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K-vector: fn of space?

  1. Apr 27, 2007 #1
    Is it possible for the k-vector to be a function of space (in the context of EM waves)? What would it imply if this was the case?
     
  2. jcsd
  3. Apr 27, 2007 #2
    Well I know that the k-vector can be related to a wave's frequency (this is called a dispersion relation). Is that what you were asking about?
     
  4. Apr 27, 2007 #3
    No... I was just curious about the spatial dependence of the k-vector (if such a thing is possible).
     
  5. Apr 28, 2007 #4

    Meir Achuz

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    If the n of the medium varied in space, then so would k.
    k=nw/c
     
  6. Apr 28, 2007 #5

    jtbell

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    In a non-planar wave (e.g. a spherical wave radiating from a pointlike source), the direction of [itex]\vec k[/itex] obviously depends on location.
     
  7. Apr 29, 2007 #6
    The equation for a spherical wave is
    [itex]e^\left(ik|\mathbf{r-r}_0|\right) [/itex]

    k doesn't depend on direction
     
  8. Apr 29, 2007 #7

    jtbell

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    That equation contains only the magnitude of the vector [itex]\vec k[/itex], whose direction is always away from the source (located at [itex]{\vec r}_0[/itex]):

    [tex]\vec k = k \frac{\vec r - {\vec r_0}}{|\vec r - {\vec r_0}|} = \left( \frac{2\pi}{\lambda} \right) \frac{\vec r - {\vec r_0}}{|\vec r - {\vec r_0}|}[/tex]
     
  9. Apr 29, 2007 #8
    I see what you're saying, but it's easier to treat k as a scalar in this case, where k has no dependence on direction.
     
  10. Apr 29, 2007 #9

    robphy

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    The wave vector can probably best thought of as
    "the gradient of the phase of the wave". Thus, one can visualize it as fields of vectors perpendicular to the wavefronts.

    (The physical quantity described by the "k-vector" is actually more naturally thought of as a "covector" (or "one-form"), but that's another story.)
     
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