Emission spectra of different materials

[JohnnyGui]

The reflectivity is apparently $\rho=1-\epsilon=.85$ and is nearly independent of wavelength. Otherwise, the radiated spectrum $M(\lambda,T)=\epsilon(\lambda) \, M_{bb}(\lambda,T)$ would not be that of a blackbody shape. i.e. if $\epsilon(\lambda)$ had hills and valleys, it would alter the shape.

Ah, so it has the same spectral shape as a blackbody but at lower energy levels for each wavelength because the energy level of each wavelength is multiplied by a fixed constant $\epsilon(\lambda)$)?

 

[Charles Link]

Ah, so it has the same spectral shape as a blackbody but at lower energy levels for each wavelength because the energy level of each wavelength is multiplied by a fixed constant $\epsilon(\lambda)$)?

Yes. I don't know myself why it is that way, but that's apparently how it is. One time, I did measure the spectrum of an incadescent lamp, and it did have very much a $T=2500 \, K$ blackbody shape. The area of the filament was unknown, so that I was unable to compute the emissivity.

 

[JohnnyGui]

Yes. I don't know myself why it is that way, but that's apparently how it is. One time, I did measure the spectrum of an incadescent lamp, and it did have very much a $T=2500 \, K$ blackbody shape. The area of the filament was unknown, so that I was unable to compute the emissivity.

But integrating the curve would yield a lower total $M$ right compared to integrating a spectral curve of a BB at the same temp, right?

There's something else I forgot to ask regarding equilibrium.
Can I say the following:
If an object at equilibrium has the same temperature as the walls $T=T_o$, this means that the object has the same approximate emissivity (i.e. absorption) for the radiation that the walls emit at $T_o$ as the emissivity that it had at its initial ambient temperature $T_{ambient}$.
For emissivity not to change too much from $T_{ambient}$ to the equilibrium $T=T_o$, they must not differ too much from each other initially.

 

[Charles Link]

In previous posts, when writing the Planck function, I used $L(\lambda, T)$ and/or $M(\lambda, T)$, but it really would have been better to write it as $L_{bb}(\lambda, T)$ and/or $M_{bb}(\lambda, T)$ to distinguish it from a greybody that has $M(\lambda,T)=\epsilon(\lambda) \, M_{bb}(\lambda, T)$. $\\$ Meanwhile, an important concept is that in an enclosed space, at temperature $T= T_{ambient}$ where everything is at thermal equilibrium, the emissivity does not matter. This is assuming no other sources $E_s(\lambda )$ are present that are not at thermal equilibrium. At thermal equilibrium in an enclosed space, everything will be at $T=T_{ambient}$. Basically $E_{incident}(\lambda)$ as well as what is emerging off of the wall is the same at thermal equilibrium, independent of the emissivity $\epsilon(\lambda)$. $E_{incident}(\lambda)$ will in fact be equal to $M_{bb}(\lambda,T_{ambient})$ at thermal equilibrium regardless of $\epsilon(\lambda)$ of that portion of the wall. And as previously mentioned, this result is also completely independent of the shape of the enclosure.

 

[JohnnyGui]

Apologies if I missed your point. I worded it poorly as well.

What I meant is if an object is exposed to a hot radiation source like the sun, the object won't necessarily reach the equilibrium temperature that one would expect according to $E_s = \sigma \cdot T^4$ because the object may have a total different absorptivity/emissivity for the wavelengths from the hot source that it's being exposed to. Therefore it can be at equilibrium at a lower $T$ for example. Just like we discussed and what I said in my statement #3 in post #225.

However, in the case of radiating walls instead of a hot source, where the temperature of the walls $T_o$ is hotter than the initial temperature of the object that is brought in, the object could eventually reach an equilibrium temperature $T$ equal to $T_o$. The fact that they're equal at equilibrium means that, although emissivity is wavelength dependent, as the temperature of the object rises to reach equilibrium, the effective emissivities at all those temperatures that the object passed by weren't enough to emit the same $M$ amount as that it receives from the walls. Such that at equilibrium $T=T_o$ and thus:
$$T^4_o \cdot \sigma \cdot \epsilon_{eff}(T_o) = T^4 \cdot \sigma \cdot \epsilon_{eff}(T)$$
The thing is, I find it to be too coincidental if the effective emissivity of the object at initial temperature (when it was brought in) was way different and that exactly at $T=T_o$ it would have the ideal emissivity that is equal to the received energy from the walls. It is therefore more likely that the emissivity of the object at $T=T_o$ (which is $\epsilon_{eff}(T_o)$) didn't change too much from the emissivity of the object at its initial temperature, when it was brought in.
Since emissivity doesn't change drastically with small wavelength/temperature change, to have a thermal equilibrium like that, the initial temperature of the object must not have been too different from the $T_o$ of the walls or else effective emissivity would have changed too much at equilibrium and one would then have a scenario similar to a hot radiating source like the sun in which the temperatures of the object and the source are different at equilibrium.

