# How to calculate radiance emitted?

Andrew Hamilton
Science 11 May '09 10:15
1. 11 May '09 10:153 edits
I want to know how to calculate the radiance emitted over a given frequency/wavelength range at a given blackbody temperature. So I tried looking it up:

http://en.wikipedia.org/wiki/Planck's_law

“…The radiance emitted over a frequency range [v1,v2] or a wavelength range [Y1,Y2] = [c / V2,c / v1] can be obtained by integrating the respective functions….” (actually it wasn’t quite “[Y1,Y2]” but it is impossible to copy and past what it does say there into this post)

-it then gives an integral equation below that but it is a very long time since I have done integral calculus and I have given up in frustration trying to understand this.

Can anyone give me an example of a specific calculation using that formula of the radiance emitted over the wavelength range [2.3nm, 4.0nm] at the blackbody temperature of 100C (thus 373.15K for this calculation) ? -if I can just see one working example then I am sure I will get it. Any help would be very much appreciate.
Baby Gauss
11 May '09 13:043 edits
Originally posted by Andrew Hamilton
I want to know how to calculate the radiance emitted over a given frequency/wavelength range at a given blackbody temperature. So I tried looking it up:

http://en.wikipedia.org/wiki/Planck's_law

“…The radiance emitted over a frequency range [v1,v2] or a wavelength range [Y1,Y2] = [c / V2,c / v1] can be obtained by integrating the respective fun ...[text shortened]... t see one working example then I am sure I will get it. Any help would be very much appreciate.
Don't calculate this integral. It can be done but if I were you I'd look it up in an integral table: http://en.wikipedia.org/wiki/Lists_of_integrals#Definite_integrals_lacking_closed-form_antiderivatives

This kind of integral can be done using these techniques: changing variables [; x = \frac{h \nu}{K T} ;], using a multiplication trick, and then resorting to the gamma function. Or you can use a the more sophisticated residue theorem.

One good way to post equations in here is using this: http://thewe.net/tex/
Kudos to Palynka.
3. 11 May '09 14:02
Thanks -but I am afraid I don’t know how to apply this (and I did look at http://en.wikipedia.org/wiki/Lists_of_integrals#Definite_integrals_lacking_closed-form_antiderivatives and http://en.wikipedia.org/wiki/Gamma_function but I don’t know exactly how to apply that to this problem)
You are obviously vastly more knowledgeable about maths than I am.
(What I need is just one working example -then I’ll get it).
Baby Gauss
11 May '09 17:54
Originally posted by Andrew Hamilton
Thanks -but I am afraid I don’t know how to apply this (and I did look at http://en.wikipedia.org/wiki/Lists_of_integrals#Definite_integrals_lacking_closed-form_antiderivatives and http://en.wikipedia.org/wiki/Gamma_function but I don’t know exactly how to apply that to this problem)
You are obviously vastly more knowledgeable about maths than I am.
(What I need is just one working example -then I’ll get it).
Just noticed this: http://en.wikipedia.org/wiki/Planck's_law#Appendix

In your case things aren't clear cut because if you have [; \nu_1 ;] and [; \nu_2 ;] and so the function you are tying to integrate doesn't have an analytical anti-derivative.So you have to resort to numerical integration. Do you run Linux or windows?
5. 11 May '09 19:193 edits
Originally posted by adam warlock
Just noticed this: http://en.wikipedia.org/wiki/Planck's_law#Appendix

In your case things aren't clear cut because if you have [; \nu_1 ;] and [; \nu_2 ;] and so the function you are tying to integrate doesn't have an analytical anti-derivative.So you have to resort to numerical integration. Do you run Linux or windows?
-I use windows.

