differential forms- $omega $ closed but not exact
$begingroup$
let be
$$ omega= |x|^{-3} left(x_1 dx_2 wedge dx_3+x_2dx_3 wedge dx_1 + x_3dx_1 wedge dx_2right) $$
and $G:= mathbb{R}^3 backslash { 0 } $
I want to prove, that $ omega$ is closed, but not exact
That $ omega $ is closed, I can prove it by looking if $ domega =0 $
But how can I prove it's not exact?
I know that a continous 2-form is exact in $G$ if there exists a 1-Form $y$ so, that $omega = dy $
how can I show there doesn't exist such $y$?
Edit:
Showing $int_S omega neq 0 $
Where $S$ is the unitsphere
Set $r=1$ and set
$$x_1= sin phi cos theta $$
$$x_2= sin phi sin theta$$
$$x_3= cos phi $$
$ theta=[0, 2pi], phi=[0, pi]$
then $ dx_1 wedge dx_2 = -sin phi cos phi dtheta wedge dphi $
$ dx_2 wedge dx_3 = -sin^2 phi cos theta dtheta wedge dphi $
$ dx_3 wedge dx_1 = -sin^2 phi sin theta d theta wedge dphi $
Putting in the equation:
$$ int_0^{2 pi} int_0^{ pi} |x| ( sin phi cos theta - sin^2 phi cos theta ~d theta wedge d phi \+ sin phi sin theta -sin^2 phi sin theta ~dtheta wedge dphi + cos phi -sin phi cos phi~ d theta wedge dphi $$
what do I put for $x$ ?
I don't know how to solve the integral
thank you for any help!
real-analysis multivariable-calculus differential-forms
edited Jan 16 at 14:16
John Hughes
65.2k24293
asked Jan 16 at 12:32
wondering1123wondering1123
14911
$endgroup$
|
show 5 more comments
$begingroup$
let be
$$ omega= |x|^{-3} left(x_1 dx_2 wedge dx_3+x_2dx_3 wedge dx_1 + x_3dx_1 wedge dx_2right) $$
and $G:= mathbb{R}^3 backslash { 0 } $
I want to prove, that $ omega$ is closed, but not exact
That $ omega $ is closed, I can prove it by looking if $ domega =0 $
But how can I prove it's not exact?
I know that a continous 2-form is exact in $G$ if there exists a 1-Form $y$ so, that $omega = dy $
how can I show there doesn't exist such $y$?
Edit:
Showing $int_S omega neq 0 $
Where $S$ is the unitsphere
Set $r=1$ and set
$$x_1= sin phi cos theta $$
$$x_2= sin phi sin theta$$
$$x_3= cos phi $$
$ theta=[0, 2pi], phi=[0, pi]$
then $ dx_1 wedge dx_2 = -sin phi cos phi dtheta wedge dphi $
$ dx_2 wedge dx_3 = -sin^2 phi cos theta dtheta wedge dphi $
$ dx_3 wedge dx_1 = -sin^2 phi sin theta d theta wedge dphi $
Putting in the equation:
$$ int_0^{2 pi} int_0^{ pi} |x| ( sin phi cos theta - sin^2 phi cos theta ~d theta wedge d phi \+ sin phi sin theta -sin^2 phi sin theta ~dtheta wedge dphi + cos phi -sin phi cos phi~ d theta wedge dphi $$
what do I put for $x$ ?
I don't know how to solve the integral
thank you for any help!
real-analysis multivariable-calculus differential-forms
edited Jan 16 at 14:16
John Hughes
65.2k24293
asked Jan 16 at 12:32
wondering1123wondering1123
14911
$endgroup$
1
$begingroup$
The first thing to try would be to integrate it on the unit sphere; if the result is not zero, then $omega$ is not exact.
$endgroup$
– Alex Provost
Jan 16 at 12:42
$begingroup$
@AlexProvost What does it mean to integrate the unit sphere?
$endgroup$
– MathOverview
Jan 16 at 12:43
$begingroup$
@MathOverview I said to "integrate it on the unit sphere", meaning the 2-form.
$endgroup$
– Alex Provost
Jan 16 at 12:44
$begingroup$
@AlexProvost But why integrate in the unitary sphere and not in $ G $?
$endgroup$
– MathOverview
Jan 16 at 12:46
1
$begingroup$
On the unit sphere, $|x| = 1$. By the way, this reduces to the standard volume form on $S^2$, so you expect the integral to yield the surface area, $4pi$.
