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Definitions

Unit vector has length 1. Orthonormal vectors are orthogonal and unit vectors.

RobJohn's suggestions for the basis in polar coordinates, here, satisfy the criteria but how can you approach this geometrically? And how to make sure they are actually orthogonal in polar coordinates?

$$\begin{align} \hat{e}_R&=\frac{(x,y,z)}{r}\\ \hat{e}_\theta&=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}\\ \hat{e}_\phi&=\frac{(-xz,-yz,r^2-z^2)}{r\sqrt{r^2-z^2}} \end{align}$$

Perhaps related

  1. Visual Ways to Remember Cross-product in $\mathbb F^3$? (third vector from two other unit vectors)

  2. Explain Dot product with Partial derivatives in Polar-coordinates (orthogonality check)

  3. Mnemonics for the Clockwiseness with Polar Coordinates

Example problem (about page 817 here)

$$\nabla\cdot\bar{F}=\left(\hat{e}_{R}\partial_{R}+\frac{1}{R}\hat{e}_{\theta}\partial_{\theta}+ \frac{1}{R\sin(\theta)} \hat{e}_{\phi}\partial_{\phi}\right)\cdot\bar{F}$$

where

$$\bar{F}=R^3 ( \cos(\phi)\sin(\theta)\bar{i}+\sin(\phi)\sin(\theta)\bar{j}+\cos(\theta)\bar{k})‌​.$$

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Usually, a unit vector is not required to be orthogonal to anything else. It sounds as if you are trying to define an orthonormal set, but you are only using a single vector. Normally, a unit vector is a vector whose length is $1$. –  robjohn Mar 26 '12 at 12:41
    
@robjohn: yes thank you. Well after learning the case 1, it is easy to learn the other cases :) –  hhh Mar 26 '12 at 18:04

1 Answer 1

up vote 1 down vote accepted

enter image description here

$\hat{e}_R$ is the unit radial vector. This is simply $(x,y,z)$ divided by its length, $r$: $\hat{e}_R=\frac{(x,y,z)}{r}$.

$\hat{e}_\theta$ is the unit vector tangent to the sphere, thus perpendicular to $(x,y,z)$, which is also perpendicular to $(0,0,1)$ since changing $\theta$ does not change $z$. Therefore, we simply need to take the cross product of $(x,y,z)$ and $(0,0,1)$ and normalize to get $\hat{e}_\theta=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}$. The sign is chosen so that the vector points counter-clockwise.

$\hat{e}_\phi$ is perpendicular to the other two, so we take the cross product and normalize: $\hat{e}_\phi=\frac{(-xz,-yz,r^2-z^2)}{r\sqrt{r^2-z^2}}$. Again, the sign is chosen to have positive $z$ component.

Thus, we get $$ \begin{align} \hat{e}_R&=\frac{(x,y,z)}{r}\\ \hat{e}_\theta&=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}\\ \hat{e}_\phi&=\frac{(-xz,-yz,r^2-z^2)}{r\sqrt{r^2-z^2}} \end{align} $$

References

  1. Source of the polar-coordinate image here from Wikipedia.
share|improve this answer
    
...could elaborate on this $\hat{e}_\theta=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}$? Which sign? When you do the cross-product, how you think it? I do the matrix thing but I find it slow... Do you think it as a Sarrus-rule here? –  hhh Mar 24 '12 at 22:01
    
@hhh: $(x,y,z)\times(0,0,1)=(y,-x,0)$ and $(0,0,1)\times(x,y,z)=(-y,x,0)$ normalizing these gives opposite vectors, so we choose the one that goes counter-clockwise. Both are perpendicular to the vectors specified. –  robjohn Mar 24 '12 at 22:46
    
The image from Wikipedia has $\theta$ and $\varphi$ switched from the way I usually see them, and from the way I have them in my answer. I will either edit my answer, or generate a new image (probably the latter so that we are copyright safe). –  robjohn Mar 26 '12 at 0:39
    
...if you generate the picture, please, do not hesitate to include the source code. I moved the clockwiseness -thing here. –  hhh Mar 26 '12 at 0:48

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