Emission spectra of different materials

[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 $\Omega=\int \frac{ dA_{projected}}{s^2}$.

 

[JohnnyGui]

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 $\Omega=\int \frac{ dA_{projected}}{s^2}$.

Ok. Two things if you don't mind.

1. The way I understood it, the $cos(\theta)^4$ factor contains a $cos(\theta)^2$ for the distance, a $cos(\theta)$ for the projected area $dA$ and a $cos(\theta)$ for the Lambertian cosine law to correct for the intensity ($I_0 \cdot cos(\theta)$), like you said in post #107. So if you say $dA$ is actually $dA_{projected}$ then one $cos(\theta)$ factor gets removed. Since $L \cdot dA_{projected} = I_0 \cdot cos(\theta)$, then another $cos(\theta)$ is not needed. This leaves a $cos(\theta)^2$ factor for the distance for which you say that the formula doesn't pick this up. But why doesn't it pick that up? Isn't the received $E$ defined by that distance?

2. If the distance and angle of each $dA$ changes as it lies more towards the edges, doesn't $dA_{projected}$ in the formula $\Omega=\int \frac{ dA_{projected}}{s^2}$ have a different value for each $dA$ so that $\Omega$ is not constant for each $d_A$?

For the above 2 points, I don't get how $E$ from each $dA$ would stay the same if distance differ significantly with each $dA$. Unless, distance from each $dA$ doesn't differ much.

 

[Charles Link]

The irradiance $E$ from each $d \Omega$ stays the same. $dE=L \, d\Omega$. Meanwhile $s$ is the distance, which is changing in the integral as a function of the location of $dA$. $\\$ Another item you mentioned was the Lambertian source factor. The intensity falls off as $cos(\theta)$ due to the intensity in a given direction $I(\theta)=L A_{projected}$ where $A_{projected}=A \cos(\theta)$. The brightness $L$ of an ideal (Lambertian) source is constant, independent of angle. $\\$ One concept that might be useful here is that you really can not determine with your eyes the distance that an object is at that has uniform brightness, regardless of its shape. The thing that allows you to focus on the object is contrasts or changes in brightness, so that you can e.g. see boundaries where the brightness changes and can focus on the boundary to determine the distance.

 

[JohnnyGui]

So sorry if I'm totally missing the point here that you're trying to make.

I understand that $L$ doesn't change with angle and distance. But if $dE = L d\Omega$ where $d\Omega = \frac{dA_{projected}}{s^2}$, and $dA_{projected}$ is changing with angle (getting smaller towards the edges) and $s^2$ is changing as a function of $dA$ (getting larger towards the edges) as you said, how can the received $dE$ from each $dA$ be the same?

 

[Charles Link]

So sorry if I'm totally missing the point here that you're trying to make.

I understand that $L$ doesn't change with angle and distance is th. But if $dE = L d\Omega$ where $d\Omega = \frac{dA_{projected}}{s^2}$, and $dA_{projected}$ is changing with angle (getting smaller towards the edges) and $s^2$ is changing as a function of $dA$ (getting larger towards the edges) as you said, how can the received $dE$ from each $dA$ be the same?

It isn't. But the $dE$ received from each solid angle $d \Omega$ covered by the source is the same. You look at the source that is spread out in front of you, and you just need to know the angular spread and the brightness to know how much energy reaches you. Take e.g. the moon in the night sky. For all practical purposes in viewing it, it could be a disc 1 ft. across and 100 ft. away. Instead, it happens to be 2000 miles across and 200,000 (approximately) miles away. The angle in radians is subtends $\Delta \theta=2000/200,000=1/100=.01$. A similar calculation can be done with the solid angle.

 

[JohnnyGui]

It isn't. But the $dE$ received from each solid angle $d \Omega$ covered by the source is the same. You look at the source that is spread out in front of you, and you just need to know the angular spread and the brightness to know how much energy reaches you. Take e.g. the moon in the night sky. For all practical purposes in viewing it, it could be a disc 1 ft. across and 100 ft. away. Instead, it happens to be 2000 miles across and 200,000 (approximately) miles away. The angle in radians is subtends $\Delta \theta=2000/200,000=1/100=.01$. A similar calculation can be done with the solid angle.

