I Emission spectra of different materials

[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, $A_s=R^2 \int\limits_{0}^{2 \pi} \int\limits_{0}^{\frac{\theta_o}{2}} sin(\theta) \, d \theta \, d \phi =2 \pi R^2(1-cos(\frac{\theta_o}{2}))$.

 

[JohnnyGui]

The problem is that your diagram is incorrect. The formula works for polar angle θ2θ2 \frac{\theta}{2} . The .5L.5L .5 L represents the ("straight line" and not "arc") distance from z=0z=0 z=0 to the edge of the circle that forms the boundary of the spherical section, where the sphere has a radius of RR R . The radius (from the z-axis) r=.5Lr=.5L r= .5 L can not be greater than RR R in the way you are using the formula. You also have the formula incorrect in your latest post. It needs sin−1(.5LR)sin−1(.5LR) sin^{-1}(\frac{.5 L}{R}) (inside the cosine) which is only equal (approximately) to .5LR.5LR \frac{.5L}{R} for small LRLR \frac{L}{R} .

Correct me if I'm missing something, but the way I see it, the formula just contains $\theta$, not how $\theta$ should be calculated. The way I chose to define $\frac{\theta}{2}$ is by saying that the angle $\frac{\theta}{2}$ in radians can be calculated by dividing its covered arc length (which I chose to be $0.5L$) by the radius $R$ of the circle. Just like when half a circle's arc has a length $Rπ$ and therefore the angle that covers it is $\frac{Rπ}{R} = π$ radians.

So the "straight line" from the z-axis that you mention in the case of an arc of length $0.5L$ is equal to $sin(\frac{0.5L}{R}) \cdot R$. So you can calculate the $\theta$ in two ways. I chose the way by using the arc length and divide that by $R$.

(Btw, with this reasoning, this says that the arc length $0.5L$ cannot exceed an angle of $0.5π$ i.e. a length of $0.5πR$)

 

[Charles Link]

Correct me if I'm missing something, but the way I see it, the formula just contains $\theta$, not how $\theta$ should be calculated. The way I chose to define $\frac{\theta}{2}$ is by saying that the angle $\frac{\theta}{2}$ in radians can be calculated by dividing its covered arc length (which I chose to be $0.5L$) by the radius $R$ of the circle. Just like when half a circle's arc has a length $Rπ$ and therefore the angle that covers it is $\frac{Rπ}{R} = π$

So the "straight line" from the z-axis that you mention in the case of an arc of length $0.5L$ is equal to $sin(\frac{0.5L}{R}) \cdot R$. So you can calculate the $\theta$ in two ways. I chose the way by using the arc length and divide that by $R$.

(Btw, with this reasoning, this says that the arc length $0.5L$ cannot exceed an angle of $0.5π$ i.e. a length of $0.5πR$)

As long as the definitions are consistent, it works. $s$ is a better letter for arc length, but there's nothing wrong with using $L$, just so long as you define it that way.

 

[JohnnyGui]

As long as the definitions are consistent, it works. $s$ is a better letter for arc length, but there's nothing wrong with using $L$, just so long as you define it that way.

So the definition of $0.5L$ being the arc length instead of the straight line from the z-axis is not the cause of A not being proportional to $R^2$. So I should be able to express the formula as:
$$A = 2πR^2 (1 - cos(\frac{0.5L}{R}))$$
Where $0.5L$ is the half arc length of the spherical cap and $\frac{0.5L}{R}$ = $\frac{\theta}{2}$ in radians.
The thing is, even if I choose $0.5L$ to be smaller than $0.5 \cdot Rπ$, the proportionality of $A$ with $R^2$ still doesn't exist. I expect that it should since energy decreases by the inverse factor of $R^2$ if it's being divided over a surface $A$ that is larger by that that same factor.

(Sorry for my stubborness on this, but I really want to understand it)

 

[Charles Link]

So the definition of $0.5L$ being the arc length instead of the straight line from the z-axis is not the cause of A not being proportional to $R^2$. So I should be able to express the formula as:
$$A = 2πR^2 (1 - cos(\frac{0.5L}{R}))$$
Where $0.5L$ is the half arc length of the spherical cap and $\frac{0.5L}{R}$ = $\frac{\theta}{2}$ in radians.
The thing is, even if I choose $0.5L$ to be smaller than $0.5 \cdot Rπ$, the proportionality of $A$ with $R^2$ still doesn't exist. I expect that it should since energy decreases by the inverse factor of $R^2$ if it's being divided over a surface $A$ that is larger by that that same factor.

(Sorry for my stubborness on this, but I really want to understand it)

For constant $\frac{ \theta_o}{2}$ the area is proportional to $R^2$. If your source radiates into the region of full cone angle $\theta_o$, which is typical of some directional type sources, (e.g. uch as an automobile headlight and/or a flashlight with a reflector that directs the beam), the area of the beam will be proportional to $R^2$ and the irradiance $E$ will fall off as $E=I/R^2$ thereby conserving power $P=\int E \, dA$, just as it should.

 

[JohnnyGui]

For constant $\frac{ \theta_o}{2}$ the area is proportional to $R^2$. If your source radiates into the region of full cone angle $\theta_o$, which is typical of some directional type sources, (e.g. uch as an automobile headlight and/or a flashlight with a reflector that directs the beam), the area of the beam will be proportional to $R^2$ and the irradiance $E$ will fall off as $E=I/R^2$ thereby conserving power $P=\int E \, dA$, just as it should.

