Emission spectra of different materials

[JohnnyGui]

Almost, but no cigar. B's radiating surface is larger by $\frac{1}{cos^2(\theta)}$ precisely accounting for the inverse square extra distance. There is no accounting for the first $cos(\theta)$ factor though, which is the projected area of the lens. For small angles $cos(\theta) \approx 1-\frac{\theta^2}{2} \approx 1$ so that we like to work with small angles $\theta$. Otherwise, it gets overly complicated mathematically, and really offers little additional instructional value.

Ah, of course. But how about the Lambertian cosine law then? Isn't that law different from the projected area of the lens and therefore the net energy should be decreased by that Lambertian factor for $B$? (A factor of $cos (\theta)$)

 

[Charles Link]

Ah, of course. But how about the Lambertian cosine law then? Isn't that law different from the projected area of the lens and therefore the net energy should be decreased by that Lambertian factor for $B$? (A factor of $cos (\theta)$)

The $\frac{1}{cos^2(\theta)}$ applies to a source that is perpendicular to the direction it is emitting, but has an extra area factor from the increased distance. If we take that source and rotate it by $\theta$ , we have the Lambertian fall-off because of the angle , but its area now increases by $\frac{1}{cos(\theta)}$. You are correct in questioning this=it actually results in a $\frac{1}{cos^3(\theta)}$ factor, but it doesn't change anything. The received irradiance is $E=L \Omega$ where $\Omega$ is the solid angle of the source measured at the observer. The one $cos(\theta )$ factor that remains unaccounted for is that this $E$ is incident on the lens at angle $\theta$.

 

[JohnnyGui]

The $\frac{1}{cos^2(\theta)}$ applies to a source that is perpendicular to the direction it is emitting, but has an extra area factor from the increased distance. If we take that source and rotate it by $\theta$ , we have the Lambertian fall-off because of the angle , but its area now increases by $\frac{1}{cos(\theta)}$. You are correct in questioning this=it actually results in a $\frac{1}{cos^3(\theta)}$ factor, but it doesn't change anything. The received irradiance is $E=L \Omega$ where $\Omega$ is the solid angle of the source measured at the observer. The one $cos(\theta )$ factor that remains unaccounted for is that this $E$ is incident on the lens at angle $\theta$.

Sorry if I'm still missing the point.
- The decrease in energy because of the increased distance by $\frac {1}{cos (\theta)^2}$ is compensated by the increased area by $cos(\theta)^2$.
- From the viewpoint of the radiating surface area, there is a $cos (\theta)$ Lambertian fall-off in intensity towards $B$. I still don't get what compensates for this.

 

[Charles Link]

Sorry if I'm still missing the point.
- The decrease in energy because of the increased distance by $\frac {1}{cos (\theta)^2}$ is compensated by the increased area by $cos(\theta)^2$.
- From the viewpoint of the radiating surface area, there is a $cos (\theta)$ Lambertian fall-off in intensity towards $B$. I still don't get what compensates for this.

To answer your last question, if you rotate the surface that is originally perpendicular to the direction you are viewing it by angle $\theta$,each area $dA$ will then radiate less by a factor of $\theta$ (the Lambertian factor), but the area contained in any solid angle $d \Omega$ of viewing will increase by $\frac{1}{cos(\theta)}$.

 

[JohnnyGui]

To answer your last question, if you rotate the surface that is originally perpendicular to the direction you are viewing it by angle $\theta$,each area $dA$ will then radiate less by a factor of $\theta$ (the Lambertian factor), but the area contained in any solid angle $d \Omega$ of viewing will increase by $\frac{1}{cos(\theta)}$.

Is the area that is originally perpendicular to B, larger by a factor of $\frac {1}{cos (\theta)^2}$, so that when it's rotated by an angle $\theta$, the total area increase is a factor of $\frac{1}{cos(\theta)^3}$?

