Is there any relation between wavelength and brightness?


by tris_d
Tags: brightness, relation, wavelength
Simon Bridge
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Nov15-12, 05:50 PM
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Quote Quote by sophiecentaur View Post
How do you intend to "Break a photon into three parts"? Before you use the term 'Photon' you should understand what it represents.
OK - I should signal better when I'm not being technically correct.

Stick to waves, wavelength, power and all the other classical ideas where they are appropriate here.
I actually didn;t need to refer to a particular model for the point I was trying to make.

The way that our colour vision (three colour analysis) works is pretty well established and 'personal' interpretations can seriously damage the understanding of newcomers to the subject.
Well called - I was too focussed on the point I was trying to make and slipped up elsewhere.
I should have talked about the receptor's response to the incoming light (simpler to have the three "signals" that a continuous frequency response) and left it at that. I was trying to convey a sense of the increased simplicity of this method and it does not help to do this if I use a complex and technical-sounding language.

Perhaps you can show me how I could have made the same points better?


---------------

Aside: parametric down conversion of photons is often described by physicists as "splitting a photon in half".
Also see Hübel H. et al. Direct generation of photon triplets using cascaded photon-pair sources Nature 466, 601–603 (29 July 2010)
... for more on how physicists understand "photon splitting". It is just not the process that happens in the eye.
tris_d
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Nov15-12, 10:58 PM
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http://www.cv.nrao.edu/course/astr534/Brightness.html

This above is a link to internet article that talks exactly about the things I want to know, it's just that some parts of it do not seem to quite fit with what I can read everywhere else. Here are some statements that do not seem to compare:




1.) The number of photons falling on the film per unit area per unit time per unit solid angle does not depend on the distance between the source and the observer.

Are they talking about intensity? Should not number of photons per unit area per unit time drop off with the square of the distance?



--//--

2.) Thus we distinguish between the brightness of the Sun, which does not depend on distance, and the apparent flux, which does.

Now they say flux depends on the distance, but is flux not the number of photons per unit area per unit time that they just previously said does not depend on the distance?



--//--

3.) Brightness is independent of distance. Brightness is the same at the source and at the detector.

I guess this is true if the light source is not point source?



--//--

4.) If a source is unresolved, meaning that it is much smaller in angular size than the point-source response of the eye or telescope observing it, its flux density can be measured but its spectral brightness cannot.

What in the world did they just say here?

Wikipedia says:
http://en.wikipedia.org/wiki/Apparent_brightness
- Note that brightness varies with distance; an extremely bright object may appear quite dim, if it is far away. Brightness varies inversely with the square of the distance.

So when I try to put everything together my interpretation is this: If light source is "resolved", that is when its focused projection has angular size greater than point source, then brightness does NOT fall off with the distance. But when the light source gets so far away that its focused projection covers no more than one pixel on the image, then it becomes "point source" or "unresolved", and then the inverse square law starts to apply in such way that the brightness DOES drop off with the square of any further distance from that point on.



--//--

5.) If a source is much larger than the point-source response, its spectral brightness at any position on the source can be measured directly, but its flux density must be calculated by integrating the observed spectral brightnesses over the source solid angle.

What is "spectral brightness" and how is it different to just "brightness"?



--//--

6.) The specific intensity or brightness is an intrinsic property of a source, while the flux density of a source also depends on the distance between the source and the observer.

How can intensity and brightness be intrinsic property of a source? Is intensity and flux not one and the same thing? -- I'd say intensity is a property of emitted light rather than a property of a light source, and that brightness is a property of an image, rather than property of either emitted light or light source itself.
sophiecentaur
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Nov16-12, 03:33 AM
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Quote Quote by Simon Bridge View Post
OK - I should signal better when I'm not being technically correct.

I actually didn;t need to refer to a particular model for the point I was trying to make.

Well called - I was too focussed on the point I was trying to make and slipped up elsewhere.
I should have talked about the receptor's response to the incoming light (simpler to have the three "signals" that a continuous frequency response) and left it at that. I was trying to convey a sense of the increased simplicity of this method and it does not help to do this if I use a complex and technical-sounding language.

Perhaps you can show me how I could have made the same points better?