That's exactly it! I don't get why I was putting a changing angle in the function. Probably a hint that I need to take a small break.

So as a fixed surface gets further away (increasing $R$), the angle $\theta$ that it makes is inversely proportional to $R$ and the energy that it receives is inversely proportional to $R^2$.

 

[JohnnyGui]

@Charles Link : So far I understand that to calculate the received energy ($P_r$) that a detector's surface $A_d$ is receiving from a radiating point source at an angle, one can use:
$$I_0 \cdot cos(\theta) \cdot \frac{A_d}{R^2} = P_r$$
The above formula basically says if a detector's surface $A_d$ at a distance $R$ is larger than $R^2$, then in that case, one would have to extrapolate the energy in a $R^2$ surface to the energy in a surface of $A_d$.

However, I noticed that one could reason this in a different way. If a surface $A_o$ at a distance $R$ is larger than $R^2$, then I would assume that it receives parts of radiant intensities that are at other angles as well. I'd therefore think that integrating the received energies from different radiant intensities at angles that cover different parts of $A_d$ would yield the same result as the above formula. Not sure if this is correct though.

 

[Charles Link]

@Charles Link : So far I understand that to calculate the received energy ($P_r$) that a detector's surface $A_d$ is receiving from a radiating point source at an angle, one can use:
$$I_0 \cdot cos(\theta) \cdot \frac{A_d}{R^2} = P_r$$
The above formula basically says if a detector's surface $A_d$ at a distance $R$ is larger than $R^2$, then in that case, one would have to extrapolate the energy in a $R^2$ surface to the energy in a surface of $A_d$.

However, I noticed that one could reason this in a different way. If a surface $A_o$ at a distance $R$ is larger than $R^2$, then I would assume that it receives parts of radiant intensities that are at other angles as well. I'd therefore think that integrating the received energies from different radiant intensities at angles that cover different parts of $A_d$ would yield the same result as the above formula. Not sure if this is correct though.

If the geometry is such as you mentioned, then the distance $R$ would not be constant either. The power received would essentially be how we computed it previously. See post # 115. I think your diagram is a good one to show how it is then computed.

 

[JohnnyGui]

If the geometry is such as you mentioned, then the distance $R$ would not be constant either. The power received would essentially be how we computed it previously. See post # 115. I think your diagram is a good one to show how it is then computed.

Right. I think the only difference is that I'm now mentioning that the radiating source is a point source (i.e. one $dA$ surface). But that won't change the relationship.
Come to think of it, doesn't that mean that in case of a larger radiating surface, just like the one in my diagram in post #115, one would have to also integrate the different radiant intensities at different angles from each $dA$ that cover different parts of the detector's surface? I understand that one doesn't need to do that and just use the following formula for each $dA$:
$$I_0 \cdot cos(\theta)^4 \cdot \frac{A_d}{R^2} = P_r$$
But what I mean is that this formula is composed of integrating different radiant intensities at different angles from each dA that cover different parts of the detector's surface.

 

[Charles Link]

Right. I think the only difference is that I'm now mentioning that the radiating source is a point source (i.e. one $dA$ surface). But that won't change the relationship.
Come to think of it, doesn't that mean that in case of a larger radiating surface, just like the one in my diagram in post #115, one would have to also integrate the different radiant intensities at different angles from each $dA$ that cover different parts of the detector's surface? I understand that one doesn't need to do that and just use the following formula for each $dA$:
$$I_0 \cdot cos(\theta)^4 \cdot \frac{A_d}{R^2} = P_r$$
But what I mean is that this formula is composed of integrating different radiant intensities at different angles from each dA that cover different parts of the detector's surface.

You got it almost completely correct: $P_r=\int \frac{I_o cos^4(\theta)}{R^2} \, dA_d$ where the integral is over the surface area of the detector.

 

[JohnnyGui]

You got it almost completely correct: $P_r=\int \frac{I_o cos^4(\theta)}{R^2} \, dA_d$ where the integral is over the surface area of the detector.

So if I understand correctly:

1. This integration is needed if a detector's surface $A_d$ is very large such that the formula $I_0 \cdot cos(\theta)^4 \cdot \frac{A_d}{R^2} = P_r$ would not be suitable?

2. That your mentioned integration $P_r=\int \frac{I_o cos^4(\theta)}{R^2} \, dA_d$ is used in the case if the radiating source can be considered a point source and the detector's surface $A_d$ is very large? Such that if the radiating source is also very large and can't be considered a point source, one would have to use:
$$P_r=\int \int \frac{I_o cos^4(\theta)}{R^2} \, dA_d dA$$
(An extra integration over every small radiating surface $dA$)

 

[Charles Link]

