Cylindrical coordinates - Orthonormal system

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Discussion Overview

The discussion revolves around the representation of cylindrical coordinates and the orthonormal system of vectors $\overrightarrow{e}_r$, $\overrightarrow{e}_{\theta}$, and $\overrightarrow{e}_z$. Participants explore how to express these vectors in terms of Cartesian unit vectors $\overrightarrow{i}$, $\overrightarrow{j}$, and $\overrightarrow{k}$, and discuss the calculation of the cross product $\overrightarrow{e}_{\theta} \times \overrightarrow{j}$ using both analytical and geometric methods.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the vector $\overrightarrow{e}_r$ can be expressed as $\overrightarrow{e}_r = \frac{x}{r}\overrightarrow{i} + \frac{y}{r}\overrightarrow{j} + \frac{z}{r}\overrightarrow{k}$, where $r$ is the radial distance.
  • Others suggest that $\overrightarrow{e}_{\theta}$ should be defined as the unit vector in the direction of how $\overrightarrow{r}$ changes with respect to $\theta$, questioning if it can be represented as $\overrightarrow{e}_{\theta} = \frac{\d {\overrightarrow r}{\theta}}{\left|\d {\overrightarrow r}{\theta}\right|}$.
  • There is a proposal that $\overrightarrow{e}_{\theta}$ can be expressed as $(-r \sin \theta)\overrightarrow{i} + (r \cos \theta)\overrightarrow{j}$, leading to a discussion on its normalization.
  • Participants discuss the expression for $\overrightarrow{e}_{\theta}$ and its relation to the Cartesian coordinates, with some suggesting it could be $-y\overrightarrow{i} + x\overrightarrow{j}$.
  • There is a calculation presented for the cross product $\overrightarrow{e}_{\theta} \times \overrightarrow{j}$, with one participant suggesting the result is $-y \overrightarrow{k}$.
  • Questions arise regarding the geometric interpretation of the cross product and how to visualize it.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the exact definitions and representations of the vectors, particularly $\overrightarrow{e}_{\theta}$. There is no consensus on the geometric approach to demonstrate the cross product.

Contextual Notes

Participants have not fully resolved the normalization of $\overrightarrow{e}_{\theta}$, and there are unresolved questions about the geometric representation of the vectors and their relationships in cylindrical coordinates.

mathmari
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Hey! :o

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)
 
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mathmari said:
Hey! :o

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)
 
I like Serena said:
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)
 
mathmari said:
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)
 
I like Serena said:
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)
I like Serena said:
Did you make a drawing? (Wondering)

No... How could I do that?? (Wondering)
 
mathmari said:
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:
CylindricalCoordinates_1001.gif
 

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