…So you have to resort to numerical integration...…

I assume you mean calculating the numerical value of a definite integral by calculating a large number of points along the curve and then joining adjacent points with short straight lines and getting an estimate of the of the numerical value of that definite integral by calculating the sum of the areas under all those short straight lines? If so, I think that would be more than good enough for my purposes.
I am extremely good at programming in java so it is easy for me to quickly make a java program to do all that tedious calculation instantly for me once I understand how to use the maths -so no problem there.

But I am having a hard time comprehending the exact physical meaning of the formulas at: http://en.wikipedia.org/wiki/Planck's_law and I don’t understand what “u(v,T)” actually means and not sure of appropriate use of the formulas. But I think a single working example of resorting to numerical integration would make me instantly comprehend it (when I did maths courses, I usually found that a single working example makes me instantly understand all that I couldn’t fathom! -like a picture saying a thousand words, I instantly 'see' all).
6. joe shmo
Strange Egg
11 May '09 19:51
I think the only way you'll "understand" is by rigorous study of quantum physics. I would think something like this has a very obscure physical meaning. Just to try to understand the mathematical derivation of the integral would require a great deal of facilitated study.
But then again i dont know that much....just throwing in my two cents
Baby Gauss
11 May '09 21:20
Originally posted by Andrew Hamilton
-I use windows.

[b]…So you have to resort to numerical integration...…

I assume you mean calculating the numerical value of a definite integral by calculating a large number of points along the curve and then joining adjacent points with short straight lines and getting an estimate of the of the numerical value of that definite integral by ...[text shortened]... all that I couldn’t fathom! -like a picture saying a thousand words, I instantly 'see' all).[/b]
I use windows.
Oh. I asked you about it because there is some mathematical software that is free and allows you do a lot of things more easily.
For purposes of ilustration: as comercial packages you things like: http://en.wikipedia.org/wiki/Mathematica and http://en.wikipedia.org/wiki/MATLAB
Some people decided to make free and/or open software that allows one to do a lot of those things too. Numerical integration being one of them:
http://en.wikipedia.org/wiki/GNU_Octave
http://en.wikipedia.org/wiki/Scilab

I haven't used any of those but from what I know they are very capable.

I assume you mean calculating the numerical value of a definite integral by calculating a large number of points along the curve and then joining adjacent points with short straight lines and getting an estimate of the of the numerical value of that definite integral by calculating the sum of the areas under all those short straight lines? If so, I think that would be more than good enough for my purposes.

I think that is integral isn't a very shaky so you can do it by straightforward triangulation but for other kind of integrals other methods have to be used. In any case here you have :~
http://www.nr.com/oldverswitcher.html

Back in the day anyone could download those books but nowadays I think that they impose some restritions. So, if you're having trouble getting the books just PM an email that I'll send you an older version with all the codes in the book too.

And now to the physics: ðŸ˜‰

[; u(\nu , T) ;] is an energy density, so it expresses the amount of energy per unit volume and it dependes in [; \nu ;] and in [; T ;]. This means that the amount an energy a black body radiates per unit volume depends not only on its temperature but also in what frequency it is emiting. So for instanve if I were to consider you a black body I'd have to take your temperature as 310 Kelvi. After this I could plot your energy spectrum by considering the function [; u(\nu , 310) ;] and if I wanted to know your total energy output I'd integrate the previous function between 0 and infinity. If I were interested in the energy some particular interval of frequencies I'd just integrate it in that interval.

And what other formulas you're having trouble in understanding?
Baby Gauss
11 May '09 21:21
Originally posted by joe shmo
I think the only way you'll "understand" is by rigorous study of quantum physics. I would think something like this has a very obscure physical meaning. Just to try to understand the mathematical derivation of the integral would require a great deal of facilitated study.
But then again i dont know that much....just throwing in my two cents
I think that for this a good grasp of integration and classifical thermodynamics would be enough. Yes there are some new things in here but with that knowledge I think that the jump can be made in an comfortable way...
9. 11 May '09 21:24
Originally posted by joe shmo
I think the only way you'll "understand" is by rigorous study of quantum physics. I would think something like this has a very obscure physical meaning. Just to try to understand the mathematical derivation of the integral would require a great deal of facilitated study.
But then again i dont know that much....just throwing in my two cents
I'm thinking Andrew Hamilton is planning to take over the world with his some type of exotic bomb.

