Laplacian in different coordinates
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Lets consider the following coordinates on $Bbb R^3$:
$$x_1=rcos(xi),x_2=rsin(xi) w_2,x_3=rsin(xi) w_3$$
with $w_2^2+w_3^2=1,r=||x||$ and $0leq xileq pi$. Further let $f$ be a function on $Bbb R^3$ depending only on $r$ and $xi$. I want to calculate the Laplacian of $f$ in the coordinates above.
In the paper I am reading this Laplacian is given but I dont get how they do it. It only says that $$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2dxi^2+r^2sin(xi)^2(dw_2^2+dw_3^2)$$ and therefore, since $Delta f=frac1sqrt gsum_k partial_k(sum_i g^iksqrt gpartial_if)$,
$$Delta f=frac1r^2fracpartialpartial rr^2fracpartialpartial rf+frac1r^2frac1sin(xi)fracpartialpartial xisin(xi)fracpartialpartial xif.$$
Could someone explain in some more detail what is happening here? What do I have to calculate? Thanks in advance for any help.
Edit:
I am able to obtain the first equation now since
$$dx_1=cos(xi)dr-rsin(xi)dxi\
dx_2=sin(xi)w_2dr+rcos(xi)w_2dxi+rsin(xi)dw_2\
dx_3=sin(xi)w_3dr+rcos(xi)w_3dxi+rsin(xi)dw_3$$
so that
$$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2dxi^2+r^2sin(xi)^2(dw_2^2+dw_3^2)$$
follows be squaring each row and using $w_2^2+w_3^2=cos(xi)^2+sin(xi)^2=1$. Now if I use the matrix $$beginpmatrix
1&0&0\
0&r^2&0\
0&0&r^2sin(xi)^2
endpmatrix$$ as matrix tensor $g=g_ijdx_iotimes dx_j$ in the new coordinates I obtain the desired result by the formula for the Laplacian above.
However I am not sure how rigorous this is since $dr,dxi,dw_2,dw_3$ don't form a basis of the cotangent bundle since we have the restriction $w_2^2+w_3^2=1$ (note that I also only added one $r^2sin(xi)^2$ term on the diagonal of $g$ despite I have it for $w_2$ and $w_3$). Do I have to write $w_2,w_3$ as $cos(t),sin(t)$ at first, so that I have the same number of variables or is there another way?
coordinate-systems differential-forms laplacian
asked Sep 9 at 10:33
Luke Mathwalker
320112
add a comment |Â
up vote
5
down vote
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Lets consider the following coordinates on $Bbb R^3$:
$$x_1=rcos(xi),x_2=rsin(xi) w_2,x_3=rsin(xi) w_3$$
with $w_2^2+w_3^2=1,r=||x||$ and $0leq xileq pi$. Further let $f$ be a function on $Bbb R^3$ depending only on $r$ and $xi$. I want to calculate the Laplacian of $f$ in the coordinates above.
In the paper I am reading this Laplacian is given but I dont get how they do it. It only says that $$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2dxi^2+r^2sin(xi)^2(dw_2^2+dw_3^2)$$ and therefore, since $Delta f=frac1sqrt gsum_k partial_k(sum_i g^iksqrt gpartial_if)$,
$$Delta f=frac1r^2fracpartialpartial rr^2fracpartialpartial rf+frac1r^2frac1sin(xi)fracpartialpartial xisin(xi)fracpartialpartial xif.$$
Could someone explain in some more detail what is happening here? What do I have to calculate? Thanks in advance for any help.
Edit:
I am able to obtain the first equation now since
$$dx_1=cos(xi)dr-rsin(xi)dxi\
dx_2=sin(xi)w_2dr+rcos(xi)w_2dxi+rsin(xi)dw_2\
dx_3=sin(xi)w_3dr+rcos(xi)w_3dxi+rsin(xi)dw_3$$
so that
$$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2dxi^2+r^2sin(xi)^2(dw_2^2+dw_3^2)$$
follows be squaring each row and using $w_2^2+w_3^2=cos(xi)^2+sin(xi)^2=1$. Now if I use the matrix $$beginpmatrix
1&0&0\
0&r^2&0\
0&0&r^2sin(xi)^2
endpmatrix$$ as matrix tensor $g=g_ijdx_iotimes dx_j$ in the new coordinates I obtain the desired result by the formula for the Laplacian above.