To let the moon look like a flat disc, a receiver must receive the same energy from every $dA$ from that source, right? Regardless of where $dA$ lies on the source. This means that according to this picture:

Every $d\Omega$ from every $dA$ in that picture (only drawn 3 here) therefore must contain the same energy, to make the moon look like a flat disc (the source's line should actually be bent since the moon is round). Is this what you mean?

 

[Charles Link]

The $d \Omega$ in the formula $dE=L \, d \Omega$ is measured from the receiver. If the solid angle increments $d \Omega$ are the same size and intercept a region of the same brightness, they will each have equal $dE=L \ d \Omega$. For uniform $L$ for a source, $E_{received}=L \, \Omega$, where $\Omega$ is the solid angle subtended by the source as measured from the receiver. $\Omega=\int \frac{d A_{projected}}{s^2}$. $\\$ For the moon, the picture I have in mind is that you start from your eye and draw an angle $\theta=.01$ radians. Spanning that .01 radians is the moon. Whether it is a mile away or 200,000 miles away really makes no difference in what you observe. (If it is a mile away, you have it be diameter .01 miles.)

 

[JohnnyGui]

The $d \Omega$ in the formula $dE=L \, d \Omega$ is measured from the receiver. If the solid angle increments $d \Omega$ are the same size and intercept a region of the same brightness, they will each have equal $dE=L \ d \Omega$. For uniform $L$ for a source, $E_{received}=L \, \Omega$, where $\Omega$ is the solid angle subtended by the source as measured from the receiver. $\Omega=\int \frac{d A_{projected}}{s^2}$. $\\$ For the moon, the picture I have in mind is that you start from your eye and draw an angle $\theta=.01$ radians. Spanning that .01 radians is the moon. Whether it is a mile away or 200,000 miles away really makes no difference in what you observe. (If it is a mile away, you have it be diameter .01 miles.)

Ah, so it's basically the received energy from an angular surface measured by the receiver, not the energy from a fixed $dA$ of the moon surface? Like this:

Such that if a source is x times further away, you'd be receiving energy from a x² times larger source surface, but since the distance is x times further, this gets compensated by a x² decrease in energy, so that the net received energy doesn't change?

 

[JohnnyGui]

@Charles Link

If my above post is (one of the) the reason(s) for the moon having the same
brightness at the edges as the center, can I say the following:

Factors that would the receiver receive less radiation from the edges of the moon:
1. The edges are further away than the center ($cos(\theta)^2$ fall-off)
2. Edges receive less light from the sun because the normals of the edge $dA's$ are deviated from the sun ($cos(\theta)$ fall off)
3. The normals of the edge $dA's$ on the moon are deviated from the receiver, so that according to Lambertian cosine law, the receiver receives less energy with a factor of $(cos(\theta))$

Factors that compensate for the above 3 factors so that the edges look as bright as the center:
1. Because the edges are further away, the receiver would look at a larger area portion towards the edges for the same solid angle ($cos(\theta)^2$)
2. Factor 1 is the case for if the source is flat. Another factor that would make the receiver look at a larger area portion for the same solid angle is the curveness of the moon since it's a sphere, like this:

3. Because the normals of the edge $dA's$ are deviated from the receiver, the projected area look smaller according to the receiver which make the receiver see the same energy coming from a smaller area. ($cos(\theta)$)

Are these factors valid?