---------------

Aside: parametric down conversion of photons is often described by physicists as "splitting a photon in half".
Also see Hübel H. et al. Direct generation of photon triplets using cascaded photon-pair sources Nature 466, 601–603 (29 July 2010)
... for more on how physicists understand "photon splitting". It is just not the process that happens in the eye.
I did sound a bit more grumpy than I meant to be - sorry.
But you have taken my point about not introducing more technical terms than necesssary - especially when it's a bit tenuous. 'Splitting photons' is not what it sounds like and it is nothing to do with the way three separate analysis filters work. If you must introduce the photon at this stage, you could just say that each sensor detects photons with a different range of energies. But I don't see how that improves on the term 'wavelength response'.
sophiecentaur
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Nov16-12, 03:43 AM
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Quote Quote by tris_d View Post
Is intensity and flux not one and the same thing?
No, they are not. Look at two lightbulbs, side by side. They have equal intensities. Turn one of them off and the remaining one has the same intensity as before but the flux reaching you has dropped to half. Move the two bulbs (both on) together or apart (not too far, or the geometry may change) and the flux from them is the same. However, if you were to superimpose the two (having two similar filaments in the same frosted envelope) then the intensity would double but the flux would be the same as having the two side by side.

You may have noticed that people are getting a bit fed up with your responses. You seem to be desperate to show that you are not wrong in your ideas, rather than willing to take on board new ones. Do you think all of the other contributors are idiots by putting things in the way they are doing? It could just possibly be you who could do something about this to resolve it.

"Don't understand" and "won't understand" are separated by a fine line.
tris_d
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Nov16-12, 04:32 AM
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Quote Quote by sophiecentaur View Post
You may have noticed that people are getting a bit fed up with your responses. You seem to be desperate to show that you are not wrong in your ideas, rather than willing to take on board new ones.
My ideas? Take on board new ideas? Huh?! What in the world are you talking about? Did you confuse me with someone else?


Do you think all of the other contributors are idiots by putting things in the way they are doing?
What are you talking about? Is this twilight zone or something?


It could just possibly be you who could do something about this to resolve it.

"Don't understand" and "won't understand" are separated by a fine line.
Understand what? Resolve what? What are you talking about? What is it you are upset about? Is that link I posted yours? Is this some joke? What is your problem?
Drakkith
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Nov16-12, 05:11 AM
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Quote Quote by tris_d View Post
1.) The number of photons falling on the film per unit area per unit time per unit solid angle does not depend on the distance between the source and the observer.

Are they talking about intensity? Should not number of photons per unit area per unit time drop off with the square of the distance?



--//--

2.) Thus we distinguish between the brightness of the Sun, which does not depend on distance, and the apparent flux, which does.

Now they say flux depends on the distance, but is flux not the number of photons per unit area per unit time that they just previously said does not depend on the distance?
I already explained this. The brightness of the Sun per unit of angular diameter does not change. Only the overall intensity, or flux does since the Sun get's smaller as you recede from it.

3.) Brightness is independent of distance. Brightness is the same at the source and at the detector.

I guess this is true if the light source is not point source?
No real light source is a true point source.

4.) If a source is unresolved, meaning that it is much smaller in angular size than the point-source response of the eye or telescope observing it, its flux density can be measured but its spectral brightness cannot.

What in the world did they just say here?
Stars are so small in telescopes that their true size cannot be seen and the appear to be point sources, so we cannot measure how bright they are per unit of angular diameter, only the flux from the whole star.

So when I try to put everything together my interpretation is this: If light source is "resolved", that is when its focused projection has angular size greater than point source, then brightness does NOT fall off with the distance. But when the light source gets so far away that its focused projection covers no more than one pixel on the image, then it becomes "point source" or "unresolved", and then the inverse square law starts to apply in such way that the brightness DOES drop off with the square of any further distance from that point on.
The overall flux from the source DOES fall off with the inverse square. Just not the brightness of the source. It's confusing terminology.


5.) If a source is much larger than the point-source response, its spectral brightness at any position on the source can be measured directly, but its flux density must be calculated by integrating the observed spectral brightnesses over the source solid angle.

What is "spectral brightness" and how is it different to just "brightness"?
I think it just means you have to measure the flux from the whole image of the source, which can be quite large if it isn't a point source.
tris_d
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Nov16-12, 06:15 AM
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Quote Quote by Drakkith View Post
The overall flux from the source DOES fall off with the inverse square. Just not the brightness of the source. It's confusing terminology.
It sure is confusing. Thanks. Some more clarification please.

