Difference between Two Vectors, Spherical Coordinates

[Higgy]

Homework Statement

I'm doing a problem that involves expressing, for two arbitrary vectors $\vec{x}$ and $\vec{x'}$,

$$|\vec{x}-\vec{x'}|$$

in spherical coordinates ($\rho,\theta,\phi$).

Homework Equations

Law of Cosines:

$$c^{2}=a^{2}+b^{2}-2ab\cos\gamma$$

where $\gamma$ is the angle between a and b.

The Attempt at a Solution

Using the law of cosines, we can write

$$|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'\cos\gamma)^{\frac{1}{2}}$$

but I can't figure out what the angle between the vectors $\gamma$ would be. I imagine a plane formed by the two vectors, and the angle being between the two vectors in that plane. How I describe this mathematically, though, is confusing. Can anyone push me in the right direction?

EDIT:
Alright, I took a new approach. I decided to express the difference in Cartesian coordinates, and then convert to spherical coordinates. In doing so, I get:

$$|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'[\sin\theta\sin\theta'(\cos\phi\cos\phi'+\sin\phi\sin\phi')+\cos\theta\cos\theta'])^{\frac{1}{2}}$$

which gives me the $\gamma$ I was looking for. It would be nice if there were an easy way to simply that, though...

 

[gabbagabbahey]

Usually, you are free to orient your coordinate system as you choose, and so a good choice is often to orient it such that $\vec{x'}$ points in the z-direction and hence $\theta'=0$ and you get a much simpler expression:

$$|\vec{x}-\vec{x'}|=\rho^{2}+(\rho ')^{2}-2\rho\rho ' cos \theta$$

 

[Higgy]

Usually, you are free to orient your coordinate system as you choose, and so a good choice is often to orient it such that $\vec{x'}$ points in the z-direction and hence $\theta'=0$ and you get a much simpler expression:

$$|\vec{x}-\vec{x'}|=\rho^{2}+(\rho ')^{2}-2\rho\rho ' cos \theta$$

That's useful a point.

In the larger problem that this is for, a rigid body (described by f(x')) is rotating around a particular axis with angular velocity $\omega$ - so I am restricted to making the z-axis parallel to $\vec{\omega}$. Hence, the most general expression is required.

 

[gabbagabbahey]

In that case, I think the most you can do to simplify is to use the trig Identity $\cos (\phi -\phi ')=\cos \phi \cos \phi '+\sin \phi \sin\phi '$

 

[Higgy]

Right! Working it out right now. Thanks.