I Emission spectra of different materials

[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.

 

[Charles Link]

For the case of uniform brightness $L$ radiating from any shape, the irradiance that reaches the receiver is simply $E=L \, \Omega$, where $\Omega$ is the solid angle measured from the receiver. There is basically a factor of $cos(\theta)$ in intensity fall-off per unit area with tilt angle $\theta$, but this is exactly compensated for because the radiating area in an incremental solid angle $d \Omega$ increases by a factor of $\frac{1}{cos(\theta) }$. The distance factor does not affect things as your diagram of post #277 explains quite clearly. $\\$ And I was glad to see you were able to follow the explanation that the moon needs to have some kind of jagged surface in order to have the nearly uniform brightness that it has.

 

[JohnnyGui]

And I was glad to see you were able to follow the explanation that the moon needs to have some kind of jagged surface in order to have the nearly uniform brightness that it has.

I was reading about this and it indeed makes sense. There's this effect also called Seeliger effect that explains why a full moon has much more than twice the brightness compared to a half moon, even though the illuminated surface ratio is 2:1. Part of it is explained by the fact that because the moon surface is jagged, the sun illuminating the moon from the side w.r.t. us would cast shadows along the illuminated surface (because of the mountains, etc.) while if the sun is directly behind us illuminating the moon in the same direction as our viewing direction, there are no shadows (or very short ones) from the mountains, which makes the moon look brighter.

Another factor that causes a higher brightness is called "coherent backscattering". This is when two (or more) light rays travel different distances beneath the moon's surface while they're being scattered, but when they come out in your direction (backscattering) the total scatter distance difference traveled among those light rays is a whole (or $n$) wavelength and therefore giving each other a constructive interference right back at your eyes.

For the case of uniform brightness LL L radiating from any shape, the irradiance that reaches the receiver is simply E=LΩE=LΩ E=L \, \Omega , where ΩΩ \Omega is the solid angle measured from the receiver. There is basically a factor of cos(θ)cos(θ) cos(\theta) in intensity fall-off per unit area with tilt angle θθ \theta , but this is exactly compensated for because the radiating area in an incremental solid angle dΩdΩ d \Omega increases by a factor of 1cos(θ)1cos(θ) \frac{1}{cos(\theta) } . The distance factor does not affect things as your diagram of post #277 explains quite clearly.

So, the distance doesn't even matter if the spherical emitter is small and near like this?

So this scenario still won't give dimmer edges?

 

[Charles Link]

So, the distance doesn't even matter if the spherical emitter is small and near like this?

So this scenario still won't give dimmer edges?

The distance isn't the important factor here. If the red surface (which is assumed to be a smooth diffuse scattering surface) is illuminated by parallel rays from the left, then it will show some dimming from the center to the edges. $\\$ And your additional reading of the moon's appearance was quite interesting.

 

[JohnnyGui]

The distance isn't the important factor here. If the red surface (which is assumed to be a smooth diffuse scattering surface) is illuminated by parallel rays from the left, then it will show some dimming from the center to the edges.

Apologies, I was talking about a diffuse black body emitter, not an illuminated sphere. Will in that case the edges in the picture still look equally bright as the center?

And your additional reading of the moon's appearance was quite interesting.

I'm glad I finally gave something in return after all your efforts XD.

 

[Charles Link]

Apologies, I was talking about a diffuse black body emitter, not an illuminated sphere. Will in that case the edges in the picture still look equally bright as the center?

The answer is yes, it will look equally bright everywhere. You might find it of interest that you also wouldn't be able to tell whether it is flat in shape or concave inward or outward.

 

[JohnnyGui]

The answer is yes, it will look equally bright everywhere. You might find it of interest that you also wouldn't be able to tell whether it is flat in shape or concave inward or outward.

Yes, this is what I indeed expected. So if distance doesn't matter regarding the received energy in a fixed solid angle, can I also say the following:

If the two $d\Omega$'s in the picture are equal in size, the power received from each solid angle would be the same?

 

[sophiecentaur]

The answer is yes, it will look equally bright everywhere. You might find it of interest that you also wouldn't be able to tell whether it is flat in shape or concave inward or outward.

This would need the emission to be taking place at the surface. Observing the Sun, you see it is a bit dark at the edges so the flux per solid angle is not actually the same in that (practical) case. (Danger - do no observe it directly!)

