I guess your surface would be a perfectly efficient mirror. Energy would be perfectly conserved with no loss to entropy.
lucida
If we're integrating over the surface of a hemisphere, is w_r always perpendicular to the surface of the sphere, making cos(theta_r) always equal to 1?
Question: what does it mean if $\rho$ is 1?
I guess your surface would be a perfectly efficient mirror. Energy would be perfectly conserved with no loss to entropy.
If we're integrating over the surface of a hemisphere, is w_r always perpendicular to the surface of the sphere, making cos(theta_r) always equal to 1?