Cylindrical Frame Field
$begingroup$
Let E be the cylindrical frame field
$E_1 = costheta U_1 + sintheta U_2, E_2 = − sintheta U_1 + costheta U_2, E_3 = U_3$
(a) Starting from the basic cylindrical equations $x = r costheta, y = r sintheta, z = z$, show that the dual 1-forms are $θ_1 = dr, θ_2 = rdtheta, θ_3 = dz$.
I started by taking the derivatives of E however, I'm not sure if that's the right first step or not. Any guidance on this please? Thanks!
geometry differential-geometry
edited Dec 2 '18 at 7:21
Tianlalu
3,08121038
asked Mar 19 '18 at 14:10
Lola Lola
156
$endgroup$
add a comment |
$begingroup$
Let E be the cylindrical frame field
$E_1 = costheta U_1 + sintheta U_2, E_2 = − sintheta U_1 + costheta U_2, E_3 = U_3$
(a) Starting from the basic cylindrical equations $x = r costheta, y = r sintheta, z = z$, show that the dual 1-forms are $θ_1 = dr, θ_2 = rdtheta, θ_3 = dz$.
I started by taking the derivatives of E however, I'm not sure if that's the right first step or not. Any guidance on this please? Thanks!
geometry differential-geometry
edited Dec 2 '18 at 7:21
Tianlalu
3,08121038
asked Mar 19 '18 at 14:10
Lola Lola
156
$endgroup$
add a comment |
$begingroup$
Let E be the cylindrical frame field
$E_1 = costheta U_1 + sintheta U_2, E_2 = − sintheta U_1 + costheta U_2, E_3 = U_3$
(a) Starting from the basic cylindrical equations $x = r costheta, y = r sintheta, z = z$, show that the dual 1-forms are $θ_1 = dr, θ_2 = rdtheta, θ_3 = dz$.
I started by taking the derivatives of E however, I'm not sure if that's the right first step or not. Any guidance on this please? Thanks!
geometry differential-geometry
edited Dec 2 '18 at 7:21
Tianlalu
3,08121038
asked Mar 19 '18 at 14:10
Lola Lola
156
$endgroup$
Let E be the cylindrical frame field
$E_1 = costheta U_1 + sintheta U_2, E_2 = − sintheta U_1 + costheta U_2, E_3 = U_3$
(a) Starting from the basic cylindrical equations $x = r costheta, y = r sintheta, z = z$, show that the dual 1-forms are $θ_1 = dr, θ_2 = rdtheta, θ_3 = dz$.
I started by taking the derivatives of E however, I'm not sure if that's the right first step or not. Any guidance on this please? Thanks!
geometry differential-geometry
geometry differential-geometry
edited Dec 2 '18 at 7:21
Tianlalu
3,08121038
asked Mar 19 '18 at 14:10
Lola Lola
156
edited Dec 2 '18 at 7:21
Tianlalu
3,08121038
asked Mar 19 '18 at 14:10
Lola Lola
156
edited Dec 2 '18 at 7:21
Tianlalu
3,08121038
edited Dec 2 '18 at 7:21
Tianlalu
3,08121038
edited Dec 2 '18 at 7:21
Tianlalu
3,08121038
3,08121038
asked Mar 19 '18 at 14:10
Lola Lola
156
asked Mar 19 '18 at 14:10
Lola Lola
156
asked Mar 19 '18 at 14:10
Lola Lola
156
156
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
One way to do it is just recalling from linear algebra that the transformation between dual basis uses the inverse of the transition matrix between the original bases. Using columns whose entries are the vectors themselves requires an extra transposition. Meaning that if $$begin{pmatrix} E_1 \ E_2 \ E_3 end{pmatrix} = begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}begin{pmatrix} U_1 \ U_2 \ U_3 end{pmatrix}$$then $$begin{pmatrix} theta_1 \ theta_2 \ theta_3 end{pmatrix} = left(begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}^{-1}right)^topbegin{pmatrix} dx\ dy \ dz end{pmatrix} = begin{pmatrix} cos theta ,dx + sintheta,dy \ -sin theta,dx + cos theta,dy \ dz end{pmatrix}.$$For free, $theta_3 = dz$. And sure enough, using $x = rcos theta$ and $y=r sin theta$ you get $$begin{align} theta_1 &= costheta(costheta ,dr - rsintheta,dtheta) + sintheta(sintheta,dr + rcostheta,dtheta) = dr, \ theta_2 &= -sintheta(costheta,dr-rsintheta,dtheta)+costheta(sintheta,dr+rcostheta,dtheta) = r,dtheta. end{align}$$