 

[Charles Link]

Is the area that is originally perpendicular to B, larger by a factor of $\frac {1}{cos (\theta)^2}$, so that when it's rotated by an angle $\theta$, the total area increase is a factor of $\frac{1}{cos(\theta)^3}$?

It looks and is greater than a factor of $\frac{1}{cos(\theta)}$ because a large part of it undergoes a huge increase in distance from the observer as it is rotated. That's why in these calculations you need to work with infinitesimal areas rather than larger areas or you will find things like this appearing that unnecessarily complicate the equations.

 

[JohnnyGui]

It looks and is greater than a factor of 1cos(θ)1cos(θ) \frac{1}{cos(\theta)} because a large part of it undergoes a huge increase in distance from the observer as it is rotated. That's why in these calculations you need to work with infinitesimal areas rather than larger areas or you will find things like this appearing that unnecessarily complicate the equations.

Please bear with me as I’m trying to illustrate this.

Suppose $A_R$ is the receiving retina surface (lens is thrown out to keep it simple) that has turned an angle $\theta$ from distance $R$ ($R$ is perpendicular to the radiating surface).

Area $A_1$, which is not iillustrated here, is the radiating area covered by the viewing angle coming from $A_R$ when $A_R$ is parallel to the radiating surface at distance $R$ (just like observer $A$ in post #310). Is it then correct to say that $A_P = \frac{A_1}{cos(\theta)^2}$?.

If this is correct, and I’d want to calculate the energy received from $A_2$ that is covered by the viewing angle coming from $A_R$, then I’d need a relationship between $A_P$ and $A_2$ or $A_2$ and $A_1$ to calculate the total energy coming from $A_2$ onto $A_R$. And this received energy from $A_2$ must be equal to the energy received from $A_1$. The question for me is first, how do I calculate the size of area $A_2$?

 

[Charles Link]

To answer your first queston, you computed $A_p$ correctly. $\\$ This is why you need to use small incremental angles in doing these calculations. If the field of view is narrow, then $A_2=\frac{A_p}{cos(\theta)}$. For a wide field of view, it is not easy to compute it.

 

[JohnnyGui]

To answer your first queston, you computed $A_p$ correctly. $\\$ This is why you need to use small incremental angles in doing these calculations. If the field of view is narrow, then $A_2=\frac{A_p}{cos(\theta)}$. For a wide field of view, it is not easy to compute it.

Ah, my illustration along with this explanation made me understand this, thanks.

If for narrow field of view $A_2 = \frac{A_P}{cos(\theta)}$, does this mean that $A_2 = \frac{A_1}{cos(\theta)^3}$?

 

[Charles Link]

Ah, my illustration along with this explanation made me understand this, thanks.

If for narrow field of view $A_2 = \frac{A_P}{cos(\theta)}$, does this mean that $A_2 = \frac{A_1}{cos(\theta)^3}$?

Correct. Very good. :) :)

 

[JohnnyGui]

Correct. Very good. :) :)

Ah, and it is this $\frac{1}{cos(\theta)^3}$ increase of the area that exactly compensates for the $cos(\theta)$ Lambertian fall-off and the $cos(\theta)^2$ energy decrease because of the increased distance w.r.t. distance $R$. I think I finally got it now :).

 

[JohnnyGui]

Correct. Very good. :) :)

One small thing; if $A_P = \frac{A_1}{cos(\theta)^2}$ because the length and width of $A_1$ are each increased by a factor $\frac{1}{cos(\theta)}$, shouldn't $A_2$ be also increased by a factor of $\frac{1}{cos(\theta)^2}$ w.r.t. $A_P$ (for narrow field of view) since the length and width of $A_P$ are again each increased by a factor of $\frac{1}{cos(\theta)}$ to get $A_2$, so that actually $A_2 = \frac{A_1}{cos(\theta)^4}$?

 

[Charles Link]

The answer is no. The angle $\theta$ is a single rotation that only affects the length in the direction that is in the plane of the rotation that is perpendicular to the line-of-sight. (e.g. if you do a horizontal rotation in the x-y plane, it does not affect the dimension in the elevation angle direction as there is no rotation in the elevation angle direction.)

 

[Charles Link]

To add to the above, $\theta$ is measured by taking one unit vector perpendicular to the surface you are observing, and taking a second unit vector along the line-of sight (pointing at you). These two vectors define the plane in which the rotation $\theta$ occurs and is measured. The angle between these two vectors is $\theta$.

 

[JohnnyGui]

To add to the above, $\theta$ is measured by taking one unit vector perpendicular to the surface you are observing, and taking a second unit vector along the line-of sight (pointing at you). These two vectors define the plane in which the rotation $\theta$ occurs and is measured. The angle between these two vectors is $\theta$.

If the field of view so large such that $A_2$ is not simply equal to $\frac{A_1}{cos(\theta)^3}$, will $A_2$ be larger than that factor such that the radiating surface from an angle would look brighter than when you look at it perpendicularly (even though the human eye is not a perfect brightness sensor)?

 

[Charles Link]

If the field of view so large such that $A_2$ is not simply equal to $\frac{A_1}{cos(\theta)^3}$, will $A_2$ be larger than that factor such that the radiating surface from an angle would look brighter than when you look at it perpendicularly (even though the human eye is not a perfect brightness sensor)?

The answer is no, because if you look at each small incremental solid angle $d \Omega$ that leads to an infinitesimal area $d A_2$ on $A_2$, it will obey the formula for the irradiance at the observer as $dE=L \, d \Omega$ , where $d \Omega$ is measured from the observer. It really is totally independent of the shape and/or distance of the surface $A_2$ or what angle it happens to be rotated at. (We are assuming the source is an ideal Lambertian radiator). The irradiance at the receiver depends only on the brightness $L$ of the surface of the source, and equal solid angles as measured from the observer give equal results.

 

[JohnnyGui]

The answer is no, because if you look at each small incremental solid angle $d \Omega$ that leads to an infinitesimal area $d A_2$ on $A_2$, it will obey the formula for the irradiance at the observer as $dE=L \, d \Omega$ , where $d \Omega$ is measured from the observer. It really is totally independent of the shape and/or distance of the surface $A_2$ or what angle it happens to be rotated at. (We are assuming the source is an ideal Lambertian radiator). The irradiance at the receiver depends only on the brightness $L$ of the surface of the source, and equal solid angles as measured from the observer give equal results.

Ah ok, I was wondering if it's possible to calculate this with $d \Omega$ is measured from the radiating surface. Please bear with me here:

Suppose we’re calculating the received energy coming from only one $dA$ of a radiating surface onto a large retinal surface $A_R$. For the following scenario:

The integration formula to calculate the received energy $P_{R1}$ in this scenario would be (as discussed):
$$\int\limits_{0}^{A_R} \frac{T^4 \cdot \sigma\cdot dA_1}{R^2 \cdot \pi} \cdot cos(\theta)^4 \cdot dA_R = P_{R1}$$
Now, for the next scenario:

We know that the infinitesimally small $dA_2$ is equal to $\frac{dA_1}{cos(\theta)^3}$. Furthermore, the perpendicular distance from $dA_2$ to $A_R$ is now a function of the length $L$ of $A_R$, such that the perpendicular distance from $dA_2$ to $A_R$ is $R – sin(\alpha) \cdot dL$. Also, since $A_R$ is large, there is now also an extra $cos(\theta)$ factor for the projected surface of $A_R$ with respect to each EM ray coming from $dA_2$. The received energy $P_{R2}$ in this case should therefore be:
$$\int\limits_{0}^{A_R} \frac{T^4 \cdot \sigma\cdot \frac{dA_1}{cos(\theta)^3}}{(R – sin(\alpha)\cdot dL)^2 \cdot \pi} \cdot cos(\theta)^4 \cdot dA_R = \int\limits_{0}^{A_R} \frac{T^4 \cdot \sigma\cdot dA_1}{(R – sin(\alpha)\cdot dL)^2 \cdot \pi} \cdot cos(\theta) \cdot dA_R = P_{R2}$$
The $cos(\theta)^4$ before simpifying the equation consists of a $cos(\theta)^2$ for the distance increase w.r.t. $R$, a $cos(\theta)$ for the Lambertian factor and a $cos(\theta)$ for the projection receiving surface of $A_R$ for each EM ray coming from $dA_2$.

