I Emission spectra of different materials

[JohnnyGui]

Yes. The radiance LL L is the surface "brightness.

Let's say an observer has moved in the x and y-axis (2 dimensional) to keep the same distance $R$ to the source. Regarding the projected area, is this schematic correct?

If this is correct, then I have 2 questions:

1. How is $x = cos(*) \cdot R$ while my schematic shows that it's $x = sin(*) \cdot R$? Perhaps I considered $x$ as your $z$ coordinate?

2. For some reason my instinct says that the size of a length when viewed from an angle should be $B$ in the below picture instead of $A$. Don't know if there is any way to explain why it's $A$ and not $B$. EDIT: Hold on, I have a hunch it's the same thing, but it looks as if the projected lengths are different.

Btw; The extra info you always add to your posts is very interesting to read. I'm not ignoring them, it's just that I'm trying to understand this first :)

 

[Charles Link]

From a previous post (#88) where I discussed the projected area of the source, your y here is that z, and y is the dimension into the paper. The source will appear to be shrunk in the x-direction so that, instead of being circular with radius $r$, it will be appear to be elliptical, and in the x direction it will appear to have a half-width of $a=rcos(\theta)$. (In the y direction it will be unaffected and appear to have a half -width of $b=r$, which is exactly what it has.) $\\$ Incidentally $z=Rcos(\theta)$ and $\frac{\sqrt{x^2+y^2}}{R}=sin(\theta)$, so you do have it correct in your figure that $x=Rsin(\theta)$, but that is really irrelevant in computing the apparent dimensions of the source as seen from angle $\theta$. $\\$ As previously mentioned, the location $R>>r$, so that the viewing angle $\theta$ is constant across the entire source. $\\$ Suggestion: Take a small circular object (like a penny) and view it from an angle off-axis. Does it look elliptical in appearance?

 

[JohnnyGui]

From a previous post (#88) where I discussed the projected area of the source, your y here is that z, and y is the dimension into the paper. The source will appear to be shrunk in the x-direction so that, instead of being circular with radius $r$, it will be appear to be elliptical, and in the x direction it will appear to have a half-width of $a=rcos(\theta)$. (In the y direction it will be unaffected and appear to have a half -width of $b=r$, which is exactly what it has.) $\\$ Incidentally $z=Rcos(\theta)$ and $\frac{\sqrt{x^2+y^2}}{R}=sin(\theta)$, so you do have it correct in your figure that $x=Rsin(\theta)$, but that is really irrelevant in computing the apparent dimensions of the source as seen from angle $\theta$. $\\$ As previously mentioned, the location $R>>r$, so that the viewing angle $\theta$ is constant across the entire source. $\\$ Suggestion: Take a small circular object (like a penny) and view it from an angle off-axis. Does it look elliptical in appearance?

Thanks, I understand now the influence of viewing angle on how the surface of the radiating source appears to be smaller, but the following stumped me:

If a blackbody with a true surface of A is sending an X amount of energy to an observer who views the blackbody at an angle, and the observer sees the blackbody having a smaller apparent surface from which he receives that X amount of energy. Shouldn't he think that it radiates MORE energy per surface unit and thus also per steradian than when he's observing it from the normal?

 

[Charles Link]

Thanks, I understand now the influence of viewing angle on how the surface of the radiating source appears to be smaller, but the following stumped me:

If a blackbody with a true surface of A is sending an X amount of energy to an observer who views the blackbody at an angle, and the observer sees the blackbody having a smaller apparent surface from which he receives that X amount of energy. Shouldn't he think that it radiates MORE energy per surface unit and thus also per steradian than when he's observing it from the normal?

The blackbody will have more energy per steradian and thereby higher intensity $I(\theta)$ when viewed from the normal because intensity $I(\theta)=LA_{effective}$. Perhaps one way of looking at the concept of radiance or brightness $L$ is to bring along your own aperture when you measure the brightness, and have that aperture be smaller than the area of the blackbody source. Place it in front of the source and view the appearance. It is not necessary to focus on the source which may be a couple inches behind the aperture. Instead. you focus on your own aperture. Sources that have the same brightness $L$ will be indistinguishable. You also will not be able to tell whether you are viewing the blackbody from on-axis or from off-axis. (Be sure and have the normal to your own aperture always pointing at you so that its area doesn't change (and always keep the distance between you and your own aperture constant.) $\\$ Alternatively, instead of viewing the source by focusing on your own aperture, you could measure the power reaching a detector a specified distance $R$ from your own aperture. Again, you could not distinguish whether you are observing from on-axis or off-axis. The power received will be the same. (Again, always have the normal to your own aperture pointing in the direction of the detector.)

 

[JohnnyGui]

Alternatively, instead of viewing the source by focusing on your own aperture, you could measure the power reaching a detector a specified distance RR R from your own aperture. Again, you could not distinguish whether you are observing from on-axis or off-axis. The power received will be the same. (Again, always have the normal to your own aperture pointing in the direction of the detector.)

Exactly, so the power will be the same from all angles. So, let's say the observer is watching the source form an angle and measures the power, he'd say that the intensity is $P$ / $A_{effective}$. Now, the observer moves towards the normal of the surface (he doesn't know he does, but he just moves in that direction) and then looks at the source. He'll still measure the same power (as you said) but he'll also see a larger surface $A$ since he's on-axis. He'll therefore calculate a smaller intensity since $P$ is the same but the surface is larger from the on-axis view.

What am I missing here?

