K-Vector Function of Space in EM Waves: Implications

[Swapnil]

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?

 

[arunma]

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?

 

[Swapnil]

No... I was just curious about the spatial dependence of the k-vector (if such a thing is possible).

 

[Meir Achuz]

If the n of the medium varied in space, then so would k.
k=nw/c

 

[jtbell]

In a non-planar wave (e.g. a spherical wave radiating from a pointlike source), the direction of \vec k obviously depends on location.

 

[christianjb]

In a non-planar wave (e.g. a spherical wave radiating from a pointlike source), the direction of \vec k obviously depends on location.

The equation for a spherical wave is
e^\left(ik|\mathbf{r-r}_0|\right)

k doesn't depend on direction

 

[jtbell]

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

\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}|}

 

[christianjb]

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

\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}|}

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.

 

[robphy]

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.)