Unit vectors

  • Thread starter queenstudy
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  • #1
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can i get some help in how i can convert from cartesian to cylindrical and spherical unit vectors and vice versa ???? thank you
 

Answers and Replies

  • #3
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i mean with proofs because i already read that website page
 
  • #4
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i mean with proofs because i already read that website page
Hmm, well, you can basically read them directly off the drawing of the coordinates.

Is there any one in particular for which you would like an explanation?
 
  • #5
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yes
in the definition of unit vectors how did we get the cartesian coordinates in terms of cylinderical coordinates
 
  • #6
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yes
in the definition of unit vectors how did we get the cartesian coordinates in terms of cylinderical coordinates
Do you mean:
[tex]
\begin{matrix}
\mathbf{\hat x} & = & \cos\phi\boldsymbol{\hat \rho}-\sin\phi\boldsymbol{\hat\phi} \\
\mathbf{\hat y} & = & \sin\phi\boldsymbol{\hat \rho}+\cos\phi\boldsymbol{\hat\phi} \\
\mathbf{\hat z} & = & \mathbf{\hat z}
\end{matrix}
[/tex]
?
 
  • #7
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yes please because we know how the opposite happens
 
  • #8
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yes please because we know how the opposite happens
The only interesting ones are [itex]\mathbf{\hat x}[/itex] and [itex]\mathbf{\hat y}[/itex].
So we're actually talking about 2-dimensional polar coordinates.


Here's one way to derive the unit vectors.

[itex]\boldsymbol{\hat \rho}[/itex] corresponds in this case with the vector from the origin to a point on the unit circle at angle [itex]\phi[/itex].

The vector [itex]\boldsymbol{\hat \rho}[/itex] can be seen as the result of a rotation of [itex]\mathbf{\hat x}[/itex] over an angle of [itex]\phi[/itex].

Or in reverse, the vector [itex]\mathbf{\hat x}[/itex] is the result of a rotation of [itex]\boldsymbol{\hat \rho}[/itex] over an angle of [itex]-\phi[/itex].
Similarly the vector [itex]\mathbf{\hat y}[/itex] is the result of a rotation of [itex]\boldsymbol{\hat \phi}[/itex] over an angle of [itex]-\phi[/itex].

We need the rotation matrix for an angle [itex]\phi[/itex] is to get what we want:
[tex]R = \begin{pmatrix}\cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix}[/tex]

Multiply [itex]R[/itex] with [itex]\boldsymbol{\hat \rho}[/itex] and [itex]\boldsymbol{\hat \phi}[/itex] and the result rolls out:
[tex]
\begin{matrix}
\mathbf{\hat x} & = & \cos\phi\boldsymbol{\hat \rho}-\sin\phi\boldsymbol{\hat\phi} \\
\mathbf{\hat y} & = & \sin\phi\boldsymbol{\hat \rho}+\cos\phi\boldsymbol{\hat\phi}
\end{matrix}
[/tex]
 
  • #9
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i know how raw and phy are in terms of x hat and y hat how did you change that to the last line im still not getting the idea??
 
  • #10
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i know how raw and phy are in terms of x hat and y hat how did you change that to the last line im still not getting the idea??
How about this:

Apparently you already know:
[tex]
\begin{matrix}
\boldsymbol{\hat \rho} & = & \cos\phi\mathbf{\hat x}+\sin\phi \mathbf{\hat y}\\
\boldsymbol{\hat\phi} & = & -\sin\phi\mathbf{\hat x}+\cos\phi \mathbf{\hat y}
\end{matrix}
[/tex]

Which is:
[tex]\begin{pmatrix} \boldsymbol{\hat \rho} \\ \boldsymbol{\hat\phi} \end{pmatrix} = \begin{pmatrix}
\cos \phi & \sin \phi \\
-\sin \phi & \cos \phi \end{pmatrix}
\begin{pmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \end{pmatrix}
[/tex]

The inverse is:

[tex]\begin{pmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \end{pmatrix} = \begin{pmatrix}
\cos \phi & -\sin \phi \\
\sin \phi & \cos \phi \end{pmatrix}
\begin{pmatrix} \boldsymbol{\hat \rho} \\ \boldsymbol{\hat\phi} \end{pmatrix}
[/tex]
 
  • #11
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ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh now i get it okay okay thank you alotttttttttttttttttttt
 

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