Imagine a surface which radiates light equally in all directions from all points

if we measure the radiance of the surface from two positions

one opposite the surface

and one at a 45 degree angle

would the measured radiance be the same?

Started by skytiger, May 05 2013 01:33 AM

61 replies to this topic

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Posted 05 May 2013 - 07:11 AM

I will confess this one left me confused for a while. **Yes, the radiance is the same**, assuming the emitter is at the same distance (it's not clear from the diagram). The reason for this is because the emitter is always at normal incidence to the detector's view angle.

There's this neat theorem called *conservation of radiance* which is really helpful in puzzling situations like these, which states that radiance measured at the emitter is the same as radiance measured at the detector, assuming no radiance is lost between emitter and detector. In this case, it becomes obvious the radiance should be the same, as in both cases the emitters are identical up to rotation (which conserves areas, distances, and basically everything you could possibly care about when working with radiance).

There are also some helpful notes under the Wikipedia Talk page for radiance, see the "cosine term" paragraph. The key point is that the cosine term in the definition of radiance does **not** use the detector's surface normal but the emitter's surface normal. Notably, the sentence "*The cosine factor in the denominator reflects the fact that the apparent size of the source goes to zero as your angle of view approaches 90°.*" highlights this. In your case, the apparent size of the source is quite clearly constant due to the nature of your setup. This can get horribly confusing especially when you see computer graphics papers such as the rendering equation which apparently use the cosine term with the detector's surface normal, but they make entirely different assumptions.

Another, perhaps more intuitive way of deducing this result is by looking at the units of radiance: watts per meter squared per steradian. Also known as power per area per solid angle. Just to be clear, "power" is the power of the emitter, "area" is the surface area **of the emitter**, and "solid angle" is the solid angle subtended by the emitter from the detector's point of view. Now, for both cases considered:

- power is constant, since the emitter is obviously still emitting the same amount of light in both cases

- area is constant, as the emitter's surface area hasn't changed

- solid angle is constant, because the emitter is held facing the detector at normal incidence in both cases

Therefore, measured radiance is the same in both cases.

That said, I could be wrong on this. Please feel free to correct me if I made a mistake.

EDIT: actually, I don't even know anymore. It's all foggy

EDIT 2: no, I think, radiance should really be interpreted as emitted intensity (power / area) per solid angle (direction). There, that's better still mighty confused though,

**Edited by Bacterius, 05 May 2013 - 11:16 AM.**

*“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”*

Posted 05 May 2013 - 01:17 PM

Radiometry is a puzzle, but using this special non-Lambertian surface makes things easier

I believe the _flux_ incident to the detector is equal in both cases

but the measured _radiance_ is 1.0 / 0.707 = 1.414 times higher for the angled detector

because from the angled detector's point of view the emitter looks smaller (it has smaller projected area)

which means you have the same amount of flux in a smaller area which increases flux density .... and therefore "brightness"

this is due to the confusing projected area term in the radiance equation:

radiance = flux / solid_angle / projected_area

and THAT is why Lambertian reflectance needs a cosine law term ... to balance out the increase in radiance due to viewing angle

that is why Lambertian surface gives constant radiance when viewing angle changes

Posted 05 May 2013 - 03:02 PM

I think part of your confusion is that you're not asking the full question radiance answers. Radiance is energy flowing at a certain point in a certain direction. (Technically differential area and solid angle, but for all practical purposes ™ it’s a point and a direction.) Radiance is always L(x,w) and you have to pick a point (which you did) _and_ a direction (which you didn't – at least not explicitly). For incident radiance imagine standing at x and looking into direction w with an extremely small fov, so small that it is just a ray. What you see then is the radiance. It doesn't make sense to ask about the radiance a point receives without specifying a single direction -- you can at most ask for the average radiance (averaged over a finite solid angle). If the receiver is left in your images, you are actually illustrating irradiance, because you are using a finite solid angle. Similarly, it doesn't make sense to ask about the radiance of a finite surface.

I think it's best to always talk about radiance with respect to a imaginary surface perpendicular to the chosen direction. In all settings I ever encountered you were free to choose the surface normal, so why not chose the easiest configuration, in which the confusing cosine term simply disappears.

Assuming the receiver is left and the emitter right in your images:

Incident radiance (say) at the center of the receiver (x) coming from the direction pointing to the center of the emitter (w) is the same in both configurations. It's also the same as the exitant radiance from the center of the emitter toward the center of the receiver.

Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal of the surface (since you asking specifically about irradiance of a concrete surface). So in this case the cosine factor is not 1 and it'll give you a smaller value.

Power collected in total by the receiver is smaller, too (obviously, if irradiance at every point is).

Posted 05 May 2013 - 03:38 PM

Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal

I do not understand exactly what you mean by this. Could you elaborate a bit?

The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.

Is this what you mean?

