Coordinate transformation: derivative of spherical coordinate with cartesian coordinate

  • Thread starter ytht100
  • Start date
  • #1
20
0
I have the following equations:
[tex]\left\{ \begin{array}{l}
x = \sin \theta \cos \varphi \\
y = \sin \theta \cos \varphi \\
z = \cos \theta
\end{array} \right.[/tex]

Assume [tex]\vec r = (x,y,z)[/tex], which is a 1*3 vector. Obviously, x, y, and z are related to each other. Now I want to calculate [tex]\frac{{\partial \vec r}}{{\partial z}}[/tex], 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:

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
20,015
11,390
[tex]r = (x,y,z)[/tex], which is a 1*3 vector. Obviously, x, y, and z are related to each other. Now I want to calculate [tex]\frac{{\partial r}}{{\partial z}}[/tex]

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
20
0
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.

1, Thank you for reminding me to write [tex]{\vec r}[/tex] as vector, I have made revisions above.
However, x, y, and z are NOT independent variables, for example [tex]{x^2} + {y^2} + {z^2} = 1[/tex]. If they are independent, the problem is very easy because [tex]\frac{{d{\vec r}}}{{dz}} = \frac{{d(x,y,z)}}{{dz}} = (0,0,1)[/tex]
2, the physical problem at my hand tells me they are NOT independent.
 
Last edited:
  • #4
334
61
which problem do you really solve? I am afraid the current statement merely does not make sense
 
  • #5
20
0
which problem do you really solve? I am afraid the current statement merely does not make sense

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
334
61
the partial derivative by its definition uses some coordinate of a coordinate system. Coordinate system consists of independent coordinates
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
20,015
11,390
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!

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
20
0
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}##

Thanks indeed! You clear up my mind!
 

Related Threads on Coordinate transformation: derivative of spherical coordinate with cartesian coordinate

Replies
5
Views
2K
  • Last Post
Replies
6
Views
1K
Replies
5
Views
8K
Replies
1
Views
2K
Replies
7
Views
2K
  • Last Post
Replies
3
Views
490
  • Last Post
Replies
1
Views
1K
Replies
3
Views
2K
Replies
24
Views
1K
Top