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Plane wave in cartesian coordinates

  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations

    ψ{x,y,z,t) = Aei(kx+ky+kz ± ωt)
    k = 2π/λ
    3. 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 = Aei(.008(cosθ +sinθ) -ωt)
     
  2. jcsd
  3. Jan 31, 2016 #2

    Simon Bridge

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    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)##.
     
  4. Jan 31, 2016 #3
    i dont see the difference in what i posted and what you posted. you posted the dot product of the propagation vector and the unit vector. isnt that i what i did?
     
  5. Jan 31, 2016 #4

    Simon Bridge

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    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##
     
  6. Jan 31, 2016 #5
    thats where my mistake is. im not sure what my k vector should be. im looking through the text right now trying to figure it out.
     
  7. Jan 31, 2016 #6

    Simon Bridge

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    Your wave vector should have magnitude ##2\pi/\lambda## and should point in the direction of propagation.
     
  8. Jan 31, 2016 #7
    i did that. thats the .008 in my solution. 2pi/700
     
  9. Jan 31, 2016 #8

    Simon Bridge

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    0.008 is the magnitude (in nm-1) - what about the direction?
     
  10. Jan 31, 2016 #9
    well if there is no z component then its headed in the x,y direction. isnt that what the initial u tells me?
     
  11. Jan 31, 2016 #10
    should there be an x and a y in front of the cos and sin, respectively.
     
  12. Jan 31, 2016 #11

    Simon Bridge

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    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.
     
  13. Jan 31, 2016 #12
    ahh ok. what i did was (cosθx, sinθy) ⋅ (x,y)

    so my x and y turned to 1's.

    thank you!!!
     
  14. Jan 31, 2016 #13

    Simon Bridge

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    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 ]
     
    Last edited: Jan 31, 2016
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