 

[Charles Link]

Yes, this is what I indeed expected. So if distance doesn't matter regarding the received energy in a fixed solid angle, can I also say the following:
View attachment 207264
If the two $d\Omega$'s in the picture are equal in size, the power received from each solid angle would be the same?

Yes, you have it correct. One thing to note that the irradiance $E=L \, d \Omega$ is the same for both. The power received per unit area by a flat detector could differ by a $cos(\theta)$ factor because the irradiance $E$ is measured in a plane normal to the direction of the small solid angle increment $d \Omega$.

 

[JohnnyGui]

Yes, you have it correct. One thing to note that the irradiance $E=L \, d \Omega$ is the same for both. The power received per unit area by a flat detector could differ by a $cos(\theta)$ factor because the irradiance $E$ is measured in a plane normal to the direction of the small solid angle increment $d \Omega$.

If you say a flat detector could receive less $E$ from the left sided $d\Omega$ (in the picture) by a factor of $cos(\theta)$ since $E$ is measured in the normal plane of $d\Omega$, doesn't this imply that if I don't look at the edges but straight normal to the flat source, I'd notice dark edges from the sides of my viewing angle because my (more or less) flat retinas are not in the normal plane of the left sided $d\Omega$. And when I move my eyes to the edges so that my retinas are normal to the left sided $d\Omega$, I'd be receiving the same $E$ from that $d\Omega$ as the normal $d\Omega$ that I was looking at previously?

This would need the emission to be taking place at the surface. Observing the Sun, you see it is a bit dark at the edges so the flux per solid angle is not actually the same in that (practical) case

I was thinking the same thing. Do you mean that because when looking at the center of the sun, there are way more emitting particles beneath the surface as compared to the edges?

(Danger - do no observe it directly!)

Too late, I have no idea what I'm typing right now.

 

[Charles Link]

If you say a flat detector could receive less $E$ from the left sided $d\Omega$ (in the picture) by a factor of $cos(\theta)$ since $E$ is measured in the normal plane of $d\Omega$, doesn't this imply that if I don't look at the edges but straight normal to the flat source, I'd notice dark edges from the sides of my viewing angle because my (more or less) flat retinas are not in the normal plane of the left sided $d\Omega$. And when I move my eyes to the edges so that my retinas are normal to the left sided $d\Omega$, I'd be receiving the same $E$ from that $d\Omega$ as the normal $d\Omega$ that I was looking at previously?

Very good. You have it correct. The eye is slightly more complicated having a lens, and the precise details of the brightness of the image at points far off-axis is beyond the scope of what we need here. In any case, if you look in the direction that is off-axis, you see the same $E$ as you stated in the last sentence.

 

[sophiecentaur]

Very good. You have it correct. The eye is slightly more complicated having a lens, and the precise details of the brightness of the image at points far off-axis is beyond the scope of what we need here. In any case, if you look in the direction that is off-axis, you see the same $E$ as you stated in the last sentence.

The varying luminosity can be seen on solar photographs too -wherever the Sun is placed on the sensor. So it's not a visual anomaly

Too late, I have no idea what I'm typing right now.

Haha.
It's worth treating fairly seriously though. Too much time looking for solar features without the right precautions actually makes people blind. Before the sextant was invented, ship's navigators always ended up blind in one eye, apparently from using the Cross Staff.

 

[JohnnyGui]

The varying luminosity can be seen on solar photographs too -wherever the Sun is placed on the sensor. So it's not a visual anomaly

You mean since the edges are "thin" compared to the center, w.r.t. your viewing direction, and therefore you'd still see darker edges?

Yes, you have it correct. One thing to note that the irradiance E=LdΩE=LdΩ E=L \, d \Omega is the same for both. The power received per unit area by a flat detector could differ by a cos(θ)cos(θ) cos(\theta) factor because the irradiance EE E is measured in a plane normal to the direction of the small solid angle increment dΩdΩ d \Omega .

I think I've finally got how brightness is independent from distance! I managed this by reasoning $L$ itself and using the following scenario (please bear with me):

The 2 factors that would make brightness differ with the angle are the observed source surface $dA$ and the eye's solid angle. Therefore, to objectify how one sees the brightness of a radiating surface $dA$ so that it can be compared to what others see, it is necessary to calculate the received energy per fixed solid angle (1 steradian) per fixed source area( 1 m²).

