A Coordinate transformation: derivative of spherical coordinate with cartesian coordinate

1. Mar 23, 2017

ytht100

I have the following equations:
$$\left\{ \begin{array}{l} x = \sin \theta \cos \varphi \\ y = \sin \theta \cos \varphi \\ z = \cos \theta \end{array} \right.$$

Assume $$\vec r = (x,y,z)$$, which is a 1*3 vector. Obviously, x, y, and z are related to each other. Now I want to calculate $$\frac{{\partial \vec r}}{{\partial z}}$$, could you please tell me if you have any hint?

I have Googled the questions a lot with different terms but can't find an answer that I am sure of. Many thanks for your attention!

Attempt 1: The problem seems related to coordinate transformation between spherical and cartesian coordinates.

Attempt 2: The problem seems related to "The Cartesian partial derivatives in spherical coordinates" shown here: http://mathworld.wolfram.com/SphericalCoordinates.html.

Last edited: Mar 23, 2017
2. Mar 23, 2017

PeroK

First, you have:

$\vec{r} = (x, y, z)$

And

$r = \sqrt{x^2 + y^2 + z^2}$

$x, y, z$ are independent variables, so are unrelated to each other. But, both $\vec{r}$ and $r$ can be treated as functions of $x, y, z$ and differentiated.

3. Mar 23, 2017

ytht100

1, Thank you for reminding me to write $${\vec r}$$ as vector, I have made revisions above.
However, x, y, and z are NOT independent variables, for example $${x^2} + {y^2} + {z^2} = 1$$. If they are independent, the problem is very easy because $$\frac{{d{\vec r}}}{{dz}} = \frac{{d(x,y,z)}}{{dz}} = (0,0,1)$$
2, the physical problem at my hand tells me they are NOT independent.

Last edited: Mar 23, 2017
4. Mar 23, 2017

zwierz

which problem do you really solve? I am afraid the current statement merely does not make sense

5. Mar 23, 2017

ytht100

I am trying to solve a ray tracing problem. That is all I can say. Other information may not appropriate on this forum. Could you please tell me which part doesn't make sense? Thank you very much!

6. Mar 23, 2017

zwierz

the partial derivative by its definition uses some coordinate of a coordinate system. Coordinate system consists of independent coordinates

7. Mar 23, 2017

PeroK

You effectively have a function of two independent variables, with $r=1$. But, to get a partial derivative wrt $z$ you must be explicit about what the other variable is. For example, you could have:

$\vec{r} = (x, \sqrt{1-x^2-z^2}, z)$ or $\vec{r} = (\sqrt{1-y^2-z^2}, y, z)$ or $\vec{r} = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ where $z = \cos{\theta}$

8. Mar 23, 2017

ytht100

Thanks indeed! You clear up my mind!