MHB Cylindrical coordinates - Orthonormal system

[mathmari]

Hey!

Using cylindrical coordinates and the orthonormal system of vectors $\overrightarrow{e}_r, \overrightarrow{e}_{\theta}, \overrightarrow{e}_z$

1. describe each of the $\overrightarrow{e}_r$, $\overrightarrow{e}_{\theta}$ and $\overrightarrow{e}_z$ as a function of $\overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k}$ and $(x, y, z)$ and
2. calculate $\overrightarrow{e}_{\theta} \times \overrightarrow{j}$ 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 cylindrical coordinates and the orthonormal system of vectors $\overrightarrow{e}_r, \overrightarrow{e}_{\theta}, \overrightarrow{e}_z$

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

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

Hi! ;)

We have the point $\overrightarrow r = x\overrightarrow{i} + y\overrightarrow{j} + z\overrightarrow{k}$.
Its spherical coordinates are $(r,\theta,\phi)$.

At the point we want to have a local orthonormal basis aligned with spherical coordinates.
The vector $\overrightarrow{e}_r$ is the unit vector with a direction that corresponds to how $\overrightarrow r$ changes if we increase $r$.
That is in the direction of $\overrightarrow r$.
Normalizing to unit length, we get:
$$\overrightarrow{e}_r = \frac x r\overrightarrow{i} + \frac y r\overrightarrow{j} + \frac z r\overrightarrow{k}$$
Alternatively we can say that:
$$\overrightarrow{e}_r = \frac{\d {\overrightarrow r}{r}}{\left|\d {\overrightarrow r}{r}\right|}$$
(Nerd)

Next is $\overrightarrow{e}_{\theta}$, which is the unit vector with a direction that corresponds to how $\overrightarrow r$ changes if we increase $\theta$.
What would its direction be? (Wondering)

 

[mathmari]

Next is $\overrightarrow{e}_{\theta}$, which is the unit vector with a direction that corresponds to how $\overrightarrow r$ changes if we increase $\theta$.
What would its direction be? (Wondering)

Is it $$\overrightarrow{e}_{\theta} = \frac{\d {\overrightarrow r}{\theta}}{\left|\d {\overrightarrow r}{\theta}\right|}$$ ?? (Wondering)

Do we have the following:

$$\overrightarrow{e}_{r}=(\cos \theta )\overrightarrow{i}+(\sin \theta )\overrightarrow{j} \\ \overrightarrow{e}_{\theta}=(-r \sin \theta )\overrightarrow{i}+(r \cos \theta )\overrightarrow{j} \\ \overrightarrow{e}_{z}=\overrightarrow{k}$$ ?? (Wondering)

So using $$x=r \cos \theta , y=r \sin \theta , z=z$$ we have the following:

$$\overrightarrow{e}_{r}=\frac{x\overrightarrow{i}+y \overrightarrow{j}}{\sqrt{x^2+y^2}} \\ \overrightarrow{e}_{\theta}=-y\overrightarrow{i}+x\overrightarrow{j} \\ \overrightarrow{e}_{z}=\overrightarrow{k}$$

Is this correct??
For the question 2. using 1. we have that $$\overrightarrow{e}_{\theta} \times \overrightarrow{j}=-y \overrightarrow{k}$$ right?? (Wondering)

But how could we show it geometrically?? (Wondering)

 

[I like Serena]

Is it $$\overrightarrow{e}_{\theta} = \frac{\d {\overrightarrow r}{\theta}}{\left|\d {\overrightarrow r}{\theta}\right|}$$ ?? (Wondering)

Do we have the following:

$$\overrightarrow{e}_{r}=(\cos \theta )\overrightarrow{i}+(\sin \theta )\overrightarrow{j} \\ \overrightarrow{e}_{\theta}=(-r \sin \theta )\overrightarrow{i}+(r \cos \theta )\overrightarrow{j} \\ \overrightarrow{e}_{z}=\overrightarrow{k}$$ ?? (Wondering)

So using $$x=r \cos \theta , y=r \sin \theta , z=z$$ we have the following:

$$\overrightarrow{e}_{r}=\frac{x\overrightarrow{i}+y \overrightarrow{j}}{\sqrt{x^2+y^2}} \\ \overrightarrow{e}_{\theta}=-y\overrightarrow{i}+x\overrightarrow{j} \\ \overrightarrow{e}_{z}=\overrightarrow{k}$$

For the question 2. using 1. we have that $$\overrightarrow{e}_{\theta} \times \overrightarrow{j}=-y \overrightarrow{k}$$ right?? (Wondering)

Is this correct??

Almost.
We still have to normalize $\overrightarrow{e}_{\theta}$ to unit length. (Wasntme)

But how could we show it geometrically?? (Wondering)

Did you make a drawing? (Wondering)

 

[mathmari]

Almost.
We still have to normalize $\overrightarrow{e}_{\theta}$ to unit length. (Wasntme)

Is it $$\overrightarrow{e}_{\theta}=-sin \theta \overrightarrow{i}+\cos \theta \overrightarrow{j}$$ ?? (Wondering)

Did you make a drawing? (Wondering)

No... How could I do that?? (Wondering)

 

[I like Serena]

Is it $$\overrightarrow{e}_{\theta}=-sin \theta \overrightarrow{i}+\cos \theta \overrightarrow{j}$$ ?? (Wondering)

Yep! (Nod)

No... How could I do that?? (Wondering)

Something like this: