I Emission spectra of different materials

[Charles Link]

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

The brightness $L$ is determined experimentally over a small solid angle $d \Omega$ simply by blocking out the other solid angles so that the receiver only looks at a very small solid angle.

 

[JohnnyGui]

The brightness $L$ is determined experimentally over a small solid angle $d \Omega$ simply by blocking out the other solid angles so that the receiver only looks at a very small solid angle.

And since $L$ is the same for each other small solid angle subtended by other $dA$'s, the receiver would measure the same $L$ but coming from the larger radiating surface area that he truly observes?

 

[JohnnyGui]

@Charles Link

The brightness LL L is determined experimentally over a small solid angle dΩdΩ d \Omega simply by blocking out the other solid angles so that the receiver only looks at a very small solid angle.

And since LLL is the same for each other small solid angle subtended by other dAdAdA's, the receiver would measure the same LLL but coming from the larger radiating surface area that he truly observes?

I think I missed something when I said this, because if a solid angle is very large (the radiating surface area covered is large), then there are $dA$ segments of that covered radiating surface that are not perpendicular to the receiving surface like you mentioned in your post #342. This means that every $dA$ from that large radiating surface would have a different $L$ right? Since there's an extra $cos(\theta)$ factor added, which therefore would need integration. Or does the $d \Omega$ get smaller accordingly when pointed at $dA$'s from the sides such that the $L$ from each $dA$ still stays the same (even if they're not perpendicular to the receiving area) and only the Irradiance gets affected?

 

[Charles Link]

@Charles LinkI think I missed something when I said this, because if a solid angle is very large (the radiating surface area covered is large), then there are $dA$ segments of that covered radiating surface that are not perpendicular to the receiving surface like you mentioned in your post #342. This means that every $dA$ from that large radiating surface would have a different $L$ right? Since there's an extra $cos(\theta)$ factor added, which therefore would need integration. Or does the $d \Omega$ get smaller accordingly when pointed at a different $dA$ from the sides such that the $L$ from each $dA$ still stays the same, even if they're not perpendicular to the receiving area?

For an ideal source, such as an ideal blackbody radiator, $L$ is constant. $L$ is clearly not constant in the case of a black and white checkerboard where the white squares could all have $L=L_1$ that might be quite large, and the dark squares could have $L=0$ .(We're working with visible light here, and not infrared). In determining $L$, the location often needs to be specified, and a large area can not be used.

 

[JohnnyGui]

So based on your post #342:

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

When it's about a large radiating surface (large solid angle), only the Irradiance of each $dA$ gets affected and not the $L$ of each $dA$?

 

[Charles Link]

So based on your post #342:
When it's about a large radiating surface (large solid angle), only the Irradiance of each $dA$ gets affected and not the $L$ of each $dA$?

If the source is ideal, yes, the brightness $L$ will be constant, but the received irradiance is lower by a factor off $\cos(\theta)$ unless you orient the receiver so that the receiving plane is perpendicular to the line of sight to the source. If you do that, the irradiance is $E=L \, d \Omega$ where $d \Omega$ is measured from the receiver.

 

[JohnnyGui]

If the source is ideal, yes, the brightness $L$ will be constant, but the received irradiance is lower by a factor off $\cos(\theta)$ unless you orient the receiver so that the receiving plane is perpendicular to the line of sight to the source. If you do that, the irradiance is $E=L \, d \Omega$ where $d \Omega$ is measured from the receiver.

I'm sorry but how does each $dA$ of one large radiating surface have the same $L$ for an observer when I just concluded in post #346 that this scenario for a fixed $dA$:

...only gives the same $L$ for $B$ and $A$ if B is perpendicular to the radiation direction?
If only one observer is looking at a large radiating surface, then that one observer would have rays coming from $dA$'s of that large surface that are not perpendicular to that observer. Since, according to the picture above, the observer $B$ needs to look perpendicularly at the same $dA$ to have the same $L$ as $A$, this means that for the one observer looking at a large surface, each $dA$ of it would give him a different $L$ since he's not perpendicular to each one of them.

 

[Charles Link]

I'm sorry but how does each $dA$ of one large radiating surface have the same $L$ for an observer when I just concluded in post #346 that this scenario for a fixed $dA$:
View attachment 210372

...only gives the same $L$ for $B$ and $A$ if B is perpendicular to the radiation direction?
If only one observer is looking at a large radiating surface, then that one observer would have rays coming from $dA$'s of that large surface that are not perpendicular to that observer. Since, according to the picture above, the observer $B$ needs to look perpendicularly at the same $dA$ to have the same $L$ as $A$, this means that for the one observer looking at a large surface, each $dA$ of it would give him a different $L$ since he's not perpendicular to each one of them.

