MHB Spherical coordinates - Orthonormal system

[mathmari]

Hey!

Using spherical coordinates and the orthonormal system of vectors $\overrightarrow{e}_{\rho}, \overrightarrow{e}_{\theta}, \overrightarrow{e}_{\phi}$

1. describe each of the $\overrightarrow{e}_{\rho}$, $\overrightarrow{e}_{\theta}$ and $\overrightarrow{e}_{\phi}$ as a function of $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ and $(x, y, z)$ and
2. calculate $\overrightarrow{e}_{\theta} \times \overrightarrow{j}$ and $\overrightarrow{e}_{\phi} \times \overrightarrow{j}$ with two ways: analytically, using (1), and geometrically.

Could you give me some hints how I could do that?? (Wondering)

 

[I like Serena]

Hey!

Using spherical coordinates and the orthonormal system of vectors $\overrightarrow{e}_{\rho}, \overrightarrow{e}_{\theta}, \overrightarrow{e}_{\phi}$

1. describe each of the $\overrightarrow{e}_{\rho}$, $\overrightarrow{e}_{\theta}$ and $\overrightarrow{e}_{\phi}$ as a function of $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ and $(x, y, z)$ and
2. calculate $\overrightarrow{e}_{\theta} \times \overrightarrow{j}$ and $\overrightarrow{e}_{\phi} \times \overrightarrow{j}$ with two ways: analytically, using (1), and geometrically.

Could you give me some hints how I could do that?? (Wondering)

Hey mathmari! :D

Easiest is to do it geometrically by drawing a vector in spherical coordinates, and deducing in which direction the vector changes if you change one of the spherical coordinates. (Thinking)

In this picture you can see how they should come out:
View attachment 4097

Or analytically by evaluating:
$$\overrightarrow{e}_{\rho} = \frac{\d {\overrightarrow r}{\rho}}{\lVert\d {\overrightarrow r}{\rho}\rVert}$$

 

[mathmari]

Or analytically by evaluating:
$$\overrightarrow{e}_{\rho} = \frac{\d {\overrightarrow r}{\rho}}{\lVert\d {\overrightarrow r}{\rho}\rVert}$$

Is it as followed?? (Wondering)

$r=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$

$x=\rho \sin \phi \cos \theta , y=\rho \sin \phi \sin \theta , z=\rho \cos \phi$

$$\overrightarrow{e}_{\rho} = \frac{\d {\overrightarrow r}{\rho}}{\lVert\d {\overrightarrow r}{\rho}\rVert}=\frac{\sin \phi \cos \theta \overrightarrow{i}+\sin \phi \sin \theta \overrightarrow{j}+\cos \phi \overrightarrow{k}}{\lVert \sin \phi \cos \theta \overrightarrow{i}+\sin \phi \sin \theta \overrightarrow{j}+\cos \phi \overrightarrow{k} \rVert}=\sin \phi \cos \theta \overrightarrow{i}+\sin \phi \sin \theta \overrightarrow{j}+\cos \phi \overrightarrow{k} \\

\overrightarrow{e}_{\theta} = \frac{\d {\overrightarrow r}{\theta}}{\lVert\d {\overrightarrow r}{\theta}\rVert}=\frac{-\rho \sin \phi \sin \theta \overrightarrow{i}+\rho \sin \phi \cos \theta \overrightarrow{j}}{\lVert -\rho \sin \phi \sin \theta \overrightarrow{i}+\rho \sin \phi \cos \theta \overrightarrow{j} \rVert}=\frac{-\rho \sin \phi \sin \theta \overrightarrow{i}+\rho \sin \phi \cos \theta \overrightarrow{j}}{\sqrt{\rho^2 \sin^2 \phi}}=\frac{-\rho \sin \phi \sin \theta \overrightarrow{i}+\rho \sin \phi \cos \theta \overrightarrow{j}}{\rho \sin \phi}=- \sin \theta \overrightarrow{i}+ \cos \theta \overrightarrow{j}\\

\overrightarrow{e}_{\phi} = \frac{\d {\overrightarrow r}{\phi}}{\lVert\d {\overrightarrow r}{\phi}\rVert}=\frac{\rho \cos \phi \cos \theta \overrightarrow{i}+\rho \cos \phi \sin \theta \overrightarrow{j}-\rho \sin \phi \overrightarrow{k}}{\lVert \rho \cos \phi \cos \theta \overrightarrow{i}+\rho \cos \phi \sin \theta \overrightarrow{j}-\rho \sin \phi \overrightarrow{k} \rVert}=\frac{\rho \cos \phi \cos \theta \overrightarrow{i}+\rho \cos \phi \sin \theta \overrightarrow{j}-\rho \sin \phi \overrightarrow{k}}{\sqrt{\rho^2}}=\frac{\rho \cos \phi \cos \theta \overrightarrow{i}+\rho \cos \phi \sin \theta \overrightarrow{j}-\rho \sin \phi \overrightarrow{k}}{\rho}=\cos \phi \cos \theta \overrightarrow{i}+\cos \phi \sin \theta \overrightarrow{j}-\sin \phi \overrightarrow{k}$$

