EM Lasing Medium Green's Function

Electricity and Magnetism Level 2

Consider electromagnetic waves polarized in the direction propagating in one dimension in an active laser medium with current density $J(x)$ . Maxwell's equation for the $z$ -component of the electric field of these waves then reads

$\left(\dfrac{d^2}{dx^2} - \kappa^2 \right) E_z (x) = J(x)$

for $\kappa$ some constant. Suppose there is a source at $x= 0$ and that there is a conducting mirror far away so that effectively $E_z (0) = 1$ and $E_z (\infty) = 0$ are the boundary conditions. Find the Green's function $G(x,y)$ for $x<y$ that will allow for determination of the electric field anywhere in space.

e^{-\kappa x} - \frac{e^{-\kappa y}}{\kappa} \sinh (\kappa x)

e^{-\kappa x} - \frac{e^{\kappa y}}{2\kappa} \cosh (\kappa x)

\frac{\sinh (\kappa y) \sinh (\kappa x)}{\kappa \sinh (\kappa y)}

\frac{\sinh (\kappa x) \sinh (\kappa y)}{\kappa \sinh (\kappa x)}

1 solution

Matt DeCross
Apr 29, 2016

The homogeneous equation is solved by the general expression:

$G(x,y) = \begin{cases} c_1 e^{\kappa x} + c_2 e^{-\kappa x}, \qquad &x<y \\ d_1 e^{\kappa x} + d_2 e^{-\kappa x}, \qquad &x>y \end{cases}.$

The boundary conditions $G(0,y) = 1, G(\infty,y) = 0$ impose the constraints on the coefficients:

$\begin{aligned} c_1 + c_2 &= 1 \\ d_1 = 0 \end{aligned}$

Solving for $c_2$ and $d_2$ and plugging into the expression for $G(x,y)$ yields:

$G(x,y) = \begin{cases} 2c_1 \sinh (\kappa x) + e^{-\kappa x}, \qquad &x<y \\ d_2 e^{-\kappa x} , \qquad &x>y \end{cases}.$

Enforcing continuity at $x=y$ gives:

$2c_1 \sinh (\kappa y) + e^{-\kappa y}= d_2 e^{-\kappa y} ,$

and requiring the derivative to jump by unity yields:

$-\kappa d_2 e^{-\kappa y } - 2c_1 \kappa \cosh (\kappa y) + \kappa e^{-\kappa y} = 1.$

Solving the two conditions for $c_1, d_1$ yields:

$\begin{aligned} c_1 &= -\frac{e^{-\kappa y}}{2 \kappa } \\ d_2 &= 1-\frac{\sinh (\kappa y)}{\kappa } \end{aligned}$

Plugging $c_1$ into $G(x,y)$ for $x<y$ , find the result:

$G(x,y) =e^{-\kappa x}-\frac{e^{-\kappa y}}{\kappa } \sinh (\kappa x) , \qquad x<y,$

as claimed.

This is wrong. The green function G(0,y)=0 is the correct selection.

Hamidreza Ramezani - 2 years, 7 months ago

No, I believe the original solution is correct. We do not have Dirichlet boundary conditions in this problem, unlike the example in the wiki this problem is used in. The boundary condition G(0,y) = 0 is not the correct boundary condition to use. In this problem we have added a source at x=0 instead of a conducting mirror, which changes the boundary conditions as given in the problem statement.

Matt DeCross - 2 years, 7 months ago

EM Lasing Medium Green's Function

1 solution

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