Integration with generalized spherical coordinates Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Change of Variable vs. Change of Coordinates.Question about some inequalityintegral of a 2-form over an oriented manifoldPolar Coordinates in $mathbb R^n$Finding the volume of a cube using spherical coordinatesComparison of the change of variable theoremWhat does it mean take the determinant of the Jacobian in: $ V_k+1 = int_M_k Biggvert det(fracpartial ypartial x) Biggvert dx$Integration Over Spherical Triangle and Change of VariableGradient in polar coordinatesShow that integral of $f$ over a manifold $M$ is independent of coordinate chart
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Integration with generalized spherical coordinates
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Change of Variable vs. Change of Coordinates.Question about some inequalityintegral of a 2-form over an oriented manifoldPolar Coordinates in $mathbb R^n$Finding the volume of a cube using spherical coordinatesComparison of the change of variable theoremWhat does it mean take the determinant of the Jacobian in: $ V_k+1 = int_M_k Biggvert det(fracpartial ypartial x) Biggvert dx$Integration Over Spherical Triangle and Change of VariableGradient in polar coordinatesShow that integral of $f$ over a manifold $M$ is independent of coordinate chart
$begingroup$
Define $varphi:(0,infty)times S^ntomathbbR^n+1$ by $varphi(t,theta)=ttheta$. Let $E$ be a subset of $mathbbR^n+1$ and let $E_rsubseteqmathbbR^n+1$ contain all points $thetain S^n$ such that $tthetain E$ for some $t>0$. Then $int_Efdx_1...dx_n+1=int_-infty^inftyint_E_rfcircvarphi r^ndrdv$, where $dv$ is the volume form of the n sphere.
I'm with the classical change of variable. For this, I need to know why the determinant of $Dvarphi=r^n$, but I don't see an easy way of seeing this.
real-analysis calculus
$endgroup$
add a comment |
$begingroup$
Define $varphi:(0,infty)times S^ntomathbbR^n+1$ by $varphi(t,theta)=ttheta$. Let $E$ be a subset of $mathbbR^n+1$ and let $E_rsubseteqmathbbR^n+1$ contain all points $thetain S^n$ such that $tthetain E$ for some $t>0$. Then $int_Efdx_1...dx_n+1=int_-infty^inftyint_E_rfcircvarphi r^ndrdv$, where $dv$ is the volume form of the n sphere.
I'm with the classical change of variable. For this, I need to know why the determinant of $Dvarphi=r^n$, but I don't see an easy way of seeing this.
real-analysis calculus
$endgroup$
$begingroup$
You will also need a constant; the Jacobian is not $r^n$, it is $C_n r^n$, where $C_n$ is the surface measure of the $n$-sphere.
$endgroup$
– Giuseppe Negro
Mar 27 at 17:06
$begingroup$
"Let $E_r subseteq Bbb R^n+1$ contain all points $theta in S^n$ such that $ttheta in E_r$ for some $t>0$". Could you please explain what this means?
$endgroup$
– Paul Sinclair
Mar 28 at 0:21
$begingroup$
@PaulSinclair sorry, there was a typo. I've fixed it.
$endgroup$
– user124910
Mar 28 at 3:36
$begingroup$
@GiuseppeNegro - that constant $C_n$ comes from the integration over $S^n$. In this case, the integration is only over $E_r$, and has yet to be performed,
$endgroup$
– Paul Sinclair
Mar 28 at 23:44
$begingroup$
Right; the Jacobian is $r^n$, not $C_nr^n$. Sorry for my mistake. Concerning the question, I prefer to see it in terms of metric tensors; $$ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This can be seen by expanding $ds^2=(d(rtheta))^2.$ Then, the volume form formula follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$. Not too rigorous, but faster for sure.
$endgroup$
– Giuseppe Negro
Mar 28 at 23:51
add a comment |
$begingroup$
Define $varphi:(0,infty)times S^ntomathbbR^n+1$ by $varphi(t,theta)=ttheta$. Let $E$ be a subset of $mathbbR^n+1$ and let $E_rsubseteqmathbbR^n+1$ contain all points $thetain S^n$ such that $tthetain E$ for some $t>0$. Then $int_Efdx_1...dx_n+1=int_-infty^inftyint_E_rfcircvarphi r^ndrdv$, where $dv$ is the volume form of the n sphere.