Your last equation needs a correction: When you go from a point source of intensity $I$ to a source that radiates over an area, you need to write it as $LA$ wheree $L$ is the brightness, so that in integral form it becomes $L \, dA$. $\\$ As much as the $cos^4(\theta)$ can be a useful result, I wouldn't use it as a universal one: e.g. in many cases, the detector faces the source, even if the detector as off at some angle $\theta$. Also in many cases, you simply measure the distance $r$ from source to detector, rather than using $z/R=cos(\theta)$. In addition, the source doesn't always have a $cos(\theta)$ intensity distribution. Instead, the intensity can in general be a function of both $\theta$ and $\phi$, and if the intensity does not vary as $\cos(\theta)$, the brightness $L$ will not be constant either, and can even vary as a function of position on the source. $\\$ The most general expressions are $P=\int \frac{I(\theta,\phi)}{r^2} \, dA_d$ for point sources, where you may need to include a $cos(\theta)$ factor if the irradiance is incident on the detector at some angle $\theta$, and $P=\int\int \frac{L \,dA\, dA_d}{r^2}$ for sources that are finite in size. In general $I=I(\theta,\phi)$ and $L=L(\theta,\phi)$. The brightness $L$ can also be a function of position $(x,y)$ on an extended source, so that more generally $L=L(x,y, \theta, \phi)$.

 

[JohnnyGui]

As much as the cos4(θ)cos4(θ) cos^4(\theta) can be a useful result, I wouldn't use it as a universal one: e.g. in many cases, the detector faces the source, even if the detector as off at some angle θθ \theta . Also in many cases, you simply measure the distance rr r from source to detector, rather than using z/R=cos(θ)z/R=cos(θ) z/R=cos(\theta) . In addition, the source doesn't always have a cos(θ)cos(θ) cos(\theta) intensity distribution. Instead, the intensity can in general be a function of both θθ \theta and ϕϕ \phi , and if the intensity does not vary as cos(θ)cos⁡(θ) \cos(\theta) , the brightness LL L will not be constant either, and can even vary as a function of position on the source. \\ The most general expressions are P=∫I(θ,ϕ)r2dAdP=∫I(θ,ϕ)r2dAd P=\int \frac{I(\theta,\phi)}{r^2} \, dA_d for point sources, where you may need to include a cos(θ)cos(θ) cos(\theta) factor if the irradiance is incident on the detector at some angle θθ \theta , and P=∫∫LdAdAdr2P=∫∫LdAdAdr2 P=\int\int \frac{L \,dA\, dA_d}{r^2} for sources that are finite in size. In general I=I(θ,ϕ)I=I(θ,ϕ) I=I(\theta,\phi) and L=L(θ,ϕ)L=L(θ,ϕ) L=L(\theta,\phi) . The brightness LL L can also be a function of position (x,y)(x,y)(x,y) on an extended source, so that more generally L=L(x,y,θ,ϕ)L=L(x,y,θ,ϕ) L=L(x,y, \theta, \phi) .

This is actually very useful to know, thanks! Helps me understand how it really works in practice.

Your last equation needs a correction: When you go from a point source of intensity II I to a source that radiates over an area, you need to write it as LALA LA wheree LL L is the brightness, so that in integral form it becomes LdALdA L \, dA .

Got it. Regarding that formula. Say $R^2$ nor $\theta$ doesn't really change with each dA, but they do change significantly with $dA_d$, can one then write the formula as:
$$P_r= A \cdot \int \frac{I_o cos^4(\theta)}{R^2}\, dA_d$$

 

[Charles Link]

This is actually very useful to know, thanks! Helps me understand how it really works in practice.
Got it. Regarding that formula. Say $R^2$ nor $\theta$ doesn't really change with each dA, but they do change significantly with $dA_d$, can one then write the formula as:
$$P_r= A \cdot \int \frac{I_o cos^4(\theta)}{R^2}\, dA_d$$

Yes, that is correct. In most cases, the detector (a photodiode) is quite small, but in problems such as heat transfer calculations, the receiving surface can be quite large. (You are also assuming the source has a $cos(\theta)$ intensity distribution, which holds for ideal blackbody radiators, but is not the case in general.)

 

[JohnnyGui]

Yes, that is correct. In most cases, the detector (a photodiode) is quite small, but in problems such as heat transfer calculations, the receiving surface can be quite large. (You are also assuming the source has a $cos(\theta)$ intensity distribution, which holds for ideal blackbody radiators, but is not the case in general.)

Great, I think I'm starting to understand how the integrals are used in this subject.

Are you familiar with the quantities based on the human's eye sensitivity? I've been reading about it and I noticed that they're basically analogues to the quantities of radiant energy that we've discusses so far, but each emitted wavelength corrected for the eye's sensitivity w.r.t. 555 nm at 1/683 Watts to give the units of Lumens. So far I've concluded the analogues as follows:

- Luminous Flux is the analogue of Radiant Flux
- Luminous Intensity (candela as unit) is the analogue of Radiant Intensity
- Emittance/Luminous Exitance is the analogue of Emissive Power
- Illumination/Illuminance (Lux as unit) is the analogue of Irrradiance
- Luminance is the analogue of Radiance

Are these correct?

 

[Charles Link]

Great, I think I'm starting to understand how the integrals are used in this subject.

Are you familiar with the quantities based on the human's eye sensitivity? I've been reading about it and I noticed that they're basically analogues to the quantities of radiant energy that we've discusses so far, but each emitted wavelength corrected for the eye's sensitivity w.r.t. 555 nm at 1/683 Watts to give the units of Lumens. So far I've concluded the analogues as follows:

- Luminous Flux is the analogue of Radiant Flux
- Luminous Intensity (candela as unit) is the analogue of Radiant Intensity
- Emittance/Luminous Exitance is the analogue of Emissive Power
- Illumination/Illuminance (Lux as unit) is the analogue of Irrradiance
- Luminance is the analogue of Radiance

Are these correct?