 

[Charles Link]

Exactly, so the power will be the same from all angles. So, let's say the observer is watching the source form an angle and measures the power, he'd say that the intensity is $P$ / $A_{effective}$. Now, the observer moves towards the normal of the surface (he doesn't know he does, but he just moves in that direction) and then looks at the source. He'll still measure the same power (as you said) but he'll also see a larger surface $A$ since he's on-axis. He'll therefore calculate a smaller intensity since $P$ is the same but the surface is larger from the on-axis view.

What am I missing here?

This is why you bring your own aperture=the area stays the same. It blocks the parts of the source whose line of sight is outside your own aperture. Right now the purpose is to measure the brightness $L$ and nothing else. To measure the intensity $I$ you need to be able to see the entire source. $\\$ (Intensity is perhaps a poor name for the quantity, because it does not refer to the intensity of the surface. Intensity $I$ refers to how much power is radiated from the whole surface per unit solid angle. It really is not "intensity" in the literal sense. It also does not refer to how intense it feels, or how much reaches a detector. That instead is given by the irradiance $E$, which is sometimes very loosely(and incorrectly) referred to as the intensity.) $\\$ To summarize something quickly, there are four radiometric quantities: $L$, $I$ , $M$, and $E$, and each of these is well defined, and represents something very specific. $I$ is called intensity, (perhaps a poor name for it, but that's what they call it.) $M$ refers to power coming off of a surface per unit area, and $E$ refers to the power per unit area across a surface or onto a surface. Of the four, $E$ and $M$ are the most closely related. (Occasionally, in some of the older literature these four quantities are referred to with the letters $N$, $J$, $W$, and $H$ respectively. In any case, the letters $L$, $I$, $M$, and $E$ are now quite standard for these quantities.) $\\$ Additional comment: These quantities are all quite useful in doing calculations such as the one with the lamp filament in the OP. It is good to use the standard terminology in the calculations, so it makes for easy reading by others.

 

[JohnnyGui]

Thanks, I think I'm starting to get it now. You've probably been trying to say this the whole time but after reading some sources, I now understand that it's about the solid angle in the observer's sphere himself, not in the sphere of the emitting source. After all, $L$ is about what the observer preceives.

Please bear with me here as I try to punch this in my head. Here's the observer again who has just move in the $x$ direction again w.r.t. the surface $A$.

As I see it, the $cos(\theta) \cdot A$ is approximately the same as its projection on the observer's sphere only in the case if the observer is very far away ($R$ is high) or if the surface $A$ is very small.

That being said, when an area of $cos(\theta) \cdot A$ is on the observer's sphere, it has a certain 3D angle $\alpha$ in the observer's sphere. From what I know, you can calculate $\alpha$ by rearranging the following formula: $cos(\theta) \cdot A = 2 \pi \cdot R^2 (1 - cos(\frac{\alpha}{2}))$

That aside, the surface $cos(\theta) \cdot A$ is part of the surface of the observer's sphere. The amount of steradians that this surface $cos(\theta) \cdot A$ contains is calculated by $\frac{cos(\theta) \cdot A}{R^2}$.
So to get the amount of energy per steradian in the observer's sphere, I'd have to divide the energy that the observer preceives by that factor. This would eventually give the amount of energy that the observer perceives per steradian (but not necessarly per steradian per m²). So:
$$\frac{E_{observed} \cdot R^2}{cos(\theta) \cdot A}=L$$
Two questions:
1. Does this all make sense? If not, where exactly did I go wrong?
2. I'm probably having a huge blackout at the moment because I'm not sure how $E_{observerd}$ is calculated first nor what quantity this is. It's the energy that the source is emitting to the eye of the observer but with a certain spatial angle of the observer's sphere himself, not of the sphere of the source. How is this energy calculated?

(Intensity is perhaps a poor name for the quantity, because it does not refer to the intensity of the surface. Intensity II I refers to how much power is radiated from the whole surface per unit solid angle. It really is not "intensity" in the literal sense. It also does not refer to how intense it feels, or how much reaches a detector. That instead is given by the irradiance EE E , which is sometimes very loosely(and incorrectly) referred to as the intensity.)

I totally agree, and that is what confused me in the first place. So the energy that a radiating surface is putting out per steradian of its own hemisphere is the Radiant Intensity, right? However, isn't Irradiance the amount of energy that the observer perceives per $m^2$ of his own surface and not per steradian of his own sphere?

 

[Charles Link]

@JohnnyGui I think you are starting to get a good handle on it. If I'm not mistaken, the brightness $L$ can be computed either from looking from the surface and how it radiates (per square m^2 per steradian), or alternatively, from the observers' reference and computing the solid angle that the source subtends,( using the irradiance $E$ at the observer), so that your calculation is correct. Keep up the good work !
$\\$ Just an additional comment: The irradiance $E_{observed}$ is often measured with a photodetector. Typically the response is linear, and the voltage from the photodetector (usually coupled to a current to voltage amplifier) is proportional to the irradiance. Typically, a photodetector can be calibrated using a calibration source of known irradiance $E_{cal}$ at a specific distance. Then $\frac{E_{observed}}{E_{cal}}=\frac{V_{observed}}{V_{cal}}$ where $V_{cal}$ is the voltage that the photodector gave for the calibration source. This allows you to compute $E_{observed}$ from the voltage $V_{observed}$ of the photodetector.

 

[JohnnyGui]

Glad I'm on the right track.

Just an additional comment: The irradiance EobservedEobserved E_{observed} is often measured with a photodetector.