Not sure if I'm making any sense here. Please correct me if I'm wrong.

 

[Charles Link]

One correction: It doesn't require an ideal emissivity to reach thermal equilibrium. When the object reaches temperature $T_{ambient}$ it is at equilibrium. The reason is at $T_{ambient}$ in an enclosed cavity, $E_s(\lambda)=M_{bb} (\lambda, T_{ambient})$. Now, writing the power as a spectral density function, $P_{absorbed} (\lambda)=A \, \epsilon (\lambda) \, E_s(\lambda)=A \, \epsilon (\lambda) \, M_{bb} (\lambda,T_{ambient})$, and $P_{reflected} (\lambda)=A \, (1-\epsilon (\lambda) ) M_{bb} (\lambda, T_{ambient})$. The power radiated $P_{radiated} (\lambda)= A \, \epsilon (\lambda) \, M(\lambda, T_{ambient})$, so that the power emerging from the wall is $P_{emerging} (\lambda)=P_{reflected} (\lambda) +P_{radiated} (\lambda)=A \, M_{bb} (\lambda, T_{ambient} )$, completely independent of any emissivities. Notice also that $P_{absorbed} (\lambda)=P_{radiated} (\lambda)$. $\\$ Regardless of the emissivity of a portion of the wall, in an enclosure at thermal equilibrium, the surface looks like a blackbody even if its emissivity is very low. This is only the case at equilibrium, but it is a very common scenario, and this concept is used in deriving the Planck function, as we previously discussed. See post #206.

 

[JohnnyGui]

One correction: It doesn't require an ideal emissivity to reach thermal equilibrium. When the object reaches temperature $T_{ambient}$ it is at equilibrium. The reason is at $T_{ambient}$ in an enclosed cavity, $E_s(\lambda)=M_{bb} (\lambda, T_{ambient})$. Now, writing the power as a spectral density function, $P_{absorbed} (\lambda)=A \, \epsilon (\lambda) \, E_s(\lambda)=A \, \epsilon (\lambda) \, M_{bb} (\lambda,T_{ambient})$, and $P_{reflected} (\lambda)=A \, (1-\epsilon (\lambda) ) M_{bb} (\lambda, T_{ambient})$. The power radiated $P_{radiated} (\lambda)= A \, \epsilon (\lambda) \, M(\lambda, T_{ambient})$, so that the power emerging from the wall is $P_{emerging} (\lambda)=P_{reflected} (\lambda) +P_{radiated} (\lambda)=A \, M_{bb} (\lambda, T_{ambient} )$, completely independent of any emissivities. Notice also that $P_{absorbed} (\lambda)=P_{radiated} (\lambda)$. $\\$ Regardless of the emissivity of a portion of the wall, in an enclosure at thermal equilibrium, the surface looks like a blackbody even if its emissivity is very low. This is only the case at equilibrium, but it is a very common scenario, and this concept is used in deriving the Planck function, as we previously discussed. See post #206.

Makes sense, so an object at its equilibrium temperature $T_{ambient}$ in an enclosed cavity absorbs and emits, as discussed, a power of:
$$P_{absorbed} = A \cdot \int M_{bb}(\lambda,T_{ambient}) \cdot \epsilon(\lambda) = T^4_{ambient} \cdot A\sigma \cdot \epsilon(T_{ambient})$$
This means that the power that is incident on the object $P_{emerging}$ must be
$$P_{emerging} = A \cdot \int M_{bb}(\lambda,T_{ambient}) = T^4_{ambient} \cdot A\sigma$$
This concludes that the walls are at the same temperature as that of the object (both $T_{ambient}$).

