Plane wave in cartesian coordinates

[nmsurobert]

Homework Statement

Provide an expression in Cartesian coordinates for a plane wave of amplitude 1 [V/m] and wavelength 700 nm propagating in u = cosθx + sinθy direction, where x and y are unit vectors along the x and y-axis and θ is the measured angle from the x axis.

Homework Equations

ψ{x,y,z,t) = Ae^{i(kx+ky+kz ± ωt)}
k = 2π/λ

The Attempt at a Solution

im not finding many good examples on this but using the plug and chug method i came up with

ψ = Ae^{i(.008(cosθ +sinθ) -ωt)}

 

[Simon Bridge]

Notice how there is no space variation in your wave?

The general expression you want is: $$\psi(\vec r) = Ae^{i(\vec k\cdot\vec r \pm \omega t)}$$ ... for Cartesian coordinates, $\vec r = (x,y,z)$ and $\vec k = (k_x,k_y,k_z)$.

 

[nmsurobert]

i don't see the difference in what i posted and what you posted. you posted the dot product of the propagation vector and the unit vector. isn't that i what i did?

 

[Simon Bridge]

i don't see the difference in what i posted and what you posted. you posted the dot product of the propagation vector and the unit vector. isn't that i what i did?

Maybe I missed it? You wrote:
$\psi = Ae^{i(.008(\cos\theta +\sin\theta) -\omega t)}$
Where is the x-y-z dependence? If you had done the dot product, wouldn't there be one?

Please write out what you got for the wave-vector $\vec k$

 

[nmsurobert]

thats where my mistake is. I am not sure what my k vector should be. I am looking through the text right now trying to figure it out.

 

[Simon Bridge]

Your wave vector should have magnitude $2\pi/\lambda$ and should point in the direction of propagation.

 

[nmsurobert]

i did that. that's the .008 in my solution. 2pi/700

 

[Simon Bridge]

0.008 is the magnitude (in nm^-1) - what about the direction?

 

[nmsurobert]

well if there is no z component then its headed in the x,y direction. isn't that what the initial u tells me?

 

[nmsurobert]

should there be an x and a y in front of the cos and sin, respectively.

 

[Simon Bridge]

That's right - the direction is the same as the direction of $\vec u$ ... since $|\vec u|=1$ you can write: $\vec k = (2\pi / \lambda )\vec u$ ...
Since $\vec u = (\cos\theta, \sin\theta, 0)$ you can write: $\vec k = \frac{2\pi}{\lambda}(\cos\theta, \sin\theta, 0)$

$\vec k\cdot\vec r = \frac{2\pi}{\lambda}(\cos\theta, \sin\theta, 0)\cdot (x,y,z) = \cdots$ ... carry out the dot product.

 

[nmsurobert]

ahh ok. what i did was (cosθx, sinθy) ⋅ (x,y)

so my x and y turned to 1's.

thank you!

 

[Simon Bridge]

Ah - then there was a notation mixup:
If we define x = (1,0,0) etc, then r = xx + yy + zz while u = cosθ x + sinθ y and the dot product proceeds correctly.
You may be used to using i-j-k for unit vectors but you can see why you don't want to do that here.

[If you were thinking that x = (x,0,0) then that's a different kind of mixup and r = x + y + z ]