$endgroup$
– Alex Provost
Jan 16 at 13:42
|
show 5 more comments
$begingroup$
let be
$$ omega= |x|^{-3} left(x_1 dx_2 wedge dx_3+x_2dx_3 wedge dx_1 + x_3dx_1 wedge dx_2right) $$
and $G:= mathbb{R}^3 backslash { 0 } $
I want to prove, that $ omega$ is closed, but not exact
That $ omega $ is closed, I can prove it by looking if $ domega =0 $
But how can I prove it's not exact?
I know that a continous 2-form is exact in $G$ if there exists a 1-Form $y$ so, that $omega = dy $
how can I show there doesn't exist such $y$?
Edit:
Showing $int_S omega neq 0 $
Where $S$ is the unitsphere
Set $r=1$ and set
$$x_1= sin phi cos theta $$
$$x_2= sin phi sin theta$$
$$x_3= cos phi $$
$ theta=[0, 2pi], phi=[0, pi]$
then $ dx_1 wedge dx_2 = -sin phi cos phi dtheta wedge dphi $
$ dx_2 wedge dx_3 = -sin^2 phi cos theta dtheta wedge dphi $
$ dx_3 wedge dx_1 = -sin^2 phi sin theta d theta wedge dphi $
Putting in the equation:
$$ int_0^{2 pi} int_0^{ pi} |x| ( sin phi cos theta - sin^2 phi cos theta ~d theta wedge d phi \+ sin phi sin theta -sin^2 phi sin theta ~dtheta wedge dphi + cos phi -sin phi cos phi~ d theta wedge dphi $$
what do I put for $x$ ?
I don't know how to solve the integral
thank you for any help!
real-analysis multivariable-calculus differential-forms
edited Jan 16 at 14:16
John Hughes
65.2k24293
asked Jan 16 at 12:32
wondering1123wondering1123
14911
$endgroup$
let be
$$ omega= |x|^{-3} left(x_1 dx_2 wedge dx_3+x_2dx_3 wedge dx_1 + x_3dx_1 wedge dx_2right) $$
and $G:= mathbb{R}^3 backslash { 0 } $
I want to prove, that $ omega$ is closed, but not exact
That $ omega $ is closed, I can prove it by looking if $ domega =0 $
But how can I prove it's not exact?
I know that a continous 2-form is exact in $G$ if there exists a 1-Form $y$ so, that $omega = dy $
how can I show there doesn't exist such $y$?
Edit:
Showing $int_S omega neq 0 $
Where $S$ is the unitsphere
Set $r=1$ and set
$$x_1= sin phi cos theta $$
$$x_2= sin phi sin theta$$
$$x_3= cos phi $$
$ theta=[0, 2pi], phi=[0, pi]$
then $ dx_1 wedge dx_2 = -sin phi cos phi dtheta wedge dphi $
$ dx_2 wedge dx_3 = -sin^2 phi cos theta dtheta wedge dphi $
$ dx_3 wedge dx_1 = -sin^2 phi sin theta d theta wedge dphi $
Putting in the equation:
$$ int_0^{2 pi} int_0^{ pi} |x| ( sin phi cos theta - sin^2 phi cos theta ~d theta wedge d phi \+ sin phi sin theta -sin^2 phi sin theta ~dtheta wedge dphi + cos phi -sin phi cos phi~ d theta wedge dphi $$
what do I put for $x$ ?
I don't know how to solve the integral
thank you for any help!
real-analysis multivariable-calculus differential-forms
real-analysis multivariable-calculus differential-forms
edited Jan 16 at 14:16
John Hughes
65.2k24293
asked Jan 16 at 12:32
wondering1123wondering1123
14911
edited Jan 16 at 14:16
John Hughes
65.2k24293
asked Jan 16 at 12:32
wondering1123wondering1123
14911
edited Jan 16 at 14:16
John Hughes
65.2k24293
edited Jan 16 at 14:16
John Hughes
65.2k24293
edited Jan 16 at 14:16
John Hughes
65.2k24293
65.2k24293
asked Jan 16 at 12:32
wondering1123wondering1123
14911
asked Jan 16 at 12:32
wondering1123wondering1123
14911
asked Jan 16 at 12:32
wondering1123wondering1123
14911
14911
1
$begingroup$
The first thing to try would be to integrate it on the unit sphere; if the result is not zero, then $omega$ is not exact.
$endgroup$
– Alex Provost
Jan 16 at 12:42
$begingroup$
@AlexProvost What does it mean to integrate the unit sphere?
$endgroup$
– MathOverview
Jan 16 at 12:43
$begingroup$
@MathOverview I said to "integrate it on the unit sphere", meaning the 2-form.
$endgroup$
– Alex Provost
Jan 16 at 12:44
$begingroup$
@AlexProvost But why integrate in the unitary sphere and not in $ G $?