I’m wondering if both integrations from each scenario would yield the same received energy so that $P_{R1} = P_{R2}$

 

[Charles Link]

When you start using a large receiver area at $A_r$, the formulas no longer work because the distance to the source is not the same at every point on the detector. Thereby, we can't say the irradiance is some number $E$ because that number changes across the receiving surface. $\\$ It may interest you that your angle $\alpha$ is equal to $\theta$.

 

[JohnnyGui]

When you start using a large receiver area at $A_r$, the formulas no longer work because the distance to the source is not the same at every point on the detector. Thereby, we can't say the irradiance is some number $E$ because that number changes across the receiving surface. $\\$ It may interest you that your angle $\alpha$ is equal to $\theta$.

Yes, but that's why I've introduced the distance to the source being a function of $dL$ according to $R – sin(\alpha) \cdot dL$. Isn't it possible to do it this way and show that $P_{R1}=P_{R2}$?
I'm indeed aware that angle $\theta = \alpha$ but since angle $\theta$ changes with each $dA_R$, I chose $\alpha$ to show that it's a fixed angle that one can use to calculate the distance to each $dA_R$.

 

[Charles Link]

You need to use an infinitesimal size receiver element to show the result. The solid angle element $d \Omega$ is measured from a point on the receiver. That solid angle for the same source will be different from another point on the receiver if the receiver is finite in size. The source won't even be at the same angle $\theta$.

 

[JohnnyGui]

You need to use an infinitesimal size receiver element to show the result. The solid angle element $d \Omega$ is measured from a point on the receiver. That solid angle for the same source will be different from another point on the receiver if the receiver is finite in size. The source won't even be at the same angle $\theta$.

Ok, but even for large receivers, shouldn't $P_{R1}=P_{R2}$ as well just like when the receiver element is infinitesimally small? Or is this wrong?

 

[JohnnyGui]

@Charles Link

Thinking this a bit through, I realized something but I need verification. Is the Radiance $L$ specifically used for receivers that have a lens in front of them (e.g. an eye or camera)? I'm thinking this because drawing solid angles coming out of the receiver instead of the source means that the receiver must have some kind of aperture/lense that allows a certain amount of light to come in on the receiving surface.

Is this all correct?

 

[Charles Link]

@Charles Link

Thinking this a bit through, I realized something but I need verification. Is the Radiance $L$ specifically used for receivers that have a lens in front of them (e.g. an eye or camera)? I'm thinking this because drawing solid angles coming out of the receiver instead of the source means that the receiver must have some kind of aperture/lense that allows a certain amount of light to come in on the receiving surface.

Is this all correct?

The question you are asking is somewhat complex, but I will attempt to answer it. For an ideal lens that is 100% transmissive, the radiance (brightness) $L$ of the image is equal to the brightness of the object. (Referring here to viewing it where the rays from the object and the image have a sufficient F# that they fill the sensor such as the eye that is viewing the object and image). The eye will create an image of the object on the retina, and similarly if you create a real image of the object by using a lens such that a focused image forms (it will be upside down) on the other side of the lens, when you view this image with your eye by focusing on it, it will appear to have exactly the same brightness $L$ as the object itself. $\\$ Meanwhile though, to use the simple formulas relating brightness $L$ to irradiance, you do not need any kind of lenses. $\\$ Hopefully the answer I gave you wasn't too complicated, and your question really seemed to be probing something that is known as the "brightness theorem" in optics, so that is how I answered it. You can google the topic for additional reading. You might find it quite interesting. $\\$ One additional item is you can write a power conservation equation, so that $P=L_1 A_1 \Omega_1=L_2 A_2 \Omega_2$ where subscript $1$ refers to the source whose light is incident on a lens and subscript $2$ refers to the real image from that lens. It can be easily shown that $A_1 \Omega_1=A_2 \Omega_2$ so that $L_1=L_2$. This is essentially what the brightness theorem tells you: The brightness $L_1$ of the object is equal to the brightness $L_2$ of the image.