Once or twice I have found it necessary to do computations with lumens and candle power, etc., but I found it a very clumsy set of conversions, and I don't have that information at my fingertips. Otherwise, yes, your comparisons look to be correct.

 

[JohnnyGui]

Once or twice I have found it necessary to do computations with lumens and candle power, etc., but I found it a very clumsy set of conversions, and I don't have that information at my fingertips. Otherwise, yes, your comparisons look to be correct.

I agree. The names that some of these quantities have are especially confusing.

I'm not sure if you can help me but there's something regarding the calculation of the Luminous flux ($\phi_v$) that has been bothering me for quite a while now which I still can't seem to grasp.

Let's say a radiating source is emitting a range of wavelengths, each wavelength at (obviously) a different amount of Watts. The amount of Lumens of one specific wavelength $\lambda_1$ is given by multiplying the amount of Watts that $\lambda_1$ is emitted at with 683 (1 Lumen is at 1/683 Watts) and then by the luminosity function $\bar y(\lambda_1)$ that indicates how sensitive that $\lambda_1$ is to the eye with respect to a wavelength of 555 nm (it's a dimensionless number, $\bar y$ at 555 nm being equal to $1$).

If the above calculation is for one specific wavelength, one would have to integrate over the whole spectrum range that the source is emitting to know the total amount of Lumens. The Wiki shows the following formula:

The $\phi_{e,\lambda}$ is the spectral radiant flux in Watts/nm.

Here's what I don't get. If a spectrum range is a continuous range of emitted wavelengths, shouldn't one have to calculate the amount of lumens (or just the energy for that matter) of each infinitesimally small wavelength increase? So for example starting from a wavelength of 500nm:
$$683.002lm/W \cdot (500nm \cdot Watts(500nm) \cdot \bar y (500nm) + 500.00001nm \cdot \bar y(500.00001) \cdot Watts(500.00001) + 500.00002nm \cdot Watts(500.00002) \cdot \bar y (500.00002nm) + ...)$$
The problem is that one could take even smaller steps in wavelength increase (0.00000...001) and I'm not sure where the limit is. The $d\lambda$ in the formula above indicates that in very small range of different wavelengths, each wavelength is emitted at the same spectral radiant flux and has the same sensitivity $\bar y$.

I have a feeling this is a general integration misunderstanding rather than a misunderstanding regarding calculating the Luminous flux.

 

[Charles Link]

The intensity spectrum $\Phi(\lambda)$ is a density function, so that $\Phi(\lambda) \, \Delta \lambda$ is the amount of energy between $\lambda$ and $\lambda + \Delta \lambda$. $\\$ You could make the analogy of letting your spectrum be represented by a football field and spreading a bunch of dirt all over it to make some hills. If you put 1000 lbs. of dirt on the field in some distribution, how much did you put at the 50 yard line? You can't give a number to that unless you know the width of the interval: e.g. 50 yds. +/- 1 yard, or 50 yards+\- 1 ft. ? There is a density of dirt that is given by pounds per yard spread across the field. If you want to know how much is at 50 yard line in an interval one foot wide, you multiply the density (in pounds per yard) by the $\Delta x$ one foot=1/3 yard. $\\$ In calculating the amount of energy in the spectrum, the summation is done with an integral, but you can also numerically evaluate the integral by using a small interval $\Delta \lambda$ and doing the computation over say 100 or 1000 points in the spectrum. The $\Delta \lambda$ is the distance between the points. You get basically the same answer if you use 100 points across the spectrum and the $\Delta \lambda$ corresponding to those 100 points, or 1000 points and a $\Delta \lambda$ 1/10 the size of the previous one. Of course, the higher resolution (more points) gives more accuracy, i.e. you need to use enough points to get an accurate answer. (You of course don't need 1 million points to get good accuracy though.)

 

[JohnnyGui]

The intensity spectrum $\Phi(\lambda)$ is a density function, so that $\Phi(\lambda) \, \Delta \lambda$ is the amount of energy between $\lambda$ and $\lambda + \Delta \lambda$. $\\$ You could make the analogy of letting your spectrum be represented by a football field and spreading a bunch of dirt all over it to make some hills. If you put 1000 lbs. of dirt on the field in some distribution, how much did you put at the 50 yard line? You can't give a number to that unless you know the width of the interval: e.g. 50 yds. +/- 1 yard, or 50 yards+\- 1 ft. ? There is a density of dirt that is given by pounds per yard spread across the field. If you want to know how much is at 50 yard line in an interval one foot wide, you multiply the density (in pounds per yard) by the $\Delta x$ one foot=1/3 yard. $\\$ In calculating the amount of energy in the spectrum, the summation is done with an integral, but you can also numerically evaluate the integral by using a small interval $\Delta \lambda$ and doing the computation over say 100 or 1000 points in the spectrum. The $\Delta \lambda$ is the distance between the points. You get basically the same answer if you use 100 points across the spectrum and the $\Delta \lambda$ corresponding to those 100 points, or 1000 points and a $\Delta \lambda$ 1/10 the size of the previous one.

