Convertion from and to spherical - cartesian

AI Thread Summary
The discussion revolves around the conversion between spherical and Cartesian coordinates, specifically focusing on the transformation equations for velocity components. The initial equations for \dot{x}, \dot{y}, and \dot{z} are provided, leading to the need for an inverse transformation for \dot{r}, \dot{θ}, and \dot{∅}. Participants emphasize that deriving the latter from the former primarily involves algebra and trigonometric identities. One user expresses frustration over the complexity of solving the equations, despite ultimately achieving the correct results. The conversation highlights the mathematical intricacies involved in coordinate transformations in physics.
M. next
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I googled it, and it says:
\dot{x}=\dot{r}sinθcos∅ + (rcosθcos∅)\dot{θ} - (rsinθsin∅)\dot{∅}
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and so on for \dot{y} & \dot{z}

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

for example they got:
\dot{r}=sinθcos∅\dot{x}+sinθsin∅\dot{y} + cosθ\dot{z}
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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

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

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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