 

[I like Serena]

Yep! (Nod)

 

[mathmari]

So, to describe each of the $\overrightarrow{e}_{\rho}$, $\overrightarrow{e}_{\theta}$ and $\overrightarrow{e}_{\phi}$ as a function of $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ and $(x, y, z)$ do we have to do the following?? (Wondering)

$r=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$

$x=\rho \sin \phi \cos \theta , y=\rho \sin \phi \sin \theta , z=\rho \cos \phi$

$\rho=\sqrt{x^2+y^2+z^2}$

$x^2+y^2=\rho^2 \sin^2 \phi \cos^2 \theta+\rho^2 \sin^2 \phi \sin^2 \theta=\rho^2 \sin^2 \phi \Rightarrow \rho \sin \phi = \sqrt{x^2+y^2}$

$\rho^2 \sin^2 \phi=x^2+y^2 \Rightarrow \rho \sin^2 \phi=\frac{x^2+y^2}{\rho} \Rightarrow \rho \sin^2 \phi=\frac{x^2+y^2}{\sqrt{x^2+y^2+z^2}}$

$z=\rho \cos \phi \Rightarrow \cos \phi=\frac{z}{\rho}=\cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}$

$$\overrightarrow{e}_{\rho} =\sin \phi \cos \theta \overrightarrow{i}+\sin \phi \sin \theta \overrightarrow{j}+\cos \phi \overrightarrow{k} =\frac{1}{\rho} \left ( \rho \sin \phi \cos \theta \overrightarrow{i}+\rho \sin \phi \sin \theta \overrightarrow{j}+\rho \cos \phi \overrightarrow{k}\right )=\frac{1}{\sqrt{x^2+y^2+z^2}} \left ( x \overrightarrow{i}+y \overrightarrow{j}+z \overrightarrow{k}\right )\\

\overrightarrow{e}_{\theta} =- \sin \theta \overrightarrow{i}+ \cos \theta \overrightarrow{j}=\frac{1}{\rho} \left (- \rho \sin \theta \overrightarrow{i}+ \rho \cos \theta \overrightarrow{j}\right )=\frac{1}{\rho \sin \phi} \left (- \rho \sin \phi \sin \theta \overrightarrow{i}+ \rho \sin \phi \cos \theta \overrightarrow{j}\right )=\frac{1}{\sqrt{x^2+y^2}} \left (- y \overrightarrow{i}+ x \overrightarrow{j}\right )\\

\overrightarrow{e}_{\phi} =\cos \phi \cos \theta \overrightarrow{i}+\cos \phi \sin \theta \overrightarrow{j}-\sin \phi \overrightarrow{k}=\frac{1}{\rho \sin \phi }\left (\cos \phi \rho \sin \phi \cos \theta \overrightarrow{i}+\cos \phi \rho \sin\phi \sin \theta \overrightarrow{j}-\rho \sin^2 \phi \overrightarrow{k}\right )=\frac{1}{\sqrt{x^2+y^2}}\left (\frac{z}{\sqrt{x^2+y^2+z^2}}x \overrightarrow{i}+\frac{z}{\sqrt{x^2+y^2+z^2}}y \overrightarrow{j}-\frac{x^2+y^2}{\sqrt{x^2+y^2+z^2}} \overrightarrow{k}\right )$$

 

[I like Serena]

So, to describe each of the $\overrightarrow{e}_{\rho}$, $\overrightarrow{e}_{\theta}$ and $\overrightarrow{e}_{\phi}$ as a function of $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ and $(x, y, z)$ do we have to do the following?? (Wondering)

That looks to be correct. (Nod)

Btw, you can basically read off the end result directly from the geometrical representation of spherical coordinates. (Nerd)

$r=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$

Shouldn't that be $\overrightarrow{\rho}$ instead of $r$? (Wondering)

 

[mathmari]

Shouldn't that be $\overrightarrow{\rho}$ instead of $r$? (Wondering)

Why? (Wondering)

 

[I like Serena]

Why? (Wondering)

Well... $r$ is defined to be a vector.
Shouldn't it have an arrow over it then? (Wondering)

Furthermore, we have $\overrightarrow r = \rho \overrightarrow e_\rho$.
It's not required, but isn't it kind of conventional to use the same symbol for the length of a vector as for the vector itself? (Wondering)