But isn't irradiance the amount of energy the observer receives but per m² of the radiating surface of $cos(\theta) \cdot A$? Since my mentioned $E_{observed}$ is the energy received by the observer by the whole surface $cos(\theta) \cdot A$, I think the relationship should be:
$$\frac{E_{observed}}{cos(\theta) \cdot A} = Irradiance$$

 

[Charles Link]

Glad I'm on the right track.
But isn't irradiance the amount of energy the observer receives but per m² of the radiating surface of $cos(\theta) \cdot A$? Since my mentioned $E_{observed}$ is the energy received by the observer by the whole surface $cos(\theta) \cdot A$, I think the relationship should be:
$$\frac{E_{observed}}{cos(\theta) \cdot A} = Irradiance$$

When the letter $E$ is used in radiometrics, like $E_{observed}$, it stands for irradiance (watts/m^2). (In other contexts it can represent an electric field, or it can represent energy, but in the present context, it represents irradiance (watts/m^2)). The letter $P$ is used for power (watts). The brightness $L=E_{observed}/\frac{Acos(\theta)}{R^2} =E_{observed}R^2/(A cos(\theta))$, where $E_{observed}=P_{observed}/A_{detector}$, but normally you can do a measurement of $E_{observed}$ as described in my previous post, and you don't need to know $A_{detector}$.

 

[JohnnyGui]

When the letter $E$ is used in radiometrics, like $E_{observed}$, it stands for irradiance (watts/m^2). (In other contexts it can represent an electric field, or it can represent energy, but in the present context, it represents irradiance (watts/m^2)). The letter $P$ is used for power (watts). The brightness $L=E_{observed}/\frac{Acos(\theta)}{R^2} =E_{observed}R^2/(A cos(\theta))$, where $E_{observed}=P_{observed}/A_{detector}$, but normally you can do a measurement of $E_{observed}$ as described in my previous post, and you don't need to know $A_{detector}$.

I noticed something but I'm not sure if this is correct. If the observer is a point, then one cannot speak of irradiance since that would require a receiving surface, not a point. One can only speak of the received energy per unit solid angle in the sphere of the point observer.
Furthermore, to speak of the amount of energy the emitted surface emits per m² of that emitting surface according to that point observer (received energy / ($cos(\theta) \cdot A$) is also not possible because this amount differs with distance from the emitting source.

However, if you consider the receiver as having a surface as well, just like the emitting source, then one would get the following scenario:

I think that in this case, one cannot speak of any quantity that has a solid angle unit in it (for example like radiant intensity) because they're not points. Except if you try to extrapolate the lines further up until they intersect. But to know the radius of that circle, one would have to know the angle at which the rays pass by the sides of the receiving surface. Since the detector you mentioned has a surface, what it does is measure the amount of energy received, by using the proportionality with the calibrated voltage like you said, and divide that amount of energy by it's own detector surface to get the irradiance in W/m². So only if the receiver is considered a surface can one speak of irradiance.

Does my reasoning make any sense?

 

[Charles Link]

I noticed something but I'm not sure if this is correct. If the observer is a point, then one cannot speak of irradiance since that would require a receiving surface, not a point. One can only speak of the received energy per unit solid angle in the sphere of the point observer.
Furthermore, to speak of the amount of energy the emitted surface emits per m² of that emitting surface according to that point observer (received energy / ($cos(\theta) \cdot A$) is also not possible because this amount differs with distance from the emitting source.

However, if you consider the receiver as having a surface as well, just like the emitting source, then one would get the following scenario:
View attachment 204547

I think that in this case, one cannot speak of any quantity that has a solid angle unit in it (for example like radiant intensity) because they're not points. Except if you try to extrapolate the lines further up until they intersect. But to know the radius of that circle, one would have to know the angle at which the rays pass by the sides of the receiving surface. Since the detector you mentioned has a surface, what it does is measure the amount of energy received, by using the proportionality with the calibrated voltage like you said, and divide that amount of energy by it's own detector surface to get the irradiance in W/m². So only if the receiver is considered a surface can one speak of irradiance.

Does my reasoning make any sense?

I think you are starting to get a good idea of all of the concepts, but it might be good to practice with a couple of calculations. Here is one that I think you could solve: $\\$ A source consists of a flat circular blackbody surface of temperature $T=1000 K$ that has area $A=.001 m^2$ and is observed at a location $\theta=60$ degrees off-axis and at a distance $R=1.0$ m from the blackbody. The receiver/detector has area $A_{detector}= .0001$ m^2, and its normal points at the center of the blackbody. Compute the radiant emittance (per unit area ) $M$ (watts/m^2) of the source (over the whole spectrum), the radiance $L$ (watts/(m^2 sr)) of the source, the intensity $I$ (watts/sr) of the source, both on-axis, and at angle $\theta=60$ degrees, and the irradiance $E$ (watts/m^2) at the detector. Finally compute the power $P_d$ incident on the detector. Also compute the total power $P$ radiated by the source. If you can successfully compute these quantities, you have a good start at understanding the different terms that arise in this type of calculation.

 

[JohnnyGui]

I think you are starting to get a good idea of all of the concepts, but it might be good to practice with a couple of calculations. Here is one that I think you could solve: \\ A source consists of a flat circular blackbody surface of temperature T=1000KT=1000K T=1000 K that has area A=.001m2A=.001m2 A=.001 m^2 and is observed at a location θ=60θ=60 \theta=60 degrees off-axis and at a distance R=1.0R=1.0 R=1.0 m from the blackbody. The receiver/detector has area Adetector=.0001Adetector=.0001 A_{detector}= .0001 m^2, and its normal points at the center of the blackbody. Compute the radiant emittance (per unit area ) MM M (watts/m^2) of the source (over the whole spectrum), the radiance LL L (watts/(m^2 sr)) of the source, the intensity II I (watts/sr) of the source, both on-axis, and at angle θ=60θ=60 \theta=60 degrees, and the irradiance EE E (watts/m^2) at the detector. Finally compute the power PdPd P_d incident on the detector. Also compute the total power PP P radiated by the source. If you can successfully compute these quantities, you have a good start at understanding the different terms that arise in this type of calculation.