I'm with the classical change of variable. For this, I need to know why the determinant of $Dvarphi=r^n$, but I don't see an easy way of seeing this.
real-analysis calculus
$endgroup$
Define $varphi:(0,infty)times S^ntomathbbR^n+1$ by $varphi(t,theta)=ttheta$. Let $E$ be a subset of $mathbbR^n+1$ and let $E_rsubseteqmathbbR^n+1$ contain all points $thetain S^n$ such that $tthetain E$ for some $t>0$. Then $int_Efdx_1...dx_n+1=int_-infty^inftyint_E_rfcircvarphi r^ndrdv$, where $dv$ is the volume form of the n sphere.
I'm with the classical change of variable. For this, I need to know why the determinant of $Dvarphi=r^n$, but I don't see an easy way of seeing this.
real-analysis calculus
real-analysis calculus
edited Mar 28 at 3:37
user124910
asked Mar 27 at 16:51
user124910user124910
1,262817
1,262817
$begingroup$
You will also need a constant; the Jacobian is not $r^n$, it is $C_n r^n$, where $C_n$ is the surface measure of the $n$-sphere.
$endgroup$
– Giuseppe Negro
Mar 27 at 17:06
$begingroup$
"Let $E_r subseteq Bbb R^n+1$ contain all points $theta in S^n$ such that $ttheta in E_r$ for some $t>0$". Could you please explain what this means?
$endgroup$
– Paul Sinclair
Mar 28 at 0:21
$begingroup$
@PaulSinclair sorry, there was a typo. I've fixed it.
$endgroup$
– user124910
Mar 28 at 3:36
$begingroup$
@GiuseppeNegro - that constant $C_n$ comes from the integration over $S^n$. In this case, the integration is only over $E_r$, and has yet to be performed,
$endgroup$
– Paul Sinclair
Mar 28 at 23:44
$begingroup$
Right; the Jacobian is $r^n$, not $C_nr^n$. Sorry for my mistake. Concerning the question, I prefer to see it in terms of metric tensors; $$ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This can be seen by expanding $ds^2=(d(rtheta))^2.$ Then, the volume form formula follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$. Not too rigorous, but faster for sure.
$endgroup$
– Giuseppe Negro
Mar 28 at 23:51
add a comment |
$begingroup$
You will also need a constant; the Jacobian is not $r^n$, it is $C_n r^n$, where $C_n$ is the surface measure of the $n$-sphere.
$endgroup$
– Giuseppe Negro
Mar 27 at 17:06
$begingroup$
"Let $E_r subseteq Bbb R^n+1$ contain all points $theta in S^n$ such that $ttheta in E_r$ for some $t>0$". Could you please explain what this means?
$endgroup$
– Paul Sinclair
Mar 28 at 0:21
$begingroup$
@PaulSinclair sorry, there was a typo. I've fixed it.
$endgroup$
– user124910
Mar 28 at 3:36
$begingroup$
@GiuseppeNegro - that constant $C_n$ comes from the integration over $S^n$. In this case, the integration is only over $E_r$, and has yet to be performed,
$endgroup$
– Paul Sinclair
Mar 28 at 23:44
$begingroup$
Right; the Jacobian is $r^n$, not $C_nr^n$. Sorry for my mistake. Concerning the question, I prefer to see it in terms of metric tensors; $$ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This can be seen by expanding $ds^2=(d(rtheta))^2.$ Then, the volume form formula follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$. Not too rigorous, but faster for sure.
$endgroup$
– Giuseppe Negro
Mar 28 at 23:51
$begingroup$
You will also need a constant; the Jacobian is not $r^n$, it is $C_n r^n$, where $C_n$ is the surface measure of the $n$-sphere.
$endgroup$
– Giuseppe Negro
Mar 27 at 17:06
$begingroup$
You will also need a constant; the Jacobian is not $r^n$, it is $C_n r^n$, where $C_n$ is the surface measure of the $n$-sphere.
$endgroup$
– Giuseppe Negro
Mar 27 at 17:06
$begingroup$
"Let $E_r subseteq Bbb R^n+1$ contain all points $theta in S^n$ such that $ttheta in E_r$ for some $t>0$". Could you please explain what this means?
