Solving for Theta and Phi in Direction of Propagation | Homework Statement

[guilesar]

Homework Statement

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?

Homework Equations

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

Components were defined in the problem.

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.

Thank you in advance!

 

[Gregg]

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

 

[gabbagabbahey]

Homework Statement

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?

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?

Homework Equations

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

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

 

[guilesar]

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

 

[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 -\omegat]

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.

 

[gabbagabbahey]

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 -\omegat]

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

what happened to the factor of 10^6?

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

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.

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

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

 

[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!