I think I'm starting to get this now. Can I say that the $\Phi(\lambda)$ in W/nm is actually the slope of the spectrum curve of a radiating source?

 

[Charles Link]

I think I'm starting to get this now. Can I say that the $\Phi(\lambda)$ in W/nm is actually the slope of the spectrum curve?

The same thing comes up in probability distribution functions and probability density functions. $F(x)=\int\limits_{- \infty}^{x} f(t) \, dt$, where $F(x)$ is the probability distribution function and $f(x)$ is the probability density function. You can take the derivative of $F(x)$, and yes, $F'(x)=f(x)$. In this case the spectral density function $\Phi(\lambda)$ corresponds to $f(x)$, and normally the equivalent of $F(x)$ is not tabulated here. $\\$ With the football field analogy, $F(x)$ would be the sum total of dirt to the left of the position $x$. With your spectrum, the function for which you want to take a derivative is not computed or tabulated. You could compute it yourself (numerically), and you would find its derivative is indeed $\Phi (\lambda)$, but in this case, the function you mentioned normally is not used in this application. They simply tabulate the spectral density function $\Phi (\lambda)$. $\\$ So to answer your question, can you say $\Phi(\lambda)$ is the slope of the spectrum curve? The answer is not really. Your idea is a good one, but they simply don't tabulate that function in this application. (Alternatively, in probability theory, the function $F(x)$ is often tabulated.)

 

[JohnnyGui]

The same thing comes up in probability distribution functions and probability density functions. $F(x)=\int\limits_{- \infty}^{x} f(t) \, dt$, where $F(x)$ is the probability distribution function and $f(x)$ is the probability density function. You can take the derivative of $F(x)$, and yes, $F'(x)=f(x)$. In this case the spectral density function $\Phi(\lambda)$ corresponds to $f(x)$, and normally the equivalent of $F(x)$ is not tabulated here. $\\$ With the football field analogy, $F(x)$ would be the sum total of dirt to the left of the position $x$. With your spectrum, the function for which you want to take a derivative is not computed or tabulated. You could compute it yourself (numerically), and you would find its derivative is indeed $\Phi (\lambda)$, but in this case, the function you mentioned normally is not used in this application. They simply tabulate the spectral density function $\Phi (\lambda)$. $\\$ So to answer your question, can you say $\Phi(\lambda)$ is the slope of the spectrum curve? The answer is not really. Your idea is a good one, but they simply don't tabulate that function in this application. (Alternatively, in probability theory, the function $F(x)$ is often tabulated.)

I think I have found an alternative way to show why $\Phi(\lambda)$ is not the slope of a spectrum curve, although I'm not sure it's the correct way to explain this. Consider the following spectrum curve that shows many watts each wavelength is emitted from a source.

Say I'd want to calculate the energy from the left side of this curve, starting from 400nm, up to the peak wavelength (around 560 nm) that is emitted at 680 Watt as shown.
If $\Phi(\lambda)$ is the slope (i.e. derivative) of this curve in Watts/nm, then integrating $\Phi(\lambda)$ would only give the difference in Watts between 400 nm and 560nm, which is 680 Watts. Correct?

So, this means that $\Phi(\lambda)$ must be the function of the curve itself since integrating the function of the curve itself would give the area beneath it.
Is this reasoning correct?

 

[Charles Link]

I think I have found an alternative way to show why $\Phi(\lambda)$ is not the slope of a spectrum curve, although I'm not sure it's the correct way to explain this. Consider the following spectrum curve that shows many watts each wavelength is emitted from a source.
View attachment 205862
Say I'd want to calculate the energy from the left side of this curve, starting from 400nm, up to the peak wavelength (around 560 nm) that is emitted at 680 Watt as shown.
If $\Phi(\lambda)$ is the slope (i.e. derivative) of this curve in Watts/nm, then integrating $\Phi(\lambda)$ would only give the difference in Watts between 400 nm and 560nm, which is 680 Watts. Correct?

So, this means that $\Phi(\lambda)$ must be the function of the curve itself since integrating the function of the curve itself would give the area beneath it.
Is this reasoning correct?

Let $A(\lambda)=\int\limits_{0}^{\lambda} \Phi(\lambda ') \, d \lambda '$, then $\frac{d A(\lambda)}{d \lambda}=\Phi(\lambda)$. $\frac{d A(\lambda)}{d \lambda }$ is of course the slope of $A(\lambda)$ vs. $\lambda$ at any point. In spectroscopy, (unlike probability theory where the integral (area under the curve) of the Gaussian distribution is often tabulated), the function $A(\lambda )$ simply isn't in widespread use. Editing: I should qualify the last statement: Many times the integration of $\Phi(\lambda)$ is performed between two wavelengths to get $P=\int\limits_{\lambda_1}^{\lambda_2} \Phi(\lambda) \, d \lambda$, but this is done without ever tabulating $A(\lambda)$. The power $P=A(\lambda_2)-A(\lambda_1)$, but $\Phi(\lambda)$ is normally tabulated, and I think I can say I have never seen a table of the function $A(\lambda)$ in any spectroscopic publication. Mathematically, it could be done this way, but it simply isn't.

