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Homework Help: Perpendicular component notation

  1. Mar 13, 2007 #1
    If I have [tex]e^{i \mathbf{q_\perp \cdot x}}[/tex] what does it mean?Specifically what does the [tex]\mathbf{q_\perp}[/tex] mean?

    thanks
     
  2. jcsd
  3. Mar 13, 2007 #2

    Dick

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    It usually means the perpendicular component of q relative to something.
     
  4. Mar 13, 2007 #3
    Yes that's what I thought. Relative to what though?

    Surely it can't mean relative to x because the dot product would imply that the term always = 0 right?
     
  5. Mar 13, 2007 #4

    Dick

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    How could I guess 'relative to what'? I'd agree it's probably not x.
     
  6. Mar 13, 2007 #5
    The equation I'm dealing with which contains this term is

    [tex]\epsilon(\mathbf{r})=\frac{i}{q_z} \int d^2 \mathbf{x} e^{i \mathbf{q_\bot \cdot x}}[\epsilon_2 e^{iq_z[H+h_2(\mathbf{x})]} - \epsilon_1 e^{iq_z h_1(\mathbf{x})}][/tex]

    I guess it could be perpendicular to r... but what difference would that make? What would it mean?
     
    Last edited: Mar 13, 2007
  7. Mar 14, 2007 #6

    dextercioby

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    It's probably perpendicular to the magnetic field intensity vector H.
     
  8. Mar 14, 2007 #7
    ahh see the H actually stands for height in this equation. :P h_x is a length also and purely a function of x.

    But hmm perpendicular to H you say... that actually makes a lot more sense to me than any of the other variables if H were a vector, unfortunately its a mean separation, so that couldn't be it could it? I mean H is measured in a particular direction but... can you use that perpendicular symbol relative to something that's not a vector but measured in a particular dimension?

    Cheers
     
  9. Mar 14, 2007 #8
    well thinking about it this is a 2 D problem using a radial or cartesian coordinate system. The radial dimensions are expressed by r and the cartesian dimensions are expressed by x= x_x + x_z.

    Saying that we are dealing with something perpendicular to r makes no sense to me in the context of the system to be honest. Since it has cartesian symmetry but no radial symmetry. Although I could be missing somthing since the equation comes from a fourier tranformation which I don't actually understand...

    (a fourier transform of the system

    [tex]\epsilon(i f, r) = \epsilon_2(i f)[/tex] when [tex]H + h_2(x) \leq z < + \infty[/tex]
    [tex]\epsilon(i f, r) = 0[/tex] when [tex]h_1(x) < z < H + h_2(x)[/tex]
    [tex]\epsilon(i f, r) = \epsilon_1(i f)[/tex] when [tex]- \infty < z \leq h_1(x)[/tex]

    )


    Saying its perpendicular to x is pointless. So I'm inclined to believe its either perpendicular to x_x or x_z. But which I don't know... :/ Nah actually though I bet if I actually understood the fourier transform I'd understand what that q is perpendicular to :/ Can anyone help please? :(
     
    Last edited: Mar 14, 2007
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