Convertion from and to spherical - cartesian

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Discussion Overview

The discussion revolves around the conversion between spherical and Cartesian coordinates, specifically focusing on the derivatives of these coordinates. Participants are exploring the algebraic manipulation required to derive the inverse transformation equations from the original equations relating the derivatives.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equations for the derivatives in spherical coordinates and questions the procedure for deriving the inverse transformation.
  • Another participant suggests that the process is primarily algebraic, involving standard trigonometric identities.
  • A different participant expresses confusion about how to derive the radial component from the Cartesian coordinates.
  • One participant mentions having solved the equations with three unknowns but notes that it required significant effort.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the best method for deriving the inverse transformation, as participants express varying levels of understanding and approaches to the problem.

Contextual Notes

Participants reference specific equations and transformations but do not clarify all assumptions or steps involved in their reasoning, leaving some aspects of the derivation unresolved.

M. next
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I googled it, and it says:
[itex]\dot{x}[/itex]=[itex]\dot{r}[/itex]sinθcos∅ + (rcosθcos∅)[itex]\dot{θ}[/itex] - (rsinθsin∅)[itex]\dot{∅}[/itex]
.
.
and so on for [itex]\dot{y}[/itex] & [itex]\dot{z}[/itex]

And then they wrote "We will also need the inverse transformation obtained by solving the equations above w.r.t [itex]\dot{r}[/itex], [itex]\dot{θ}[/itex], and [itex]\dot{∅}[/itex]

for example they got:
[itex]\dot{r}[/itex]=sinθcos∅[itex]\dot{x}[/itex]+sinθsin∅[itex]\dot{y}[/itex] + cosθ[itex]\dot{z}[/itex]
.
.
and so on, my question how did they deduced the latter from the former equations?

Here's the link if you want to see them clearer:
http://www.physics.sc.edu/~yar/Phys701_2009/homework/hw9_solutions.pdf

Just what's the procedure?
Thanks
 
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Hi M. next! :smile:
M. next said:
… for example they got:
[itex]\dot{r}[/itex]=sinθcos∅[itex]\dot{x}[/itex]+sinθsin∅[itex]\dot{y}[/itex] + cosθ[itex]\dot{z}[/itex]

Just what's the procedure?

It's just algebra (plus standard trigonometric identities) …

for example, the r' component of the RHSs is r'(sin2θcos2∅ + sin2θsin2∅ + cos2θ) = r' :wink:
 
Hmm.. I mean how do we get r' from (x', y', z')?
 
I solved three equations with three unknowns and it worked but after so much effort! Thanks anyways Tim.
 

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