 

[JohnnyGui]

Let $A(\lambda)=\int\limits_{0}^{\lambda} \Phi(\lambda ') \, d \lambda '$, then $\frac{d A(\lambda)}{d \lambda}=\Phi(\lambda)$. $\frac{d A(\lambda)}{d \lambda }$ is of course the slope of $A(\lambda)$ vs. $\lambda$ at any point. In spectroscopy, (unlike probability theory where the integral (area under the curve) of the Gaussian distribution is often tabulated), the function $A(\lambda )$ simply isn't in widespread use. Editing: I should qualify the last statement: Many times the integration of $\Phi(\lambda)$ is performed between two wavelengths to get $P=\int\limits_{\lambda_1}^{\lambda_2}$, but this is done without ever tabulating $A(\lambda)$. The above $P=A(\lambda_2)-A(\lambda_1)$, but $\Phi(\lambda)$ is normally tabulated, and I think I can say I have never seen a table of the function $A(\lambda)$ in any spectroscopic publication.

So if I understand correctly $A(\lambda)$ is the total energy in a chosen wavelength range of the spectrum curve shown in my previous post. And $A(\lambda)$ has the derivatie $\Phi(\lambda)$ which is the function of the spectrum curve? (apologies if I misunderstood your post)

 

[Charles Link]

So if I understand correctly $A(\lambda)$ is the total energy in a specific wavelength range of the spectrum curve shown in my previous post that has the derivatie $\Phi(\lambda)$ which is the function of the spectrum curve? (apologies if I misunderstood your post)

Starting at the left end of the curve, yes, $A (\lambda)$ is the area under the curve up to (and including) wavelength $\lambda$. It is analogous mathematics that distance $s(t)=\int\limits_{0}^{t} v(t') \, dt'$ and $\frac{d s(t)}{dt}=v(t)$. If you have a graph of the velocity $v$ vs. time $t$, the area under the curve is the distance traveled. Meanwhile on a graph of $s$ vs. $t$, $v=\frac{ds}{dt}$ is the slope of that graph at any point.

 

[JohnnyGui]

Starting at the left end of the curve, yes, $A (\lambda)$ is the area under the curve up to (and including) wavelength $\lambda$.

Great, because this leads me to the root of my problem XD. If $\Phi(\lambda)$ is the function of the spectrum curve, then how does it have units of Watts per nanometre? Shouldn't it give only Watts for a chosen wavelength? The per nanometre part kind of implies that it's a derivative of the spectrum curve while it isn't.

 

[Charles Link]

Starting at the left end of the curve, yes, $A (\lambda)$ is the area under the curve up to (and including) wavelength $\lambda$.

Great, because this leads me to the root of my problem XD. If $\Phi(\lambda)$ is the function of the spectrum curve, then how does it have units of Watts per nanometre? Shouldn't it give only Watts for a chosen wavelength? The per nanometre part kind of implies that it's a derivative of the spectrum curve while it isn't.

Please read the edited version of my previous post. $\\$ To answer the nanometer question, if the wavelength is in nanometers, then $\Phi(\lambda)$ has units of watts per nanometer. Over a 50 nm interval, say from 550-600 nm, it will integrate correctly if you take the height of the curve at 575 nm (assuming it is a uniform height) and multiply by 50. Numerically, you could do the following $P=\Phi(550) \, 5 \, nm+\Phi(555) \, 5 \, nm +\Phi(560) \, 5 \,nm +... +\Phi(595) \, 5 \, nm$. Alternatively, at higher resolution $P=\Phi(550) \, 1 \, nm +\Phi(551) \, 1 \, nm+... +\Phi(599) \, 1 \, nm$. You'll basically get the same answer both ways with the higher resolution possibly providing increased accuracy. $\\$ Meanwhile, yes, $\Phi(\lambda)$ is the derivative of $A(\lambda)$, but they don't call $A(\lambda)$ the spectrum curve (they call $\Phi(\lambda)$ the spectral curve) =they normally don't give $A(\lambda)$ any name=again, mathematically, they could have gone the route you are asking about, but it simply has never been presented that way. $\\$ Additional item of interest is the Planck blackbody spectral function $L(\lambda, T)$ back in post # 81. You may find it of interest that this function also has units of watts/.../per unit wavelength. If a graph is displayed with wavelength in nm, then it has units of watts/.../nm. Other than the extra units of watts/(m^2 sr nm), this function is very much like the function $\Phi(\lambda)$.

 

[JohnnyGui]

Please read the edited version of my previous post. $\\$ To answer the nanometer question, if the wavelength is in nanometers, then $\Phi(\lambda)$ has units of watts per nanometer. Over a 50 nm interval, say from 550-600 nm, it will integrate correctly if you take the height of the curve at 575 nm (assuming it is a uniform height) and multiply by 50. Numerically, you could do the following $P=\Phi(550) \, 5 \, nm+\Phi(555) \, 5 \, nm +\Phi(560) \, 5 \,nm +... +\Phi(595) \, 5 \, nm$. Alternatively, at higher resolution $P=\Phi(550) \, 1 \, nm +\Phi(551) \, 1 \, nm+... +\Phi(599) \, 1 \, nm$. You'll basically get the same answer both ways with the higher resolution possibly providing increased accuracy. $\\$ Meanwhile, yes, $\Phi(\lambda)$ is the derivative of $A(\lambda)$, but they don't call $A(\lambda)$ the spectrum curve (they call $\Phi(\lambda)$ the spectral curve) =they normally don't give $A(\lambda)$ any name=again, mathematically, they could have gone the route you are asking about, but it simply has never been presented that way. $\\$ Additional item of interest is the Planck spectral function $L(\lambda, T)$ back in post # ... You may find it of interest that this function also has units of watts/.../per unit wavelength. If a graph is displayed with wavelength in nm, then it has units of watts/.../nm.

