How to calculate radiance emitted?

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.

Originally posted by Andrew Hamilton
…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.
..…

Suppose there is a blackbody that is radiating out photons and suppose in a snapshot of time there are 10^10 photons in a particula ...[text shortened]... rds but I am not good at explaining what I mean here)

-obviously I am missing something here.[/b]

The blackbody radiation spectrum is derived from the limiting case of considering a box with edge L and taking the limit as L goes to infinity. In this limit, the possible wavelengths also form a continuum (and the chance of a particular wavelength being emitted is indeed zero), but in the case of finite L, the spectrum is discrete and single wavelengths can have a finite probability.

Originally posted by Andrew Hamilton
…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.
..…

Suppose there is a blackbody that is radiating out photons and suppose in a snapshot of time there are 10^10 photons in a particula ...[text shortened]... rds but I am not good at explaining what I mean here)

-obviously I am missing something here.[/b]

You are talking about the photons as if they had definite frequencies. They don´t, real particles come as wave packets, so there is a blurring of the wavelength even when only a single photon is involved. This means that you are getting confused about measure, essentially you are integrating a density function (photon number density vs frequency), although it is possible to do the integral consistently even if photons have definite frequency using delta functions, from a real physics point of view you don´t have to. Each photon has a probability density of having a wavelength and the sum of these is what you are integrating.

Originally posted by Andrew Hamilton
…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.
..…

Suppose there is a blackbody that is radiating out photons and suppose in a snapshot of time there are 10^10 photons in a particula ...[text shortened]... rds but I am not good at explaining what I mean here)

-obviously I am missing something here.[/b]

But that's a common thing whenever you're calculating the probabilities of a continuous variable.

A common example: what is the probability of you having the height that you have? It's 0!!! But you do have the height that you have don't you. 😉

Mathematically speaking: if you want to calculate the probability of a continuous variable you have to do it by calculating an integral.

Let's make things concrete by giving a specific example: The height of people in the world. You define the probability density [; \rho ;] and then if you want to know the probability for someone having a height in a given interval [; (h_1 , h_2) ;] you just calculate the integral [; \int_{h_1}^{h_2} \rho ;]. Now if you want to know the probability of having a specific height the integral just is : [; \int_{h_1}^{h_1} \rho ;] and the integral is calculated just in point it amounts to 0!!!

So now you know that just because have a 0 probability it doesn't mean that the event doesn't happen.
But the big problem isn't that. The problem is that in a black body situation it doesn't make much sense to keep a count of the number of photons because they change. In technical jargon: it isn't a quantum number.

Originally posted by adam warlock
But that's a common thing whenever you're calculating the probabilities of a continuous variable.

A common example: what is the probability of you having the height that you have? It's 0!!! But you do have the height that you have don't you. 😉

Mathematically speaking: if you want to calculate the probability of a continuous variable you have to d ...[text shortened]... r of photons because they change. In technical jargon: it isn't a quantum number.

Well, it doesn't happen. A black body in the limiting case of L to infinity will never emit a photon of, say, exactly 544 nm.

Originally posted by KazetNagorra
Well, it doesn't happen. A black body in the limiting case of L to infinity will never emit a photon of, say, exactly 544 nm.

This is incorrect. That an event has probability zero of happening doesn't mean it cannot happen, only that it almost surely won't happen. Only the reverse is true (an event that cannot happen has necessarily probability zero of happening) .

Originally posted by Palynka
This is incorrect. That an event has probability zero of happening doesn't mean it cannot happen, only that it almost surely won't happen. Only the reverse is true (an event that cannot happen has necessarily probability zero of happening) .

I didn't say it cannot happen, just that it won't happen.

Originally posted by KazetNagorra
I didn't say it cannot happen, just that it won't happen.

Well, that's imprecise too, because events of probability zero may happen. They almost surely won't happen.

Originally posted by Palynka
Well, that's imprecise too, because events of probability zero may happen. They almost surely won't happen.

Hmm no, in this case it's absolutely certain it won't happen.

But then again, this whole perfect black body is rather artificial since it doesn't occur in nature.

Originally posted by KazetNagorra
Hmm no, in this case it's absolutely certain it won't happen.

But then again, this whole perfect black body is rather artificial since it doesn't occur in nature.

Then you don't understand what probability zero means. That's fine. You're not alone.

Originally posted by Palynka
Then you don't understand what probability zero means. That's fine. You're not alone.