Thanks for the exercise!

- The radiant emittance would be $M = 1000^4\cdot \sigma = 56703,73 W/m^2$ and the total power $P$ by the source $56703,73 \cdot 0.001 = 56,70 W$.

- The radiance of the source is $L = \frac{M}{\pi}$ and therefore $\frac{56703,73}{\pi} = 18049,36 W/(m^2 \cdot sr)$ and its intensity on-axis would be $I = LA = 18049,36 \cdot 0.001 = 18,05 W/sr$ while the off-axis intensity at $\theta = 60$ would be $18,05 \cdot cos(60) = 10,61 W/sr$.

- Regarding the power $P_d$ incident on the detector, if the source emits 10,61 W per steradian (1² m²) at 60 degrees towards the detector, and the detector has a surface of 0.0001 m², then I'd think that it will receive $10,61 W \cdot 0.0001 = 0.001061 W$.
The irradiance would then be $P_d / A_{detector} = 0.001061 / 0.0001 = 10,61 W/m^2$ which is the same as the radiant intensity since $R= 1$.

Is this correct?

 

[Charles Link]

Thanks for the exercise!

- The radiant emittance would be $M = 1000^4\cdot \sigma = 56703,73 W/m^2$ and the total power $P$ by the source $56703,73 \cdot 0.001 = 56,70 W$.

- The radiance of the source is $L = \frac{M}{\pi}$ and therefore $\frac{56703,73}{\pi} = 18049,36 W/(m^2 \cdot sr)$ and its intensity on-axis would be $I = LA = 18049,36 \cdot 0.001 = 18,05 W/sr$ while the off-axis intensity at $\theta = 60$ would be $18,05 \cdot cos(60) = 10,61 W/sr$.

- Regarding the power $P_d$ incident on the detector, if the source emits 10,61 W per steradian (1² m²) at 60 degrees towards the detector, and the detector has a surface of 0.0001 m², then I'd think that it will receive $10,61 W \cdot 0.0001 = 0.001061 W$.
The irradiance would then be $P_d / A_{detector} = 0.001061 / 0.0001 = 10,61 W/m^2$ which is the same as the radiant intensity since $R= 1$.

Is this correct?

Very good. I have one small correction: cos(60)=.500 (and not .573 ). One other comment is that you should compute the irradiance $E_d$ at the detector before you compute the power onto the detector. $E_d=\frac{I}{R^2}$ and then $P_d=E_d A_d$.

 

[JohnnyGui]

Very good. I have one small correction: cos(60)=.500 (and not .573 ). One other comment is that you should compute the irradiance $E_d$ at the detector before you compute the power onto the detector. $E_d=\frac{I}{R^2}$ and then $P_d=E_d A_d$.

Great. One question though; we have now calculated the radiant intensity of an emitting source because its surface is so small that it can be considered a point source. But how is radiant intensity then calculated when the emitting source has a much larger surface and is therefore emitting multiple "hemispheres" (each dA having 1 hemisphere)? Is the radiant intensity then equal to the $I$ per dA multiplied by $\frac{A}{dA}$?

 

[Charles Link]

Great. One question though; we have now calculated the radiant intensity of an emitting source that is considered a point source, which is possible since its surface is very small. But how is radiant intensity then calculated when the emitting source has a much larger surface and is therefore emitting multiple "hemispheres" (each dA having 1 hemisphere)? Is the radiant intensity then equal to the $I$ per dA multiplied by $\frac{A}{dA}$?

It is then done with a surface integral over the emitting surface as seen at a given location/observation point. If the location is nearby, the angle $\theta$ and distance $r$ will not be constant over the entire surface, so that it can be a somewhat detailed calculation. The integral for the irradiance $E$ would be $E=\int \frac{Lcos(\theta) }{r^2} dA$. $\\$ In many cases, the brightness $L$ is a constant, independent of angle and/or location on the surface. For cases where $r$ also stays (approximately) constant over the whole surface, and $cos(\theta )$ is also constant, this becomes $E=\frac{LA cos(\theta)}{r^2}$.

 

[Charles Link]

@JohnnyGui Back in post #87, you had a diagram of the irradiance at surfaces S,S2, and S3. It now might be a good time to do a calculation with that configuration: First of all, can you see that the irradiance $E$ will be proportional to $cos(\theta)^4$? The reason for this is that the intensity goes as $cos(\theta)$, the distance $r$ is such that $z=rcos(\theta)$ (where z is the vertical distance to the plane containing S,S2, and S3), so that $\frac{1}{r^2}=\frac{cos^2(\theta)}{z^2}$, and finally the irradiance onto the small section of surface, (e.g. S3), will be the irradiance normal to the surface $E_{normal}$ multiplied by $cos(\theta)$. $\\$ The problem is to compute the power radiated by a source of brightness $L$ and area $A$, which is $P=LA \pi$, and show that this is the power received by the surface in the entire plane containing S, S2, and S3. Try doing a surface integral of $\int E \, dA$ in a polar coordinate system. If you get stuck, I'd be glad to show you the result. $\\$ And one hint for the calculation: In the polar coordinate system where $R$ is the radial distance in the plane, $\frac{R}{z}=tan(\theta)$. Instead of integrating over $R \, dR \, d \phi$ where $R$ goes from $0$ to $+\infty$ , you can do the ($R$) integration over $\theta$ from $0$ to $\pi/2$, where again, you use the substitution $R=z \, tan(\theta)$.

