# Lasers Homework 1

1. We know that the $\mathbf{D}$ and $\mathbf{H}$ fields are defined as $\mathbf{D}=\epsilon_0 \mathbf{E}+\mathbf{P}$ and $\mathbf{H}=\mathbf{B}/\mu_0 -\mathbf{M}$, and the bound current density is given by $\mathbf{J_s}=\nabla \times \mathbf{M} +d\mathbf{P}/d t$. Starting with the Maxwell equation $\nabla \times \mathbf{H} =\mathbf{J_f}+d\mathbf{D}/d t$.

$∇×B=μ_0 (J_t+ϵ_0 ∂E/∂t)$ with $J_{total}=J_{bound}+J_{free}$

$∇×B=μ_0 [(J_b+J_f )+∂(ϵ_0 E)/∂t]$ with $B=μ_0 H+M$

$∇×(μ_0 (H+M))=μ_0 [(∇×M+∂P/∂t)+J_f+∂(D-P)/∂t]$

$μ_0 [(∇×H)+(∇×M) ]=μ_0 [(∇×M)+∂P/∂t+J_f+∂D/∂t-∂P/∂t]$

$∇×H=(J_f+∂D/∂t)$

2. A linear medium is defined as each of the three Cartesian components of the electric displacement vector $D$ (“response”) is linearly dependent on the three Cartesian components of the electric field $E$ (“cause”). That is $D_i=\sum_i \epsilon_{i j} E_j$, where $\epsilon_{i j}$ is a tensor and is independent of $E$. in general $E$ and $D$ are in different directions. For a linear medium please prove that if the direction of $E$ is fixed in space, then the direction of $D$ is also fixed in space.

Linear medium means $D_i=∑_j ϵ_{i j} E_j$ with $i,j=1,2,3$ (or $i,j=(x,y,z)$)

$E ⃗=E_x*x ̂+E_y*y ̂+E_z*z ̂ $ is fixed,means $E_x$, $E_y$, $E_z$ are fixed

$D ⃗=D_x*x ̂+D_y*y ̂+D_z*z ̂=x ̂*∑_j ϵ_{(1,j)} E_j+y ̂*∑_j ϵ_{(2,j)} E_j+z ̂*∑_k ϵ_{(3,j)} E_j$

But since all $E_j$ are fixed and $ϵ_{i j}$ is a tensor (it means $ϵ_{i j}$ are constant numbers) it follows that

all sums $∑_j ϵ_{(1,j)} E_j ,∑_j ϵ_{(2,j)} E_j ,…$ are fixed.

Therefore $D ⃗$ is fixed because $x ̂,y ̂,z ̂$ are versors (unit vectors in the x,y,z directions).

3. As we know, an infinitesimal change of the energy density of an electric field is given by $\delta U=E*\delta D$. please find the expression od $U_e$ for a linear medium in terms of $E$. Note that a linear medium is defined as $D_i=\sum_j \epsilon_{i j} E_j$ where $\epsilon_{i j}$ is tensor and is independent of $E$

$δ=d$ so that $dU=E ⃗*dD ⃗$

$dU=(E_x*x ̂+E_y*y ̂+E_z*z ̂ )*(dD_x*x ̂+dD_y*y ̂+dD_z*z ̂)$

$dU=E_x*dD_x+E_y*dD_y+E_z*dD_z$

but $D_x=∑_j ϵ_{(1,j)} E_j ,…$ so that

$dU=E_x*∑_j ϵ_{(1,j)} dE_j+E_y*∑_j ϵ_{(2,j)} dE_j+ E_z*∑_j ϵ_{(3,j)} dE_j$

Integrating to the left and to the right we get

$U=((ϵ_{11} E_x^2)/2+ϵ_{12} E_x E_y+ϵ_{13} E_x E_z )+$

$+(ϵ_{21} E_y E_x+(ϵ_{22} E_y^2)/2+ϵ_{23} E_y E_z )+$

$+(ϵ_{31} E_z E_x+ϵ_{32} E_z E_y+(ϵ_{33} E_z^2)/2)$

But usually for tensors $ϵ_{i j}=ϵ_{ji}$ (if they are real) so that

$U=1/2*∑_i ϵ_{ii} E_i^2 +2(ϵ_{12} E_x E_y+ϵ_{13} E_x E_y+ϵ_{23} E_y E_z )$

$U=1/2*∑_i ϵ_{ii} E_i^2+2∑_(i? j) ϵ_{i j} E_i E_j$

In case the dielectric constant is a diagonal tensor (all $ϵ_{i j}=0$ for $i? j$) then

$U=1/2 ϵE^2$

4. Suppose we have an isotropic but nonlinear medium whose constitutive equation is given by $D=\alpha E+\beta E^2$, where $\alpha$ and $\beta$ are scalar constants. Please find the expression od $U_e$ in terms of $E$.

Isotropic medium with

$D=αE+βE^2$ so that $dD=α*dE+β*d(E^2)=α*dE+2β(E*dE)$

$dU=E*dD=E*(α*dE)+E*[2β(E*dE)]$

$U=α∫ E*dE+2β∫ E^2*dE=(αE^2)/2+(2βE^3)/3$