Explaining Spherical Coordinates and Coordinate Vectors

[jhosamelly]

Homework Statement

(a) For spherical coordinates, show that $\hat{\theta}$ points along the negative z-axis if $\theta$ = 90°.
(b) If $\phi$ also equals 90°, in what direction are $\hat{r}$ and $\hat{\phi}$?

Homework Equations

The Attempt at a Solution

can i just explain this in words.. like

for a.
since we are rotating about the z axis , if theta is 90°, theta will automatically point to negative z-axis

 

[jhosamelly]

Can someone help me with this one?? I'm really having a problem solving this.

 

[jambaugh]

You'll probably need to express these coordinate vectors as functions of r,theta, phi or at least parallel vectors. That is take the partial derivative of the position vector $\vec{r}$ with respect to theta and plug in theta = 90deg, likewise with (b).

 

[jhosamelly]

You'll probably need to express these coordinate vectors as functions of r,theta, phi or at least parallel vectors. That is take the partial derivative of the position vector $\vec{r}$ with respect to theta and plug in theta = 90deg, likewise with (b).

... is that something like this??

$\frac{\partial \hat{r}}{\partial \hat{\theta}}$ = r cos $\phi$$\frac{\partial sin \theta}{\partial {\theta}}$ $\hat{i}$ + r sin $\phi$ $\frac{\partial sin \theta}{\partial {\theta}}$ $\hat{j}$ + r $\frac{\partial cos \theta}{\partial {\theta}}$ $\hat{k}$

= $r cos \phi (cos \theta) \hat{i} + r sin \phi cos \theta \hat{j} - r sin \theta \hat{k}$

so if i plug in 90° only -r$\hat{k}$ will be left since cos 90° = 0

how thus that prove anything?

 

[jambaugh]

... is that something like this??

$\frac{\partial \hat{r}}{\partial \hat{\theta}}$ = r cos $\phi$$\frac{\partial sin \theta}{\partial {\theta}}$ $\hat{i}$ + r sin $\phi$ $\frac{\partial sin \theta}{\partial {\theta}}$ $\hat{j}$ + r $\frac{\partial cos \theta}{\partial {\theta}}$ $\hat{k}$

= $r cos \phi (cos \theta) \hat{i} + r sin \phi cos \theta \hat{j} - r sin \theta \hat{k}$

so if i plug in 90° only -r$\hat{k}$ will be left since cos 90° = 0

how thus that prove anything?

? ? ? It shows the intended conclusion. $$\vec{\theta} = \frac{\partial \vec{r}}{\partial \theta}$$
$$\hat{\theta} = \frac{\vec{\theta}}{|\vec{\theta}|} = -\hat{k}$$ at $\theta=90^o$.

 

[jhosamelly]

? ? ? It shows the intended conclusion. $$\vec{\theta} = \frac{\partial \vec{r}}{\partial \theta}$$
$$\hat{\theta} = \frac{\vec{\theta}}{|\vec{\theta}|} = -\hat{k}$$ at $\theta=90^o$.

ow.. hahahaha.. I don't know that equation.. hahaha! i mean i know an equation something like that but we only use it to get unit vector.. Is that similar? Thanks for the help )) much appreciated

 

[jambaugh]

ow.. hahahaha.. I don't know that equation.. hahaha! i mean i know an equation something like that but we only use it to get unit vector.. Is that similar? Thanks for the help )) much appreciated

Yes, in general coordinates, say (u,v,w) you can define the local unit vectors $\hat{u},\hat{v},\hat{w}$ as the unit vectors in the directions of increase of the corresponding coordinate. They will be what you get when you normalize each of:
$$\frac{\partial \vec{r}}{\partial u}, \frac{\partial \vec{r}}{\partial v},\text{ and }\frac{\partial \vec{r}}{\partial w}.$$

Note these unit vectors vary with coordinates so be careful when differentiating vectors expressed in terms of these. Also for general coordinates don't assume they are orthogonal to each other. However cylindrical, polar, and spherical coordinates are orthogonal coordinate systems so in these cases you do get a local orthonormal basis.

 

[jhosamelly]

Yes, in general coordinates, say (u,v,w) you can define the local unit vectors $\hat{u},\hat{v},\hat{w}$ as the unit vectors in the directions of increase of the corresponding coordinate. They will be what you get when you normalize each of:
$$\frac{\partial \vec{r}}{\partial u}, \frac{\partial \vec{r}}{\partial v},\text{ and }\frac{\partial \vec{r}}{\partial w}.$$

Note these unit vectors vary with coordinates so be careful when differentiating vectors expressed in terms of these. Also for general coordinates don't assume they are orthogonal to each other. However cylindrical, polar, and spherical coordinates are orthogonal coordinate systems so in these cases you do get a local orthonormal basis.

last question.. you can just answer yes or no.

for cylindrical coordinates if I'm looking for the direction of $\hat{r}$ should i do

$\frac{\partial \hat{r}}{\partial \rho}$ then normalize? like what i did earlier?

 

[jambaugh]

last question.. you can just answer yes or no.

for cylindrical coordinates if I'm looking for the direction of $\hat{r}$ should i do

$\frac{\partial \hat{r}}{\partial \rho}$ then normalize? like what i did earlier?

Yes, but understand in the context of the problem whether you are being asked to find the unit coordinate vector or the normalized position vector. That is to say the coordinate vector is the unit vector pointing away from the z axis while the normalized position vector always points away from the origin in any coordinate system.

Since we use $\vec{r}$ for the position vector even when we aren't using cylindrical or polar coordinates the notation can be ambiguous. That's why sometimes we write $\hat{\mathbf{e}}_r$ instead of $\hat{r}$ (and likewise for other coordinates) to indicate the unit vector in the direction of increasing coordinate.

 

[jhosamelly]

Yes, but understand in the context of the problem whether you are being asked to find the unit coordinate vector or the normalized position vector. That is to say the coordinate vector is the unit vector pointing away from the z axis while the normalized position vector always points away from the origin in any coordinate system.

Since we use $\vec{r}$ for the position vector even when we aren't using cylindrical or polar coordinates the notation can be ambiguous. That's why sometimes we write $\hat{\mathbf{e}}_r$ instead of $\hat{r}$ (and likewise for other coordinates) to indicate the unit vector in the direction of increasing coordinate.

I didn't want to take much of your time... but thanks for explaining further.. I understand it better now! Thank you so much! ))