$endgroup$
– MathOverview
Jan 16 at 12:46
1
$begingroup$
On the unit sphere, $|x| = 1$. By the way, this reduces to the standard volume form on $S^2$, so you expect the integral to yield the surface area, $4pi$.
$endgroup$
– Alex Provost
Jan 16 at 13:42
|
show 5 more comments
1
$begingroup$
The first thing to try would be to integrate it on the unit sphere; if the result is not zero, then $omega$ is not exact.
$endgroup$
– Alex Provost
Jan 16 at 12:42
$begingroup$
@AlexProvost What does it mean to integrate the unit sphere?
$endgroup$
– MathOverview
Jan 16 at 12:43
$begingroup$
@MathOverview I said to "integrate it on the unit sphere", meaning the 2-form.
$endgroup$
– Alex Provost
Jan 16 at 12:44
$begingroup$
@AlexProvost But why integrate in the unitary sphere and not in $ G $?
$endgroup$
– MathOverview
Jan 16 at 12:46
1
$begingroup$
On the unit sphere, $|x| = 1$. By the way, this reduces to the standard volume form on $S^2$, so you expect the integral to yield the surface area, $4pi$.
$endgroup$
– Alex Provost
Jan 16 at 13:42
1
1
$begingroup$
The first thing to try would be to integrate it on the unit sphere; if the result is not zero, then $omega$ is not exact.
$endgroup$
– Alex Provost
Jan 16 at 12:42
$begingroup$
The first thing to try would be to integrate it on the unit sphere; if the result is not zero, then $omega$ is not exact.
$endgroup$
– Alex Provost
Jan 16 at 12:42
$begingroup$
@AlexProvost What does it mean to integrate the unit sphere?
$endgroup$
– MathOverview
Jan 16 at 12:43
$begingroup$
@AlexProvost What does it mean to integrate the unit sphere?
$endgroup$
– MathOverview
Jan 16 at 12:43
$begingroup$
@MathOverview I said to "integrate it on the unit sphere", meaning the 2-form.
$endgroup$
– Alex Provost
Jan 16 at 12:44
$begingroup$
@MathOverview I said to "integrate it on the unit sphere", meaning the 2-form.
$endgroup$
– Alex Provost
Jan 16 at 12:44
$begingroup$
@AlexProvost But why integrate in the unitary sphere and not in $ G $?
$endgroup$
– MathOverview
Jan 16 at 12:46
$begingroup$
@AlexProvost But why integrate in the unitary sphere and not in $ G $?
$endgroup$
– MathOverview
Jan 16 at 12:46
1
1
$begingroup$
On the unit sphere, $|x| = 1$. By the way, this reduces to the standard volume form on $S^2$, so you expect the integral to yield the surface area, $4pi$.
$endgroup$
– Alex Provost
Jan 16 at 13:42
$begingroup$
On the unit sphere, $|x| = 1$. By the way, this reduces to the standard volume form on $S^2$, so you expect the integral to yield the surface area, $4pi$.
$endgroup$
– Alex Provost
Jan 16 at 13:42
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Switch to spherical coordinates and integrate $omega$ on the unit sphere $r = 1$. Let $theta,phi$ denote the azimuth and polar angles, respectively; then $$dx = cos theta sin phi , dr - sin theta sin phi , dtheta + costheta cos phi , dphi$$
$$ dy = sin theta sin phi , dr + cos theta sinphi , dtheta + sin theta cos phi , dphi $$
$$dz = cos phi , dr - sinphi , dphi$$
and after some algebra we find that $omega = sinphi , dphi wedge dtheta$. Hence $$int_{S^2} omega = int_0^{2pi} int_0^pi sin phi , dphi , dtheta = 2cdot 2pi = 4pi.$$
This is the expected answer, namely the surface area of the sphere, as $omega$ restricted to $S^2$ is precisely the volume form induced by that of $mathbb R^3$. And since the integral is not zero, $omega$ cannot be exact, lest Stoke's theorem be violated.
answered Jan 16 at 17:32
Alex ProvostAlex Provost
15.6k22351
$endgroup$
add a comment |
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$begingroup$
Switch to spherical coordinates and integrate $omega$ on the unit sphere $r = 1$. Let $theta,phi$ denote the azimuth and polar angles, respectively; then $$dx = cos theta sin phi , dr - sin theta sin phi , dtheta + costheta cos phi , dphi$$
$$ dy = sin theta sin phi , dr + cos theta sinphi , dtheta + sin theta cos phi , dphi $$
$$dz = cos phi , dr - sinphi , dphi$$
and after some algebra we find that $omega = sinphi , dphi wedge dtheta$. Hence $$int_{S^2} omega = int_0^{2pi} int_0^pi sin phi , dphi , dtheta = 2cdot 2pi = 4pi.$$
This is the expected answer, namely the surface area of the sphere, as $omega$ restricted to $S^2$ is precisely the volume form induced by that of $mathbb R^3$. And since the integral is not zero, $omega$ cannot be exact, lest Stoke's theorem be violated.
