I think the rest of the energy is absorbed or transform to other format of energy
jifengy
How does the notation fr(wi->w0) work? Do we input two values wi and w0 into some fr equation then get out a value, or does the arrow indicate something else?
enzyme
@jifengy I think $f_r (\omega_i \rightarrow \omega_o)$ tells us what fraction of the incoming light from direction $\omega_i$ ends up outputted in direction $\omega_o$. So we end up with a distribution over the possible directions.
FeiFeiFei
I think the basis of Helmholtz reciprocity is the reversibility of the optical path that we probably learned in middle school. Each "micro" light path is reversible so that overall it turns out to hold Helmholtz reciprocity in the macro view.
oadrian96
is there a reason why the outgoing distribution is a lobe and not a cone? why do the rays rejoin at the tip?
aylu
@oadrian96 I think it is representing the intensity of the distribution (the magnitude of light in that direction) rather than the resulting cone of light
Azure
Not exactly sure why Helmholtz reciprocity should hold; since there's like one-way windows, where light can pass through one way but not the other. I feel like something similar could happen for scattering, where light would bounce in a certain direction one way but not for the reverse...
I think the rest of the energy is absorbed or transform to other format of energy
How does the notation fr(wi->w0) work? Do we input two values wi and w0 into some fr equation then get out a value, or does the arrow indicate something else?
@jifengy I think $f_r (\omega_i \rightarrow \omega_o)$ tells us what fraction of the incoming light from direction $\omega_i$ ends up outputted in direction $\omega_o$. So we end up with a distribution over the possible directions.
I think the basis of Helmholtz reciprocity is the reversibility of the optical path that we probably learned in middle school. Each "micro" light path is reversible so that overall it turns out to hold Helmholtz reciprocity in the macro view.
is there a reason why the outgoing distribution is a lobe and not a cone? why do the rays rejoin at the tip?
@oadrian96 I think it is representing the intensity of the distribution (the magnitude of light in that direction) rather than the resulting cone of light
Not exactly sure why Helmholtz reciprocity should hold; since there's like one-way windows, where light can pass through one way but not the other. I feel like something similar could happen for scattering, where light would bounce in a certain direction one way but not for the reverse...