Emissivity of a flat surface varies with zenith angle

[Callisto]

If the emissivity of a flat surface varies with zenith angle according to

e=E*cos(theta)
where E is the emissivity at zenith.
Would this surface radiate isotropically?

I think that because the emissivity varies then the emitted radiation varies accordingly so the energy measured from any fixed point and at any angle would not be the same so therefore the surface would be an anisotropic source of radiation.

Does this sound like a fair argument?
anybody care to correct me?

Callisto

 

[Antiphon]

If the emissivity of a flat surface varies with zenith angle according to

e=E*cos(theta)
where E is the emissivity at zenith.
Would this surface radiate isotropically?

I think that because the emissivity varies then the emitted radiation varies accordingly so the energy measured from any fixed point and at any angle would not be the same so therefore the surface would be an anisotropic source of radiation.

Does this sound like a fair argument?
anybody care to correct me?

Callisto

No, the surface is an isotropic radiator. The cos law means the projected
area of the surface is being reduced. Any deviation from the cos variation
means it's not an isotropic radiator.

 

[sir-pinski]

, your argument is correct. The emissivity of a flat surface varying with zenith angle means that the surface is not emitting radiation equally in all directions, therefore it is not radiating isotropically. This can be observed by measuring the radiation emitted from different angles, which would show variations in the amount of energy being emitted. An isotropic source of radiation emits energy equally in all directions, regardless of the angle of measurement. So, in this case, the surface would be considered an anisotropic source of radiation.