answered Dec 2 '18 at 8:53
Ivo TerekIvo Terek
45.4k952141
$endgroup$
add a comment |
$begingroup$
Considering directly Cylindrical/polar coordinates for an arc $ds$ the three differentials are, :
along $z$ direction $=dz$,
helical arc component along circumference direction $r, d theta $, and along the radius $dr.$
answered Dec 2 '18 at 9:29
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
One way to do it is just recalling from linear algebra that the transformation between dual basis uses the inverse of the transition matrix between the original bases. Using columns whose entries are the vectors themselves requires an extra transposition. Meaning that if $$begin{pmatrix} E_1 \ E_2 \ E_3 end{pmatrix} = begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}begin{pmatrix} U_1 \ U_2 \ U_3 end{pmatrix}$$then $$begin{pmatrix} theta_1 \ theta_2 \ theta_3 end{pmatrix} = left(begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}^{-1}right)^topbegin{pmatrix} dx\ dy \ dz end{pmatrix} = begin{pmatrix} cos theta ,dx + sintheta,dy \ -sin theta,dx + cos theta,dy \ dz end{pmatrix}.$$For free, $theta_3 = dz$. And sure enough, using $x = rcos theta$ and $y=r sin theta$ you get $$begin{align} theta_1 &= costheta(costheta ,dr - rsintheta,dtheta) + sintheta(sintheta,dr + rcostheta,dtheta) = dr, \ theta_2 &= -sintheta(costheta,dr-rsintheta,dtheta)+costheta(sintheta,dr+rcostheta,dtheta) = r,dtheta. end{align}$$
answered Dec 2 '18 at 8:53
Ivo TerekIvo Terek
45.4k952141
$endgroup$
add a comment |
$begingroup$
One way to do it is just recalling from linear algebra that the transformation between dual basis uses the inverse of the transition matrix between the original bases. Using columns whose entries are the vectors themselves requires an extra transposition. Meaning that if $$begin{pmatrix} E_1 \ E_2 \ E_3 end{pmatrix} = begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}begin{pmatrix} U_1 \ U_2 \ U_3 end{pmatrix}$$then $$begin{pmatrix} theta_1 \ theta_2 \ theta_3 end{pmatrix} = left(begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}^{-1}right)^topbegin{pmatrix} dx\ dy \ dz end{pmatrix} = begin{pmatrix} cos theta ,dx + sintheta,dy \ -sin theta,dx + cos theta,dy \ dz end{pmatrix}.$$For free, $theta_3 = dz$. And sure enough, using $x = rcos theta$ and $y=r sin theta$ you get $$begin{align} theta_1 &= costheta(costheta ,dr - rsintheta,dtheta) + sintheta(sintheta,dr + rcostheta,dtheta) = dr, \ theta_2 &= -sintheta(costheta,dr-rsintheta,dtheta)+costheta(sintheta,dr+rcostheta,dtheta) = r,dtheta. end{align}$$
answered Dec 2 '18 at 8:53
Ivo TerekIvo Terek
45.4k952141
$endgroup$
add a comment |
$begingroup$