However I am not sure how rigorous this is since $dr,dxi,dw_2,dw_3$ don't form a basis of the cotangent bundle since we have the restriction $w_2^2+w_3^2=1$ (note that I also only added one $r^2sin(xi)^2$ term on the diagonal of $g$ despite I have it for $w_2$ and $w_3$). Do I have to write $w_2,w_3$ as $cos(t),sin(t)$ at first, so that I have the same number of variables or is there another way?
coordinate-systems differential-forms laplacian
asked Sep 9 at 10:33
Luke Mathwalker
320112
1
Well, the problem, as you correctly pointed out, is that $r$, $xi$, $w_1$ and $w_2$ are not coordinates (they are four and $mathbb R^3$ is three-dimensional). The proper way to do this is to write $w_1=cos t$ and $w_2=sin t$. Nevertheless the expression you gave for $ds^2$ in terms of these functions is correct (it does not matter if the differential $dw_1$ and $dw_2$ are not independent).
– Brightsun
Sep 14 at 19:26
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Lets consider the following coordinates on $Bbb R^3$:
$$x_1=rcos(xi),x_2=rsin(xi) w_2,x_3=rsin(xi) w_3$$
with $w_2^2+w_3^2=1,r=||x||$ and $0leq xileq pi$. Further let $f$ be a function on $Bbb R^3$ depending only on $r$ and $xi$. I want to calculate the Laplacian of $f$ in the coordinates above.
In the paper I am reading this Laplacian is given but I dont get how they do it. It only says that $$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2dxi^2+r^2sin(xi)^2(dw_2^2+dw_3^2)$$ and therefore, since $Delta f=frac1sqrt gsum_k partial_k(sum_i g^iksqrt gpartial_if)$,
$$Delta f=frac1r^2fracpartialpartial rr^2fracpartialpartial rf+frac1r^2frac1sin(xi)fracpartialpartial xisin(xi)fracpartialpartial xif.$$
Could someone explain in some more detail what is happening here? What do I have to calculate? Thanks in advance for any help.
Edit:
I am able to obtain the first equation now since
$$dx_1=cos(xi)dr-rsin(xi)dxi\
dx_2=sin(xi)w_2dr+rcos(xi)w_2dxi+rsin(xi)dw_2\
dx_3=sin(xi)w_3dr+rcos(xi)w_3dxi+rsin(xi)dw_3$$
so that
$$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2dxi^2+r^2sin(xi)^2(dw_2^2+dw_3^2)$$
follows be squaring each row and using $w_2^2+w_3^2=cos(xi)^2+sin(xi)^2=1$. Now if I use the matrix $$beginpmatrix
1&0&0\
0&r^2&0\
0&0&r^2sin(xi)^2
endpmatrix$$ as matrix tensor $g=g_ijdx_iotimes dx_j$ in the new coordinates I obtain the desired result by the formula for the Laplacian above.
However I am not sure how rigorous this is since $dr,dxi,dw_2,dw_3$ don't form a basis of the cotangent bundle since we have the restriction $w_2^2+w_3^2=1$ (note that I also only added one $r^2sin(xi)^2$ term on the diagonal of $g$ despite I have it for $w_2$ and $w_3$). Do I have to write $w_2,w_3$ as $cos(t),sin(t)$ at first, so that I have the same number of variables or is there another way?
coordinate-systems differential-forms laplacian
asked Sep 9 at 10:33
Luke Mathwalker
320112
Lets consider the following coordinates on $Bbb R^3$:
$$x_1=rcos(xi),x_2=rsin(xi) w_2,x_3=rsin(xi) w_3$$
with $w_2^2+w_3^2=1,r=||x||$ and $0leq xileq pi$. Further let $f$ be a function on $Bbb R^3$ depending only on $r$ and $xi$. I want to calculate the Laplacian of $f$ in the coordinates above.
In the paper I am reading this Laplacian is given but I dont get how they do it. It only says that $$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2dxi^2+r^2sin(xi)^2(dw_2^2+dw_3^2)$$ and therefore, since $Delta f=frac1sqrt gsum_k partial_k(sum_i g^iksqrt gpartial_if)$,
$$Delta f=frac1r^2fracpartialpartial rr^2fracpartialpartial rf+frac1r^2frac1sin(xi)fracpartialpartial xisin(xi)fracpartialpartial xif.$$
Could someone explain in some more detail what is happening here? What do I have to calculate? Thanks in advance for any help.
