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.