I suggest a Chemical Engineering degree first, might be truely helpful!

I agree with Shmo, it is difficult for someone even with the physics background to truely understand such a thing. Also, the link you provided gives the integral at the bottom of the page. There is also a link at the bottom of the wikipedia page that takes you to a web site that will calculate the energy density for you:

http://www.vias.org/simulations/simusoft_blackbody.html
10. Palynka
Upward Spiral
12 May '09 02:003 edits
For numerical integration, there are quite a number of methods that you can read up on the web.

If you want a quick and rough one, just go with a basic rectangle rule. This just divides the function in n-1 intervals, draws a rectangle centered on the image of the midpoint and just sums them up over the whole function.

This is the brutish approach, but it's easy to code so you gain time there.

For example, integrate 1.3^x between 1 and 5, for comparison. This integral should be equal to 1/ln(1.3)*(1.3^5-1.3) = 9.1969

If you code (this is Octave/Matlab but very readable) simply:

sum = 0;
intervalsize = 1/50000;

% now with a loop, keep adding each rectangle (evaluated at its midpoint) to the sum variable.
for i = 1:intervalsize: (5-intervalsize);
sum = sum+intervalsize*1.3^(i+0.5*intervalsize);
end;

Now calling your variable sum should give you the result.

This is not very parsimonious, but I wrote it to be as readable as possible.

If your function is reasonably well-behaved, this should work and all you need to do to increase your accuracy is to decrease the interval size (obviously at the cost of computational speed). This might not work very well with some functional forms, though, particularly if the derivatives go to infinity as you approach one of the bounds (or in between).

If you need more advanced methods let me know.
11. 12 May '09 09:331 edit
Originally posted by mlprior
I'm thinking Andrew Hamilton is planning to take over the world with his some type of exotic bomb.

I suggest a Chemical Engineering degree first, might be truely helpful!

I agree with Shmo, it is difficult for someone even with the physics background to truely understand such a thing. Also, the link you provided gives the integral at the bottom of t ...[text shortened]... alculate the energy density for you:

http://www.vias.org/simulations/simusoft_blackbody.html
…I'm thinking Andrew Hamilton is planning to take over the world with his some type of exotic bomb. ..…

-yes, or he could be trying to calculate the probable longest wavelengths an ‘alternate solar cell’ should be designed to convert to electrical energy assuming the solar cell’s temperature doesn’t go over, say, 100C in natural conditions even in a hot climate.

…http://www.vias.org/simulations/simusoft_blackbody.html..…

Thanks for that ðŸ™‚
I have downloaded this onto my desktop and and used it and it is pretty good!
I think this may have inadvertently given me what I am looking for which is a method of estimating the radiance emitted from a blackbody at given temperature between a given wavelength range -I could just get an estimate simply by reading it off this graph! So I may not have to do any computations myself after all.
12. 12 May '09 10:265 edits
I am not sure if I understand the physical meaning of “energy density” in the context of a blackbody:

I understand “energy density” is the energy within in a unit volume of space (presumably joules per cubic meter).
Ok -but where is this volume in the context of a blackbody? -I assume it to be in the blackbody itself? -so, if all the inner walls/surfaces in the interior of a sealed empty box (containing no air i.e. containing a vacuum only) is at temperature 100C then the “energy density” of a particular wavelength Y of radiation is the amount of energy of the sum of energys in all the photons that have that EXACT particular wavelength Y per cubic meter of volume of empty space within the box and at a single point in time?

-if so, then this is what confuses me about this; the range of possible wavelengths that can exist at a given temperature is infinite thus the number of photons that are EXACTLY at a particular wavelength Y must be about zero hence the energy density of wavelength Y in the box is zero!