answered Jan 16 at 17:32
Alex ProvostAlex Provost
15.6k22351
$endgroup$
add a comment |
$begingroup$
Switch to spherical coordinates and integrate $omega$ on the unit sphere $r = 1$. Let $theta,phi$ denote the azimuth and polar angles, respectively; then $$dx = cos theta sin phi , dr - sin theta sin phi , dtheta + costheta cos phi , dphi$$
$$ dy = sin theta sin phi , dr + cos theta sinphi , dtheta + sin theta cos phi , dphi $$
$$dz = cos phi , dr - sinphi , dphi$$
and after some algebra we find that $omega = sinphi , dphi wedge dtheta$. Hence $$int_{S^2} omega = int_0^{2pi} int_0^pi sin phi , dphi , dtheta = 2cdot 2pi = 4pi.$$
This is the expected answer, namely the surface area of the sphere, as $omega$ restricted to $S^2$ is precisely the volume form induced by that of $mathbb R^3$. And since the integral is not zero, $omega$ cannot be exact, lest Stoke's theorem be violated.
answered Jan 16 at 17:32
Alex ProvostAlex Provost
15.6k22351
$endgroup$
add a comment |
$begingroup$
Switch to spherical coordinates and integrate $omega$ on the unit sphere $r = 1$. Let $theta,phi$ denote the azimuth and polar angles, respectively; then $$dx = cos theta sin phi , dr - sin theta sin phi , dtheta + costheta cos phi , dphi$$
$$ dy = sin theta sin phi , dr + cos theta sinphi , dtheta + sin theta cos phi , dphi $$
$$dz = cos phi , dr - sinphi , dphi$$
and after some algebra we find that $omega = sinphi , dphi wedge dtheta$. Hence $$int_{S^2} omega = int_0^{2pi} int_0^pi sin phi , dphi , dtheta = 2cdot 2pi = 4pi.$$
This is the expected answer, namely the surface area of the sphere, as $omega$ restricted to $S^2$ is precisely the volume form induced by that of $mathbb R^3$. And since the integral is not zero, $omega$ cannot be exact, lest Stoke's theorem be violated.
answered Jan 16 at 17:32
Alex ProvostAlex Provost
15.6k22351
$endgroup$
Switch to spherical coordinates and integrate $omega$ on the unit sphere $r = 1$. Let $theta,phi$ denote the azimuth and polar angles, respectively; then $$dx = cos theta sin phi , dr - sin theta sin phi , dtheta + costheta cos phi , dphi$$
$$ dy = sin theta sin phi , dr + cos theta sinphi , dtheta + sin theta cos phi , dphi $$
$$dz = cos phi , dr - sinphi , dphi$$
and after some algebra we find that $omega = sinphi , dphi wedge dtheta$. Hence $$int_{S^2} omega = int_0^{2pi} int_0^pi sin phi , dphi , dtheta = 2cdot 2pi = 4pi.$$
This is the expected answer, namely the surface area of the sphere, as $omega$ restricted to $S^2$ is precisely the volume form induced by that of $mathbb R^3$. And since the integral is not zero, $omega$ cannot be exact, lest Stoke's theorem be violated.
answered Jan 16 at 17:32
Alex ProvostAlex Provost
15.6k22351
edited Jan 16 at 20:54
answered Jan 16 at 17:32
Alex ProvostAlex Provost
15.6k22351
answered Jan 16 at 17:32
Alex ProvostAlex Provost
15.6k22351
answered Jan 16 at 17:32
Alex ProvostAlex Provost
15.6k22351
15.6k22351
add a comment |
add a comment |
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$begingroup$
The first thing to try would be to integrate it on the unit sphere; if the result is not zero, then $omega$ is not exact.
$endgroup$
– Alex Provost
Jan 16 at 12:42
$begingroup$
@AlexProvost What does it mean to integrate the unit sphere?
$endgroup$
– MathOverview
Jan 16 at 12:43
$begingroup$
@MathOverview I said to "integrate it on the unit sphere", meaning the 2-form.
$endgroup$
– Alex Provost
Jan 16 at 12:44
$begingroup$
@AlexProvost But why integrate in the unitary sphere and not in $ G $?
$endgroup$
– MathOverview
Jan 16 at 12:46
1
$begingroup$
On the unit sphere, $|x| = 1$. By the way, this reduces to the standard volume form on $S^2$, so you expect the integral to yield the surface area, $4pi$.
$endgroup$
– Alex Provost
Jan 16 at 13:42