Therefore, if the observing eye has a surface of $dA_o$, then receiver $A$ would measure a total received power of $I_0\cdot \frac{dA_o}{R^2} = P_A$. Receiver $A$ would thus say that if $dA$ is emitting at distance $R$ an energy of $I_0\cdot \frac{dA_o}{R^2} = P_A$ on a surface of $dA_o$, that the energy per steradian per $1$ m² radiating source would be
$$I_0 \cdot \frac{dA_o}{R^2} \cdot \frac{R^2}{dA_o} \cdot \frac{1}{dA} = \frac{I_0}{dA} = L$$
Receiver $B$ would receive a total power of $I_0 \cdot \frac{dA_o}{R^2} \cdot cos(\theta)^3 = P_B$ because of Lambert's cosine law and the increase in distance. B would reason that per steradian, this power would be $I_0 \cdot \frac{dA_o}{R^2} \cdot cos(\theta)^3 \cdot \frac{R^2}{dA_o \cdot cos(\theta)^2} = P / Sr$ (intensity). B would furthermore see that this energy is coming from a surface of $dA \cdot cos(\theta)$. Per $1$ m² this would give :
$$I_0 \cdot \frac{dA_o}{R^2} \cdot cos(\theta)^3 \cdot \frac{R^2}{dA_o \cdot cos(\theta)^2} \cdot \frac{1}{dA \cdot cos(\theta)} = \frac{I_0}{dA} = L$$
Therefore, the brightness is the same for both A and B. Apologies for taking so long to finally understand this. I tend to take things step by step to guarantee I fully understand something.

 

[JohnnyGui]

@Charles Link : Is there actually another intrinsic physical explanation, other than "looking at a smaller projected source area and therefore the same radiance", that explains why we would see the same brightness? In other words;

Why would our retinas register the same brightness even though the total received energy as well as the photon density that bombards the retinas are both less when they're coming from a $dA$ at an angle? (regardless of the observed source area being smaller by a factor of $cos(\theta)$). I mean, isn't photon density the intrinsic factor that determines how the retina measures brightness instead of observing a smaller projected source area?

 

[Charles Link]

@Charles Link : Is there actually another intrinsic physical explanation, other than "looking at a smaller projected source area and therefore the same radiance", that explains why we would see the same brightness? In other words;

Why would our retinas register the same brightness even though the total received energy as well as the photon density that bombards the retinas are both less when they're coming from a $dA$ at an angle? (regardless of the observed source area being smaller by a factor of $cos(\theta)$). I mean, isn't photon density the intrinsic factor that determines how the retina measures brightness instead of observing a smaller projected source area?

A very good question...You need to keep the solid angle of all sources the same for this kind of comparison. The area on the retina $dA_r$ that gets imaged is $dA_r=f^2 \, d \Omega$, where $d \Omega$ is the solid angle subtended by the source as measured from the eye, and $f$ is the focal length of the lens of the eye. $\\$ (In the focal plane, for small angles $d \theta$ , we can write $dx=f \, d \theta$ where $dx$ is the image size on the retina (which is in the focal plane) and $d \theta$ is the range of angles covered by the incoming rays. From the center of the object, essentially the rays are parallel and on-axis and image at the center of the focal plane, and if the rays come in from the edge of the object all parallel but at some angle $d \theta$ , they will come to focus in the focal plane (retina) at distance $dx =f \, d \theta$ from the center. Thereby $dx=f \, d \theta$ and $dA_r=f^2 \, d \Omega$). $\\$ If you look at an area $dA_r$ of the retina, for sources of equal brightness, it receives the same number of photons per second. $\\$ Additional item, in case you didn't know it: Whether it is a camera lens or the lens of your eye, it creates a focused image of the scene in the image plane, which for a camera is the focal plane array of pixels, and for your eye, it is the retina. The focusing is such that $\frac{1}{f}=\frac{1}{b}+\frac{1}{m}$ where $b$ is the object distance and $m$ is the image distance. For faraway objects=large $b$, $m \approx f$, and the focused image occurs in the focal plane. For objects at closer distances, a refocusing is necessary. $\\$ Additional item: Oftentimes, there is too much stray light and you don't see the image, but try seeing the image of a light bulb with a magnifying glass: Place the (lit up) light bulb about 4 ft. from the magnifying glass, and look for the focused image using a sheet of paper as the screen. Your focused image should appear on the screen about 4" from the magnifying glass for a typical magnifying glass with $f=4"$.

 

[JohnnyGui]

...You need to keep the solid angle of all sources the same for this kind of comparison.