 

[Charles Link]

The moon isn't exactly perfectly uniform brightness across it as it is illuminated by the sun, but as I understand it, it is more nearly uniform in brightness than was originally expected. For a spherical body receiving parallel rays from a source, the edges would be expected to be much dimmer than near the center because it doesn't receive the same irradiance per unit area of surface. (If the moon were a blackbody radiator with uniform brightness, the radiant emittance $M$ would be uniform. Since it is a reflective type radiator, the radiant emittance (per unit area) $M$ will be proportional to the irradiance per unit area $E$ that it receives from the sun. This will clearly be reduced going out from the center to the edges. If I understand it correctly, this allowed scientists to forecast that the moon's surface must be somewhat jagged and not completely smooth.) $\\$ In any case, for a spherical radiator with uniform brightness $L$, the irradiance $E$ received by a receiver is $E=L \, \Omega$ , where $\Omega$ is the solid angle subtended by the spherical radiator, (e.g. the moon), as measured from the receiver. In this case $\Omega=\frac{\pi R_{moon}^2}{s^2}$ (almost precisely). $\\$ Additional comments: The moon is 93 million miles from the sun, so an extra distance of travel of 1000 miles to the edges is not going to make any difference. The thing that should make a difference though is that the angle of incidence is almost a glancing angle at the edges. This means that the irradiance per unit area is very low unless there is a jagged surface like small mountains present in which case the light even at the edges might strike the surface at nearly normal incidence.

 

[JohnnyGui]

The moon isn't exactly perfectly uniform brightness across it as it is illuminated by the sun, but as I understand it, it is more nearly uniform in brightness than was originally expected. For a spherical body receiving parallel rays from a source, the edges would be expected to be much dimmer than near the center because it doesn't receive the same irradiance per unit area of surface. (If the moon were a blackbody radiator with uniform brightness, the radiant emittance $M$ would be uniform. Since it is a reflective type radiator, the radiant emittance (per unit area) $M$ will be proportional to the irradiance per unit area $E$ that it receives from the sun. This will clearly be reduced going out from the center to the edges. If I understand it correctly, this allowed scientists to forecast that the moon's surface must be somewhat jagged and not completely smooth.) $\\$ In any case, for a spherical radiator with uniform brightness $L$, the irradiance $E$ received by a receiver is $E=L \, \Omega$ , where $\Omega$ is the solid angle subtended by the spherical radiator, (e.g. the moon), as measured from the receiver. In this case $\Omega=\frac{\pi R_{moon}^2}{s^2}$ (almost precisely). $\\$ Additional comments: The moon is 93 million miles from the sun, so an extra distance of travel of 1000 miles to the edges is not going to make any difference. The thing that should make a difference though is that the angle of incidence is almost a glancing angle at the edges. This means that the irradiance per unit area is very low unless there is a jagged surface like small mountains present in which case the light even at the edges might strike the surface at nearly normal incidence.

I see, so basically the factors that I mentioned that make the edges look brighter aren't really enough to compensate for the factors that make the edges dimmer. It needs an extra factor; the jagged surface you mentioned.

I'm trying to visualize how, in case of a smooth Lambertian spherical black body emitter (not a reflector like the moon), the edges would have the same brightness as the center. I'm not sure how to draw and explain this with the solid angle coming from the receiver instead of one of the edge $dA's$.

All I can say is the following about a smooth Lambertian spherical emitter:
1. For a spherical black body emitter, the normals of the $dA's$ at the edges are deviated from the receiver. But the $cos(\theta)$ fall-off in intensity (Lambert's law) is compensated by the smaller projected area of those $dA's$ w.r.t. the receiver.
2. You mentioned that an extra 1000 miles of distance to the edges in case of a very far black body emitter won't make a difference. If the emitter is near, the distance difference will matter. In that case, with the same solid angle, you'd be looking at a larger surface regarding the edge that is further away. But this is in the case of a flat emitting surface. Since it's a sphere, per the same solid angle, you'd be even looking at a larger surface area at the edge according to the posted picture:

But now, there are 3 factors that cause a brighter edge and just one factor that compensates for one of them. The net result would be that the edges should be brighter than the center.

If this is wrong (probably is), perhaps it's better to show me a picture how the factors that reduce brightness from the spherical edges are exactly compensated by the ones that increases their brightness.