# Homework Help: Finding Theta and Phi

1. Sep 21, 2009

### guilesar

1. The problem statement, all variables and given/known data
The direction of propagation is defined as ex=sin(theta)cos(phi), ey=sin(theta)sin(phi), ez=cos(theta). What are the values of theta and phi characterizing the direction of propagation?

2. Relevant equations
We have a point in space with the coordinates (r,theta,phi)=(2,3,4)

Components were defined in the problem.
3. The attempt at a solution
Well, seeing as how the point is being designated by these numbers, and according to the spherical notation these numbers should represent theta and phi, I would think the answer would be theta=3 and phi=4. But that seems way too easy... so any guidance would be appreciated.

2. Sep 21, 2009

### Gregg

what does ex,ey,ez mean. Use LaTeX!

3. Sep 21, 2009

### gabbagabbahey

This statement makes no sense to me. The equations $\mathbf{\hat{e}}_x=\sin\theta\cos\phi\mathbf{\hat{e}}_r$, $\mathbf{\hat{e}}_y=\sin\theta\sin\phi\mathbf{\hat{e}}_{\theta}$,
and $\mathbf{\hat{e}}_z=\cos\theta\mathbf{\hat{e}}_{\phi}$ simply relate the Cartesian unit vectors $\{\mathbf{\hat{e}}_x,\mathbf{\hat{e}}_y,\mathbf{\hat{e}}_z\}$ to the spherical unit vectors, $\{\mathbf{\hat{e}}_{r},\mathbf{\hat{e}}_{\theta},\mathbf{\hat{e}}_{\phi}\}$

They do not define the direction of propagation of a particle or wave. However, any direction can be written in terms of either set of unit vectors.

What was the exact wording on your original problem statement?

This seems an unlikely way to give you a position vector....are you sure they don't mean $(x,y,z)=(2,3,4)$?

4. Sep 21, 2009

### guilesar

sorry, those are just elementary vectors. They could easily just be said x,y,z. I don't know what LaTeX is...

5. Sep 21, 2009

### guilesar

This is just the second part of a four part problem. The original problem stated exactly as it was on my homework is:

A plane wave propagating in a given medium is expressed as

u(x,y,z,t)=u0exp[i(2x+3y+4z)*106-i1015t]

a. Find the given unit vector for the direction of propagation

A plane wave is generally described by:
expi[k*r -$$\omega$$t]

So the k vector is: (2,3,4) or k*r: 2x+3y+4z

the direction is (2x+3y+4z)/ (22+32+42) =2/(29)1/2x+3/(29)1/2y+4/(29)1/2.

The problem I originally stated is exactly what my worksheet says for part B.

6. Sep 21, 2009

### gabbagabbahey

what happened to the factor of $10^6$?

You should really use bolded letters, or something in order to denote vectors...Writing,

$$\mathbf{\hat{k}}=\frac{2}{\sqrt{29}}\mathbf{\hat{e}}_x+\frac{3}{\sqrt{29}}\mathbf{\hat{e}}_y+\frac{4}{\sqrt{29}}\mathbf{\hat{e}}_z$$

makes your intended meaning much clearer.

Okay, it seems like they are telling you (in a very confusing manner!), that the x, y, and z-components of the direction of propagation are $\sin\theta\cos\phi$, $\sin\theta\sin\phi$ and $\cos\theta$ respectively. If so, then you basically have:

$$\mathbf{\hat{k}}=\frac{2}{\sqrt{29}}\mathbf{\hat{e}}_x+\frac{3}{\sqrt{29}}\mathbf{\hat{e}}_y+\frac{4}{\sqrt{29}}\mathbf{\hat{e}}_z=\sin\theta\cos\phi\mathbf{\hat{e}}_x+\sin\theta\sin\phi\mathbf{\hat{e}}_y+\cos\theta\mathbf{\hat{e}}_z$$

and you are asked to solve for $\theta$ and $\phi$

7. Sep 21, 2009

### guilesar

Yeah. I don't "forum" much so I just learned what Latex is tonight. But I'll work on it :)

Thank you very very much!!