Edit:
I am able to obtain the first equation now since
$$dx_1=cos(xi)dr-rsin(xi)dxi\
dx_2=sin(xi)w_2dr+rcos(xi)w_2dxi+rsin(xi)dw_2\
dx_3=sin(xi)w_3dr+rcos(xi)w_3dxi+rsin(xi)dw_3$$
so that
$$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2dxi^2+r^2sin(xi)^2(dw_2^2+dw_3^2)$$
follows be squaring each row and using $w_2^2+w_3^2=cos(xi)^2+sin(xi)^2=1$. Now if I use the matrix $$beginpmatrix
1&0&0\
0&r^2&0\
0&0&r^2sin(xi)^2
endpmatrix$$ as matrix tensor $g=g_ijdx_iotimes dx_j$ in the new coordinates I obtain the desired result by the formula for the Laplacian above.
However I am not sure how rigorous this is since $dr,dxi,dw_2,dw_3$ don't form a basis of the cotangent bundle since we have the restriction $w_2^2+w_3^2=1$ (note that I also only added one $r^2sin(xi)^2$ term on the diagonal of $g$ despite I have it for $w_2$ and $w_3$). Do I have to write $w_2,w_3$ as $cos(t),sin(t)$ at first, so that I have the same number of variables or is there another way?
coordinate-systems differential-forms laplacian
coordinate-systems differential-forms laplacian
asked Sep 9 at 10:33
Luke Mathwalker
320112
asked Sep 9 at 10:33
Luke Mathwalker
320112
edited Sep 13 at 22:40
asked Sep 9 at 10:33
Luke Mathwalker
320112
asked Sep 9 at 10:33
Luke Mathwalker
320112
asked Sep 9 at 10:33
Luke Mathwalker
320112
320112
1
Well, the problem, as you correctly pointed out, is that $r$, $xi$, $w_1$ and $w_2$ are not coordinates (they are four and $mathbb R^3$ is three-dimensional). The proper way to do this is to write $w_1=cos t$ and $w_2=sin t$. Nevertheless the expression you gave for $ds^2$ in terms of these functions is correct (it does not matter if the differential $dw_1$ and $dw_2$ are not independent).
– Brightsun
Sep 14 at 19:26
add a comment |Â
1
Well, the problem, as you correctly pointed out, is that $r$, $xi$, $w_1$ and $w_2$ are not coordinates (they are four and $mathbb R^3$ is three-dimensional). The proper way to do this is to write $w_1=cos t$ and $w_2=sin t$. Nevertheless the expression you gave for $ds^2$ in terms of these functions is correct (it does not matter if the differential $dw_1$ and $dw_2$ are not independent).
– Brightsun
Sep 14 at 19:26
1
1
Well, the problem, as you correctly pointed out, is that $r$, $xi$, $w_1$ and $w_2$ are not coordinates (they are four and $mathbb R^3$ is three-dimensional). The proper way to do this is to write $w_1=cos t$ and $w_2=sin t$. Nevertheless the expression you gave for $ds^2$ in terms of these functions is correct (it does not matter if the differential $dw_1$ and $dw_2$ are not independent).
– Brightsun
Sep 14 at 19:26
Well, the problem, as you correctly pointed out, is that $r$, $xi$, $w_1$ and $w_2$ are not coordinates (they are four and $mathbb R^3$ is three-dimensional). The proper way to do this is to write $w_1=cos t$ and $w_2=sin t$. Nevertheless the expression you gave for $ds^2$ in terms of these functions is correct (it does not matter if the differential $dw_1$ and $dw_2$ are not independent).
– Brightsun
Sep 14 at 19:26
add a comment |Â
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Well, the problem, as you correctly pointed out, is that $r$, $xi$, $w_1$ and $w_2$ are not coordinates (they are four and $mathbb R^3$ is three-dimensional). The proper way to do this is to write $w_1=cos t$ and $w_2=sin t$. Nevertheless the expression you gave for $ds^2$ in terms of these functions is correct (it does not matter if the differential $dw_1$ and $dw_2$ are not independent).
– Brightsun
Sep 14 at 19:26