Question, is it possible for an object in an enclosed cavity to be at equilibrium while having a different temperature than the walls? As discussed, the object at $T_{ambient}$ receives a $P_{absorbed}= T^4_{ambient} \cdot A\sigma \cdot \epsilon(T_{ambient})$. However, is it possible that the object is receiving that exact amount of $P_{absorbed}$ but caused by a different temperature of the walls plus the object having a different emissivity for the wavelengths that those walls emit at that different temperature? So that:
$$T_{walls}^4 \cdot A\sigma \cdot \epsilon(T_{walls}) = T^4_{ambient} \cdot A\sigma \cdot \epsilon(T_{ambient}) = P_{absorbed}$$
Where $\epsilon(T_{walls})$ and $T_{walls}$ are different from $\epsilon(T_{ambient})$ and $T_{ambient}$ respectively. The equation would be the same as:
$$A \int M_{bb}(\lambda, T_{walls}) \cdot \epsilon(\lambda) = A \int M_{bb}(\lambda, T_{ambient}) \cdot \epsilon(\lambda) = P_{absorbed}$$
Is this possible or not?

 

[Charles Link]

Your very last equation answers your question. If $T_{walls}>T_{ambient}$, then $M_{bb}(\lambda, T_{walls})>M_{bb}(\lambda,T_{ambient})$ for all $\lambda$, and thereby these two integrals could not be equal. A similar statement applies with the inequality reversed for $T_{walls}<T_{ambient}$. The only way the two integrals can be equal is if $T_{walls}=T_{ambient}$.

 

[JohnnyGui]

Your very last equation answers your question. If $T_{walls}>T_{ambient}$, then $M_{bb}(\lambda, T_{walls})>M_{bb}(\lambda,T_{ambient})$ for all $\lambda$, and thereby these two integrals could not be equal. A similar statement applies with the inequality reversed for $T_{walls}<T_{ambient}$. The only way the two integrals can be equal is if $T_{walls}=T_{ambient}$.

I get what you're saying. But the reason I find this possible is because $\epsilon(\lambda)$ differs with wavelength. So looking at the blackbody spectra at different temperatures:

Looking at the red line of $300K$, the emissivity $\epsilon(\lambda)$ of the object could be for example 1 for the wavelengths corresponding to that $300K$ line while it could be very low for the wavelengths corresponding to the $10000K$ line. Since the $10000K$ line has a higher total energy when integrated but is compensated by the low emissivity, there is a possibility that the total absorbed energy from that $10000K$ line at a low emissivity is equal to the total absorbed energy from the $300K$ line at a high emissivity. (I see that the $10000K$ line also covers the wavelengths from the 300K line but emissivity can also change due to a higher temperature itself so that it is lower for those $300K$ wavelengths at $10000$).

Isn't this possible?

 

[Charles Link]

I get what you're saying. But the reason I find this possible is because $\epsilon(\lambda)$ differs with wavelength. So looking at the blackbody spectra at different temperatures:
View attachment 206732
Looking at the red line of $300K$, the emissivity $\epsilon(\lambda)$ of the object could be for example 1 for the wavelengths corresponding to that $300K$ line while it could be very low for the wavelengths corresponding to the $10000K$ line. Since the $10000K$ line has a higher total energy when integrated but is compensated by the low emissivity, there is a possibility that the total absorbed energy from that $10000K$ line at a low emissivity is equal to the total absorbed energy from the $300K$ line at a high emissivity.

Isn't this possible?

The answer is no, it is not possible. In your last equation of post #247, every single term $\epsilon (\lambda) M_{bb} (\lambda, T_{walls})$ of the integral on the left hand side of the equation is greater than the corresponding term $\epsilon(\lambda) M_{bb} (\lambda , T_{ambient})$ on the right hand side of the equation for $T_{walls}>T_{ambient}$. The emissivity $\epsilon (\lambda)$ is the same for both terms. For the same $\lambda$, we don't have two different $\epsilon (\lambda)$.

 

[JohnnyGui]

The answer is no, it is not possible. In your last equation of post #247, every single term $\epsilon (\lambda) M_{bb} (\lambda, T_{walls})$ of the integral on the left hand side of the equation is greater than the corresponding term $\epsilon(\lambda) M_{bb} (\lambda , T_{ambient})$ on the right hand side of the equation for $T_{walls}>T_{ambient}$. The emissivity $\epsilon (\lambda)$ is the same for both terms. For the same $\lambda$, we don't have two different $\epsilon (\lambda)$.

