Dot product of basis vectors in orthogonal coordinate systems

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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 [itex] \hat R \cdot \hat r [/itex] 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: [itex] (r,\phi,z) [/itex]
Spherical: [itex] (R, \phi, \theta ) [/itex]

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

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

So lets say [itex] \theta = 0[/itex], then [itex] \theta_{rR} [/itex] would be 90 degrees right?

Now say [itex] \theta = 45\,\,deg [/itex], then [itex] \theta_{rR} [/itex] would be 45 degrees right?

Now say [itex] \theta = 135 \,\,deg [/itex], then does [itex] \theta_{rR} [/itex] equal 45 degrees?

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

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

  • #2
quasar987
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It's probably easier to go at this problem from the point of view [itex]\vec{A}\cdot \vec{B}=A_xB_x+A_yB_y[/itex] rather than [itex]\vec{A}\cdot \vec{B}=ABcos(\theta_{AB})[/itex]

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

*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.
 
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  • #3
HallsofIvy
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The [itex]\hat r[/itex], in cylindrical coordinates, is just the projection of the [itex]\hat R[/itex], in spherical coordinates, projected onto the xy-plane. Since [itex]\phi[/itex] is the "co-latitude", the angle [itex]\hat R[/itex] makes with the z-axis, the angle you want, between [itex]\hat R[/itex] and [itex]\hat r[/itex] is the complement of that [itex]\frac{\pi}{2}-\phi[/itex].
 
  • #4
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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 :smile: I had every intention of working on it last night, but sleep got the best of me. Good looking out though :approve:

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

I ended up with,

[tex] \hat R = \hat r \sin \theta + \hat z \cos \theta [/tex]
(in cylindrical coordinates)

Thus, the dot product [itex] \hat R \cdot \hat r = \sin \theta [/itex]

HallsofIvy: I appreciate the help :smile: I think what you are saying is a confirmation of what I found. Does everything look ok?
 
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  • #5
quasar987
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[itex]\sin\theta[/itex] is my answer too. (Or in cartesian coordinates, [itex]\sqrt{x^2+y^2}/\sqrt{x^2+y^2+z^2}[/itex])

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

[tex] \hat R = \hat r \sin \theta + \hat z \cos \theta [/tex]
(in cylindrical coordinates)
I'm not sure this is correct. In my understanding of the term "coordinates", what you wrote is [itex] \hat R [/itex] in spherical coordinates in the [itex]\hat{r},\hat{\phi},\hat{z}[/itex] (cylindrical) basis.

But I could easily be wrong! Anyone can comment on that?
 
  • #6
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quasar987 said:
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 [itex] \theta [/itex] is in use for the basis vector.
 

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