Integration of closed 2-form over $S^2$ independent of radius
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I have this 2-form on $mathbb{R}^3 setminus left{0 right}$:
$$ sigma = frac{ x_1 dx_2 wedge dx_3 - x_2 dx_1 wedge dx_3 + x_3 dx_1 wedge dx_2}{|| x||^3}$$ where $(x_1, x_2, x_3)$ are the standard coordinates on $mathbb{R}^3$ and $||x|| = sqrt{x_1^2 + x_2^2 + x_3^2}$ is the norm. I have shown that $d sigma = 0$ i.e. that the two-form is closed.
I also computed the integral $$ int_{S^2 (r)} j_r^{*} sigma $$ where $j_r$ is the inclusion of the two-sphere $S^2 (r)$ into $mathbb{R}^3 setminus left{0 right}$.
I did this in the standard way by defining $$ F : mathbb{R}^3 setminus left{0 right} rightarrow S^2 (r) : (x_1, x_2, x_3) mapsto (r sin phi cos theta, r sin phi sin theta, r cos phi)$$ where $0 leq phi leq pi$ and $0 leq theta leq 2 pi$ and then working with the pullback $F^{*}$.
I got the result $$ int_{S^2 (r)} j_r^{*} sigma = 4 pi. $$
I think this is correct, because the problem asks to me argue why the result of the integration is independent of the radius . I should use for this the fact that $sigma$ is a closed two-form.
However, I don't know the answer to this, and was hoping for some help. Why is the integral independent of the radius, given that the two-form is closed? Someone has an explanation for this, maybe using Stoke's?
integration differential-geometry differential-forms
asked Jan 2 at 11:29
KamilKamil
2,05721649
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add a comment |
$begingroup$
I have this 2-form on $mathbb{R}^3 setminus left{0 right}$:
$$ sigma = frac{ x_1 dx_2 wedge dx_3 - x_2 dx_1 wedge dx_3 + x_3 dx_1 wedge dx_2}{|| x||^3}$$ where $(x_1, x_2, x_3)$ are the standard coordinates on $mathbb{R}^3$ and $||x|| = sqrt{x_1^2 + x_2^2 + x_3^2}$ is the norm. I have shown that $d sigma = 0$ i.e. that the two-form is closed.
I also computed the integral $$ int_{S^2 (r)} j_r^{*} sigma $$ where $j_r$ is the inclusion of the two-sphere $S^2 (r)$ into $mathbb{R}^3 setminus left{0 right}$.
I did this in the standard way by defining $$ F : mathbb{R}^3 setminus left{0 right} rightarrow S^2 (r) : (x_1, x_2, x_3) mapsto (r sin phi cos theta, r sin phi sin theta, r cos phi)$$ where $0 leq phi leq pi$ and $0 leq theta leq 2 pi$ and then working with the pullback $F^{*}$.
I got the result $$ int_{S^2 (r)} j_r^{*} sigma = 4 pi. $$
I think this is correct, because the problem asks to me argue why the result of the integration is independent of the radius . I should use for this the fact that $sigma$ is a closed two-form.
However, I don't know the answer to this, and was hoping for some help. Why is the integral independent of the radius, given that the two-form is closed? Someone has an explanation for this, maybe using Stoke's?
integration differential-geometry differential-forms
asked Jan 2 at 11:29
KamilKamil
2,05721649
$endgroup$
3
$begingroup$
Apply Stokes to the $3$-dimensional "annulus" bounded by two spheres of radii $r_1$ and $r_2$ centred at the origin.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:41
add a comment |
$begingroup$
I have this 2-form on $mathbb{R}^3 setminus left{0 right}$:
$$ sigma = frac{ x_1 dx_2 wedge dx_3 - x_2 dx_1 wedge dx_3 + x_3 dx_1 wedge dx_2}{|| x||^3}$$ where $(x_1, x_2, x_3)$ are the standard coordinates on $mathbb{R}^3$ and $||x|| = sqrt{x_1^2 + x_2^2 + x_3^2}$ is the norm. I have shown that $d sigma = 0$ i.e. that the two-form is closed.