You're right, the 10000K line also covers the wavelengths from 300K at a higher energy level. It might be farfetched but what if emissivity also changes with temperature itself so that it is specifically low for the wavelengths from 300K when it's at 10000K?

 

[Charles Link]

You're right, the 10000K line also covers the wavelengths from 300K at a higher energy level. It might be farfetched but what if emissivity also changes with temperature itself so that it is specifically low for the wavelengths from 300K when it's at 10000K?

Very far-fetched. What you are proposing is that since $\epsilon=\epsilon(\lambda,T)$ that the two $\epsilon(\lambda)$ are not necessarily the same for the left integral and the right integral (for $T=T_{walls}$ on the left and $T=T_{ambient}$ on the right), that in an extreme case of emissivity varying with the temperature, that it might be possible. Mathematically, perhaps, yes, but in a real physical system, it's not going to do that. This would basically be a system for which you reach equilibrium without having thermal (temperature) equilibrium. It just doesn't happen that way.

 

[JohnnyGui]

Very far-fetched. What you are proposing is that since $\epsilon=\epsilon(\lambda,T)$ that the two $\epsilon(\lambda)$ are not necessarily the same for the left integral and the right integral (for $T=T_{walls}$ on the left and $T=T_{ambient}$ on the right), that in an extreme case of emissivity varying with the temperature, that it might be possible. Mathematically, perhaps, yes, but in a real physical system, it's not going to do that. This would basically be a system for which you reach equilibrium without having thermal (temperature) equilibrium. It just doesn't happen that way.

Got it. So in case of an enclosed cavity in which $E = \int M(\lambda,T) \cdot \epsilon(\lambda) \cdot d\lambda$, an equilibrium can only occur when $T = T_{walls}$.
I noticed two other scenarios and I was wondering if these statements are correct about them.

1. If the $E$ on an object decreases with distance (unlike in the case of an enclosed cavity), then the object would have a lower equilibrium temperature than the radiating blackbody source. However, when 2 objects with different $\epsilon(\lambda)$ (i.e. black and white) are placed and both receiving the same $E$, then they will both have the same equilibrium temperature as each other, but different from the blackbody source.

2. If the source is not a black body and emitting only certain wavelengths, then an object receiving radiation from it can have a different equilibrium temperature than the source. When 2 objects with different $\epsilon(\lambda)$'s are exposed to that non-black body radiation (i.e. black and white), they can both differ in equilibrium temperature from each other.

 

[Charles Link]

(1) is incorrect. They absorb $\int \epsilon(\lambda) E(\lambda) \, d \lambda$, which in general will be different for both. You could expect the black object to normally absorb more visible, and if they only radiate in the IR, (if they aren't very warm), the black object will normally have the higher temperature, especially if their emissivities in the IR are similar.

 

[JohnnyGui]

(1) is incorrect. They absorb $\int \epsilon(\lambda) E(\lambda) \, d \lambda$, which in general will be different for both. You could expect the black object to normally absorb more visible, and if they only radiate in the IR, (if they aren't very warm), the black object will normally have the higher temperature, especially if their emissivities in the IR are similar.

I just found out that it was indeed incorrect and was about to change it XD.

1) But in case of an enclosed cavity, according to the equation, a white and black object would both have the same equilibrium temperature, as the walls, right? Apologies if I still got this wrong.
2) If statement 2 is correct, and a black object gets warmer in the sun than a white one, doesn't that mean that the sun is not a black body? Because it is considered as one.

 

[Charles Link]

I just found out that it was indeed incorrect and was about to change it XD.

1) But in case of an enclosed cavity, according to the equation, a white and black object would both have the same equilibrium temperature, as the walls, right? Apologies if I still got this wrong.
2) If statement 2 is correct, and a black object gets warmer in the sun than a white one, doesn't that mean that the sun is not a black body? Because it is considered as one.

The answer to (2) is that the sun is an isolated source that disrupts whatever equilibrium would be present without it. It has nearly a blackbody output of $T=6000$K, but there is no enclosure with all of the walls also at $T=6000$ K radiating like the sun. There is a big difference between an isolated source radiating like a blackbody and being inside a blackbody enclosure at thermal equilibrium.