$endgroup$
– Paul Sinclair
Mar 28 at 0:21
$begingroup$
"Let $E_r subseteq Bbb R^n+1$ contain all points $theta in S^n$ such that $ttheta in E_r$ for some $t>0$". Could you please explain what this means?
$endgroup$
– Paul Sinclair
Mar 28 at 0:21
$begingroup$
@PaulSinclair sorry, there was a typo. I've fixed it.
$endgroup$
– user124910
Mar 28 at 3:36
$begingroup$
@PaulSinclair sorry, there was a typo. I've fixed it.
$endgroup$
– user124910
Mar 28 at 3:36
$begingroup$
@GiuseppeNegro - that constant $C_n$ comes from the integration over $S^n$. In this case, the integration is only over $E_r$, and has yet to be performed,
$endgroup$
– Paul Sinclair
Mar 28 at 23:44
$begingroup$
@GiuseppeNegro - that constant $C_n$ comes from the integration over $S^n$. In this case, the integration is only over $E_r$, and has yet to be performed,
$endgroup$
– Paul Sinclair
Mar 28 at 23:44
$begingroup$
Right; the Jacobian is $r^n$, not $C_nr^n$. Sorry for my mistake. Concerning the question, I prefer to see it in terms of metric tensors; $$ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This can be seen by expanding $ds^2=(d(rtheta))^2.$ Then, the volume form formula follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$. Not too rigorous, but faster for sure.
$endgroup$
– Giuseppe Negro
Mar 28 at 23:51
$begingroup$
Right; the Jacobian is $r^n$, not $C_nr^n$. Sorry for my mistake. Concerning the question, I prefer to see it in terms of metric tensors; $$ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This can be seen by expanding $ds^2=(d(rtheta))^2.$ Then, the volume form formula follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$. Not too rigorous, but faster for sure.
$endgroup$
– Giuseppe Negro
Mar 28 at 23:51
add a comment |
2 Answers
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$begingroup$
Now that the meaning of $E_r$ is cleared up, I can say with certainty that your iterated integral formula makes no sense. There is a small problem that the integrals are backwards from the differentials. ($r$ does not vary over $E_r$, nor does $v$ vary over $(-infty, infty)$.)
But the real issue is that the region of integration for iterated integral does not match $E$ at all. To give the correct expression, you need to define another collection of sets $$E(theta) = t in Bbb R : ttheta in E, t > 0quadforall theta in E_r$$
(I didn't want to use $E_theta$ because $E(theta)$ depends on $theta$, while the previously established "$E_r$" does not depend on $r$.) Then the actual formula would be
$$int_EfdV=int_E_rint_E(theta)fcircvarphi r^n,dr,dv$$
If you turned it around: Let $E_r subseteq S^n$ and let $E = ttheta : theta in E_r, t > 0$ (i.e., $E$ is the union of all rays passing through $E_r$), then you could justify the equation
$$int_EfdV=int_0^inftyint_E_rfcircvarphi r^n,dv,dr$$
since $E(theta) = (0,infty)$ for all $theta$ in this case. But there is no way to justify integrating $r$ over $(-infty, infty)$.
As for your differential question, informally: $dr$ is the direction perpendicular to $rS^n$, the sphere of radius $r$, so the volume element at radius $r$ is $dr$ times the "$n$-area" $dw$ on $rS^n$. But $rS^n$ is just $S^n$ magnified by a factor of $r$, so every one of its $n$ independent directions is scaled by $r$, and therefore, its differential area is $r^n$ times as big as the corresponding differential area $dv$ on $S^n$.
To make that handwaving rigorous, define an orthonormal coordinate system on $S^n$. For instance, $(theta_1, ..., theta_n)$, where $$x_0 = sintheta_1 ... sintheta_n\x_1 = costheta_1sin theta_2 ... sin theta_n\x_2 = costheta_2sin theta_3 ... sin theta_n\vdots\x_n = cos theta_n$$
And therefore $$varphi(r,theta_1, ..., theta_n) = (rsintheta_1 ... sintheta_n, rcostheta_1sin theta_2 ... sin theta_n, ..., rcostheta_n)$$
Now you have something concrete to base your Hessian calculation on.