 

[Charles Link]

@JohnnyGui One additional hint for the above problem: When $R=z \, tan(\theta)$, $dR=z\, sec^2(\theta)\, d \theta$. (Note that the $d \phi$ integration is separate and just gives a factor of $2 \pi$.)

 

[JohnnyGui]

Back in post #87, you had a diagram of the irradiance at surfaces S,S2, and S3. It now might be a good time to do a calculation with that configuration: First of all, can you see that the irradiance EE E will be proportional to cos(θ)4cos(θ)4 cos(\theta)^4 ? The reason for this is that the intensity goes as cos(θ)cos(θ) cos(\theta) , the distance rr r is such that z=rcos(θ)z=rcos(θ) z=rcos(\theta) (where z is the vertical distance to the plane containing S,S2, and S3), so that 1r2=cos2(θ)z21r2=cos2(θ)z2 \frac{1}{r^2}=\frac{cos^2(\theta)}{z^2} , and finally the irradiance onto the small section of surface, (e.g. S3), will be the irradiance normal to the surface EnormalEnormal E_{normal} multiplied by cos(θ)cos(θ) cos(\theta) . \\ The problem is to compute the power radiated by a source of brightness LL L and area AA A , which is P=LAπP=LAπ P=LA \pi , and show that this is the power received by the surface in the entire plane containing S, S2, and S3. Try doing a surface integral of ∫EdA∫EdA \int E \, dA in a polar coordinate system. If you get stuck, I'd be glad to show you the result. \\ And one hint for the calculation: In the polar coordinate system where RR R is the radial distance in the plane, Rz=tan(θ)Rz=tan(θ) \frac{R}{z}=tan(\theta) . Instead of integrating over RdRdϕRdRdϕ R \, dR \, d \phi where RR R goes from 00 0 to +∞+∞ +\infty , you can do the (RR R ) integration over θθ \theta from 00 0 to π/2π/2 \pi/2 , where again, you use the substitution R=ztan(θ)R=ztan(θ) R=z \, tan(\theta) .

Apologies for the late reply. Your read my mind; after solving your problem in post #102 I indeed went by myself and tried to understand your mentioned proportionality of $E$ with $cos(\theta)^4$ in my post #87. The factors that you mentioned (Lambert's cosine law, orientation of the detector's surface $A_d$ and increase in distance) are indeed exactly the ones that I concluded myself and led me to deduce that it's proportional with $cos(\theta)^4$.

Your problem is an interesting one, although I'm not very good at integrals (I just finally understood the integration of your post #79!). What I do realize, I think, is that the formula for the irradiance $E$ for the surfaces S, S2 and S3 or any other surface in that plane is:
$$\frac{I_0 \cdot cos(\theta)^4}{R^2} = E$$
So, I guess that this means that I need to prove that:
$$\int \frac{I_0 \cdot cos(\theta)^4}{R^2} \cdot dA = LA\pi = P$$
This is an integration in just one plane so I'd need to integrate this for 1 other axis as well. So there must be a double integration here. Am I taking the right first steps regarding this?

 

[Charles Link]

Apologies for the late reply. Your read my mind; after solving your problem in post #102 I indeed went by myself and tried to understand your mentioned proportionality of $E$ with $cos(\theta)^4$ in my post #87. The factors that you mentioned (Lambert's cosine law, orientation of the detector's surface $A_d$ and increase in distance) are indeed exactly the ones that I concluded myself and led me to deduce that it's proportional with $cos(\theta)$.

Your problem is an interesting one, although I'm not very good at integrals (I just finally understood the integration of your post #79!). What I do realize, I think, is that the formula for the irradiance $E$ for the surfaces S, S2 and S3 (which I'll call $A_d$ as in the detector surface) or any other surface in that plane is:
$$\frac{I_0 \cdot cos(\theta)^4}{R^2} = E$$
So, I guess that this means that I need to prove that:
$$\int \frac{I_0 \cdot cos(\theta)^4}{R^2} \cdot dA = LA\pi = P$$
This is an integration of just one plane so I'd need to integrate this for other coordinates as well. So there must be at least a double integration here. Am I taking the right first steps regarding this?

So far so good. $R$ is the distance between the two planes. There are two ways that you can proceed on this one to do the integral: $\\$ 1) Let $r$ and $\phi$ be the polar coordinates in the plane. The differential area $dA=r \, dr \, d \phi$. Integrating over $\phi$ will give you a factor $2 \pi$. You also have $\frac{r}{R}=tan(\theta)$. Given $r=R \, tan(\theta)$, you get $dr=R \, sec^2(\theta) \, d \theta$, and integrating for $r$ from $0$ to $+\infty$ is equivalent to integrating $\theta$ from $0$ to $\frac{\pi}{2}$. That is the simple way. $\\$ 2) An alternative route, which also is not difficult is to see that $cos^2(\theta)=\frac{1}{sec^2(\theta)}$ and since $sec^2(\theta)=tan^2(\theta)+1$, $sec^2(\theta)=(\frac{r}{R})^2 +1$. (And $cos^4(\theta)=\frac{1}{(sec^2(\theta))^2}$ ). You can simply evaluate the $r$ integral in closed form where $r$ goes from $0$ to $+ \infty$. (In this alternative method, the integral over $\phi$ also gives $2 \pi$ just as in the previous case.)

