# A problem on the paraxial wave equation?

1. Nov 28, 2009

### jeebs

Hi,
I have this electrodynamics problem sheet on the paraxial approximation, and I am not getting very far with it. It starts off talking about a laser beam travelling in the z-direction, and says that a scalar wave has the form F(r,w)eiwt.

The first part of the question ends with me proving that the homogeneous wave equation

$$\nabla^2 \Psi = \frac{1}{c^2} \frac{\partial^2 t}{\partial t^2}$$

takes the form

$$(\nabla^2 + k^2)F(\vec{r},w) = 0$$, which was fairly straight forward.

I am then told to substitute F(r,w) = G(r,w)e-ikz, rewrite the wave equation in terms of G, and apply the paraxial approximation:

$$2ik\frac{\partial G}{\partial z} >> \frac{\partial^2 G}{\partial z^2}$$

to get the "paraxial wave equation". So, here's what I have tried...

$$\frac{\partial G}{\partial z} = ikF(\vec{r},w)e^i^k^z = ikG(\vec{r},w)$$

$$\frac{\partial^2 G}{\partial z^2} = -k^2F(\vec{r},w)e^i^k^z = -k^2G(\vec{r},w)$$

hence $$2ik\frac{\partial G}{\partial z} = -2k^2F(\vec{r},w)e^i^k^z = -2k^2G(\vec{r},w)$$.

I also said that $$\nabla^2G(\vec{r},w)e^-^i^k^z = -k^2G(\vec{r},w)e^-^i^k^z$$ (rewriting my wave equation in terms of G)

ie. $$\nabla^2G(\vec{r},w) = -k^2G(\vec{r},w)$$

or $$\nabla^2G(\vec{r},w) = \frac{\partial^2 G}{\partial z^2}$$.

(so the second derivatives with respect to x and y are zero, don't know if this has any significance?).

This is essentially as far as I have made it with this question so far. I am not really sure what to do with that so-called paraxial approximation, because when you stick the first and second G derivatives into it and cancel, you get 2>>1, which doesn't really make sense. All I really said beyond this was that

$$2ik\frac{\partial G}{\partial z} >> \nabla^2 G$$, and I don't really see how that helps.

Can anyone offer any suggestions on how to proceed here? I'd really appreciate it.
Thanks.

Last edited: Nov 28, 2009
2. Nov 28, 2009

### gabbagabbahey

You need to remember that both $F(\textbf{r},t)$ and $G(\textbf{r},t)$ are functions of the position vector $\textbf{r}$ (and hence $x$, $y$ and $z$)....so when you are taking spatial derivatives of the product of either of these functions with say, $e^{ikz}$, you need to use the product rule.

For example,

$$\frac{\partial G}{\partial z} =\frac{\partial}{\partial z}\left(F(\textbf{r},\omega)e^{ikz}\right)= ikF(\textbf{r},\omega)e^{ikz}+\frac{\partial G}{\partial z}e^{ikz}\neq ikF(\textbf{r},\omega)e^{ikz}$$

3. Nov 28, 2009

### jdwood983

No. $G$ is a function $x$, $y$ and $z$, so the second derivatives with respect to $x$ and $y$ should not be zero.

4. Nov 29, 2009

### jeebs

Ah of course, can't believe I missed that... thanks guys.