I also computed the integral $$ int_{S^2 (r)} j_r^{*} sigma $$ where $j_r$ is the inclusion of the two-sphere $S^2 (r)$ into $mathbb{R}^3 setminus left{0 right}$.
I did this in the standard way by defining $$ F : mathbb{R}^3 setminus left{0 right} rightarrow S^2 (r) : (x_1, x_2, x_3) mapsto (r sin phi cos theta, r sin phi sin theta, r cos phi)$$ where $0 leq phi leq pi$ and $0 leq theta leq 2 pi$ and then working with the pullback $F^{*}$.
I got the result $$ int_{S^2 (r)} j_r^{*} sigma = 4 pi. $$
I think this is correct, because the problem asks to me argue why the result of the integration is independent of the radius . I should use for this the fact that $sigma$ is a closed two-form.
However, I don't know the answer to this, and was hoping for some help. Why is the integral independent of the radius, given that the two-form is closed? Someone has an explanation for this, maybe using Stoke's?
integration differential-geometry differential-forms
asked Jan 2 at 11:29
KamilKamil
2,05721649
$endgroup$
I have this 2-form on $mathbb{R}^3 setminus left{0 right}$:
$$ sigma = frac{ x_1 dx_2 wedge dx_3 - x_2 dx_1 wedge dx_3 + x_3 dx_1 wedge dx_2}{|| x||^3}$$ where $(x_1, x_2, x_3)$ are the standard coordinates on $mathbb{R}^3$ and $||x|| = sqrt{x_1^2 + x_2^2 + x_3^2}$ is the norm. I have shown that $d sigma = 0$ i.e. that the two-form is closed.
I also computed the integral $$ int_{S^2 (r)} j_r^{*} sigma $$ where $j_r$ is the inclusion of the two-sphere $S^2 (r)$ into $mathbb{R}^3 setminus left{0 right}$.
I did this in the standard way by defining $$ F : mathbb{R}^3 setminus left{0 right} rightarrow S^2 (r) : (x_1, x_2, x_3) mapsto (r sin phi cos theta, r sin phi sin theta, r cos phi)$$ where $0 leq phi leq pi$ and $0 leq theta leq 2 pi$ and then working with the pullback $F^{*}$.
I got the result $$ int_{S^2 (r)} j_r^{*} sigma = 4 pi. $$
I think this is correct, because the problem asks to me argue why the result of the integration is independent of the radius . I should use for this the fact that $sigma$ is a closed two-form.
However, I don't know the answer to this, and was hoping for some help. Why is the integral independent of the radius, given that the two-form is closed? Someone has an explanation for this, maybe using Stoke's?
integration differential-geometry differential-forms
integration differential-geometry differential-forms
asked Jan 2 at 11:29
KamilKamil
2,05721649
asked Jan 2 at 11:29
KamilKamil
2,05721649
asked Jan 2 at 11:29
KamilKamil
2,05721649
asked Jan 2 at 11:29
KamilKamil
2,05721649
asked Jan 2 at 11:29
KamilKamil
2,05721649
2,05721649
3
$begingroup$
Apply Stokes to the $3$-dimensional "annulus" bounded by two spheres of radii $r_1$ and $r_2$ centred at the origin.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:41
add a comment |
3
$begingroup$
Apply Stokes to the $3$-dimensional "annulus" bounded by two spheres of radii $r_1$ and $r_2$ centred at the origin.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:41
3
3
$begingroup$
Apply Stokes to the $3$-dimensional "annulus" bounded by two spheres of radii $r_1$ and $r_2$ centred at the origin.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:41
$begingroup$
Apply Stokes to the $3$-dimensional "annulus" bounded by two spheres of radii $r_1$ and $r_2$ centred at the origin.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:41
add a comment |
0
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$begingroup$
Apply Stokes to the $3$-dimensional "annulus" bounded by two spheres of radii $r_1$ and $r_2$ centred at the origin.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:41