 

[JohnnyGui]

So far so good. RR R is the distance between the two planes. There are two ways that you can proceed on this one to do the integral: \\ 1) Let rr r and ϕϕ \phi be the polar coordinates in the plane. The differential area dA=rdrdϕdA=rdrdϕ dA=r \, dr \, d \phi . Integrating over ϕϕ \phi will give you a factor 2π2π 2 \pi . You also have rR=tan(θ)rR=tan(θ) \frac{r}{R}=tan(\theta) . Given r=Rtan(θ)r=Rtan(θ) r=R \, tan(\theta) , you get dr=Rsec2(θ)dθdr=Rsec2(θ)dθ dr=R \, sec^2(\theta) \, d \theta , and integrating for rr r from 00 0 to +∞+∞ +\infty is equivalent to integrating θθ \theta from 00 0 to π2π2 \frac{\pi}{2} . That is the simple way. \\ 2) An alternative route, which also is not difficult is to see that cos2(θ)=1sec2(θ)cos2(θ)=1sec2(θ) cos^2(\theta)=\frac{1}{sec^2(\theta)} and since sec2(θ)=tan2(θ)+1sec2(θ)=tan2(θ)+1 sec^2(\theta)=tan^2(\theta)+1 , sec2(θ)=(rR)2+1sec2(θ)=(rR)2+1 sec^2(\theta)=(\frac{r}{R})^2 +1 . (And cos4(θ)=1(sec2(θ))2cos4(θ)=1(sec2(θ))2 cos^4(\theta)=\frac{1}{(sec^2(\theta))^2} ). You can simply evaluate the rr r integral in closed form where rr r goes from 00 0 to +∞+∞ + \infty .

Great, I'll delve deeper into this. But I noticed something that is colliding with my reasoning.

If I understand you correctly, basically what you're saying is that if you straighten out that orange radiation arc (that I drew in post #87) into a straight line that overlap the plane of surfaces S, S2 and S3, you'd have the same total energy $P$ as in the orange radiation arc. If this is really what you meant, then integrating the irradiance $E$ over $dA$ in which you take the increasing distance into account (a proportionality with $cos(\theta)^2$) would not add up pieces of energy from the orange radiation arc, but add up energy pieces from other larger radiation arcs in which the total energy $P$ is spread differently (over a larger arc). The result is not calculating the total energy $P$ from the orange radiation arc itself but adding up energy per $dA$ from larger radiation arcs.

Here's a schematic of what I mean. When taking the increasing distance into account for surfaces S, S2 and S3, then you'll be calculating irradiances of the light blue projected surfaces from 3 different radiation arcs in which the total energy is spread differently.

An example is shown saying that the irradiance $E$ in the light blue projected surface of $S2$ (on the red radiation arc) is not the same as the pink projected surface of $S2$ (on the orange arc). Their formulas are respectively $E = \frac{I_0 \cdot cos(\theta)^4}{R^2}$ and $E = \frac{I_0 \cdot cos(\theta)^2}{R^2}$, so a factor of $cos(\theta)^2$ difference like I mentioned (factor of increasing distance).

What I think is the correct way to calculate the total energy $P$ of the orange radiation arc, is if you take out the factor of increasing distance ($cos(\theta)^2$) out of the integral so that you'd be adding up energy pieces of the orange radiation arc itself, like this:

So that one would have to integrate and solve:
$$\int \frac{I_0 \cdot cos(\theta)^2}{R^2} \cdot dA = LA\pi = P$$
Apologies if all this reasoning is wrong. Please correct me if I am.

 

[Charles Link]

The answer is, yes, we do want to compute it as in your first diagram above at positions (arcs) of different irradiance. We want the observation points to cover an entire plane at a height a distance $R$ ($h$ might be a better letter) above the original plane. $\\$ Your second diagram will not give the correct answer as written, and the reason is that coming off the source gives one factor of $cos(\theta)$, and although the irradiance onto the detector is a second factor of $cos(\theta)$, that is because the solid angle $\Delta \Omega$ covered by the detector is $\Delta \Omega=\frac{A_d}{R^2} cos(\theta)$. $P_d=I_o cos(\theta) \Delta \Omega$. If you want to compute it on the orange arc, you need to keep the detector on the surface of the sphere rather than level with the plane of the emitting surface. Otherwise, there is not a one-to-one correspondence between $dA$ in the integral and $A_d$ of the detector. $\\$ In this second case, $R$ is the arc radius. In the first case, $R$ is the distance to the plane above. In the second case, $dA=R^2 sin(\theta) \, d \theta \, d \phi$ and the integral is on the surface of the hemisphere. In the first case, $dA=r \, dr \, d \phi$, and the integral covers the plane at height $R$ above the plane of the source. (Notice the use of $r$. It represents the radial distance in the plane in polar coordinates. And that is "polar" coordinates in the plane.) The integral over the plane (which is the first diagram in your previous post) of $P=\int \frac{I_o}{R^2} cos^4 (\theta) \, dA$ ,(where $R$ is the height above the emitting surface), should also give $P=I_o \, \pi$, and it does.

 

[JohnnyGui]

The answer is, yes, we do want to compute it as in your first diagram above at positions (arcs) of different irradiance. We want the observation points to cover an entire plane at a height a distance $R$ ($h$ might be a better letter) above the original plane. $\\$ Your second diagram will not give the correct answer as written, and the reason is that coming off the source gives one factor of $cos(\theta)$, and although the irradiance onto the detector is a second factor of $cos(\theta)$, that is because the solid angle $\Delta \Omega$ covered by the detector is $\Delta \Omega=\frac{A_d}{R^2} cos(\theta)$. $P_d=I_o cos(\theta) \Delta \Omega$. If you want to compute it on the orange arc, you need to keep the detector on the surface of the sphere rather than level with the plane of the emitting surface. Otherwise, there is not a one-to-one correspondence between $dA$ in the integral and $A_d$ of the detector. $\\$ In this second case, $R$ is the arc radius. In the first case, $R$ is the distance to the plane above. In the second case, $dA=R^2 sin(\theta) \, d \theta \, d \phi$ and the integral is on the surface of the hemisphere. In the first case, $dA=r \, dr \, d \phi$, and the integral covers the plane at height $R$ above the plane of the source. (Notice the use of $r$. It represents the radial distance in the plane in polar coordinates. And that is "polar" coordinates in the plane.) The integral over the plane (which is the first diagram in your previous post) of $P=\int \frac{I_o}{R^2} cos^4 (\theta) \, dA$ ,(where $R$ is the height above the emitting surface), should also give $P=I_o \, \pi$, and it does.