I think your analogy with the velocity-time diagram made me find the culprit. My spectrum graph shown in post #140 has a y-axis in Watts, and not in Watts/nanometre. Hence if you integrate a function that has Watts/nm as units ($\Phi(\lambda)$) you'll merely get the difference in Watts between the 2 chosen wavelengths in my Watts vs $\lambda$ spectrum curve in post #140.

So with your mentioned example of integrating $\Phi(\lambda)$ between $550nm-600nm$, in a Watts vs $\lambda$ graph this would give give the following answer:

However, in a graph of Watts/nm vs $\lambda$, integrating $\Phi(\lambda)$ would give the area under that Watts/nm vs $\lambda$ graph.

So my question is; if I'm using a Watts vs $\lambda$ graph (like in my post #140) instead of a Watts/nm vs $\lambda$ graph, and I want to calculate the energy within a specific wavelength range using that graph, shouldn't I integrate a function other than $\Phi(\lambda)$ to get the area under that Watts vs $\lambda$ graph? Is that the step by step numerical calculation that you were talking about in your post #145?

 

[Charles Link]

I think your analogy with the velocity-time diagram made me find the culprit. My spectrum graph shown in post #140 has a y-axis in Watts, and not in Watts/nanometre. Hence if you integrate a function that has Watts/nm as units ($\Phi(\lambda)$) you'll merely get the difference in Watts between the 2 chosen wavelengths in my Watts vs $\lambda$ spectrum curve in post #140.

So with your mentioned example of integrating $\Phi(\lambda)$ between $550nm-600nm$, in a Watts vs $\lambda$ graph this would give give the following answer:
View attachment 205866

However, in a graph of Watts/nm vs $\lambda$, integrating $\Phi(\lambda)$ would give the area under that Watts/nm vs $\lambda$ graph.

So my question is; if I'm using a Watts vs $\lambda$ graph (like in my post #140) instead of a Watts/nm vs $\lambda$ graph, and I want to calculate the energy within a specific wavelength range using that graph, shouldn't I integrate a function other than $\Phi(\lambda)$ to get the area under that Watts vs $\lambda$ graph? Is that the step by step numerical calculation that you were talking about in your post #145?

I do believe you have most of it figured out. Good work ! $\\$ In your post #140, that graph is clearly mislabeled on the y-axis and it should read watts/nm. Sometimes, the per nm designation is understood, but it really belongs in the label on the y-axis. $\\$ Just an additional comment: With these spectral curves, you always compute the area under the curve=you don't take a difference such as $N-X$. In that particular instance, I think you are trying to do something that is incorrect. These spectral curves are $\Phi(\lambda)$ type curves in all cases (e.g. per unit wavelength). They are not $A(\lambda)$ type curves. If you were showing a graph of $A(\lambda)$ vs. $\lambda$ in post 146, then what you did is correct, but again, it simply is never done that way in spectroscopy. $\\$ (And to say it again=it is a common technique in probability theory to tabulate the integrated curve $A(\lambda)$. The two letters they typically use are $F(\lambda)$ (for the integrated), and $f(\lambda)$ for what they call the probability density function, but the spectroscopists do it without the $A(\lambda)$).

 

[JohnnyGui]

I do believe you have most of it figured out. Good work ! $\\$ In your post #140, that graph is clearly mislabeled on the y-axis and it should read watts/nm. Sometimes, the per nm designation is understood, but it really belongs in the label on the y-axis.

This has been bothering me for a long while and I can't believe this was actually the culprit. Thanks a lot for explaining this!

Let's say I'm very stubborn and I want to use the graph in my post #140 with just Watts on the y-axis vs $\lambda$ to calculate the energy between 550nm-600nm. Should I read off the amount of Watts for each wavelength within 550-600 nm and add them all together? Or would this still give an incorrect answer?

 

[Charles Link]

This has been bothering me for a long while and I can't believe this was actually the culprit. Thanks a lot for explaining this!

Let's say I'm very stubborn and I want to use the graph in my post #140 with just Watts on the y-axis vs $\lambda$ to calculate the energy between 550nm-600nm. Should I read off the amount of Watts for each wavelength within 550-600 nm and add them all together? Or would this still give an incorrect answer?