The brightness $L$ is independent of how you observe it. The formula $E=L \, d \Omega$ only holds if the receiver is aimed at the surface. If you keep the receiver fixed in position at angle $\theta$ but turn it horizontal with the same receiver area $dA_r$, you get a power incident on that surface $P=L \, d \Omega \, dA_r \, cos(\theta)$ and the irradiance onto the receiver surface is $E=L \, d \Omega \, cos(\theta)$.

 

[JohnnyGui]

The brightness $L$ is independent of how you observe it. The formula $E=L \, d \Omega$ only holds if the receiver is aimed at the surface. If you keep the receiver fixed in position at angle $\theta$ but turn it horizontal with the same receiver area $dA_r$, you get a power incident on that surface $P=L \, d \Omega \, dA_r \, cos(\theta)$ and the irradiance onto the receiver surface is $E=L \, d \Omega \, cos(\theta)$.

My bad, I think I got it. Let me write $P$ and $d \Omega$ all out for the following scenario:

Here, $dA_1 = dA_2$ so that the field of view must be $\alpha_1 > \alpha_2$. The receiver $dA_R$ is only perpendicular to $dA_1$, not to $dA_2$.
To calculate $L_1$ for $dA_1$ by writing the formula for $P$ and $d \Omega$ all out, I'd have:
$$L_1=(\frac{\sigma T^4 \cdot dA_1}{\pi} \cdot \frac{dA_R}{R^2}) \cdot \frac{R^2}{dA_1} \cdot \frac{1}{dA_R} = \frac{I_0}{dA_1}$$
To calculate $L_2$ for $dA_2$, in which the $P$ contains a $cos(\theta)^4$ in the numerator for increased distance, projection of $dA_R$ and Lambert's cosine law. And in which the $d \Omega$ contains a $cos(\theta)^2$ for the increased steradian area and a $cos(\theta)$ for the projection of $dA_2$, both in the denominator, along with an extra $cos(\theta)$ since you're dividing by the projection of $dA_R$. This gives:
$$L_2=(\frac{\sigma T^4 \cdot dA_2}{\pi} \cdot \frac{dA_R \cdot cos(\theta)^4}{R^2}) \cdot \frac{R^2}{dA_2 \cdot cos(\theta)^3} \cdot \frac{1}{dA_R \cdot cos(\theta)} = \frac{I_0}{dA_2}$$
So both $dA_1$ and $dA_2$ have the same $L$ since they have an equal surface area as said earlier.

Sorry for missing this, I keep on forgetting certain $cos(\theta)$ factors every time I write this down in my head.

 

[Charles Link]

Very good. I have one minor correction/suggestion. In your equations for $L_1$ and $L_2$, your final result can be simplified. If you simply do the algebra, you get $L_1=\frac{\sigma T^4}{\pi}$, and likewise $L_2=\frac{\sigma T^4}{\pi}$. (This is a very well-known result for a blackbody). In any case, very good. :) :)

 

[JohnnyGui]

Very good. I have one minor correction/suggestion. In your equations for $L_1$ and $L_2$, your final result can be simplified. If you simply do the algebra, you get $L_1=\frac{\sigma T^4}{\pi}$, and likewise $L_2=\frac{\sigma T^4}{\pi}$. (This is a very well-known result for a blackbody). In any case, very good. :) :)

Great, I indeed forgot to mention that $M=\pi L$. Thanks for the verification.

Something else I noticed regarding the difference between a receiving surface $A_R$ having an aperture or not.

- Say $A_R$ is increasing its distance away from the radiating surface $A$. Since its field of view is not covering the whole radiating surface area (yet), this means that $A_R$ would measure the same brightness regardless of its increasing distance. However, there is a limit at a certain distance where the field of view would cover a larger area than the actual radiating surface area. From that point on, increasing the distance would make the radiating surface look less bright.

- However, if $A_R$ does not have an aperture, then this means that it's already receiving power from the whole radiating surface at its initial distance. Increasing distance of $A_R$ without an aperture would make the radiating surface look less bright for $A_R$ immediately.

Are these 2 statements correct?