Hmm, I'm actually surprised regarding this. Isn't the relationship $P=I_0\pi$ caused by Lambert's cosine law? If so, shouldn't that cosine law be independent from the distance $R$ from the emitting source such that you can calculate this relationship even if you don't change the distance $R$?
You showed in post #79 that $P=\int \int I(\theta) dΩ$ should also give $P=I_0\pi$.

 

[Charles Link]

This one is an exercise to show mathematical consistency. In the steady state, if the source puts out $P=5$ watts in a $I(\theta)=I_o cos(\theta)$ pattern, we have already shown that the power $P$ crossing the hemisphere at radius $R$ is 5 watts. The additional thing we are attempting to show here is that the power $P$ crossing the entire plane at a distance $z=R$ above our emittting source will also be 5 watts. The irradiance onto a surface $dA$ in this plane is found to obey $E=E_o cos^4(\theta) =\frac{I_o}{R^2}cos^4(\theta)$ where $E_o$ is the irradiance onto the surface $dA$ located at $\theta=0$ (on-axis). We know there is no energy loss or build-up anywhere, so if energy is conserved, the power crossing the entire plane should be $P= 5$ watts. The integral computations shows this is indeed the case.

 

[JohnnyGui]

This one is an exercise to show mathematical consistency. In the steady state, if the source puts out P=5P=5 P=5 watts in a I(θ)=Iocos(θ)I(θ)=Iocos(θ) I(\theta)=I_o cos(\theta) pattern, we have already shown that the power PP P crossing the hemisphere at radius RR R is 5 watts. The additional thing we are attempting to show here is that the power PP P crossing the entire plane at a distance z=Rz=R z=R above our emittting source will also be 5 watts. The irradiance onto a surface dAdA dA in this plane is found to obey E=Eocos4(θ)=IoR2cos4(θ)E=Eocos4(θ)=IoR2cos4(θ) E=E_o cos^4(\theta) =\frac{I_o}{R^2}cos^4(\theta) where EoEo E_o is the irradiance onto the surface dAdA dA located at θ=0θ=0 \theta=0 (on-axis). We know there is no energy loss or build-up anywhere, so if energy is conserved, the power crossing the entire plane should be P=5P=5 P= 5 watts. The integral computations shows this is indeed the case.

Ah, I got it now! Basically what it's saying is that, if a hemisphere with a fixed radius $R$ is giving a total output of $P = I_0 \cdot \pi$, then bringing this relation to a plane above the source along with corrections for the orientation angle and changing distances should still give this total power according to the same relationship since no energy gets lost.

Correct me if I'm wrong, but I noticed that the same relationship of $E = E_o \cdot cos(\theta)^4$ can be applied for a scenario in which both the surfaces of the emitting source ($A$) and the detector ($A_d$) are large surfaces, without the detector surface moving in a plane like S, S2 and S3. Instead, each $dA$ of the emitting source has a different angle to the detector's surface for which the same relationship $E = E_o \cdot cos(\theta)^4$ applies for each $dA$.

Here's a schematic (each $dA$ is emitting two lines that cover the surface $A_d$):

 

[Charles Link]

@JohnnyGui It looks like you are very much on the right track. :) :) In general, most measurements are done where the geometry is such that both the source and the detector can be treated approximately as points (i.e. the angle $\theta$ doesn't change with position on the source), but yes, the calculation above can be applied to the completely general case. One example that might have widespread application is the heat transfer between radiating and absorbing surfaces.

 

[JohnnyGui]

@JohnnyGui It looks like you are very much on the right track. :) :) In general, most measurements are done where the geometry is such that both the source and the detector can be treated approximately as points (i.e. the angle $\theta$ doesn't change with position on the source, but yes, the calculation above can be applied to the completely general case. One example that might have widespread application is the heat transfer between radiating and absorbing surfaces.

I'm glad I'm reasoning this the correct way :). I often tend to think of more difficult scenarios (in this case, larger surfaces) just to understand the more detailed calculations. Thanks for verifying all this.

I've thought of yet another (different) problem. I'm trying to deduce the inverse square relationship between energy and distance, but through the change in the solid angle as a fixed surface $A$ gets further away from an emitter (perpendicularly to the emitter's plane). I know the relationship of:
$$A = 2πR^2 (1 - cos(\frac{\theta}{2}))$$
The $\frac{\theta}{2}$ can be written as the $tan^{-1}$ function of half the length of the surface $A$ ($L$) divided by the distance $R$. So that:
$$A = 2πR^2 (1 - cos(tan^{-1}(\frac{0.5L}{R})))$$
What I'd expect is, that the factor by which $A$ increases in the formula as $R$ increases, should be inversely proportional to the amount of energy in a fixed surface that moves further away by that same $R$ increase. And at the same time, that the amount of energy should be inversely proportional to the square by which the corresponding $R's$ have increased.

