Hodgman

very interesting point about the area of the emitter that is visible being greater

however it just confirms the radiance of 1.414

here I show the radiance calculation from both points of view, and the result is 1.414 for both:

Your idea that the solid angle changes is wrong ...

As the angle becomes more oblique to the wall on the left the solid angle the detector subtends will decrease. The solid angle will reach it's maximum when the detector and emitter are parallel.

The solid angle the detector subtends is constant in all my diagrams

because the distance between detector and emitter (center points) is equal

and the detector is always normal to the light

In my first post the solid angle the emitter subtends does decrease with angle

But in the post above the solid angles subtended by both emitter and detector are constant

In the first post the emitter area is constant, in the last post the emitter *projected* area is constant

Either way the result is the same, radiance increases with viewing angle

because radiance = flux / solid_angle / projected_area

as the solid_angle decreases radiance must increase

as the projected_area decreases radiance must increase

as the flux increases radiance must increase

In my diagram above you can measure the plane angle (depicting the solid angle) with a protractor and see it is constant

If the solid angle is constant then the area on the left wall will increase as the angle goes further oblique.

That only makes sense if you let the emitter area increase unbounded but physical devices are bounded in area so it doesn't make sense to think in those terms.

-= Dave

Exactly

as the emitter area increases the flux increases which means the measured radiance *increases*

And now you conveniently forget mentioning the cosine term in the denominator which you've been arguing about for the past two days. Funny how that works, isn't it?

Let's go with the diagram you posted and walk through it step by step:

Emitter is on the left, detector is on the right. For the diagrams on the left, we are considering radiance emitted towards the detector, and for the diagrams on the right, we are considering radiance received by the detector (they should, of course, be the same). Assume the detector and emitter have respective areas Ad and Ae and unit distance from each other. Let P be the total power emitted by the emitter (in every direction over its entire surface). Now:

Now, for calculating radiance emitted by a surface into a given solid angle:

-- ? is the angle (with respect to the emitter's normal) at which the radiance is being emitted

-- d²? is the radiant flux emitted by the surface (this is proportional to the solid angle being emitted into)

-- d? is the solid angle being emitted into

-- dA is a surface patch on the emitter's surface

And, for calculating radiance received by a detector from a given solid angle:

-- ? is the angle (with respect to the detector's normal) at which the radiance is being received

-- d²? is the radiant flux received by the detector (again, proportional to the solid angle where the light is being received from)

-- d? is the solid angle being received from

-- dA is a surface patch on the detector's surface

Now, the top-left diagram (radiance measured as it exits the emitter towards the detector):

-- ? is 0 // straightforward

-- d²? is equal to P * (Ad / Ae) // power over surface Ae projected into cross-sectional area Ad - note projected area = real area since we're at normal incidence

-- d? is equal to Ad // fine, we see the detector at normal incidence so its solid angle subtends over its entire area

-- dA is equal to Ae // straightforward

--> L = (P * (Ad / Ae)) / (Ae * Ad) = P / Ae^2

Bottom-left diagram (radiance measured as it exits the emitter towards the detector):

-- ? is equal to 45 degrees // angle made with the emitter's surface normal

-- d²? is equal to P * ((Ad * cos(?)) / Ae) // power over surface area Ae projected into cross-sectional area Ad * cos(?)

-- d? is equal to Ad // the solid angle is the same as the detector is always facing the emitter at normal incidence

-- dA is equal to Ae // straightforward

--> L = (P * ((Ad * cos(?)) / Ae)) / (Ae * Ad * cos(?)) = P / Ae^2

Top-right diagram (radiance measured as it falls on the detector):

-- ? is 0 // straightforward

-- d²? is equal to P * (Ad / Ae) // power over cross-sectional area Ae

-- d? is equal to Ae // fine, we see the emitter at normal incidence so its solid angle subtends over its entire area

-- dA is equal to Ad // straightforward

--> L = (P * (Ad / Ae)) / (Ad * Ae) = P / Ae^2

Bottom-right diagram (radiance measured as it falls on the detector):

-- ? is equal to 0 degrees // angle made with the detector's surface normal

-- d²? is equal to P * ((Ad * cos(45)) / Ae) // same reasoning as with bottom-left, power over surface area Ae projected into cross-sectional area Ad * cos(?)

-- d? is equal to Ae * cos(45) // the projected area of the emitter, no problem

-- dA is equal to Ad // straightforward

--> L = (P * ((Ad * cos(45)) / Ae)) / (Ad * Ae * cos(45) * cos(?)) = P / Ae^2

And we see the radiance is constant with view angle, as expected. It is also dependent on the emitter's surface area and power - of course, a higher power leads to larger radiance and a larger surface area leads to a smaller radiance as the emitter subtends a larger solid angle). And it doesn't go to infinity unless the emitter has no area which is not physical (and radiance is then meaningless). And it is of course independent of detector surface area, which only comes into play when you consider irradiance at the detector (where you multiply the radiance by Ad * cos(?) to obtain the irradiance)

Worth noting P already depends on the emitter's surface area (since a higher surface area leads to higher power) so in reality radiance is proportional to 1 / Ae if you keep the emitter's power per unit area (also known as intensity) constant, which is generally the case.

Your "non-lambertian geometry" concept is meaningless. "Lambertian" doesn't refer to any form of "geometry", at least not to do with aiming detectors and emitters at various angles with respect to each other. Lambert's cosine law simply makes an observation that at grazing angles, the flux tends to zero, and that is true in general:

It may vary for various materials (that is where the BRDF comes in - the Lambertian model is simply a base case for the notion of BRDF) but it is completely unphysical for a surface to be able to emit constant flux (or, in general, flux which does not decrease at least as fast as cos(?) with view angle) at grazing angles, as that would mean it has infinite power. That is what you have been assuming all along, I believe. It just does not happen for any emitter which has area.

I would nevertheless like to thank you for making this thread as it has really challenged (and improved, in many ways) my understanding of radiance, radiant flux, etc..

Your interpretation is "radiance goes to infinity as view angle tends to 90 degrees". Yes or no?

Do you understand this diagram? Yes or no? Can you see why radiance is independent of view angle? (in this case, I mean. with a BRDF it is obviously not constant)