next up previous contents index
Next: Degenerate Eigenvalues Up: The Helmholtz Equation Previous: The Helmholtz Equation   Contents   Index


Cartesian versus Polar Coordinates

Relative to the standard rectilinear Cartesian coordinates Helmholtz's equation has the form

$\displaystyle -\nabla^2\psi\equiv -\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}
{\partial y^2}\right)\psi = k^2\psi\,.
$

If one uses
$\displaystyle x$ $\displaystyle =$ $\displaystyle r\cos\theta$  
$\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta$  

to make a transition to polar coordinates, the Helmholtz equation assumes the form

$\displaystyle -\nabla^2\psi\equiv -\left(\frac{1}{r}~\frac{\partial}{\partial r...
...al r}+\frac{1}{r^2}~\frac{\partial^2}{\partial\theta}
\right)\psi = k^2\psi\,.
$

(Nota bene: To show that the Laplacian relative to polars has the form indicated, it is easiest to actually start with that polar expression and then use the above coordinate transformation to recover the Cartesian expression for $ \nabla^2$ . Going the other way is, of course, equivalent but takes a little extra work.)

Given these two representations of the Laplacian $ \nabla^2$ , how do their eigenfunctions compare and how are they related?

This is a very fruitful question to ask, because in answering it, we shall not only obtain a deep and thorough understanding of waves on the flat Euclidean plane, but also develop the framework for dealing with waves on a sphere as well as with waves in three dimensional Euclidean space.

Figure: An instantaneous plane wave consists of a set of parallel phase fronts, the isograms of the phase function. Its gradient, which is perpendicular to these isograms, is the wave propagation vector $ \vec k$ .
\begin{figure}\centering\epsfig{file=fig_planewave.eps}\end{figure}

Plane wave solutions play a key role in the development. Thus we must have a natural and precise way of identifying them relative to Cartesian as well as polar coordinates.

The solutions to

$\displaystyle \frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2} +
k^2\psi = 0
$

are the ``plane wave'' solutions

$\displaystyle e^{i\vec k\cdot\vec x} = e^{i(k_x x+k_y y)}~~\qquad~~\textrm{(relative to
Cartesians).}
$

Such a solution is characterized by its wave propogation vector

$\displaystyle \vec k = (k_x,k_y)\,.
$

The polar representation of this vector,

$\displaystyle \vec k$ $\displaystyle = (k\cos\alpha ,k\sin\alpha )$    

where


$\displaystyle k^2$ $\displaystyle =k^2_x+k^2_y~~,$    

is appropriate relative to polar coordinates. The wave propagation vector $ \vec k$ is the gradient of the phase for a plane wave solution. This phase has the form
$\displaystyle \textrm{phase}$ $\displaystyle \equiv$ $\displaystyle k_xx +k_y y~~\qquad~~\qquad~~\textrm{(relative to
Cartesians)}$  
  $\displaystyle =$ $\displaystyle kr (\cos\alpha\cos\theta +\sin\alpha\sin\theta )$  
  $\displaystyle =$ $\displaystyle kr \cos (\alpha -\theta )~~\qquad~~\qquad~~\textrm{(relative to
polars)}\,.$  

Consequently,

$\displaystyle e^{i\vec k\cdot x} = e^{ikr\cos (\alpha -\theta )}~~\qquad~~\textrm{(relative
to polars)\,.}
$

Thus relative to polar coordinates, a plane wave is represented by the magnitude $ k$ , and the direction angle $ \alpha $ of its propagation vector $ \vec k$ .


next up previous contents index
Next: Degenerate Eigenvalues Up: The Helmholtz Equation Previous: The Helmholtz Equation   Contents   Index
Ulrich Gerlach 2007-04-05