 

[JohnnyGui]

The answer to (2) is that the sun is an isolated source that disrupts whatever equilibrium would be present without it. It has nearly a blackbody output of $T=6000$K, but there is no enclosure with all of the walls also at $T=6000$ K radiating like the sun. There is a big difference between an isolated source radiating like a blackbody and being inside a blackbody enclosure at thermal equilibrium.

Ah, this is basically what you corrected me for in your post #254. Two objects with different $\epsilon(\lambda)$ will have different equilibrium temperatures from a black body if $E$ differs with distance. They will have the same equilibrium temperature if they're inside a blackbody enclosure.

 

[JohnnyGui]

@Charles Link :The link you gave me for drawing the Planck function is great. I was playing a bit with it and I noticed something.

Consider the y-axis as $E$ instead of $L$. I then considered a blackbody source at 600K and a receiver at a certain distance from that source, such that the $E$ that it receives is half of what it would receive if it's at a distance of 1m away. If I consider that the receiver has an emissivity $\epsilon(\lambda)$ of $1$ for wavelengths > 15 µm, and $0$ for < 15µm, then to know the equilibrium temperature that the receiver would have, one could seek a blackbody curve at a certain temperature that overlaps the $600K \cdot 0.5$ curve at > 15µm and up. Like this:

The yellow line is the $E$ at 1m away from the blackbody source of 600K. The red line is what the receiver receives (half of $E$) because of the distance. The blue line is a blackbody curve at 430K. As shown, the blue line overlaps the received $E$ (red line) starting from 15 µm (green vertical line) and up. If I'm not wrong, this means that $A \cdot 430^4 \cdot \sigma \cdot \epsilon(\lambda)$ would give the same received energy as what the receiver would calculate from the red line, since any energy below 15 µm does not matter for the receiver.

Am I making sense?

 

[Charles Link]

You need to match $\int \epsilon (\lambda) E_s(\lambda) \, d \lambda$ with $\int M(\lambda) \, d \lambda=\int \epsilon(\lambda) M_{bb} (\lambda, T)$ to get equilibrium. Those two curves from the graph may look like the numbers are all equal for $\lambda>15 \, \mu m$, but that is not the case. Anyway, glad you are finding the website in the "link" of interest. :) (The graph really needs to be expanded for $\lambda >15 \, \mu m$ to get a closer look at it, but those two curves will not match for the whole interval $\lambda > 15 \, \mu m$ ).

 

[JohnnyGui]

You need to match $\int \epsilon (\lambda) E_s(\lambda) \, d \lambda$ with $\int M(\lambda) \, d \lambda=\int \epsilon(\lambda) M_{bb} (\lambda, T)$ to get equilibrium. Those two curves from the graph may look like the numbers are all equal for $\lambda>15 \, \mu m$, but that is not the case. Anyway, glad you are finding the website in the "link" of interest. :) (The graph really needs to be expanded for $\lambda >15 \, \mu m$ to get a closer look at it, but those two curves will not match for the whole interval $\lambda > 15 \, \mu m$ ).

Ah, I should've indeed matched integrals since I'm using $\epsilon(\lambda)$. So the $\int \epsilon (\lambda) E_s(\lambda) \, d \lambda$ should be performed on the red line and $\int \epsilon(\lambda) M_{bb} (\lambda, T)$ is done on the blue line, right?

 

[JohnnyGui]

@Charles Link

Sorry for bringing this up again (probably having a blackout again), but I noticed something regarding 2 objects with different $\epsilon(\lambda)$ getting radiation from a far away black body source (thus not inclosed in a black body).

As discussed, for an enclosed cavity, $P_{absorbed} = P_{radiated}$ and is equal to
$$A \cdot \int M_{bb}(\lambda, T_{walls}) d\lambda \cdot \epsilon(\lambda) = A \cdot \int M_{bb}(\lambda, T_{object}) d\lambda \cdot \epsilon(\lambda)$$
Since the $\epsilon(\lambda)$ is the same on both sides, any object with any $\epsilon(\lambda)$ will have the same equilibrium temperature (black or white)

However, when it comes to 2 objects exposed to a $P_{absorbed}$ dependent on the distance from a spherical blackbody source, then $P_{absorbed} = P_{radiated}$ would be equal to:
$$\frac{A}{4\pi R^2} \cdot \int M_{bb}(\lambda, T_{bb}) d\lambda \cdot \epsilon(\lambda) = A \int M_{bb}(\lambda, T_{object}) d\lambda \cdot \epsilon(\lambda)$$
Where $R$ is the distance to the centre of the blackbody. This equation merely has the $\frac{1}{4\pi R^2}$ factor on the left side but still contain the same $\epsilon(\lambda)$ on both sides as well. I'm aware that objects with different $\epsilon(\lambda)$'s have different equilibrium temperatures when receiving power from a BB that is dependent on distance, but this equation implies that, since $\epsilon(\lambda)$ is on both sides as well, any object with any $\epsilon(\lambda)$ will have the same equilibrium temperature (but different from the BB source because of the $\frac{1}{4\pi R^2}$ factor).

