Dot product of basis vectors in orthogonal coordinate systems

[FrogPad]

I'm doing a series of questions right now that is basically dealing with the dot and cross products of the basis vectors for cartesian, cylindrical, and spherical coordinate systems.

I am stuck on $\hat R \cdot \hat r$ right now.

I'll try to explain my work, and the problem I am running into as well as I can. It's somewhat difficult to do since I can't draw pictures, but...

Ok, first the notation being used is:

Cylindrical: $(r,\phi,z)$
Spherical: $(R, \phi, \theta )$

First, the dot product of any two unit vectors, say $\hat A$ and $\hat B$ will be: $\hat A \cdot \hat B = \cos \theta_{AB}$

So my trouble is finding the angle between $\hat r$ and $\hat R$. I know that $\phi$ does not affect this angle, so it really comes down to $\theta$.

So let's say $\theta = 0$, then $\theta_{rR}$ would be 90 degrees right?

Now say $\theta = 45\,\,deg$, then $\theta_{rR}$ would be 45 degrees right?

Now say $\theta = 135 \,\,deg$, then does $\theta_{rR}$ equal 45 degrees?

I'm not sure how to deal with measuring the angle between the two vectors when $\theta$ causes $\hat R$ to point below the x-y plane.

I hope this makes sense. Any help would be nice. Thanks :)

 

[quasar987]

It's probably easier to go at this problem from the point of view $\vec{A}\cdot \vec{B}=A_xB_x+A_yB_y$ rather than $\vec{A}\cdot \vec{B}=ABcos(\theta_{AB})$

In the partcular case $$\hat{R}\cdot \hat{r}$$, try expressing $$\hat{R}$$ in cylindrical coordinates*. If you can do that, then the dot product will be obviously just the r-component of $$\hat{R}$$!

*If you struggle too long on this, it is because you're not using the easiest approach.. I'll give you a hint if you need one.

 

[HallsofIvy]

The $\hat r$, in cylindrical coordinates, is just the projection of the $\hat R$, in spherical coordinates, projected onto the xy-plane. Since $\phi$ is the "co-latitude", the angle $\hat R$ makes with the z-axis, the angle you want, between $\hat R$ and $\hat r$ is the complement of that $\frac{\pi}{2}-\phi$.

 

[FrogPad]

Thanks quasar987 for pointing me in the right direction. I didn't really struggle on the problem. I actually posted it, went up to the store and got some food, ate it and fell asleep I had every intention of working on it last night, but sleep got the best of me. Good looking out though

I ended up solving it (I think?) by finding the scalar projection of $\hat R$ on the x-y plane in the direction of $\hat r$, and the $\hat z$ component along the cylinder with another projection in the direction of $\hat z$.

I ended up with,

$$\hat R = \hat r \sin \theta + \hat z \cos \theta$$
(in cylindrical coordinates)

Thus, the dot product $\hat R \cdot \hat r = \sin \theta$

HallsofIvy: I appreciate the help I think what you are saying is a confirmation of what I found. Does everything look ok?

 

[quasar987]

$\sin\theta$ is my answer too. (Or in cartesian coordinates, $\sqrt{x^2+y^2}/\sqrt{x^2+y^2+z^2}$)

Though if you want an honest answer to "Does everything look ok?", I'd answer that when you wrote...

$$\hat R = \hat r \sin \theta + \hat z \cos \theta$$
(in cylindrical coordinates)

I'm not sure this is correct. In my understanding of the term "coordinates", what you wrote is $\hat R$ in spherical coordinates in the $\hat{r},\hat{\phi},\hat{z}$ (cylindrical) basis.

But I could easily be wrong! Anyone can comment on that?

 

[FrogPad]

I'm not sure this is correct. In my understanding of the term "coordinates", what you wrote is LaTeX graphic is being generated. Reload this page in a moment. in spherical coordinates in the LaTeX graphic is being generated. Reload this page in a moment. (cylindrical) basis.

I'm very glad you expressed the concern in my wording. I see where you are coming from. It is definitely not accurate for me to say "in cylindrical coordinates" while the variable $\theta$ is in use for the basis vector.