Emissivity e varies with zenith angle according to e = E*cos(theta)

[Callisto]

If the emissivity e varies with zenith angle according to e = E*cos(theta) where E is the emissivity normal to the surface. Would this surface be an isotropic source of radiation?

Well, since e varies with angle then the flux density must vary accordingly so the surface would radiate anistropically.

Anybody disagree?
Callisto

 

[selfAdjoint]

This cosine law is like the one for intensity of incoming solar radiation, and that's a cylinder, so I agree, anisotropic.

 

[Bystander]

"Anisotropic" for a point on a surface; integrate intensity at any point above an infinite plane surface, and it's isotropic.