 

[JohnnyGui]

Thanks for the extensive explanation. So I take it that $L$ can also be calculated without a receiving surface having any lenses.

One additional item is you can write a power conservation equation, so that P=L1A1Ω1=L2A2Ω2P=L1A1Ω1=L2A2Ω2 P=L_1 A_1 \Omega_1=L_2 A_2 \Omega_2 where subscript 11 1 refers to the source whose light is incident on a lens and subscript 22 2 refers to the real image from that lens. It can be easily shown that A1Ω1=A2Ω2A1Ω1=A2Ω2 A_1 \Omega_1=A_2 \Omega_2 so that L1=L2L1=L2 L_1=L_2 . This is essentially what the brightness theorem tells you: The brightness L1L1 L_1 of the object is equal to the brightness L2L2 L_2 of the image.

Is the $\Omega_1$ the solid angle coming from the radiated source and subtended by the lens and $\Omega_2$ the solid angle coming from the lens and subtended by the radiating source?I noticed something regarding a question in another thread about Radiance posted here .
The statement says that if a $1m^2$ receiving surface $A_R$ on Earth receives 1050 W from the sun, such that $E= 1050W/m^2$, the brightness $L$ would be $15.4 MW/m^2/sr$. I was able to calculate this, but the way I calculated it gave me a different idea about what Radiance really means.

Here's how I drew it in my mind to calculate this:

I assumed that one steradian would be the surface area of 1 Astronomical unit (AU) squared (calculated in meters) since the sun is 1 AU away from us. I then simply calculated how much larger that 1 steradian of $1 AU^2$ is with respect to the sun's disc area $A_{sun}$; which is $\frac{1 AU^2}{A_{sun}}$. Multiplying this ratio by the received energy of 1050W would indeed give me approximately $15.4MW/m^2/sr$.

This makes me think that $L$ is simply the amount of energy $A_R$ would receive per $1m^2$ of that $A_R$, if the sun had an area of 1 steradian (distance to the sun squared). So it's basically the Irradiance but then coming from a radiating area of $s^2$ (distance squared).
Is this correct? If not, why does this thought still give me the correct answer of $15.4MW/m^2/sr$?

 

[Charles Link]

A little algebra gives the result that you are trying to determine whether your method is logical: Irradiance $E=L \, \Omega=L \, (A_{sun}/s^2)$ so that $L=E (s^2/A_{sun})$. You are essentially calling $s^2=A_o$ and asking is $L=E (A_o/A_{sun})$?, and the answer is yes, but the algebra is really a better route than to try to introduce any new concept.

 

[JohnnyGui]

A little algebra gives the result that you are trying to determine whether your method is logical: Irradiance $E=L \, \Omega=L \, (A_{sun}/s^2)$ so that $L=E (s^2/A_{sun})$. You are essentially calling $s^2=A_o$ and asking is $L=E (A_o/A_{sun})$?, and the answer is yes, but the algebra is really a better route than to try to introduce any new concept.

Ah of course, didn't come to my mind to prove it like this.

But the thing is, this new concept makes me understand $L$ a whole lot better. The descrption of $L$ being "the amount of energy you receive on $m^2$ of your receiving surface from a radiating source area of 1 steradian" somehow makes more sense to me. The weird part is that I'm using here the $m^2$ of the receiving surface while the other description of $L$ (which I had difficulty understanding the meaning of it) says that the $m^2$ in the units is the $m^2$ area of the radiating source. I guess this is what you meant in your post #338 regarding the power conservation equation?

