Question

How to arrive at the following from given \( x = r\sin \theta \cos \phi, y = r\sin \theta \sin \phi, z=r\cos\theta \)
\[ \begin{bmatrix}
A_x\\
A_y\\
A_z
\end{bmatrix}
= \begin{bmatrix}
\sin \theta \cos \phi & \cos \theta \cos \phi & -\sin\phi\\
\sin \theta \sin \phi & \cos \theta \sin \phi & \cos\phi\\
\cos\theta & -\sin\theta & 0
\end{bmatrix}
\begin{bmatrix}
A_r\\
A_\theta\\
A_\phi
\end{bmatrix} \]

Also how show that
\[ \begin{bmatrix}
\hat i\\
\hat j\\
\hat k
\end{bmatrix}
= \begin{bmatrix}
\sin \theta \cos \phi & \cos \theta \cos \phi & -\sin\phi\\
\sin \theta \sin \phi & \cos \theta \sin \phi & \cos\phi\\
\cos\theta & -\sin\theta & 0
\end{bmatrix}
\begin{bmatrix}
\hat e_r\\
\hat e_\theta\\
\hat e_\phi
\end{bmatrix} \]
How to change \( (a,b,c) \) into spherical polar coordinates and \( (r ,\theta, \phi) \) into Cartesian coordinates using this matrix?

Answer 1

did you not drop an r from some parts of your matrix?

Answer 2

\[x=r\sin\theta\cos\phi\qquad\qquad y=r\sin\theta\sin\phi\qquad\qquad z=r\cos\theta\]

\[x_r=\sin\theta\cos\phi\qquad x_\theta=r\cos\theta\cos\phi\qquad x_\phi=-r\sin\theta\sin\phi\]\[y_r=\sin\theta\sin\phi\qquad y_\theta=r\cos\theta\sin\phi\qquad y_\phi=r\sin\theta\cos\phi\]\[z_r=\cos\theta\qquad\qquad z_\theta=-r\sin \theta\qquad\qquad z_\phi=0\]

\[\frac{\partial(x,y,z)}{\partial ( r,\theta, \phi)}=\left|\begin{matrix} x_r&x_\theta&x_\phi\\{y}_{r}&y_{\theta }&y_{\phi}\\{z}_{r}&{ z}_{\theta }&{z}_{\phi}\end{matrix}\right|\]

Answer 3

does this answer you question @hash.nuke
do you need help calculating the Jacobian determinant?

Answer 4

http://www.web-formulas.com/Math_Formulas/Linear_Algebra_Transform_from_Cartesian_to_Spherical_Coordinate.aspx
@UnkleRhaukus I don't this is Determinant of Jacobian

Answer 5

pardon?

Answer 6

I don't know what that site is @hash.nuke , but @UnkleRhaukus has given the correct Jacobian
http://en.wikipedia.org/wiki/Spherical_coordinate_system#Cartesian_coordinates

Answer 7

No ... that was just transformation rule for unit vectors. As there is no linear relation between curvilinear coordinate system and rectangular coordinate system, the transformation cannot be represented by matrix.
I have come to conclude that.

Answer 8

\[\frac{\partial(x,y,z)}{\partial ( r,\theta, \phi)}=\left|\begin{matrix} x_r&x_\theta&x_\phi\\{y}_{r}&y_{\theta }&y_{\phi}\\{z}_{r}&{ z}_{\theta }&{z}_{\phi}\end{matrix}\right|\]
\[=x_r\left((y_\theta\times z_\phi)-(y_\phi\times z_\theta)\right)-x_\theta\left((y_r\times z_\phi)-(y_\phi \times z_r)\right)+x_\phi\left((y_r\times z_\theta)-(y_\theta \times z_r)\right)\]\[=x_r\left((r\cos\theta\sin\phi\times 0)-(r\sin\theta\cos\phi\times -r\sin\theta )\right)\]\[ \qquad\qquad-x_\theta\left((\sin\theta\sin\phi\times 0)-(r\sin\theta\cos\phi \times \cos\theta)\right)\]\[\qquad\qquad+x_\phi\left((\sin\theta\sin\phi\times -r\sin\theta )-(r\cos\theta\sin\phi \times \cos\theta)\right)\]
\[=x_r\left(0+r^2\sin^2\theta\cos\phi)\right)-x_\theta\left(0-r\sin\theta\cos\phi\cos\theta\right)-x_\phi\left(r\sin^2\theta\sin\phi+r\cos^2\theta\sin\phi\right)\]\[=x_rr^2\sin^2\theta\cos\phi+x_\theta r\sin\theta\cos\phi\cos\theta-x_\phi r\sin\phi\]\[=\sin\theta\cos\phi\ r^2\sin^2\theta\cos\phi+r\cos\theta\cos\phi\ r\sin\theta\cos\phi\cos\theta+r\sin\theta\sin\phi r\sin\phi\]\[=r^2\sin\theta\left(\sin^2\theta\cos^2\phi-\cos^2\theta\cos^2\phi+\sin^2\phi\right)\]\[=r^2\sin\theta\left((\sin^2\theta+\cos^2\theta)\cos^2\phi+\sin^2\phi\right)\]\[=r^2\sin\theta\left(\sin^2\phi+\cos^2\phi\right)\]\[{=r^2\sin\theta}\]\[\text{---------}\]