$endgroup$
add a comment |
$begingroup$
I prefer to see this in terms of metric tensors; $$tag1ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This formula can be quickly proved by expanding $ds^2=(d(rtheta))^2=(dr, theta+r,dtheta)^2, $ using that $thetacdot dtheta=0$, because $thetacdot theta=1$.
Once (1) has been proved, the volume form formula immediately follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$.
Not too rigorous, maybe, but faster.
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Now that the meaning of $E_r$ is cleared up, I can say with certainty that your iterated integral formula makes no sense. There is a small problem that the integrals are backwards from the differentials. ($r$ does not vary over $E_r$, nor does $v$ vary over $(-infty, infty)$.)
But the real issue is that the region of integration for iterated integral does not match $E$ at all. To give the correct expression, you need to define another collection of sets $$E(theta) = t in Bbb R : ttheta in E, t > 0quadforall theta in E_r$$
(I didn't want to use $E_theta$ because $E(theta)$ depends on $theta$, while the previously established "$E_r$" does not depend on $r$.) Then the actual formula would be
$$int_EfdV=int_E_rint_E(theta)fcircvarphi r^n,dr,dv$$
If you turned it around: Let $E_r subseteq S^n$ and let $E = ttheta : theta in E_r, t > 0$ (i.e., $E$ is the union of all rays passing through $E_r$), then you could justify the equation
$$int_EfdV=int_0^inftyint_E_rfcircvarphi r^n,dv,dr$$
since $E(theta) = (0,infty)$ for all $theta$ in this case. But there is no way to justify integrating $r$ over $(-infty, infty)$.
As for your differential question, informally: $dr$ is the direction perpendicular to $rS^n$, the sphere of radius $r$, so the volume element at radius $r$ is $dr$ times the "$n$-area" $dw$ on $rS^n$. But $rS^n$ is just $S^n$ magnified by a factor of $r$, so every one of its $n$ independent directions is scaled by $r$, and therefore, its differential area is $r^n$ times as big as the corresponding differential area $dv$ on $S^n$.
To make that handwaving rigorous, define an orthonormal coordinate system on $S^n$. For instance, $(theta_1, ..., theta_n)$, where $$x_0 = sintheta_1 ... sintheta_n\x_1 = costheta_1sin theta_2 ... sin theta_n\x_2 = costheta_2sin theta_3 ... sin theta_n\vdots\x_n = cos theta_n$$
And therefore $$varphi(r,theta_1, ..., theta_n) = (rsintheta_1 ... sintheta_n, rcostheta_1sin theta_2 ... sin theta_n, ..., rcostheta_n)$$
Now you have something concrete to base your Hessian calculation on.
$endgroup$
add a comment |
$begingroup$
Now that the meaning of $E_r$ is cleared up, I can say with certainty that your iterated integral formula makes no sense. There is a small problem that the integrals are backwards from the differentials. ($r$ does not vary over $E_r$, nor does $v$ vary over $(-infty, infty)$.)
But the real issue is that the region of integration for iterated integral does not match $E$ at all. To give the correct expression, you need to define another collection of sets $$E(theta) = t in Bbb R : ttheta in E, t > 0quadforall theta in E_r$$
(I didn't want to use $E_theta$ because $E(theta)$ depends on $theta$, while the previously established "$E_r$" does not depend on $r$.) Then the actual formula would be
$$int_EfdV=int_E_rint_E(theta)fcircvarphi r^n,dr,dv$$
If you turned it around: Let $E_r subseteq S^n$ and let $E = ttheta : theta in E_r, t > 0$ (i.e., $E$ is the union of all rays passing through $E_r$), then you could justify the equation
$$int_EfdV=int_0^inftyint_E_rfcircvarphi r^n,dv,dr$$
since $E(theta) = (0,infty)$ for all $theta$ in this case. But there is no way to justify integrating $r$ over $(-infty, infty)$.
As for your differential question, informally: $dr$ is the direction perpendicular to $rS^n$, the sphere of radius $r$, so the volume element at radius $r$ is $dr$ times the "$n$-area" $dw$ on $rS^n$. But $rS^n$ is just $S^n$ magnified by a factor of $r$, so every one of its $n$ independent directions is scaled by $r$, and therefore, its differential area is $r^n$ times as big as the corresponding differential area $dv$ on $S^n$.