 

[Charles Link]

Normally I like to work from the source, rather than from the observer, and write $I=L \, A$. The next step is to employ the inverse square law to get irradiance $E=I/s^2=LA/s^2$. $\\$ Alternatively, it can be viewed from the observer as $E=L \, \Omega$ where $\Omega=A/s^2$, but this second method is usually not my preferred way of calculating the irradiance. $\\$ A couple additional comments on the surface area $A_o$ for $\Omega=1 \, sr$: It should be remembered that for large solid angles, this area $A_o=s^2$ is the surface area on the surface of a sphere of radius $s$. In addition, the formula $E=L \, \Omega$ is not completely valid for large solid angles. The formula $E= L \, \Omega$ works for relatively small solid angles $\Omega$ on-axis, and upon going off-axis, there is a factor of $\cos(\theta)$ on the irradiance because the receiver is not perpendicular to the line-of-sight to the source. In general $E=\int L \cos(\theta) \, d \Omega$ and is only $E=\int L \, d \Omega = L \, \int d \Omega =L \, \Omega$ for sources covering a small solid angle $\Omega$ on-axis.

 

[JohnnyGui]

A couple additional comments on the surface area AoAo A_o for Ω=1srΩ=1sr \Omega=1 \, sr : It should be remembered that for large solid angles, this area Ao=s2Ao=s2 A_o=s^2 is the surface area on the surface of a sphere of radius ss s . In addition, the formula E=LΩE=LΩ E=L \, \Omega is not completely valid for large solid angles. The formula E=LΩE=LΩ E= L \, \Omega works for relatively small solid angles ΩΩ \Omega on-axis, and upon going off-axis, there is a factor of cos(θ)cos⁡(θ) \cos(\theta) on the irradiance because the receiver is not perpendicular to the line-of-sight to the source. In general E=∫Lcos(θ)dΩE=∫Lcos⁡(θ)dΩ E=\int L \cos(\theta) \, d \Omega and is only E=∫LdΩ=L∫dΩ=LΩE=∫LdΩ=L∫dΩ=LΩ E=\int L \, d \Omega = L \, \int d \Omega =L \, \Omega for sources covering a small solid angle ΩΩ \Omega on-axis.

Ah, thanks for reminding me this.

I'd like some verification again regarding calculating $L$ from the receiver’s perspective ($A_R$)

Let’s say the solid angle $d \Omega$ subtended by a partial source surface $A$ is small enough so that $\frac{A_P}{cos(\theta)}=A$. And receiver $dA_R$ is receiving a power $P_R$ from $A$ that is a distance $s$ away.

Does this mean that $d \Omega = \frac{A \cdot cos(\theta)}{s^2}$?
And therefore $L = P_R \cdot \frac{s^2}{A \cdot cos(\theta)} \cdot \frac{1}{dA_R}$?

 

[Charles Link]

Very good. Your calculation is completely correct assuming the receiver element $dA_r$ faces the source. Otherwise, if $dA_r$ is horizontal, you will get a second factor of $\cos(\theta)$ in the denominator.

 

[JohnnyGui]

Very good. Your calculation is completely correct assuming the receiver element $dA_r$ faces the source. Otherwise, if $dA_r$ is horizontal, you will get a second factor of $\cos(\theta)$ in the denominator.

Thanks! I indeed forgot to put the $cos(\theta)$ in the denominator for a horizontal $dA_R$. So it should indeed be $L=P_R \cdot \frac{s^2}{A \cdot cos(\theta)^2 \cdot dA_R}$