 

[Charles Link]

The brightness needs to be measured with an aperture that limits the field of view. Averaging a reading over a larger area where part of the area is not radiating is not a measure of the brightness of the source. In the case of doing this with something like a checkerboard surface (with black and white squares) , it would give you an average brightness level that could be considered a low resolution measurement that misses the finer detail.

 

[JohnnyGui]

The brightness needs to be measured with an aperture that limits the field of view. Averaging a reading over a larger area where part of the area is not radiating is not a measure of the brightness of the source. In the case of doing this with something like a checkerboard surface (with black and white squares) , it would give you an average brightness level that could be considered a low resolution measurement that misses the finer detail.

I found a quote from a link that shows what I meant:
"To make this clearer, consider the brightness of the Sun as seen from different planets. From the Earth, the Sun has an angular diameter of about 1/2 degree. If we look at a tiny portion of the surface, say a 1" (1 arcsecond) square area, we will measure a certain brightness. From Jupiter, at a distance of 5.2 AU from the Sun, however, the Sun has an angular diameter of 1/2 degree/5.2 ~ 0.1 degree. Its flux, and so its magnitude, will be correspondingly smaller, yet if we again look at a 1" square portion of the surface we find that it has the same brightness"

What do they mean here with brightness? Shouldn't that be the Radiance here since it doesn't change with distance?

 

[Charles Link]

I found a quote from a link that shows what I meant:
"To make this clearer, consider the brightness of the Sun as seen from different planets. From the Earth, the Sun has an angular diameter of about 1/2 degree. If we look at a tiny portion of the surface, say a 1" (1 arcsecond) square area, we will measure a certain brightness. From Jupiter, at a distance of 5.2 AU from the Sun, however, the Sun has an angular diameter of 1/2 degree/5.2 ~ 0.1 degree. Its flux, and so its magnitude, will be correspondingly smaller, yet if we again look at a 1" square portion of the surface we find that it has the same brightness"

What do they mean here with brightness? Shouldn't that be the Radiance here since it doesn't change with distance?

"Brightness" and radiance are the same thing. I believe when brightness is used to describe stars, it has a different definition and refers to the total amount of light that is received. Here we are using $L$ as radiance=brightness.

 

[JohnnyGui]

"Brightness" and radiance are the same thing. I believe when brightness is used to describe stars, it has a different definition and refers to the total amount of light that is received. Here we are using $L$ as radiance=brightness.

You're right about how they describe brightness when it comes to stars. Further down in the link it says: "If we define brightness as the flux through that 1" square area,..". Isn't that basically the received energy $P$ through that 1'' square area?
If so, does that mean that, according to that quote I quoted in my previous post, the received energy $P$ through the 1'' square does not change with distance as long as the Sun does not appear smaller than that 1'' square?

 

[Charles Link]

You're right about how they describe brightness when it comes to stars. Further down in the link it says: "If we define brightness as the flux through that 1" square area,..". Isn't that basically the received energy $P$ through that 1'' square area?
If so, does that mean that, according to that quote I quoted in my previous post, the received energy $P$ through the 1'' square does not change with distance as long as the Sun does not appear smaller than that 1'' square?

You are mixing up what they are saying. In the case of the stars, their receiver is "one square". They are not using a receiver with a small "one square" aperture a couple of meters in front of a small receiver to limit the viewing angle of the receiver to measure brightness like we would if we experimentally measure it from a wall or a small portion of the sun or moon.

 

[JohnnyGui]

You are mixing up what they are saying. In the case of the stars, their receiver is "one square". They are not using a receiver with a small "one square" aperture a couple of meters in front of a small receiver to limit the viewing angle of the receiver to measure brightness like we would if we experimentally measure it from a wall or a small portion of the sun or moon.

Ah ok, but since they're defining brightness as the flux on that 1 arcsecond area, isn't this basically the received energy $P$ on that 1 arcsecond area? If so, the link then says that $P$ does not decrease with distance as long as the Sun does not appear smaller than area. Is this actually correct?

 

[Charles Link]

Ah ok, but since they're defining brightness as the flux on that 1 arcsecond area, isn't this basically the received energy $P$ on that 1 arcsecond area? If so, the link then says that $P$ does not decrease with distance as long as the Sun does not appear smaller than area. Is this actually correct?