Consider some finite time interval. In this finite time interval, a black body will emit a finite amount of photons. Consider one emitted photon, which has some random wavelength. In the decimal representation of the wavelength, the probability of one decimal corresponding is 10%, at least in the limit of a large numer of decimals behind the comma (so that the actual physics is irrelevant). So the probability of all decimals corresponding to some finite value is simply the limit of 0.1^x as x goes to infinity, in other words, zero. So zero photons of wavelength 788 nm (or whatever) are emitted in this finite time interval. Now take the limit as the time goes to infinity - still zero. It will never happen.

Perhaps not very mathematically rigorous, but that's basically what the integral representation of the probability distribution means.

Originally posted by KazetNagorra
Consider some finite time interval. In this finite time interval, a black body will emit a finite amount of photons. Consider one emitted photon, which has some random wavelength. In the decimal representation of the wavelength, the probability of one decimal corresponding is 10%, at least in the limit of a large numer of decimals behind the comma (so t ...[text shortened]... s, but that's basically what the integral representation of the probability distribution means.

Jesus, man, what I'm saying is that there is a difference between something having probability zero and never happening. How does your sequence converging to probability zero help you?

Please read up on what almost surely means in probability.

Edit - Besides, if you take the limit to infinity, then the distribution of photon wavelengths ever emitted will converge to its probability distribution, which means that the set of photons with exactly 544 nm will be non-empty (as long as 544 nm is in the support).

Originally posted by KazetNagorra
Well, it doesn't happen. A black body in the limiting case of L to infinity will never emit a photon of, say, exactly 544 nm.

Yawn...

But humor me on this one please: why do we analyze the case L=0 and then take the limit L to infinity when we are studying a black body?

Originally posted by KazetNagorra
Consider some finite time interval. In this finite time interval, a black body will emit a finite amount of photons. Consider one emitted photon, which has some random wavelength. In the decimal representation of the wavelength, the probability of one decimal corresponding is 10%, at least in the limit of a large numer of decimals behind the comma (so t ...[text shortened]... s, but that's basically what the integral representation of the probability distribution means.

Consider some finite time interval. In this finite time interval, a black body will emit a finite amount of photons
Why is the number of photons emitted finite?

Consider one emitted photon, which has some random wavelength. In the decimal representation of the wavelength, the probability of one decimal corresponding is 10%, at least in the limit of a large numer of decimals behind the comma (so that the actual physics is irrelevant). So the probability of all decimals corresponding to some finite value is simply the limit of 0.1^x as x goes to infinity, in other words, zero. So zero photons of wavelength 788 nm (or whatever) are emitted in this finite time interval. Now take the limit as the time goes to infinity - still zero. It will never happen.
I'm fighting really hard to make sense of this. Do you think you can be more clear?

Perhaps not very mathematically rigorous, but that's basically what the integral representation of the probability distribution means.
No. It isn't just a case of hand-waving it is a case of missing everything that there is to miss about a given subject.

Thanks to all for the help; I still struggling hard with some of the concepts but I think I have got what I want from this thread. But I now have found something new that confuses me:

I have heard of wave packets but looked it up to revise them:

http://www.statemaster.com/encyclopedia/Wave-packet

“…the energy of light packets is a discrete function of frequency:

E = nhf

The energy, E, is given as an integer, n, multiple of Planck's constant, h, and frequency, f. ….”

AND:

http://209.85.229.132/search?q=cache:ys2HjnDlgawJ:ualr.edu/planetarium/physics/davis/sddsecondsemester/CH30.pdf+wave+packets+%22E+%3D+nhf%22&cd=4&hl=en&ct=clnk&gl=uk&lr=lang_en

“…Planck's relation was E = nhf
….
and energy was emitted from an atom as a transition between two different energy states (i.e., different n's). …”


Now this confuses me because I was always taught that the energy of a photon is given simply by:

E = hf

Which I understand perfectly.
But now I see it written as E = nhf where n can be any positive whole number!
I understand perfectly that the energy of an emitted photon from an atom is determined by transition between two different energy states (i.e., different n's) but I assumed that the difference in potential energy between the two different energy states in question is all that determines the frequency AND therefore the energy of the admitted photon?

So what confuses me about E = nhf is that the “n” in it seems to imply to me that you could have two photons with identical frequency f and yet for one n=1 so that its energy is E = hf but for the other n=2 so that its energy is E = 2hf !!! How could TWO different photons with the SAME frequency have DIFFERENT energies?
Please could someone tell me what I am missing here.