So taking an example of $0.5L=27$, the ratio of the 2 calculated surfaces $A$ at $R = 5$ and $R = 10$ is $3.191965$. I'd therefore think that the amount of energy in a fixed surface that moves from $R=5$ to $R=10$ should be decreased by a factor of $3.191965$. However, the corresponding $R$'s have increased by a factor of $2$ and therefore the amount of energy in a fixed surface should be decreased by a factor of $2^2=4$ instead of $3.191965$.

I don't get why the inverse of the factor by which $A$ increases does not represent the amount by which energy should decrease in a fixed surface.

 

[Charles Link]

The above needs a couple of corrections=perhaps this will be helpful: The formula $A_s=2 \pi R^2(1-cos(sin^{-1}(\frac{r}{R}) )$ works best for smaller $r$, so that $sin^{-1}(\frac{r}{R})=\frac{r}{R}$ (approximately), and $cos(\frac{r}{R})=1-\frac{1}{2} (\frac{r}{R})^2$ in which case $A_s=\pi r^2$. (approximately). The area $A_s$ is actually an area on a spherical surface, but for small $\frac{r}{R}$, simple flat geometry equations apply. Trying to apply the formula for $r=27$ and $R=5$ and $R=10$ really distorts the meaning of the formula.

 

[JohnnyGui]

The above needs a couple of corrections=perhaps this will be helpful: The formula $A_s=2 \pi R^2(1-cos(tan^{-1}(\frac{r}{R}) )$ works best for smaller $r$, so that $tan^{-1}(\frac{r}{R})=\frac{r}{R}$ (approximately), and $cos(\frac{r}{R})=1-\frac{1}{2} (\frac{r}{R})^2$ in which case $A_s=\pi r^2$. (approximately). The area $A_s$ is actually an area on a spherical surface, but for small $\frac{r}{R}$, simple flat geometry equations apply. Trying to apply the formula for $r=27$ and $R=5$ and $R=10$ really distorts the meaning of the formula.

Ah, this crossed my mind but I have a hunch that there is a larger cause that breaks the relation of $A ∝ R^2$ in the formula.

Let's throw out the $tan^{1}$ function and call the $L$ now the length of the spherical cap itself so that the spherical cap is the receiving surface $A$:

The angle in radians that $0.5L$ makes is therefore equal to $\frac{0.5L}{R}$. This angle should represent the exact length of the receiving surface $A$, which is the the spherical cap. Now when I use the formula for the same values: $0.5L = 27$ and calculate the spherical cap surface $A$ at $R=5$ and $R=10$:
$$A = 2πR^2 (1 - cos(\frac{0.5L}{R}))$$
I'd get that $A$ has increased by a factor of $20.849$ while the $R^2$ has increased by a factor of $4$. So even if the angle represents the exact length $L$ of the receiving surface it's still not showing that energy decreases by the same factor that $R^2$ increases with.

What is now the issue in this case?

 

[Charles Link]

Ah, this crossed my mind but I have a hunch that there is a larger cause that breaks the relation of $A ∝ R^2$ in the formula.

Let's throw out the $tan^{1}$ function and call the $L$ now the length of the spherical cap itself so that the spherical cap is the receiving surface $A$:
View attachment 205231
The angle in radians that $0.5L$ makes is therefore equal to $\frac{0.5L}{R}$. This angle should represent the exact length of the receiving surface $A$, which is the the spherical cap. Now when I use the formula for the same values: $0.5L = 27$ and calculate the spherical cap surface $A$ at $R=5$ and $R=10$:
$$A = 2πR^2 (1 - cos(\frac{0.5L}{R}))$$
I'd get that $A$ has increased by a factor of $20.849$ while the $R^2$ has increased by a factor of $4$. So even if the angle represents the exact length $L$ of the receiving surface it's still not showing that energy decreases by the same factor that $R^2$ increases with.

What is now the issue in this case?

The problem is that your diagram is incorrect. The formula works for polar angle $\frac{\theta}{2}$. The $.5 L$ represents the ("straight line" and not "arc") distance from $z=0$ to the edge of the circle that forms the boundary of the spherical section, where the sphere has a radius of $R$. The radius (from the z-axis) $r= .5 L$ can not be greater than $R$ in the way you are using the formula. You also have the formula incorrect in your latest post. It needs $sin^{-1}(\frac{.5 L}{R})$ (inside the cosine) which is only equal (approximately) to $\frac{.5L}{R}$ for small $\frac{L}{R}$. $\\$ [Note: I made an error in post #118 which I corrected. The formula correctly reads $A_s=2 \pi R^2(1-cos(sin^{-1}(\frac{r}{R})))$, with $sin^{-1}(\frac{r}{R})$ , and not $tan^{-1}(\frac{r}{R})$. (Compare to the quote in post #119 and you'll see where I corrected it.) Also, I see in post #117, you have it incorrectly as $tan^{-1}(\frac{.5L}{R})$.] $\\$ In any case, when $r=R$, $sin^{-1}(1)=\frac{\pi}{2}$, (90 degrees), and $cos(\frac{\pi}{2})=0$ so that $A_s=2 \pi R^2$, which is a hemisphere. $\\$ I think you will find with the correct formula, and correct use of the formula, that it is completely consistent. $\\$ Additional item: You may ask, how is it possible to memorize all of the details about this formula? And the answer is, you don't=it is readily derived so that whenever a question comes up about it, you simply derive it: Using spherical coordinates, $A_s=R^2 \int\limits_{0}^{2 \pi} \int\limits_{0}^{\frac{\theta_o}{2}} sin(\theta) \, d \theta \, d \phi =2 \pi R^2(1-cos(\frac{\theta_o}{2}))$.