Can anyone explain this to me? -I am obviously seriously missing something here.
Baby Gauss
12 May '09 12:08
Originally posted by Andrew Hamilton
I am not sure if I understand the physical meaning of “energy density” in the context of a blackbody:

I understand “energy density” is the energy within in a unit volume of space (presumably joules per cubic meter).
Ok -but where is this volume in the context of a blackbody? -I assume it to be in the blackbody itself? -so, if all the inner walls ...[text shortened]... ox is zero!

Can anyone explain this to me? -I am obviously seriously missing something here.
Th energy density is not about the volume of the black body but about the volume you are considering.

If you have a black body radiating energy you can know the energy it is radiating in a given wavelength for cubic meter. So it's not about what's radiating but to where it is radiating.

if so, then this is what confuses me about this; the range of possible wavelengths that can exist at a given temperature is infinite thus the number of photons that are EXACTLY at a particular wavelength Y must be about zero hence the energy density of wavelength Y in the box is zero!
No, no, no. Why do you say that since we have a dense set of possible values of wavelengths we must have 0 photons for each wavelength? This is a wrong conclusion but nevertheless I need to know how you arrived at it.

We indeed consider all possible wavelengths but the concept of total photons for wavelength isn't a well defined one. Remember that photons aren't conserved in this situation.
14. 12 May '09 18:43
Originally posted by Palynka
For numerical integration, there are quite a number of methods that you can read up on the web.

If you want a quick and rough one, just go with a basic rectangle rule. This just divides the function in n-1 intervals, draws a rectangle centered on the image of the midpoint and just sums them up over the whole function.

This is the brutish approach, bu ...[text shortened]... u approach one of the bounds (or in between).

If you need more advanced methods let me know.
This function is pretty well-behaved, so it should be no problem.
15. 12 May '09 19:112 edits
Originally posted by adam warlock
Th energy density is not about the volume of the black body but about the volume you are considering.

If you have a black body radiating energy you can know the energy it is radiating in a given wavelength for cubic meter. So it's not about what's radiating but to where it is radiating.

[b]if so, then this is what confuses me about this; the range ...[text shortened]... velength isn't a well defined one. Remember that photons aren't conserved in this situation.
…Why do you say that since we have a dense set of possible values of wavelengths we must have 0 photons for each wavelength? This is a wrong conclusion but nevertheless I need to know how you arrived at it.
..…[/b]

Suppose there is a blackbody that is radiating out photons and suppose in a snapshot of time there are 10^10 photons in a particular cubic meter volume that all originated from the blackbody. Now the photon there that happens to have the longest wavelength (out of all the photons currently there) may be, say, 40000nm in wavelength while the shortest may be, say 400nm in wavelength. So we have a range of wavelengths but this range can be split up into many smaller ranges such as the range from 1000nm to 1001nm and each of those small ranges would have less photons existing in that smaller wavelength range in that specified cubic meter than the 4000nm to 40000nm range and each of those smaller ranges such as from 1000nm to 1001nm can be split into many even smaller ranges such as from 1000.01nm to 1000.02nm and each of those small ranges would have less photons existing in that wavelength range in that specified cubic meter than the 1000nm to 1001nm range and so on until you get to the range like from 1000.00000000000001nm to 1000.0000000000002nm where you can calculate the probability of a particular photon with a wavelength existing in that tiny range in a particular cubic meter to be small despite there being, say, 10^10 photons in that particular cubic meter. Thus considering the mathematical limits that are implied here, I conclude that the probability of there being a photon in that particular cubic meter with a wavelength of EXACTLY 1000.0000000000000155555555nm to be virtually zero because there are an infinite number of numbers between 1000.00000000000001 and 1000.0000000000002.

(sorry for using a ridiculous number of words but I am not good at explaining what I mean here)

-obviously I am missing something here.