Posted 05 May 2013 - 05:06 PM

I do not understand exactly what you mean by this. Could you elaborate a bit?

The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.

Is this what you mean?

Yes, that is correct. The incident beam would be smaller due to the cosine factor, hence irradiance decreases by a corresponding amount. That is not radiance, though, but irradiance.

Assuming the receiver is left and the emitter right in your images:

That was my assumption as well. Perhaps this was not the expected interpretation?

**Edited by Bacterius, 05 May 2013 - 05:40 PM.**

Posted 05 May 2013 - 06:03 PM

The emitter is on the left

and the receivers are on the right

the blue represents the solid angle subtended by the receivers

the magenta shows the area and the projected area of the emitter

they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver

Posted 05 May 2013 - 07:13 PM

The emitter is on the left

and the receivers are on the right

the blue represents the solid angle subtended by the receivers

the magenta shows the area and the projected area of the emitter

they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver

Well in that case, yes, radiance is lower. Think about what happens when you aim the emitter perpendicular to the receiver, the projected area goes to zero and no light moves towards the receiver so radiance is zero, as expected. So a lower emitter projected area for the emitter -> lower radiance, by the factor you stated in your first post.

The blue illustration in the diagram was confusing, though. And it's important to not confuse the cosine term in the emitter's projected area with the cosine term for the irradiance's incidence area, they are not the same!

Posted 06 May 2013 - 12:33 AM

Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal

I do not understand exactly what you mean by this. Could you elaborate a bit?

The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.

Is this what you mean?

That's what I meant. Although it wasn't quite correct of me to speak of 'the' cosine factor when a finite solid angle is involved, because there are many cosine factors; one for each direction in the direction bundle that makes up the solid angle. But each factor is smaller in (b) than the corresponding factor in (a), so it's save to say that the irradiance will be lower due to 'the' cosine factor.

I have to disagree, in my opinion you'll approximately get the same results.

The emitter is on the left

and the receivers are on the right

the blue represents the solid angle subtended by the receivers

the magenta shows the area and the projected area of the emitter

they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver

Well in that case, yes, radiance is lower. Think about what happens when you aim the emitter perpendicular to the receiver, the projected area goes to zero and no light moves towards the receiver so radiance is zero, as expected. So a lower emitter projected area for the emitter -> lower radiance, by the factor you stated in your first post.

The blue illustration in the diagram was confusing, though. And it's important to not confuse the cosine term in the emitter's projected area with the cosine term for the irradiance's incidence area, they are not the same!

- Radiance is the same. You look up from the receiver with in a single direction. Either you see the emitter, then you get the radiance it emits, or you don't see the emitter, then you get zero. The orientation of the emitter doesn't matter since it's defined to emit equally in all directions.

- Irradiance is lower in b than in a. This time not because of the cosine factor(s), but because of the smaller solid angle.

- Power also lower.

Posted 06 May 2013 - 01:15 AM

I have to disagree, in my opinion you'll approximately get the same results.

- Radiance is the same. You look up from the receiver with in a single direction. Either you see the emitter, then you get the radiance it emits, or you don't see the emitter, then you get zero. The orientation of the emitter doesn't matter since it's defined to emit equally in all directions.

- Irradiance is lower in b than in a. This time not because of the cosine factor(s), but because of the smaller solid angle.

- Power also lower.

Yes, I agree with everything you say. Good point about the orientation of the emitter.

Posted 06 May 2013 - 02:04 AM

emitter doesn't matter since it's defined to emit equally in all directions

Yes, but it's defined to emit LIGHT equally in all directions

LIGHT <> RADIANCE

If the incident radiance was the same in both cases the surface would be Lambertian ... but it isn't

From wiki:

In optics, **Lambert's cosine law** says that the radiant intensity or luminous intensity observed from an ideal diffusely reflecting surface or ideal diffuse radiator is directly proportional to the cosine of the angle θ between the observer's line of sight and the surface normal

My surface above is NOT Lambertian as the radiant intensity is not proportional to the cosine ... it is constant

In the radiance equation we are dividing by cosine, as the view angle approaches 90º the radiance approaches infinity ...

Posted 06 May 2013 - 02:14 AM

In the radiance equation we are dividing by cosine, as the view angle approaches 90º the radiance approaches infinity ...

The cosine is counter-balanced by the other terms in the formula (namely radiant flux). Radiance clearly doesn't tend to infinity as the angle approaches grazing incidence. See the Wikipedia talk page for details, someone has asked the same question.

Posted 06 May 2013 - 03:25 AM

I strongly disagree

The talk article says "The radiant flux for physical sources falls off at least as fast with angle as cos(?)"

"for physical sources" he is saying that _in real life_ it never happens ...

That is why I created the non-Lambertian geometry above

the surface is not a "physical source" it is constructed deliberately to isolate the cosine / projected area term in the radiance equation

and expose it for what it is - highly confusing!