How must this formula be corrected to show that objects with different $\epsilon(\lambda)$'s will have different equilibrium temperatures when exposed to a far away BB source? Or can this only be deduced from looking at the spectral curves and that the equation is not always reliable?

 

[Charles Link]

If emissivity $\epsilon(\lambda)=1$ or some other constant independent of wavelength, it becomes a very simple calculation using a couple $\sigma T^4$ 's, since $\int\limits_{0}^{+\infty} M_{bb} (\lambda,T) \, d \lambda=\sigma T^4$. When $\epsilon$ is a function of wavelength, it becomes more complicated, and there is no simple way to solve for $T$ without knowing the function $\epsilon(\lambda)$. $\\$ Note: In the integral on the left hand side of the equation, you also need to include a factor $A_s$ for the area of the source. And additional correction: $E=LA_s/R^2=MA_s/(\pi R^2)$ and not $4 \pi$.

 

[JohnnyGui]

If emissivity $\epsilon(\lambda)=1$ or some other constant independent of wavelength, it becomes a very simple calculation using a couple $\sigma T^4$ 's, since $\int\limits_{0}^{+\infty} M_{bb} (\lambda,T) \, d \lambda=\sigma T^4$. When $\epsilon$ is a function of wavelength, it becomes more complicated, and there is no simple way to solve for $T$ without knowing the function $\epsilon(\lambda)$. $\\$ Note: In the integral on the left hand side of the equation, you also need to include a factor $A_s$ for the area of the source. And additional correction: $E=LA_s/R^2=MA_s/(\pi R^2)$ and not $4 \pi$.

Of course. I skipped the fact that $\epsilon(\lambda)$ is part of the integration itself so that it's not possible to cancel them on both sides of the equation. So it's indeed necessary to either know the function of $\epsilon(\lambda)$ or look at the curves to calculate the equilibrium temperature.

I indeed forgot the $A_s$. One thing though, if the black body is a sphere (like the sun) shouldn't the radiating area $A_s$ be divided by 2 since the object is exposed to 1 half of the spherical black body?

 

[Charles Link]

Of course. I skipped the fact that $\epsilon(\lambda)$ is part of the integration itself so that it's not possible to cancel them on both sides of the equation. So it's indeed necessary to either know the function of $\epsilon(\lambda)$ or look at the curves to calculate the equilibrium temperature.

I indeed forgot the $A_s$. One thing though, if the black body is a sphere (like the sun) shouldn't the radiating area $A_s$ be divided by 2 since the object is exposed to 1 half of the spherical black body?

For the sun, $A_s$ is the projected area which is $A_s=\pi R^2$. It is not $2 \pi R^2$. The irradiance $E=L A_s/s^2=MA_s/(\pi s^2)$. This is a common mistake among novices, where they often want to use $2 \pi R^2$ for such a problem. $\\$ In an alternative method of solution, $P=M 4 \pi R^2=E (4 \pi s^2)$ so that $E=M R^2/s^2=L \pi R^2/s^2$.

 

[JohnnyGui]

For the sun, $A_s$ is the projected area which is $A_s=\pi R^2$. It is not $2 \pi R^2$. The irradiance $E=L A_s/s^2=MA_s/(\pi s^2)$. This is a common mistake among novices, where they often want to use $2 \pi R^2$ for such a problem. $\\$ In an alternative method of solution, $P=M 4 \pi R^2=E (4 \pi s^2)$ so that $E=M R^2/s^2=L \pi R^2/s^2$.

Can the use of $A_s = \pi R^2$ instead of $2\pi R^2$ be explained as follows:

Towards the edges of a radiating hemispherical body of area $A_s = 2\pi R^2$, its $dA_s$'s will deviate more and more from the receiver's direction which results in a stronger energy decrease per angle increase from the receiver, compared to the energy decrease per angle from a flat circular radiating body. The net result is that the effective radiating area would therefore be equal to a flat radiating circle with the same radius $R$.