Light intensity is number of photons per unit area per unit time?

Light flux is number of photons per unit area per unit time, or what?

Stars are so small in telescopes that their true size cannot be seen and the appear to be point sources, so we cannot measure how bright they are per unit of angular diameter, only the flux from the whole star.
How flux relates to brightness?
sophiecentaur
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Nov16-12, 10:17 AM
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Quote Quote by tris_d View Post
My ideas? Take on board new ideas? Huh?! What in the world are you talking about? Did you confuse me with someone else?

What are you talking about? Is this twilight zone or something?

Understand what? Resolve what? What are you talking about? What is it you are upset about? Is that link I posted yours? Is this some joke? What is your problem?
No confusion. Just read your responses to the last few replies and then read the contents of them again. They are not wrong.
tris_d
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Nov16-12, 12:49 PM
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Quote Quote by sophiecentaur View Post
No confusion. Just read your responses to the last few replies and then read the contents of them again. They are not wrong.
#12
Oh man! This is not simple. -- Thank you all, I'm chewing on it.
Something wrong with this post?

#15
And if we swap eyes, maybe you would see it's purple what you previously called green.
Something wrong with this post?

#20
This above is a link to internet article that talks exactly about the things I want to know..
Something wrong with this post?



By the way, can you answer this:

Light intensity is number of photons per unit area per unit time?

Light flux is number of photons per unit area per unit time, or what?
Drakkith
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Nov16-12, 01:56 PM
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Let's look at an example. Let's say that the Sun puts 1,000 photons per second onto a sensor of 100 pixels at the focal point of a telescope here on Earth. Then we take this sensor and move it twice as far away from the Sun as it was. The Sun now puts 250 photons per second onto the sensor. BUT, in both cases, each pixel that receives light receives the same amount of photons per second. The reason that there are 1/4 as many photons hitting the sensor is that the image formed at the focal plain is half the size as before in both the X and Y directions. So the surface area of this image at the focal plane is 1/4 what it used to be and only 1/4 as many pixels are even hit by light from the Sun. So it was originally 1,000 photons over 100 pixels is 10 photons/second/pixel. Now its 250 photons over 25 pixels, which is still 10 photons/second/pixel! This also means that the TOTAL amount of photons falling onto the aperture of the telescope has fallen from 1,000 to 250, so as you can see in both the focused and unfocused case the RADIANT FLUX, the photons per second, has decreased to 1/4 just by doubling the distance.

Now, what about brightness? I am not familiar enough to figure out which of the many units (See here) to use, so I will have to explain it my way again instead. Let's say that brightness is the number of photons coming from an angular section of the sky, as that seems to be the only way it makes sense.

Let's say I measured the number of photons per second coming from an area of the Sun that is 15 arcminutes x 15 arcminutes. So the area would be 225 arcminutes. Since the Sun was putting a total 1,000 photons per second onto the sensor, and 15x15 arcminutes is 1/4 the size of the Sun from the Earth (the Sun is 30 arcminutes across), the number of photons per second from this area is 250.

Now, I move the telescope twice as far away. How many photons per second to I now get from this same 15x15 arcminutes? Well, if the Sun has had it's dimensions halved, it is now 15 arcminutes across instead of 30. Which means that the area is 1/4 what it was, which means the whole Sun now fits in this 15x15 area! And if the Sun was putting a TOTAL of 250 photons per second onto our sensor earlier, then it must still be doing the same thing now since we are at the same distance. So even though I've moved twice as far away and the total light from the Sun has decreased to 1/4 what it was, I still have the same amount of light coming from the same angular area of sky. (Note that I've simplified the explanation by using the area of a square, not a circle. However the result is the same.) So the BRIGHTNESS, which I mean as the number of photons per area of sky, is exactly the same. Note that this also happens if we move CLOSER to the Sun. At half the distance to the Sun the light is quadrupled, but the image of the Sun is now 4 times as large! So 4,000 photons, over 400 pixels is still 10 photons/second/pixel!

But what about far away stars? Here we run into an issue. My telescope focuses the light down to a point called an airy disc. Let's say I'm measuring 500 nm light. With an aperture of 250 mm and a focal length of 1,000 mm my telescope will focus 500 nm light down to a spot that is 4.88 microns in diameter. But, what if my star image is even smaller than that? Like, much smaller? Well, in that case we treat the star as a "point source". At this point we cannot measure the brightness of the star, only the total FLUX. If we know the size of the star and it's distance we could calculate the brightness, however we cannot measure it.
tris_d
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Nov16-12, 03:33 PM
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Quote Quote by Drakkith View Post
Let's look at an example...
Thank you Drakkith, that's great! Let me chew on it.

Just one thing in the mean time, what are the units of light flux?
Drakkith
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Nov16-12, 04:38 PM
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Quote Quote by tris_d View Post
Thank you Drakkith, that's great! Let me chew on it.

Just one thing in the mean time, what are the units of light flux?
If you mean Radiant Flux, that would be Watts. Hopefully I haven't butchered optical terminology too badly in my example lol.
tris_d
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Nov16-12, 04:45 PM
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Quote Quote by Drakkith View Post
Let's look at an example. Let's say that the Sun puts 1,000 photons per second onto a sensor of 100 pixels at the focal point of a telescope here on Earth. Then we take this sensor and move it twice as far away from the Sun as it was. The Sun now puts 250 photons per second onto the sensor. BUT, in both cases, each pixel that receives light receives the same amount of photons per second. The reason that there are 1/4 as many photons hitting the sensor is that the image formed at the focal plain is half the size as before in both the X and Y directions. So the surface area of this image at the focal plane is 1/4 what it used to be and only 1/4 as many pixels are even hit by light from the Sun. So it was originally 1,000 photons over 100 pixels is 10 photons/second/pixel. Now its 250 photons over 25 pixels, which is still 10 photons/second/pixel! This also means that the TOTAL amount of photons falling onto the aperture of the telescope has fallen from 1,000 to 250, so as you can see in both the focused and unfocused case the RADIATIVE FLUX, the photons per second, has decreased to 1/4 just by doubling the distance.
Great. I think you could/should make an internet article out of this stuff. It's fairly simple, but not very obvious if you don't consider there is a lens in between and that focusing is involved, to which I was completely oblivious previously.


Now, what about brightness? I am not familiar enough to figure out which of the many units (See here) to use, so I will have to explain it my way again instead. Let's say that brightness is the number of photons coming from an angular section of the sky, as that seems to be the only way it makes sense.

Let's say I measured the number of photons per second coming from an area of the Sun that is 15 arcminutes x 15 arcminutes. So the area would be 225 arcminutes. Since the Sun was putting a total 1,000 photons per second onto the sensor, and 15x15 arcminutes is 1/4 the size of the Sun from the Earth (the Sun is 30 arcminutes across), the number of photons per second from this area is 250.

Now, I move the telescope twice as far away. How many photons per second to I now get from this same 15x15 arcminutes? Well, if the Sun has had it's dimensions halved, it is now 15 arcminutes across instead of 30. Which means that the area is 1/4 what it was, which means the whole Sun now fits in this 15x15 area! And if the Sun was putting a TOTAL of 250 photons per second onto our sensor earlier, then it must still be doing the same thing now since we are at the same distance. So even though I've moved twice as far away and the total light from the Sun has decreased to 1/4 what it was, I still have the same amount of light coming from the same angular area of sky. (Note that I've simplified the explanation by using the area of a square, not a circle. However the result is the same.) So the BRIGHTNESS, which I mean as the number of photons per area of sky, is exactly the same. Note that this also happens if we move CLOSER to the Sun. At half the distance to the Sun the light is quadrupled, but the image of the Sun is now 4 times as large! So 4,000 photons, over 400 pixels is still 10 photons/second/pixel!
Would you agree that instead of defining brightness as a measure of "number of photons per area in the sky" would be better to say it is a measure of "number of photons per pixel", so that brightness be a property of an image rather than property of the light itself?


But what about far away stars? Here we run into an issue. My telescope focuses the light down to a point called an airy disc. Let's say I'm measuring 500 nm light. With an aperture of 250 mm and a focal length of 1,000 mm my telescope will focus 500 nm light down to a spot that is 4.88 microns in diameter. But, what if my star image is even smaller than that? Like, much smaller? Well, in that case we treat the star as a "point source". At this point we cannot measure the brightness of the star, only the total FLUX. If we know the size of the star and it's distance we could calculate the brightness, however we cannot measure it.
Are most of the stars in our galaxy point light sources? Are most of the other galaxies point light sources? -- When you say "500 nm light" you refer to angular size of "airy disc"? I guess the size of airy disc depends on magnification, so how do you know it's 500nm and not 200nm, that is how do you know you focused it properly (if this question makes sense)? -- How do we measure the flux? You say we can not measure the brightness in case of point light sources, but would not image itself of such point light source be a measure of its brightness?
sophiecentaur
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Nov16-12, 04:46 PM
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Quote Quote by tris_d View Post

Light intensity is number of photons per unit area per unit time?

Light flux is number of photons per unit area per unit time, or what?
If you insist on using photons in your arguments the everything changes with frequency and that just doesn't help anyone. That's the whole point of using Watts - it works for any combination of wavelengths. Your attempted definition will not work for a mixture of frequencies.
Why bring photons into this at all? All this stuff was done and dusted before anyone came up with E=hf
sophiecentaur
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Nov16-12, 04:59 PM
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Quote Quote by Drakkith View Post
Let's look at an example. . . . .
I am having great difficulty in seeing why you keep bringing photons, pixels and peculiarities of imaging systems into this particular question. I realise the bottom line for a practical astronomer is looking at resolvable, bright enough images but is all this fundamental to the actual question?
You will have read my reservations about the use of photons on account of the variable energy. Watts were good enough for all the original calculations on this stuff.

(You know I have a general aversion to explanations of things that bring in Photons when their actual nature is not specified initially. It is a potentially risky process and the raw results are suspect. Surely it isn't any harder to consider light as a continuum for basic optics than ignoring the fact that a massive object consists of atoms, in the context of mechanics)
tris_d
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Nov16-12, 05:15 PM
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Quote Quote by sophiecentaur View Post
I am having great difficulty in seeing why you keep bringing photons, pixels and peculiarities of imaging systems into this particular question. I realise the bottom line for a practical astronomer is looking at resolvable, bright enough images but is all this fundamental to the actual question?
You will have read my reservations about the use of photons on account of the variable energy. Watts were good enough for all the original calculations on this stuff.

(You know I have a general aversion to explanations of things that bring in Photons when their actual nature is not specified initially. It is a potentially risky process and the raw results are suspect. Surely it isn't any harder to consider light as a continuum for basic optics than ignoring the fact that a massive object consists of atoms, in the context of mechanics)
I'm interested in brightness, and I believe we established brightness is a property of an image rather than property of the light source or light itself. Therefore, since the image is digital object rather than analog, since it is a collection of discrete pixels, I think we also need to quantize the light so we can talk about the relation between the image and the light, and then we could perhaps define brightness as the number of photons per pixel.

That's where my original question came from. I was wondering if there is anything else beside the number of photons per pixel, like wavelength, that would define the brightness of a pixel.
Drakkith
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Nov16-12, 05:16 PM
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Why wouldn't it be fundamental to the question? I showed that before and after the optical system the results are the same. I am merely putting an optical system in because I am far more familiar with the workings of optics, sensors, and photons than I am waves and energy per area and such. I know we can't treat light as little particles traveling through space, but unless I've gravely misunderstood something I think my explanation works out the same either way. Please correct me if I'm wrong on something.
tris_d
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Nov16-12, 05:32 PM
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Quote Quote by Drakkith View Post
Why wouldn't it be fundamental to the question? I showed that before and after the optical system the results are the same. I am merely putting an optical system in because I am far more familiar with the workings of optics, sensors, and photons than I am waves and energy per area and such. I know we can't treat light as little particles traveling through space, but unless I've gravely misunderstood something I think my explanation works out the same either way. Please correct me if I'm wrong on something.
I like the way you explained it, that's how it naturally comes to me to think about it. I think we understand each other just fine now. And if there is nothing wrong about it, then I'd prefer we stick with photons and pixels. Individual photons can be considered to have wavelength and frequency, I guess, so we could include those concepts as well, and that I believe should then cover the whole subject and explain all the related phenomena.


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