What happens at 90° is not described in the radiance equation ... it is due to occlusion ... there is no radiance because you can not see the surface

The key to understanding radiance is understanding what happens when the viewing angle _approaches_ 90° ... the radiance approaches infinity

This is due to flux density:

**Edited by skytiger, 06 May 2013 - 03:59 AM.**

Posted 06 May 2013 - 07:11 AM

Here is another gamedev topic that agrees with me (very messily though ...)

http://www.gamedev.net/topic/533397-questions-about-radiance/

quoting:

That's right. Radiance gives the transmitted radiant power through the given solid angle **per unit of projected area**. That is why you need to divide by the projected area given by *cos(theta)*dA*. At a grazing angle the projected area approaches zero, hence the radiant power **per unit of projected area** approaches infinity.

Posted 06 May 2013 - 10:05 AM

I can't quite follow your argument with this strange 'non-Lambertian geometry', but if you define your emitter in such a way that it emits inifinite radiance at grazing angles then you'll see infinite radiance when you look at it at grazing angles. But this has nothing to do with the general definition of radiance. It is also not a very useful definition.

Anyway, I just wanted to say two things for other people who may be reading this:

1. Bacterius is absolutely right about the cosine factor.

2. Incident radiance at a point x from a direction w has nothing to with whether the surface on which x lies is Lambertian or not. Incident radiance is called incident radiance because it's measured before it interacts with the surface. It is also idependent of the surface orientation, as described above.

Anyway, I just wanted to say two things for other people who may be reading this:

1. Bacterius is absolutely right about the cosine factor.

2. Incident radiance at a point x from a direction w has nothing to with whether the surface on which x lies is Lambertian or not. Incident radiance is called incident radiance because it's measured before it interacts with the surface. It is also idependent of the surface orientation, as described above.

Posted 06 May 2013 - 12:06 PM

I don't think you understand anything I have said

It is an isotropic surface - it radiates intensity equally in all directions

Lambertian surfaces do *not* radiate intensity equally in all directions - they radiate intensity proportional to cosine(viewing_angle)

Radiance depends on viewing angle

As the viewing angle increases the projected area of the emitter decreases and the radiance increases

When you COMBINE Lambertian reflectance's cosine term with radiance's 1/cosine term you get CONSTANT RADIANCE

You seem to believe that both radiance *and* intensity are constant!

your point that "radiance is independent of surface orientation" is simply completely wrong

radiance can ONLY be measured with respect to surface orientation!

**Edited by skytiger, 06 May 2013 - 01:26 PM.**

Posted 06 May 2013 - 12:08 PM

To reach agreement on this point we need to agree:

(1) that incident power is the same for both detectors

(2) that solid angle subtended by the detector is the same for both detectors

(3) that the area of the emitter does not change

(4) that the projected area of the emitter is smaller for the angled detector (by factor of 0.707)

(5) that radiance = flux / solid_angle / projected_area MUST be greater for the angled detector (by factor of 1.0 / 0.707 = 1.414)

Follow the worked example:

radiance = flux / solid_angle / projected_area

Given flux of 1.0 and solid_angle of 1.0:

For my isotropic surface:

non-angled

radiance = flux / solid_angle / 1.0 = 1.0

angled

radiance = flux / solid_angle / 0.707 = 1.414 // radiance varies

*If* it was Lambertian: (the flux would be reduced due to cosine term)

non-angled

radiance = flux / solid_angle / 1.0 = 1.0

angled

radiance = flux * 0.707 / solid_angle / 0.707 = 1.0 // radiance is constant

Posted 06 May 2013 - 12:49 PM

I think I understood some of it, but some things elude me. And I'm probably not alone.I don't think you understand anything I have said

One last remark, again, mostly for the grinning and/or confused bystanders:your point that "radiance is independent of surface orientation" is simply completely wrong

radiance can ONLY be measured with respect to surface orientation!

Indeed, you can measure radiance only with respect to surface orientation. It's even better: you can choose any surface orientation you like. The point is you'll always get the same result. If you tilt the surface normal of the _imaginary_ surface away from the direction in which you're probing for radiance the _irradiance_ will get lower. Less energy per unit area. Because of the tilt. Now the cute little cosine factor in the denominator comes in, counterbalances this (exactly like Bacterius said), and brings the result back up again. So, the surface orientation you choose for 'measuring' radiance doesn't matter. In other words: radiance is independent of surface orientation.

When you go from irradiance to radiance you do two things:

1. You look only in a single direction instead of the whole hemisphere (that's dw)

2. You go to projected area from oriented (or real) area, in order to detatch yourself from any concrete surfaces (that's cos(t))

Posted 06 May 2013 - 01:55 PM

Thanks for your posts

Can you please point out which parts of my above post are wrong?

I have carefully reread everything that has been said and still think I'm right ...

(I also have read many articles that seem to agree with my point of view)