Thanks for the detailed explanation! One question regarding keeping the solid angle the same. Is it correct to say the following:

If one has a fixed viewing angle, for example the human eye (150 degrees). And that eye first looks perpendicularly at a radiating surface, and after that looks at it from an angle, doesn't this mean that the solid angle $\frac{dA_{projected}}{s^2}$ will always be different, regardless of the viewing angle being fixed? So in other words, one can not keep the solid angle fixed unless you change the viewing angle?

 

[Charles Link]

I don't understand your question. Basically, a given amount of solid viewing angle $d \Omega$ corresponds to a given amount of area on the retina $dA_r$ with $d \Omega= \frac{dA_r}{f^2}$. If you mean that it will use a different portion of the area on the retina to see it if you view it at an angle, the answer is yes, and the response of those receptors could be different. Basically the retina is like the array of pixels of a digital camera that are in the focal plane of the lens. I think different parts of the retina even have fewer pixels per unit area, but for the eye, you aren't generating quantitative brightness information.

 

[JohnnyGui]

I don't understand your question. Basically, a given amount of solid viewing angle $d \Omega$ corresponds to a given amount of area on the retina $dA_r$ with $d \Omega= \frac{dA_r}{f^2}$. If you mean that it will use a different portion of the area on the retina to see it if you view it at an angle, the answer is yes, and the response of those receptors could be different. Basically the retina is like the array of pixels of a digital camera that are in the focal plane of the lens. I think different parts of the retina even have fewer pixels per unit area, but for the eye, you aren't generating quantitative brightness information.

Apologies, I meant it more generally. So in the following picture:

If viewing angle $\alpha_A$ = $\alpha_B$ (same angle, in degrees), then this means that the projected area (dashed line) that $B$ sees is the same as the area that A sees. I'm taught that $B$ would see this projected area at a larger distance than $A$ would see his surface by $\frac{s_A}{cos(\theta)} = s_B$. However, from the drawing and from geometry, the distance from $B$ to the dashed line looks to be the same as the distance from $A$ to his own surface, $s_A = s_B$.

If the drawing is incorrect and $B$ is seeing the projected area at a distance of $\frac{s_A}{cos(\theta)} = s_B$, then it should be drawn like this instead:

If this is correct, then that means that the solid angle of $A$ subtended by his own surface is different from the solid angle of B subtended by the projected surface, regardless of the viewing angle $\alpha$ being the same for both.

Is this right?

 

[Charles Link]

B in some ways thinks he is seeing the projected area, but he is actually seeing the surface which is farther away and with an increased area. He really can't tell the difference between the two if there is nothing of contrast to focus on. And remember, solid angle $\Omega=\int \frac{dA_{projected}}{s^2}$. $s$ is different for the different parts of $dA_{prrojected}$. You can't compute the solid angle properly in your diagram if $s$ changes throughout the diagram. Also, your diagram is unclear on what you want to call $s$ and $A_{projected}$. In the formula that I just wrote with the integral, the $dA_{projected}$ and $s$ are measured right to that portion of the radiating surface. You can of course even measure $\Omega$ from your eye or from any surface that intercepts that $\Omega$, but you need to be precise, or you will get inconsistencies in the results. The formula are all very consistent.

 

[JohnnyGui]

B in some ways thinks he is seeing the projected area, but he is actually seeing the surface which is farther away and with an increased area. He really can't tell the difference between the two if there is nothing of contrast to focus on.

Ah ok, but if one wants to calculate B's radiance using his solid angle $d\Omega_B$ he should use $$d\Omega_B = \frac{A_{projected}}{(\frac{s_A}{cos(\theta)})^2}$$
Where $s_A$ is the distance of A to his own surface, right?

 

[Charles Link]

Ah ok, but if one wants to calculate B's radiance using his solid angle $d\Omega_B$ he should use $$d\Omega_B = \frac{A_{projected}}{(\frac{s_A}{cos(\theta)})^2}$$
Where $s_A$ is the distance of A to his own surface, right?

The angle $\theta$ here doesn't stay constant throughout the calculation. $s_A$ is the perpendicular distance to the second surface, but since you have an extended angle of viewing, $\theta$ will not be constant. Also, $A_{projected}$, if you are using this formula, which has an increased distance $s=s_A/cos(\theta)$, needs to come from the location of that point on the surface. (The $A_{projected}$ will be larger if you measure it at a location which is farther away, as is necessary to keep the ratio $\Omega=\frac{A_{projected}}{s^2}$ constant for constant $\Omega$).