To make that handwaving rigorous, define an orthonormal coordinate system on $S^n$. For instance, $(theta_1, ..., theta_n)$, where $$x_0 = sintheta_1 ... sintheta_n\x_1 = costheta_1sin theta_2 ... sin theta_n\x_2 = costheta_2sin theta_3 ... sin theta_n\vdots\x_n = cos theta_n$$
And therefore $$varphi(r,theta_1, ..., theta_n) = (rsintheta_1 ... sintheta_n, rcostheta_1sin theta_2 ... sin theta_n, ..., rcostheta_n)$$
Now you have something concrete to base your Hessian calculation on.
$endgroup$
add a comment |
$begingroup$
Now that the meaning of $E_r$ is cleared up, I can say with certainty that your iterated integral formula makes no sense. There is a small problem that the integrals are backwards from the differentials. ($r$ does not vary over $E_r$, nor does $v$ vary over $(-infty, infty)$.)
But the real issue is that the region of integration for iterated integral does not match $E$ at all. To give the correct expression, you need to define another collection of sets $$E(theta) = t in Bbb R : ttheta in E, t > 0quadforall theta in E_r$$
(I didn't want to use $E_theta$ because $E(theta)$ depends on $theta$, while the previously established "$E_r$" does not depend on $r$.) Then the actual formula would be
$$int_EfdV=int_E_rint_E(theta)fcircvarphi r^n,dr,dv$$
If you turned it around: Let $E_r subseteq S^n$ and let $E = ttheta : theta in E_r, t > 0$ (i.e., $E$ is the union of all rays passing through $E_r$), then you could justify the equation
$$int_EfdV=int_0^inftyint_E_rfcircvarphi r^n,dv,dr$$
since $E(theta) = (0,infty)$ for all $theta$ in this case. But there is no way to justify integrating $r$ over $(-infty, infty)$.
As for your differential question, informally: $dr$ is the direction perpendicular to $rS^n$, the sphere of radius $r$, so the volume element at radius $r$ is $dr$ times the "$n$-area" $dw$ on $rS^n$. But $rS^n$ is just $S^n$ magnified by a factor of $r$, so every one of its $n$ independent directions is scaled by $r$, and therefore, its differential area is $r^n$ times as big as the corresponding differential area $dv$ on $S^n$.
To make that handwaving rigorous, define an orthonormal coordinate system on $S^n$. For instance, $(theta_1, ..., theta_n)$, where $$x_0 = sintheta_1 ... sintheta_n\x_1 = costheta_1sin theta_2 ... sin theta_n\x_2 = costheta_2sin theta_3 ... sin theta_n\vdots\x_n = cos theta_n$$
And therefore $$varphi(r,theta_1, ..., theta_n) = (rsintheta_1 ... sintheta_n, rcostheta_1sin theta_2 ... sin theta_n, ..., rcostheta_n)$$
Now you have something concrete to base your Hessian calculation on.
$endgroup$
Now that the meaning of $E_r$ is cleared up, I can say with certainty that your iterated integral formula makes no sense. There is a small problem that the integrals are backwards from the differentials. ($r$ does not vary over $E_r$, nor does $v$ vary over $(-infty, infty)$.)
But the real issue is that the region of integration for iterated integral does not match $E$ at all. To give the correct expression, you need to define another collection of sets $$E(theta) = t in Bbb R : ttheta in E, t > 0quadforall theta in E_r$$
(I didn't want to use $E_theta$ because $E(theta)$ depends on $theta$, while the previously established "$E_r$" does not depend on $r$.) Then the actual formula would be
$$int_EfdV=int_E_rint_E(theta)fcircvarphi r^n,dr,dv$$
If you turned it around: Let $E_r subseteq S^n$ and let $E = ttheta : theta in E_r, t > 0$ (i.e., $E$ is the union of all rays passing through $E_r$), then you could justify the equation
$$int_EfdV=int_0^inftyint_E_rfcircvarphi r^n,dv,dr$$
since $E(theta) = (0,infty)$ for all $theta$ in this case. But there is no way to justify integrating $r$ over $(-infty, infty)$.