In the case of the sun, the 1 arc second is measured from the receiver. It basically means to put a small aperture a large distance (e.g. a meter or so) in front of a small receiver to limit the viewing angle. The 1 arc second is not measured from the sun. Incidentally, when measuring the brightness of the sun or moon, the aperture you use would normally be much larger than 1 arc second. $\\$ Additional item: In reading the "link", they use the distance 1" (1 arc second) very loosely. 1 arc second is 1/3600 of a degree. (With one degree =1/57.3 radians (approximately)). The sun is 93,000,000 miles away, so that in looking at a 1 arc second portion of the surface, that distance would not be 1" but rather about 450 miles across. They were trying to describe it in simple terms, but calling it 1" was very inaccurate. They would be better off just to call it a small portion, but even the word "small" is relative. In this case "small" means about 450 miles across. $\\$ Editing this: I looked at this further, and the angular measurement of one arc minute is often designated as 1' and one arc second as 1". So they were not referring to inches at all, and I stand corrected, they were not using 1" loosely=they meant 1 arc second and not 1 inch.

 

[JohnnyGui]

In the case of the sun, the 1 arc second is measured from the receiver. It basically means to put a small aperture a large distance (e.g. a meter or so) in front of a small receiver to limit the viewing angle. The 1 arc second is not measured from the sun. Incidentally, when measuring the brightness of the sun or moon, the aperture you use would normally be much larger than 1 arc second. $\\$ Additional item: In reading the "link", they use the distance 1" (1 arc second) very loosely. 1 arc second is 1/3600 of a degree. (With one degree =1/57.3 radians (approximately)). The sun is 93,000,000 miles away, so that in looking at a 1 arc second portion of the surface, that distance would not be 1" but rather about 450 miles across. They were trying to describe it in simple terms, but calling it 1" was very inaccurate. They would be better off just to call it a small portion, but even the word "small" is relative. In this case "small" means about 450 miles across. $\\$ Editing this: I looked at this further, and the angular measurement of one arc minute is often designated as 1' and one arc second as 1". So they were not referring to inches at all, and I stand corrected, they were not using 1" loosely=they meant 1 arc second and not 1 inch.

Yes, regarding the arc second this is correct. I indeed pictured it as a receiver having an aperture in front of it so that the field of view angle 1'' is coming out of the receiver. What I was wondering in such a scenario is if the received power $P$ through that aperture would indeed stay constant, independent of distance, until the whole sun itself would have a smaller viewing angle than 1''. Does $P$ truly stay constant until then?

 

[Charles Link]

Yes, regarding the arc second this is correct. I indeed pictured it as a receiver having an aperture in front of it so that the field of view angle 1'' is coming out of the receiver. What I was wondering in such a scenario is if the received power $P$ through that aperture would indeed stay constant, independent of distance, until the whole sun itself would have a smaller viewing angle than 1''. Does $P$ truly stay constant until then?

If you were to take your measurement apparatus out to Jupiter and farther from the sun to increase the distance, the answer is yes, other than the phenomenon of limb darkening that was previously mentioned in this thread. Let me see if I can find that post=yes, posts # 287 and # 291 by @sophiecentaur .

 

[JohnnyGui]

If you were to take your measurement apparatus out to Jupiter and farther from the sun to increase the distance, the answer is yes, other than the phenomenon of limb darkening that was previously mentioned in this thread. Let me see if I can find that post=yes, posts # 287 and # 291 by @sophiecentaur .

Ah ok, got it. Let's see if I understand this formula-wise:

So let's say on Earth the receiver $A_R$ is a distance $R$ from the sun and with a field of view of 1'' this would cover an area $A_1$ of the sun. The received energy $P_{R1}$ would be (let's do an integration for the sake of accuracy):
$$\int \limits_{0}^{A_1} L \cdot \frac{dA \cdot cos(\theta)^3}{R^2} \cdot A_R \cdot cos(\theta) = P_{R1}$$
Where the fraction $\frac{dA \cdot cos(\theta)^3}{R^2} = d \Omega$. I’m writing it out to integrate over $dA$ instead of over $d \Omega$.
Now, if the receiver $A_R$ increases its distance x times (standing on Jupiter) so that the new distance is $R \cdot x$, then the newly covered area of the Sun $A_2$ by the same field of view 1'' would be larger by a factor of $x^2$ so that the new source area $A_1 \cdot x^2 = A_2$. The integration for the received energy $P_R2$ should then be done for a surface area of $A_1 \cdot x^2$:
$$\int \limits_{0}^{A_1 x^2} L \cdot \frac{dA \cdot cos(\theta)^3}{x^2R^2} \cdot A_R \cdot cos(\theta) = P_{R2}$$
Since $P_R$ is independent of distance. Does this mean that both integrations should give the same value so that $P_{R2} = P_{R1}$?