When they make what is really a mistake with their units on the y-axis, you need to question whether the numbers such as 600 and 700 watts/nm are correct, but using those numbers, when integrating this source, the answer that you get it that it has about 70,000 watts of radiated power. (I estimated this by taking 500 watts/nm at the shoulders near the peak multiplied by a width of about $\Delta \lambda=100$ nm, (for wavelength $\lambda$ from 600 nm to 700 nm) and then observing the entire area under the curve might be about 35% more). It general, I would recommend finding a book that gets the units correct, but it still made for a useful diagram. $\\$ Note: In computing these integrals numerically, it is much easier to use a lower resolution , e.g. $\Delta \lambda=10 \, nm$ or even $\Delta \lambda =100 \, nm$. The summation you do is $\sum\limits_{i} \Phi(\lambda_i) \, \Delta \lambda$. (At higher resolution=smaller $\Delta \lambda$, there are more $\Phi(\lambda_i)$ terms in the sum. At low resolution (e.g. $\Delta \lambda=100 \, nm$ ), you might have i=1 to 10. At very high resolution (e.g. $\Delta \lambda=1 \, nm$), you would then have i=1 to 1000, etc.) As I showed in post # 145, you can do it at various resolutions, but there is no reason to use $\Delta \lambda=1 \, nm$. When the spectral curve has little structure, you can get a very accurate answer with a low resolution summation. If you followed the explanation of how I estimated 70,000 watts, I basically computed the integral at a resolution of $\Delta \lambda=100 \, nm$. If you compute it at $\Delta \lambda =1 \, nm$ resolution, it might take you 30 minutes or more by hand ,(of course you can simply use a spreadsheet and let the computer do the work), and most likely you will still get an answer for the area under the entire curve of about 70,000 watts.

 

[JohnnyGui]

When they make what is really a mistake with their units on the y-axis, you need to question whether the numbers such as 600 and 700 watts/nm are correct, but using those numbers, when integrating this source, the answer that you get it that it has about 70,000 watts of radiated power. (I estimated this by taking 500 watts/nm at the shoulders near the peak multiplied by a width of about Δλ=100Δλ=100 \Delta \lambda=100 nm, (for wavelength λλ \lambda from 600 nm to 700 nm) and then observing the entire area under the curve might be about 35% more). It general, I would recommend finding a book that gets the units correct, but it still made for a useful diagram. \\ Note: In computing these integrals numerically, it is much easier to use a lower resolution , e.g. Δλ=10nmΔλ=10nm \Delta \lambda=10 \, nm or even Δλ=100nmΔλ=100nm \Delta \lambda =100 \, nm . The summation you do is ∑iΦ(λi)Δλ∑iΦ(λi)Δλ \sum\limits_{i} \Phi(\lambda_i) \, \Delta \lambda . (At higher resolution=smaller ΔλΔλ \Delta \lambda , there are more Φ(λi)Φ(λi) \Phi(\lambda_i) terms in the sum. At low resolution (e.g. Δλ=100nmΔλ=100nm \Delta \lambda=100 \, nm ), you might have i=1 to 10. At very high resolution (e.g. Δλ=1nmΔλ=1nm \Delta \lambda=1 \, nm ), you would then have i=1 to 1000, etc.) As I showed in post # 145, you can do it at various resolutions, but there is no reason to use Δλ=1nmΔλ=1nm \Delta \lambda=1 \, nm . When the spectral curve has little structure, you can get a very accurate answer with a low resolution summation. If you followed the explanation of how I estimated 70,000 watts, I basically computed the integral at a resolution of Δλ=100nmΔλ=100nm \Delta \lambda=100 \, nm . If you compute it at Δλ=1nmΔλ=1nm \Delta \lambda =1 \, nm resolution, it might take you 30 minutes or more by hand ,(of course you can simply use a spreadsheet and let the computer do the work), and most likely you will still get an answer for the area under the entire curve of about 70,000 watts.

You mentioned here that this computation has to be done for a graph that has Watts/nm on the y-axis (a $\Phi(\lambda)$ vs $\lambda$ curve) right?
I might have misunderstood you, but I meant calculating the energy if the graph is a $A(\lambda)$ vs $\lambda$ curve (i.e. Watts on the y-axis instead of Watts/nm).

I might be blacking out again but here's what I find colliding:

For a $\Phi(\lambda)$ vs $\lambda$ curve that has Watts/nm on the y-axis, you'd have to integrate $\Phi(\lambda)$ for a certain wavelength range to calculate the energy. Just like integrating a $v$ vs $t$ curve between $t_1-t_2$ would give the distance $s$ traveled between $t_1 - t_2$ which is the distance $s$ at $t_2$ minus distance at $t_1$ in a $s$ vs $t$ curve, integration of $\Phi(\lambda)$ between $\lambda_1 - \lambda_2$ in a $\Phi(\lambda)$ vs $\lambda$ curve would give the energy between $\lambda_1-\lambda_2$ which is the energy at $\lambda_2$ minus the energy of $\lambda_1$ in a $A(\lambda)$ curve, if I understand this correctly.

However, say I now have a $A(\lambda)$ vs $\lambda$ curve which has Watts on the y-axis and I want to calculate the energy from that curve between $\lambda_1-\lambda_2$. In this case, my instinct would say that I'd have to add up every amount of Watts corresponding to each wavelength between $\lambda_1-\lambda_2$ (with a high resolution for examle). Notice that this would NOT give the same answer as when I integrate $\Phi{\lambda}$ in a $\Phi(\lambda)$ curve between $\lambda_1 - \lambda_2$, since that would merely give the difference of the energy between $\lambda_1$ and $\lambda_2$ in a $A(\lambda)$ curve.

You probably have been trying to explain me this the whole time, but I don't really get how I should calculate the amount of energy from a $A(\lambda)$ vs $\lambda$ curve to get the same amount of energy as integrating a $\Phi(\lambda)$ vs $\lambda$ curve.