Spherical coordinates - Orthonormal system

In summary, The conversation discussed using spherical coordinates and the orthonormal system of vectors $\overrightarrow{e}_{\rho}, \overrightarrow{e}_{\theta}, \overrightarrow{e}_{\phi}$ to describe each vector as a function of $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ and $(x, y, z)$. Analytical and geometrical approaches were used to calculate $\overrightarrow{e}_{\theta} \times \overrightarrow{j}$ and $\overrightarrow{e}_{\phi} \times \overrightarrow{j}$. The conversation also touched on the notation for vectors, specifically whether or not to include an arrow over the symbol for a vector.
  • #1
mathmari
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Hey! :eek:

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)
 
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  • #2
mathmari said:
Hey! :eek:

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:
$$\overrightarrow{e}_{\rho} = \frac{\d {\overrightarrow r}{\rho}}{\lVert\d {\overrightarrow r}{\rho}\rVert}$$
 

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  • #3
I like Serena said:
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}$$
 
  • #4
Yep! (Nod)
 
  • #5
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 )$$
 
  • #6
mathmari said:
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)
 
  • #7
I like Serena said:
Shouldn't that be $\overrightarrow{\rho}$ instead of $r$? (Wondering)

Why? (Wondering)
 
  • #8
mathmari said:
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)
 

FAQ: Spherical coordinates - Orthonormal system

1. What are spherical coordinates and how are they different from Cartesian coordinates?

Spherical coordinates are a system used to describe the location of a point in three-dimensional space. They use three coordinates: radius, azimuth angle, and polar angle. The main difference from Cartesian coordinates is that the position of a point is described using angles instead of distances along x, y, and z axes.

2. What is the purpose of using an orthonormal system in spherical coordinates?

An orthonormal system in spherical coordinates ensures that the basis vectors at any point are perpendicular to each other and have a unit length. This simplifies calculations and makes it easier to interpret geometric properties of a point.

3. How do you convert between spherical and Cartesian coordinates?

To convert from spherical coordinates (r, θ, φ) to Cartesian coordinates (x, y, z), we use the following equations: x = r*sin(θ)*cos(φ), y = r*sin(θ)*sin(φ), z = r*cos(θ). To convert from Cartesian coordinates to spherical coordinates, we use the equations: r = √(x² + y² + z²), θ = arctan(√(x² + y²)/z), and φ = arctan(y/x).

4. Can spherical coordinates be used to describe any point in three-dimensional space?

Yes, spherical coordinates can be used to describe any point in three-dimensional space as long as the origin and axes are properly defined.

5. What are some common applications of spherical coordinates in science and engineering?

Spherical coordinates are commonly used in fields such as physics, astronomy, and engineering. They are particularly useful for describing the motion of objects in three-dimensional space, such as the movement of planets and satellites. They are also used in navigation systems, computer graphics, and 3D modeling.

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