Convertion from and to spherical - cartesian

  • Thread starter M. next
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  • #1
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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]
.
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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]
.
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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 [Broken]

Just what's the procedure?
Thanks
 
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Answers and Replies

  • #2
tiny-tim
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Hi M. next! :smile:
… 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:
 
  • #3
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Hmm.. I mean how do we get r' from (x', y', z')?
 
  • #4
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I solved three equations with three unknowns and it worked but after so much effort! Thanks anyways Tim.
 

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