Suppose we have the following equation 1: $$\tag{1} A_G(x,y,z) = \frac{A_1}{q(z)} e^{-ik \frac{x^2 + y^2}{2q(z)}} $$ where $$ q(z) = z+iz_0 $$ and $i$ is equal to $\sqrt{-1}$.

Suppose we have another equation 2 (where $X(.)$, $Y(.)$, and $Z(z)$ are real functions): $$\tag{2} A(x,y,z) = X( \sqrt{2} \frac{x}{W(z)})Y(\sqrt{2} \frac{y}{W(z)})e^{iZ(z)}A_G(x,y,z) $$

Lastly, we have equation 3: $$ \Delta _T A(x,y,z) - 2ik \frac{ \partial A}{ \partial z} =0 \tag{3}$$

where $\Delta _T=\partial_{xx}+\partial_{yy}$ is the transverse Laplacian operator.

I need to show that substituting equation (2) into equation (3), given that equation (1) is also a solution of equation (3), will produce the following equation: $$ \frac{1}{X} ( \frac{\partial ^{2}X}{\partial u^{2}} - 2u \frac{\partial X}{\partial u}) + \frac{1}{Y} ( \frac{\partial ^{2}Y}{\partial v^{2}} - 2v \frac{\partial Y}{\partial v}) + kW^{2}(z) \frac{\partial Z}{\partial z}=0 $$ where $$ u=\frac{\sqrt2x}{W(z)} $$ and $$ v=\frac{\sqrt2y}{W(z)} $$ Can somebody give me some pointers as to how to begin here?

  • $\begingroup$ You could begin by labelling with (1), (2), and (3) the equations you refer to by those labels. Then you could tell us what all those $\delta$-symbols are - are they supposed to be $\partial$, partial derivatives? And what's $\Delta$? And is $i$ the same as $j$ the same as $\sqrt{-1}$? And if all of this comes from some book, you could give the name and page number, or a link to a scan, or something. $\endgroup$ Sep 30, 2012 at 5:03
  • $\begingroup$ Another question - where do the $u$ and $v$ in your target come from? They are nowhere to be found in the other equations. $\endgroup$ Sep 30, 2012 at 5:05
  • $\begingroup$ All valid criticisms. I have addressed most of them now, I think. Thanks. $\endgroup$ Sep 30, 2012 at 14:02
  • $\begingroup$ Good. What's $r$? (As in $A(r)$, when $A$ has been defined as a function of 3 variables, not 1). My advice would be to give it to a good computer algebra package. $\endgroup$ Sep 30, 2012 at 23:01
  • $\begingroup$ I have fixed this up as well, and added a clarification to the Laplace operator definition. Which computer algebra package would you recommend? $\endgroup$ Oct 1, 2012 at 13:25

1 Answer 1


This is not as difficult as it might look at first. The calculation is long, but when you don't forget anything, it works fine.

First, you need to find all the derivatives in Eq. (3). It's important to remember, that also $X$ and $Y$ are functions of $z$, so that the $z$ derivative gives $i XYe^{iZ}A_G\partial_z Z + XYe^{iZ}\partial_z A_G + Ye^{iZ}A_G\partial_z X + Xe^{iZ}A_G \partial_z Y$. Second derivatives with respect to $x$ and $y$ each give three terms, proportional to $\partial_{xx}X$, $\partial_x X\partial_x A_G$, $\partial_{xx}A_G$ (similar for $y$). Plugging all these into (3) and using that $A_G$ fulfilles (3) as well, you can get rid of terms with $\partial_{xx}A_G$, $\partial_{yy}A_G$ and $\partial_zA_G$. Next, using $\partial_xA_G = -ikxA_G/q$, you can divide the equation by $A_G$ and $e^{iZ}$, so that you should arrive at \begin{equation} Y(\partial_{xx}X-\frac{2ikx}{q}\partial_xX-2ik\partial_zX)+X(\partial_{yy}Y-\frac{2iky}{q}\partial_yY-2ik\partial_zY)+2kXY\partial_zZ = 0. \end{equation}

Multiplying this by $W^2/(2XY)$ you directly get the $Z$ term. To obtain the terms with $X$ and $Y$, you need to explicitly make the derivatives, so that \begin{eqnarray} \partial_z X(\sqrt{2}x/W(z)) &=& -X'\frac{\sqrt{2}x}{W^2}\partial_zW \\ \partial_x X(\sqrt{2}x/W(z)) &=& \frac{\sqrt{2}}{W} X' \\ \partial_{xx} X(\sqrt{2}x/W(z)) &=& \frac{2}{W^2} X'', \end{eqnarray} denoting $X'$, $X''$ first and second derivatives with respect to $u = \sqrt{2}x/W$. If you now express the derivative $\partial_z W$ and use the relations that bind the parameters of the Gaussian beam, you can arrive at the final equation.

  • $\begingroup$ Hey man, thanks for the answer. Just a quick question - Should the line: "$Y(\partial_{xx}X-\frac{2ikx}{q}\partial_xX-2ik\partial_zX)+X(\partial_{yy}Y- \frac{2iky}{q}\partial_yY-2ik\partial_zY)+2kXY\partial_zZ = 0$" actually be "$\frac{1}{X}(\partial_{xx}X-\frac{2ikx}{q}\partial_xX-2ik\partial_zX)+\frac{1}{Y}(\partial_{yy}Y-\frac{2iky}{q}\partial_yY-2ik\partial_zY)+2kXY\partial_zZ = 0$", or am I mistaken? $\endgroup$ Oct 2, 2012 at 18:43
  • $\begingroup$ No, the terms $1/X(...)$, $1/Y(...)$ you get after you divide the equation by $XY$, which is in the next step. $\endgroup$ Oct 2, 2012 at 18:58
  • $\begingroup$ Ah, I see now. But how do we then end up getting rid of the $\frac{W^{2}}{2}$ term that appears outside the $1/X$ and $1/Y$ functions? $\endgroup$ Oct 2, 2012 at 19:07
  • $\begingroup$ Those cancel with the terms you get when you express the derivative with respect to $x$ as derivative with respect to $u$. $\endgroup$ Oct 2, 2012 at 19:34
  • $\begingroup$ I see. Thanks a lot man, really appreciate it. If I had five more reputation, I'd mark your answer as useful too. $\endgroup$ Oct 2, 2012 at 19:38

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