Transformation matrix derivation problem

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Fryderyk
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Homework Statement



As part of an assignment, I need to derive a transformation matrix to convert a vector in cartesian coordinates to spherical coordinates.

Homework Equations



What I've got so far is:

For an arbitrary vector V,

[tex] \textbf{V}=\left[\begin {array}{ccc}V_{x}&V_{y}&V_{z}\end{array}\right]\left[\begin{array}{c}\textbf{i}\\\textbf{j}\\\textbf{k}\end{array}\right] = \left[\begin{array}{ccc}V_{R}&V_{\theta}&V_{\phi}\end{array}\right]\left[\begin{array}{c}\textbf{e}_{R}\\\textbf{e}_{\theta}\\\textbf{e}_{\phi}\end{array}\right][/tex]

Knowing that VR is the component of V in the eR direction, then:

[tex] V_{R}=aV_{x}+bV_{y}+cV_{z}[/tex]

where a, b and c are some transformation coefficients to be derived from geometry.

Similarly,

[tex] V_{\theta}=dV_{x}+eV_{y}+fV_{z}[/tex]
and
[tex] V_{\phi}=gV_{x}+hV_{y}+iV_{z}[/tex]

From this, a transformation matrix T exists such that:

[tex] \left[\begin{array}{ccc}V_{x}&V_{y}&V_{z}\end{array}\right]\textit{T}=\left[\begin{array}{ccc}aV_{x}+bV_{y}+cV_{z}&dV_{x}+eV_{y}+fV_{z}&gV_{x}+hV_{y}+iV_{z}\end{array}\right]=\left[\begin{array}{ccc}V_{R}&V_{\theta}&V_{\phi}\end{array}\right][/tex]

The Attempt at a Solution



My problem is finding this matrix T.
It's obvious from inspection that:

[tex] T=\left[\begin{array}{ccc}a&d&g\\b&e&h\\c&f&i\end{array}\right][/tex]

But, I'm not sure that just stating 'it's obvious that' will be sufficient.
I'm wondering if and how the below equation can be rearranged and solved to give T.

[tex] \left[\begin{array}{ccc}V_{x}&V_{y}&V_{z}\end{array}\right]\textit{T}=\left[\begin{array}{ccc}V_{R}&V_{\theta}&V_{\phi}\end{array}\right][/tex]

After this point, finding the coefficients shouldn't be a problem.

Thank you
 
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Welcome to PF!

Hi Fryderyk! Welcome to PF! :smile:

(have a theta: θ and a phi: φ :wink:)
Fryderyk said:
My problem is finding this matrix T.
It's obvious from inspection that:

[tex] T=\left[\begin{array}{ccc}a&d&g\\b&e&h\\c&f&i\end{array}\right][/tex]

But, I'm not sure that just stating 'it's obvious that' will be sufficient.
I'm wondering if and how the below equation can be rearranged and solved to give T.

[tex] \left[\begin{array}{ccc}V_{x}&V_{y}&V_{z}\end{array}\right]\textit{T}=\left[\begin{array}{ccc}V_{R}&V_{\theta}&V_{\phi}\end{array}\right][/tex]

Yes, you're really just stating the obvious …

Finding T and finding a b c etc are the same thing.

Try it this way … you have two sets of axes, with angles between them, and you first need to find the angles …

do the two-dimensional case first (it's easier! :wink:) …

er and eθ are at angle θ to i and j, so the matrix is … ? :smile:
After this point, finding the coefficients shouldn't be a problem.

:smile: :smile:
 
hi, I am new to PF and i have a very similer problem to this. I am also having trouble finding the T matrix. so far i have found:
R= (x^2+y^2+z^2)^1/2
theta= sin-1(y/(x^2+y^2)^1/2)
phi =??

hope some one can help...
thanks
 
Welcome to PF!

Hi demha! Welcome to PF! :smile:

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
demha said:
hi, I am new to PF and i have a very similer problem to this. I am also having trouble finding the T matrix. so far i have found:
R= (x^2+y^2+z^2)^1/2
theta= sin-1(y/(x^2+y^2)^1/2)
phi =??

You seem to be trying to find a matrix that transforms "global" coordinates.

A matrix is linear, and only gives linear transformations, but Cartesian to polar vectors isn't linear.

The matrix T isn't intended to convert vectors which start at the origin … it's only for converting "local" vectors at a particular point (x,y,z) = (r,θ,φ) … so you're converting from i j and k at that point to er eθ and eφ at that point.

For that, you don't need √ or sin-1, you just use the angles. :wink:

As I said to Fryderyk, try the two-dimensional case first. :smile: