# I Emission spectra of different materials

1. Sep 7, 2017

### JohnnyGui

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?

2. Sep 7, 2017

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.

3. Sep 8, 2017

### JohnnyGui

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?

4. Sep 8, 2017

"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.

5. Sep 8, 2017

### JohnnyGui

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?

6. Sep 8, 2017

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.

7. Sep 8, 2017

### JohnnyGui

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?

8. Sep 8, 2017

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.

Last edited: Sep 8, 2017
9. Sep 8, 2017

### JohnnyGui

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?

10. Sep 8, 2017

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 .

Last edited: Sep 8, 2017
11. Sep 8, 2017

### JohnnyGui

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 $P_{R2} = P_{R1}$?

Last edited: Sep 8, 2017