[Math] how to change polar coordinate into cartesian coordinate using transformation matrix

coordinate systems

I would like to change $(3,4,12)$ in $xyz$ coordinate to spherical coordinate using the following relation
enter image description here
It is from the this link. I do not understand the significance of this matrix (if not for coordinate transformation) or how it is derived. Also please check my previous question building transformation matrix from spherical to cartesian coordinate system. Please I need your insight on building my concept.

Thank you.
EDIT::
I understand that $ \left [ A_x \sin \theta\cos \phi \hspace{5 mm} A_y \sin \theta\sin\phi \hspace{5 mm} A_z\cos\theta\right ]$ gives $A_r$ but how is other coordinates $ (A_\theta, A_\phi)$ equal to their respective respective rows from Matrix multiplication?

Best Answer

The transformation from Cartesian to polar coordinates is not a linear function, so it cannot be achieved by means of a matrix multiplication.

Related Solutions

Transforming Unit Vectors from Cartesian to Cylindrical Coordinates

A far more simple method would be to use the gradient.
Lets say we want to get the unit vector $\boldsymbol { \hat e_x } $. What we then do is to take $\boldsymbol { grad(x) } $ or $\boldsymbol { ∇x } $.
This; $\boldsymbol ∇ $, is the nabla-operator. It is a vector containing each partial derivative like this...

$\boldsymbol { ∇= ( \frac {∂} {∂x}, \frac {∂} {∂y}, \frac {∂} {∂z}) } $

When we take the gradient of x we get this...

$\boldsymbol { ∇x= ( \frac {∂x} {∂x}, \frac {∂x} {∂y}, \frac {∂x} {∂z})=(1,0,0)=\hat e_x } $

To get the unit vector of $\boldsymbol x$ in cylindrical coordinate system we have to rewrite $x$ in the form of $\boldsymbol {r_c}$ and $\boldsymbol {\phi}$.

$\boldsymbol {x= r_c cos(x) } $

Now you have to use the more general definition of nabla ($\boldsymbol ∇ $).
Lets say we have a curve-linear coordinate system where the position vector is defined like this...

$\boldsymbol {\vec r = u_1 \hat e_{u1} + u_2 \hat e_{u2} + u_3 \hat e_{u3}} $

... Then the nabla operator for that coordinate system is as follows...

$\boldsymbol { ∇ = \frac {1}{h_1} \frac {∂}{∂u_1} \hat e_{u1} + \frac {1}{h_2} \frac {∂}{∂u_2} \hat e_{u1} + \frac {1}{h_3} \frac {∂}{∂u_3} \hat e_{u1}} $

"$\boldsymbol { h_n } $" is the scale factor to the variable "$\boldsymbol { u_n } $". The scale-factor is defined as: $\boldsymbol {h_n = \frac {\partial \vec r}{\partial u_n}}$
For cylindrical coordinates the position vector is defined as: $\boldsymbol {\vec r = r_c \hat e_{rc} + z \hat e_z }$
With some simple math we can get the scale factors and they are...

$\boldsymbol {h_{rc} = 1 \ \ ,\ h_{\phi} = r_c \ \ ,\ h_z = 1}$

We already know that in cylindircal cooridnates $\boldsymbol x $ is defined as $\boldsymbol {x = r_c cos(x)}$, so now we can get the gradient.

$\boldsymbol { ∇x = ∇(r_c cos(x))= \frac {\partial (r_c \cos(x))}{\partial r_c} \hat e_{rc} + \frac {1}{r_c} \frac {\partial (r_c \cos(x))}{\partial \phi} \hat e_{\phi} + \frac {\partial (r_c cos(x))}{\partial z} \hat e_{z}}$

The result from this gradient is then...

$\boldsymbol {\hat e_{x} = \cos(\phi)\hat e_{rc} - \sin(\phi) \hat e_{\phi}}$

When the same method is applied to $\boldsymbol y$, where $\boldsymbol {y = r_c sin(\phi) }$, we get with ease that...

$\boldsymbol {\hat e_{y} = sin(\phi)\hat e_{rc} + \cos(\phi) \hat e_{\phi}}$

Hope it helped!
I also hope the use of $\boldsymbol \phi $ instead of $\boldsymbol \theta $ and $\boldsymbol {r_c} $ instead of $\boldsymbol \rho $ wasn't to confusing. As a physics student I am more used to the $\boldsymbol {(r_c,\phi,z)}$ standard for cylindrical coordinates.

[Math] building transformation matrix from spherical to cartesian coordinate system

I can partially answer this. I believe your first matrix is not the correct general transformation matrix for cartesian to spherical coordinates because you are missing factors of $\rho$ (the radial coordinate), as well as some other incorrect pieces. So it is not clear what you are trying to show.

If you are trying to derive the general transformation matrix from spherical to cartesian, it is: $$\begin{bmatrix} A_x\\ A_y\\ A_z \end{bmatrix} = \begin{bmatrix} \sin \theta \cos \phi & \rho \cos \theta \cos \phi & -\rho \sin \theta \sin\phi\\ \sin \theta \sin \phi & \rho \cos \theta \sin \phi & \rho \sin\theta \cos\phi\\ \cos\theta & -\rho \sin\theta & 0 \end{bmatrix} \begin{bmatrix} A_r\\ A_\theta\\ A_\phi \end{bmatrix} $$ This matrix is formed from the derivative matrix in the following way: $ \sum_{(i,j)} \frac {\partial x^i} {\partial u^j} $ where $$x^1 = \rho\sin \theta \cos \phi, x^2 = \rho\sin \theta \sin \phi, x^3=\rho\cos\theta$$ and $u^1 = \rho, u^2 = \theta, u^3 = \phi $ (Note: the superscripts are indices, not exponents). If you write out all the derivatives, you get this matrix:

$$ \begin{bmatrix} A_x\\ A_y\\ A_z \end{bmatrix} = \begin{bmatrix} \frac {\partial (\rho \sin \theta \cos \phi)} {\partial \rho} & \frac {\partial (\rho \sin \theta \cos \phi)} {\partial \theta} & \frac {\partial (\rho \sin \theta \cos \phi)} {\partial \phi}\\ \frac {\partial (\rho \sin \theta \sin \phi)} {\partial \rho} & \frac {\partial (\rho \sin \theta \sin \phi)} {\partial \theta} & \frac {\partial (\rho \sin \theta \sin \phi)} {\partial \phi}\\ \frac {\partial (\rho \cos \theta)} {\partial \rho} & \frac {\partial (\rho \cos \theta)} {\partial \theta} & \frac {\partial (\rho \cos \theta)} {\partial \phi} \end{bmatrix} \begin{bmatrix} A_r\\ A_\theta\\ A_\phi \end{bmatrix} $$ ...which evaluates to the correct general transformation matrix I listed above.

Best Answer

Related Solutions

Transforming Unit Vectors from Cartesian to Cylindrical Coordinates

[Math] building transformation matrix from spherical to cartesian coordinate system

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