As for your differential question, informally: $dr$ is the direction perpendicular to $rS^n$, the sphere of radius $r$, so the volume element at radius $r$ is $dr$ times the "$n$-area" $dw$ on $rS^n$. But $rS^n$ is just $S^n$ magnified by a factor of $r$, so every one of its $n$ independent directions is scaled by $r$, and therefore, its differential area is $r^n$ times as big as the corresponding differential area $dv$ on $S^n$.
To make that handwaving rigorous, define an orthonormal coordinate system on $S^n$. For instance, $(theta_1, ..., theta_n)$, where $$x_0 = sintheta_1 ... sintheta_n\x_1 = costheta_1sin theta_2 ... sin theta_n\x_2 = costheta_2sin theta_3 ... sin theta_n\vdots\x_n = cos theta_n$$
And therefore $$varphi(r,theta_1, ..., theta_n) = (rsintheta_1 ... sintheta_n, rcostheta_1sin theta_2 ... sin theta_n, ..., rcostheta_n)$$
Now you have something concrete to base your Hessian calculation on.
answered Mar 29 at 2:19
Paul SinclairPaul Sinclair
20.9k21543
20.9k21543
add a comment |
add a comment |
$begingroup$
I prefer to see this in terms of metric tensors; $$tag1ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This formula can be quickly proved by expanding $ds^2=(d(rtheta))^2=(dr, theta+r,dtheta)^2, $ using that $thetacdot dtheta=0$, because $thetacdot theta=1$.
Once (1) has been proved, the volume form formula immediately follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$.
Not too rigorous, maybe, but faster.
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add a comment |
$begingroup$
I prefer to see this in terms of metric tensors; $$tag1ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This formula can be quickly proved by expanding $ds^2=(d(rtheta))^2=(dr, theta+r,dtheta)^2, $ using that $thetacdot dtheta=0$, because $thetacdot theta=1$.
Once (1) has been proved, the volume form formula immediately follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$.
Not too rigorous, maybe, but faster.
$endgroup$
add a comment |
$begingroup$
I prefer to see this in terms of metric tensors; $$tag1ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This formula can be quickly proved by expanding $ds^2=(d(rtheta))^2=(dr, theta+r,dtheta)^2, $ using that $thetacdot dtheta=0$, because $thetacdot theta=1$.
Once (1) has been proved, the volume form formula immediately follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$.
Not too rigorous, maybe, but faster.
$endgroup$
I prefer to see this in terms of metric tensors; $$tag1ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This formula can be quickly proved by expanding $ds^2=(d(rtheta))^2=(dr, theta+r,dtheta)^2, $ using that $thetacdot dtheta=0$, because $thetacdot theta=1$.
Once (1) has been proved, the volume form formula immediately follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$.
Not too rigorous, maybe, but faster.
edited Apr 8 at 7:16
community wiki
2 revs
Giuseppe Negro
add a comment |
add a comment |
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$begingroup$
You will also need a constant; the Jacobian is not $r^n$, it is $C_n r^n$, where $C_n$ is the surface measure of the $n$-sphere.
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– Giuseppe Negro
Mar 27 at 17:06
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"Let $E_r subseteq Bbb R^n+1$ contain all points $theta in S^n$ such that $ttheta in E_r$ for some $t>0$". Could you please explain what this means?
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– Paul Sinclair
Mar 28 at 0:21
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@PaulSinclair sorry, there was a typo. I've fixed it.
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– user124910
Mar 28 at 3:36
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@GiuseppeNegro - that constant $C_n$ comes from the integration over $S^n$. In this case, the integration is only over $E_r$, and has yet to be performed,
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– Paul Sinclair
Mar 28 at 23:44
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Right; the Jacobian is $r^n$, not $C_nr^n$. Sorry for my mistake. Concerning the question, I prefer to see it in terms of metric tensors; $$ds^2=dr^2 + r^2dtheta^2, $$where $ds^2$ is the metric tensor of $mathbb R^n+1$, and $dtheta^2$ is the metric tensor of $mathbb S^n$. This can be seen by expanding $ds^2=(d(rtheta))^2.$ Then, the volume form formula follows from $$dx=sqrtgdrdtheta, $$ where $g$ is the determinant of $ds^2$. Not too rigorous, but faster for sure.
$endgroup$
– Giuseppe Negro
Mar 28 at 23:51