Let's say distance $R$ is the perpedicular distance between $dA_R$ and the horizontal plane of the radiating source, such that $\frac{R}{cos(\theta)} = s$.
If the power $P_R$ comes from $d \Omega$ that subtends a radiating area of $A$, can I express $P_R$ as:
$$P_R = \int \limits_{0}^{A} \frac{T^4 \cdot \sigma \cdot dA_R}{\pi \cdot R^2} \cdot cos(\theta_{dA})^4 \cdot dA$$
Combining this $P_R$ expression with the formula for $L$ from my previous post #343 gives:
$$L = \frac{R^2}{A \cdot dA_R \cdot cos(\theta)^4} \cdot \int \limits_{0}^{A} \frac{T^4 \cdot \sigma \cdot dA_R}{\pi \cdot R^2} \cdot cos(\theta_{dA})^4 \cdot dA$$
Which simplifies to:
$$L=\frac{T^4 \cdot \sigma}{A \cdot \pi \cdot cos(\theta)^4} \cdot \int \limits_{0}^{A} cos(\theta_{dA})^4 \cdot dA$$
I have the feeling I'm close to concluding $\frac{I}{A} = L$ but the problem here is, if $\theta_{dA}$ doesn't differ much with each $dA$, then the integration part collapses to $cos(\theta)^4 \cdot A$ and the whole formula will be $L = \frac{T^4 \cdot \sigma}{\pi} = I_0$ which is incorrect.

Edit: Nevermind, found out that I forgot that it's $\frac{T^4 \cdot \sigma \cdot A}{\pi} = I_0$ such that the last formula indeed shows $L = I / A$ if the integration collapses to $cos(\theta)^4 \cdot A$. So my concluded formula of $L$ indeed shows that one can interchange between the perspective of the receiver as well as the perspective ofthe radiating source, if I understand correctly.

 

[JohnnyGui]

@Charles Link

From my previous post I noticed, as we discussed, that the $cos(\theta)^4$ both in the numerator and the denominator cancel each other out so that looking at the same $A$ from any angle would give the same Radiance $L$.

This made me realize something about 2 different scenarios;:

Left picture
If the radiating surface area is fixed and 2 observers are looking at that fixed radiating surface area, then $B$'s field of view $\alpha_B$ is smaller than $\alpha_A$ (let's say because B has a smaller iris/aperture). Nevertheless, they should still measure the same Radiance $L$?

Right picture
And when the field of view angles are the same for both $A$ and $B$ ($\alpha_A = \alpha_B$), such that each observer would cover a different radiating surface ($A_B > A_A$) then the measured $L$ would still be the same for both observers.

Are these 2 statements correct?

 

[Charles Link]

@Charles Link

From my previous post I noticed, as we discussed, that the $cos(\theta)^4$ both in the numerator and the denominator cancel each other out so that looking at the same $A$ from any angle would give the same Radiance $L$.

This made me realize something about 2 different scenarios;:
View attachment 210141 View attachment 210143

Left picture
If the radiating surface area is fixed and 2 observers are looking at that fixed radiating surface area, then $B$'s field of view $\alpha_B$ is smaller than $\alpha_A$ (let's say because B has a smaller iris/aperture). Nevertheless, they should still measure the same Radiance $L$?

Right picture
And when the field of view angles are the same for both $A$ and $B$ ($\alpha_A = \alpha_B$), such that each observer would cover a different radiating surface ($A_B > A_A$) then the measured $L$ would still be the same for both observers.

Are these 2 statements correct?

The answer is precisely the same brightness $L$, provided the surface is a Lambertian radiator. An ideal blackbody surface is a Lambertian radiator. $\\$ There are many sources that have some directional dependence so that when viewed from a different angle you get totally different results. A good example of this is a flashlight, where if you are in the center of the beam the whole reflector is illuminated, but outside the central beam, all you see is a lit filament without any contribution from the reflector. In this case the reflector has a different brightness when viewed from an angle.

 

[JohnnyGui]

The answer is precisely the same brightness $L$, provided the surface is a Lambertian radiator. An ideal blackbody surface is a Lambertian radiator. $\\$ There are many sources that have some directional dependence so that when viewed from a different angle you get totally different results. A good example of this is a flashlight, where if you are in the center of the beam the whole reflector is illuminated, but outside the central beam, all you see is a lit filament without any contribution from the reflector. In this case the reflector has a different brightness when viewed from an angle.

