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## Main Question or Discussion Point

What is equivalent to the unit k (vector in cartesian coords) in spherical coordinates? And why?

z=rcos(t)

z=rcos(t)

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What is equivalent to the unit k (vector in cartesian coords) in spherical coordinates? And why?

z=rcos(t)

z=rcos(t)

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if the vector is in three dimensions, one more variable(of spherical) is required to define j.pivoxa15 said:What is equivalent to the unit j (vector in cartesian coords) in spherical coordinates? And why?

z=rcos(t)

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dav2008

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If you are asking what the unit vectors in spherical coordinates are then see: http://mathworld.wolfram.com/SphericalCoordinates.html (Equations 18, 19, 20)

Also http://www.engin.brown.edu/courses/en3/Notes/Vector_Web2/Vectors6a/Vectors6a.htm

Also http://www.engin.brown.edu/courses/en3/Notes/Vector_Web2/Vectors6a/Vectors6a.htm

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In cartesian, it is (0,0,1). What is it in spherical (0,0,what)?

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arildno

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For the most common choice of spherical polar coordinates,

[tex]x=r\sin\phi\cos\theta,y=r\sin\phi\sin\thea,z=r\cos\phi,0\leq{r},0\leq\phi\leq\pi,0\leq\theta\leq{2}\pi[/itex]

we have the following unit vetors relations:

[tex]\vec{i}_{r}=\sin\phi(\cos\theta\vec{i}+\sin\theta\vec{j})+\cos\phi\vec{k}[/tex]

[tex]\vec{i}_{\phi}=\frac{\partial}{\partial\phi}\vec{i}_{r}=\cos\phi(\cos\theta\vec{i}+\sin\theta\vec{j})-\sin\phi\vec{k}[/tex]

[tex]\vec{i}_{\theta}=\frac{1}{\sin\phi}\frac{\partial}{\partial\theta}\vec{i}_{r}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]

Solving for the Cartesian unit vectors we gain, in particular:

[tex]\vec{k}=\cos\phi\vec{i}_{r}-\sin\phi\vec{i}_{\phi}[/tex]

That is of course equal to the coordinate transformation:

[tex](0,0,1)\to(\cos\phi,0,\sin\phi)[/tex]

In order to find the correct expressions for the other two unit Cartesian vectors, utilize the intermediate result:

[tex]\sin\phi\vec{i}_{r}+\cos\phi\vec{i}_{\phi}=\vec{i}_{\hat{r}}=\cos\theta\vec{i}+\sin\theta\vec{j}[/itex]

[itex]\vec{i}_{\hat{r}},\vec{i}_{\theta}[/itex] are polar coordinate vectors in the horizontal plane.

[tex]x=r\sin\phi\cos\theta,y=r\sin\phi\sin\thea,z=r\cos\phi,0\leq{r},0\leq\phi\leq\pi,0\leq\theta\leq{2}\pi[/itex]

we have the following unit vetors relations:

[tex]\vec{i}_{r}=\sin\phi(\cos\theta\vec{i}+\sin\theta\vec{j})+\cos\phi\vec{k}[/tex]

[tex]\vec{i}_{\phi}=\frac{\partial}{\partial\phi}\vec{i}_{r}=\cos\phi(\cos\theta\vec{i}+\sin\theta\vec{j})-\sin\phi\vec{k}[/tex]

[tex]\vec{i}_{\theta}=\frac{1}{\sin\phi}\frac{\partial}{\partial\theta}\vec{i}_{r}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]

Solving for the Cartesian unit vectors we gain, in particular:

[tex]\vec{k}=\cos\phi\vec{i}_{r}-\sin\phi\vec{i}_{\phi}[/tex]

That is of course equal to the coordinate transformation:

[tex](0,0,1)\to(\cos\phi,0,\sin\phi)[/tex]

In order to find the correct expressions for the other two unit Cartesian vectors, utilize the intermediate result:

[tex]\sin\phi\vec{i}_{r}+\cos\phi\vec{i}_{\phi}=\vec{i}_{\hat{r}}=\cos\theta\vec{i}+\sin\theta\vec{j}[/itex]

[itex]\vec{i}_{\hat{r}},\vec{i}_{\theta}[/itex] are polar coordinate vectors in the horizontal plane.

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