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?