 

[Charles Link]

Can the use of $A_s = \pi R^2$ instead of $2\pi R^2$ be explained as follows:

Towards the edges of a radiating hemispherical body of area $A_s = 2\pi R^2$, its $dA_s$'s will deviate more and more from the receiver's direction which results in a stronger energy decrease per angle increase from the receiver, compared to the energy decrease per angle from a flat circular radiating body. The net result is that the effective radiating area would therefore be equal to a flat radiating circle with the same radius $R$.

This is actually a result of the brightness theorem and a Lambertian radiator. Even though parts of (most of) the round object are being viewed at an angle from the normal to the surface, the radiating surface appears to have equal brightness $L$ everywhere, and there is no perceived roundness of the surface as seen by the viewer. The round object looks like a flat circular disc.

 

[JohnnyGui]

This is actually a result of the brightness theorem and a Lambertian radiator. Even though parts of (most of) the round object are being viewed at an angle from the normal to the surface, the radiating surface appears to have equal brightness $L$ everywhere, and there is no perceived roundness of the surface as seen by the viewer. The round object looks like a flat circular disc.

There's something small bothering me regarding this, as when I read the same thing about a dome.
I understand that the radiance $L$ from each $dA$ would be the same, even when the $dA$ is near the edges, because the Lambertian cosine law gets compensated by the projected $dA \cdot cos(\theta)$ w.r.t. the receiver.
But there's also the distance factor that influences the received energy by the receiver, since distance of each $dA$ to the receiver changes as it is near the edges. For a dome or hemisphere, this distance factor changes more aggressively towards the edges than for a flat surface for example. The result is that the receiver would still measure a different energy coming from the sides because of that distance change from each $dA$. I can't see how the receiver would therefore still see the same "brightness" from those edges.

All what I stated above can of course be negated if the distance change from each $dA$ is neglectable (very far or small radiating object), but I'm not sure whether this is the exact cause why a receiver would see the same brightness from each $dA$ or if there's another reason that I'm missing.

 

[Charles Link]

The irradiance received from a source of brightness $L$ that spans a solid angle $\Omega$ is given by $E=L \, \Omega$. The solid angle is given by $\Omega=A/s^2$. The result is the distance really doesn't matter=the important factor is how much of the viewing field (in solid angle) is covered by the source. $\\$ Editing: Note: To be more precise $\Omega=\int \frac{dA}{s^2}$.

 

[JohnnyGui]

Note: To be more precise $\Omega=\int \frac{dA}{s^2}$.

So if I understand correctly, if the distance $s$ for each $dA$ changes significantly towards the edges so that for a $dA$ at the source's edge the irradiance is $L \cdot \frac{dA \cdot cos(\theta)^4}{s^2}= E$, the $cos(\theta)^4$ would still not matter for the integral so that;
$$\Omega=\int \frac{dA}{s^2} cos(\theta)^4 = \frac{dA}{s^2}$$
?

 

[Charles Link]

So if I understand correctly, if the distance $s$ for each $dA$ changes significantly towards the edges so that for a $dA$ at the source's edge the irradiance is $L \cdot \frac{dA \cdot cos(\theta)^4}{s^2}= E$, the $cos(\theta)^4$ would still not matter for the integral so that;
$$\Omega=\int \frac{dA}{s^2} cos(\theta)^4 = \frac{dA}{s^2}$$
?

For this $\Omega=\int \frac{dA}{s^2}$, $s$ is the total distance and doesn't pick up a factor of $s=\frac{s_o}{cos(\theta)}$. Meanwhile, the surface element $dA$ in the integral is a projected surface area and is not the actual $dA$. (It already contains one factor of $cos(\theta)$). The final $cos(\theta)$ factor in the $cos^4(\theta)$ result is from the irradiance $E$ not being perpendicular to the area that it encounters at the receiver. This factor isn't required because the $E=L \, \Omega$ doesn't include this result. Thereby, the $cos^4(\theta)$ isn't not in the formula $\Omega=\int \frac{dA}{s^2}$, but the one part that perhaps should be corrected is the formula should read $\Omega=\int \frac{ dA_{projected}}{s^2}$.