One way to do it is just recalling from linear algebra that the transformation between dual basis uses the inverse of the transition matrix between the original bases. Using columns whose entries are the vectors themselves requires an extra transposition. Meaning that if $$begin{pmatrix} E_1 \ E_2 \ E_3 end{pmatrix} = begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}begin{pmatrix} U_1 \ U_2 \ U_3 end{pmatrix}$$then $$begin{pmatrix} theta_1 \ theta_2 \ theta_3 end{pmatrix} = left(begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}^{-1}right)^topbegin{pmatrix} dx\ dy \ dz end{pmatrix} = begin{pmatrix} cos theta ,dx + sintheta,dy \ -sin theta,dx + cos theta,dy \ dz end{pmatrix}.$$For free, $theta_3 = dz$. And sure enough, using $x = rcos theta$ and $y=r sin theta$ you get $$begin{align} theta_1 &= costheta(costheta ,dr - rsintheta,dtheta) + sintheta(sintheta,dr + rcostheta,dtheta) = dr, \ theta_2 &= -sintheta(costheta,dr-rsintheta,dtheta)+costheta(sintheta,dr+rcostheta,dtheta) = r,dtheta. end{align}$$
answered Dec 2 '18 at 8:53
Ivo TerekIvo Terek
45.4k952141
$endgroup$
One way to do it is just recalling from linear algebra that the transformation between dual basis uses the inverse of the transition matrix between the original bases. Using columns whose entries are the vectors themselves requires an extra transposition. Meaning that if $$begin{pmatrix} E_1 \ E_2 \ E_3 end{pmatrix} = begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}begin{pmatrix} U_1 \ U_2 \ U_3 end{pmatrix}$$then $$begin{pmatrix} theta_1 \ theta_2 \ theta_3 end{pmatrix} = left(begin{pmatrix} cos theta & sin theta & 0 \ -sin theta & cos theta & 0 \ 0 & 0 & 1end{pmatrix}^{-1}right)^topbegin{pmatrix} dx\ dy \ dz end{pmatrix} = begin{pmatrix} cos theta ,dx + sintheta,dy \ -sin theta,dx + cos theta,dy \ dz end{pmatrix}.$$For free, $theta_3 = dz$. And sure enough, using $x = rcos theta$ and $y=r sin theta$ you get $$begin{align} theta_1 &= costheta(costheta ,dr - rsintheta,dtheta) + sintheta(sintheta,dr + rcostheta,dtheta) = dr, \ theta_2 &= -sintheta(costheta,dr-rsintheta,dtheta)+costheta(sintheta,dr+rcostheta,dtheta) = r,dtheta. end{align}$$
answered Dec 2 '18 at 8:53
Ivo TerekIvo Terek
45.4k952141
answered Dec 2 '18 at 8:53
Ivo TerekIvo Terek
45.4k952141
answered Dec 2 '18 at 8:53
Ivo TerekIvo Terek
45.4k952141
answered Dec 2 '18 at 8:53
Ivo TerekIvo Terek
45.4k952141
45.4k952141
add a comment |
add a comment |
$begingroup$
Considering directly Cylindrical/polar coordinates for an arc $ds$ the three differentials are, :
along $z$ direction $=dz$,
helical arc component along circumference direction $r, d theta $, and along the radius $dr.$
answered Dec 2 '18 at 9:29
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
$begingroup$
Considering directly Cylindrical/polar coordinates for an arc $ds$ the three differentials are, :
along $z$ direction $=dz$,
helical arc component along circumference direction $r, d theta $, and along the radius $dr.$
answered Dec 2 '18 at 9:29
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
$begingroup$
Considering directly Cylindrical/polar coordinates for an arc $ds$ the three differentials are, :
along $z$ direction $=dz$,
helical arc component along circumference direction $r, d theta $, and along the radius $dr.$
answered Dec 2 '18 at 9:29
NarasimhamNarasimham
20.6k52158
$endgroup$
Considering directly Cylindrical/polar coordinates for an arc $ds$ the three differentials are, :
along $z$ direction $=dz$,
helical arc component along circumference direction $r, d theta $, and along the radius $dr.$
answered Dec 2 '18 at 9:29
NarasimhamNarasimham
20.6k52158
answered Dec 2 '18 at 9:29
NarasimhamNarasimham
20.6k52158
answered Dec 2 '18 at 9:29
NarasimhamNarasimham
20.6k52158
answered Dec 2 '18 at 9:29
NarasimhamNarasimham
20.6k52158
20.6k52158
add a comment |
add a comment |
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