Thanks, I tried it and I was indeed able to calculate and prove the same $L$ for A and B in all scenarios and that a Lambertian radiator is needed to measure the same $L$. :)

A question arises when the radiating surface #A## is too large though. Let's take this scenario for example:

As stated in my previous post, for a small radiating surface, $L=P_R \cdot \frac{R^2}{A \cdot cos(\theta)^4 \cdot dA_R}$. Where $P_R = \frac{T^4 \cdot \sigma \cdot dA_R \cdot A}{\pi \cdot R^2} \cdot cos(\theta)^4$

However, if the radiating surface $A$ is large, then an integration is needed which would imply the following large formula:
$$L = \int \limits_{0}^{A} \frac{T^4 \cdot \sigma \cdot dA_R}{\pi \cdot R^2} \cdot cos(\theta_{dA})^4 \cdot dA \cdot \int \limits_{0}^{A}\frac{R^2}{dA_R \cdot cos(\theta_{dA})^4} \cdot \frac{1}{dA}$$
Do both integrations in the formula still cancel each other out so that $L=\frac{I_0}{A}$?

 

[Charles Link]

Thanks, I tried it and I was indeed able to calculate and prove the same $L$ for A and B in all scenarios and that a Lambertian radiator is needed to measure the same $L$. :)

A question arises when the radiating surface #A## is too large though. Let's take this scenario for example:
View attachment 210224
As stated in my previous post, for a small radiating surface, $L=P_R \cdot \frac{R^2}{A \cdot cos(\theta)^4 \cdot dA_R}$.
Where $P_R = \frac{T^4 \cdot \sigma \cdot dA_R \cdot A}{\pi \cdot R^2} \cdot cos(\theta)^4$

However, if the radiating surface $A$ is large, then an integration is needed which would imply the following large formula:
$$L = \int \limits_{0}^{A} \frac{T^4 \cdot \sigma \cdot dA_R}{\pi \cdot R^2} \cdot cos(\theta_{dA})^4 \cdot dA \cdot \int \limits_{0}^{A}\frac{R^2}{dA_R \cdot cos(\theta_{dA})^4} \cdot \frac{1}{dA}$$
Do both integrations in the formula still cancel each other out so that $L=\frac{I_0}{A}$?

It would be poor mathematics to split the numerator and denominator on the right side of the equation into two integrals. It would not tell you anything extra by doing this except give you an averaged brightness level over the surface. e.g. what happens in the case that $L$ is not uniform everywhere? One term in the integral is essentially the irradiance and the other term a solid angle, but also, remember that the irradiance formula $E=L \Omega$ only works for near on-axis because the receiver needs to point in the direction of the source. $\\$ Thereby, in this case, it is far better to work with small areas and not attempt to integrate to a larger area. You want to determine the brightness $L$ at a location, and your small area formula does exactly this. You don't want to average it over a larger radiating surface.

 

[JohnnyGui]

It would be poor mathematics to split the numerator and denominator on the right side of the equation into two integrals. It would not tell you anything extra by doing this except give you an averaged brightness level over the surface. e.g. what happens in the case that $L$ is not uniform everywhere? One term in the integral is essentially the irradiance and the other term a solid angle, but also, remember that the irradiance formula $E=L \Omega$ only works for near on-axis because the receiver needs to point in the direction of the source. $\\$ Thereby, in this case, it is far better to work with small areas and not attempt to integrate to a larger area. You want to determine the brightness $L$ at a location, and your small area formula does exactly this. You don't want to average it over a larger radiating surface.

This is what bothers me a bit because, how would one, without integrating, be able to prove that an observer is measuring the same $L$ if the covered radiating surface is very large? We discussed that neither angle nor the field of view width changes $L$ even though the covered radiating surface would be larger because of that larger angle or larger field of view. And how can one then speak of the same $L$ or the same $E$ if the power that the receiver receives is not uniform along the receiving surface?