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- Thread starter queenstudy
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http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

That is, the conversions of the vectors, as well as the conversions of the unit vectors.

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i mean with proofs because i already read that website page

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

Is there any one in particular for which you would like an explanation?

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in the definition of unit vectors how did we get the cartesian coordinates in terms of cylinderical coordinates

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Do you mean:

in the definition of unit vectors how did we get the cartesian coordinates in terms of cylinderical coordinates

[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]

?

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yes please because we know how the opposite happens

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The only interesting ones are [itex]\mathbf{\hat x}[/itex] and [itex]\mathbf{\hat y}[/itex].yes please because we know how the opposite happens

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]

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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]

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ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh now i get it okay okay thank you alotttttttttttttttttttt

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