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jeebs
Nov28-09, 05:09 PM
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.

gabbagabbahey
Nov28-09, 10:22 PM
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}

jdwood983
Nov28-09, 10:25 PM
so the second derivatives with respect to x and y are zero, don't know if this has any significance?.


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

jeebs
Nov29-09, 08:21 AM
Ah of course, can't believe I missed that... thanks guys.