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#ActualCryZe

Posted 26 February 2013 - 07:49 AM

we just use the fresnel term for LDotN

Not quite. If you assume that your surface is perfectly flat, your microfacet distribution, as you mentioned, returns 1 where N=H and 0 where N!=H, and thus your BRDF only returns a value other than 0, when N=H. But H is still calculated from L and V. So LDotH and VDotH are still the same (because that's always the case, since H is the halfway vector). But your BRDF requires H to be N to return a value other than 0. But if that's the case, LDotH is the same as LDotN, and VDotH is the same as VDotN. But since LDotH and VDotH are the same, LDotN and VDotN are the same as well. And thus, Helmholtz reciprocity still applies in this case.

No. All non-magnetic, non-optically active, linear (i.e. ordinary) light-matter interaction must obey Helmholtz reciprocity, no matter how many reflections and scatterings the light undergoes. It also applies in any ordinary participating medium (e.g. subsurface scattering) but that is generally approximated as well.

Yeah, I don't know. Something is really weird in the situation Hodgman pointed out

#3CryZe

Posted 26 February 2013 - 07:46 AM

we just use the fresnel term for LDotN

Not quite. If you assume that your surface is perfectly flat, your microfacet distribution, as you mentioned, returns 1 where N=H and 0 where N!=H, and thus your BRDF only returns a value other than 0, when N=H. But H is still calculated from L and V. So LDotH and VDotH are still the same (because that's always the case, since H is the halfway vector). But your BRDF requires H to be N to return a value other than 0. But if that's the case, LDotH is the same as LDotN, and VDotH is the same as VDotN. But since LDotH and VDotH are the same, LDotN and VDotN are the same as well. And thus, Helmholtz reciprocity still applies in this case.

No. All non-magnetic, non-optically active, linear (i.e. ordinary) light-matter interaction must obey Helmholtz reciprocity, no matter how many reflections and scatterings the light undergoes. It also applies in any ordinary participating medium (e.g. subsurface scattering) but that is generally approximated as well.

Yeah, but I can't find any other explanation, other than "diffuse is too much of an approximation of subsurface scattering, to fully work".

#2CryZe

Posted 26 February 2013 - 07:44 AM

we just use the fresnel term for LDotN

Not quite. If you assume that your surface is perfectly flat, your microfacet distribution, as you mentioned, returns 1 where N=H and 0 where N!=H, and thus your BRDF only returns a value other than 0, when N=H. But H is still calculated from L and V. So LDotH and VDotH are still the same (because that's always the case, since H is the halfway vector). But your BRDF requires H to be N to return a value other than 0. But if that's the case, LDotH is the same as LDotN, and VDotH is the same as VDotN. But since LDotH and VDotH are the same, LDotN and VDotN are the same as well. And thus, Helmholtz reciprocity still applies in this case.

#1CryZe

Posted 26 February 2013 - 07:43 AM

we just use the fresnel term for LDotN

Not quite. If you assume that your surface is perfectly flat, your microfacet distribution, as you mentioned returns 1 where N=H and 0 where N!=H, and thus your BRDF only returns a value other than 0, when N=H. But H is still calculated from L and V. So LDotH and VDotH are still the same (because that's always the case, since H is the halfway vector). But your BRDF requires H to be N to return a value other than 0. But if that's the case, LDotH is the same as LDotN, and VDotH is the same as VDotN. But since LDotH and VDotH are the same, LDotN and VDotN are the same as well. And thus, Helmholtz reciprocity still applies in this case.

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