23 at 20:54
Kenny WongKenny Wong
20.1k21442
20.1k21442
$begingroup$
I couldn't expect a better answer. You nailed it completely. Also, thank you for suggesting the viewpoint of volume forms; that makes for much slicker computations.
$endgroup$
– Giuseppe Negro
Apr 23 at 22:10
add a comment |
$begingroup$
I couldn't expect a better answer. You nailed it completely. Also, thank you for suggesting the viewpoint of volume forms; that makes for much slicker computations.
$endgroup$
– Giuseppe Negro
Apr 23 at 22:10
$begingroup$
I couldn't expect a better answer. You nailed it completely. Also, thank you for suggesting the viewpoint of volume forms; that makes for much slicker computations.
$endgroup$
– Giuseppe Negro
Apr 23 at 22:10
$begingroup$
I couldn't expect a better answer. You nailed it completely. Also, thank you for suggesting the viewpoint of volume forms; that makes for much slicker computations.
$endgroup$
– Giuseppe Negro
Apr 23 at 22:10
add a comment |
$begingroup$
The Jacobians of the two functions aren't equal by the chain rule.
In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$
answered Apr 23 at 21:04
George DewhirstGeorge Dewhirst
1,86025
$endgroup$
$begingroup$
I accepted the other answer, but this one is also correct and useful, as I remarked in the "Solution" section, at the end of the question. Thank you.
$endgroup$
– Giuseppe Negro
Apr 24 at 11:27
add a comment |
$begingroup$
The Jacobians of the two functions aren't equal by the chain rule.
In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$
answered Apr 23 at 21:04
George DewhirstGeorge Dewhirst
1,86025
$endgroup$
$begingroup$
I accepted the other answer, but this one is also correct and useful, as I remarked in the "Solution" section, at the end of the question. Thank you.
$endgroup$
– Giuseppe Negro
Apr 24 at 11:27
add a comment |
$begingroup$
The Jacobians of the two functions aren't equal by the chain rule.
In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$
answered Apr 23 at 21:04
George DewhirstGeorge Dewhirst
1,86025
$endgroup$
The Jacobians of the two functions aren't equal by the chain rule.
In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$
answered Apr 23 at 21:04
George DewhirstGeorge Dewhirst
1,86025
answered Apr 23 at 21:04
George DewhirstGeorge Dewhirst
1,86025
answered Apr 23 at 21:04
George DewhirstGeorge Dewhirst
1,86025
answered Apr 23 at 21:04
George DewhirstGeorge Dewhirst
1,86025
1,86025
$begingroup$
I accepted the other answer, but this one is also correct and useful, as I remarked in the "Solution" section, at the end of the question. Thank you.
$endgroup$
– Giuseppe Negro
Apr 24 at 11:27
add a comment |
$begingroup$
I accepted the other answer, but this one is also correct and useful, as I remarked in the "Solution" section, at the end of the question. Thank you.
$endgroup$
– Giuseppe Negro
Apr 24 at 11:27
$begingroup$
I accepted the other answer, but this one is also correct and useful, as I remarked in the "Solution" section, at the end of the question. Thank you.
$endgroup$
– Giuseppe Negro
Apr 24 at 11:27
$begingroup$
I accepted the other answer, but this one is also correct and useful, as I remarked in the "Solution" section, at the end of the question. Thank you.
$endgroup$
– Giuseppe Negro
Apr 24 at 11:27
add a comment |
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