- #1

- 75

- 0

Ex = Ax*cos(Kx*x)*sin(Ky*y)*sin(Kz*z)*exp(iwt)

Ey = Ax*sin(Kx*x)*cos(Ky*y)*sin(Kz*z)*exp(iwt)

Ez = Ax*sin(Kx*x)*sin(Ky*y)*cos(Kz*z)*exp(iwt)

satisfies the wave equation in 3-d. I can do this. We also need to show that they satisfy the condition that div(E)=0 (I think this is because there is no charge density in the box) provided that there are two independent polarisations for light.

just doing algebra I can get the expression (Kx*Ax + Ky*Ay + Kz*Az) = 0 , but I'm not sure how this connects to light having two independent polarisations (or in fact what it means for light to have two independent polarisiations). any hlpful hints would be appreciated.

Sachi