How to calculate radiance emitted?

Originally posted by Andrew Hamilton
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:

[b]http://www.statemaster.com/encyclopedia/Wave-packet

“…the energy of light packets is a dis ...[text shortened]... ME frequency have DIFFERENT energies?
Please could someone tell me what I am missing here.[/b]

n represents the number of photons. So if n=2 you have two photons transiting.

Originally posted by Palynka
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 emi ...[text shortened]... of photons with exactly 544 nm will be [b]non-empty (as long as 544 nm is in the support).[/b]

Well, Jesus man, tell me what you mean by "almost surely".

Originally posted by KazetNagorra
Well, Jesus man, tell me what you mean by "almost surely".

It's a well-defined concept in probability theory.

http://en.wikipedia.org/wiki/Almost_surely

Emphasis on this part.
http://en.wikipedia.org/wiki/Almost_surely#Throwing_a_dart

Originally posted by Palynka
It's a well-defined concept in probability theory.

http://en.wikipedia.org/wiki/Almost_surely

Emphasis on this part.
http://en.wikipedia.org/wiki/Almost_surely#Throwing_a_dart

That's basically what I said. If you keep throwing darts at that square, a dart will never exactly hit the diagonal, even though in retrospect the point that the dart landed on had zero probability.

Originally posted by adam warlock
n represents the number of photons. So if n=2 you have two photons transiting.

oh right! 😛 I knew I was missing something. Odd that I couldn’t find any websites that stated what “n” is given it is simply the number of photons rather than some property of a single photon.

Originally posted by Andrew Hamilton
oh right! 😛 I knew I was missing something. Odd that I couldn’t find any websites that stated what “n” is given it is simply the number of photons rather than some property of a single photon.

Well, if you did have a single photon with energy E = nhv, the frequency would simply be nv.

Originally posted by KazetNagorra
That's basically what I said. If you keep throwing darts at that square, a dart will never exactly hit the diagonal, even though in retrospect the point that the dart landed on had zero probability.

Pay attention. Every dart throw results in an event that had probability zero.

Therefore, events with probability zero are not events that cannot or won't happen.

Originally posted by KazetNagorra
That's basically what I said. If you keep throwing darts at that square, a dart will never exactly hit the diagonal, even though in retrospect the point that the dart landed on had zero probability.

That's nothing of what you said.

Originally posted by Palynka
Pay attention. Every dart throw results in an event that had probability zero.

Therefore, events with probability zero are not events that cannot or won't happen.

You can start throwing those darts, give me a call when you hit the diagonal.

Originally posted by KazetNagorra
You can start throwing those darts, give me a call when you hit the diagonal.

It's funny how a so-called student of quantum physics can be so ignorant about basic probability theory. This is 101 stuff in any university.

Originally posted by Palynka
It's funny how a so-called student of quantum physics can be so ignorant about basic probability theory. This is 101 stuff in any university.

I think you just misunderstood what I said (or I worded it badly).

Originally posted by KazetNagorra
I think you just misunderstood what I said (or I worded it badly).

Nice cop-out, but considering you "tried" to correct adam when he said:

So now you know that just because have a 0 probability it doesn't mean that the event doesn't happen.

then I don't know how it's possible that I misinterpreted you.

Originally posted by adam warlock
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?

Bump for Kazet!

And please read up on probability cause you really need it.

Originally posted by Palynka
Nice cop-out, but considering you "tried" to correct adam when he said:

So now you know that just because have a 0 probability it doesn't mean that the event doesn't happen.

then I don't know how it's possible that I misinterpreted you.

I don't see the conflict - if you regard a black body at t = 0 and wait for a photon to be emitted at exactly 744 nm it won't happen. The wavelength of an emitted photon will contain an infinite string of random numbers in decimal representation and even though the infinite string of all zeros is not any less likely than any other particular infinite string it's safe to say a black body will never emit a 744 nm photon (or "almost sure", if you insist).

Originally posted by adam warlock
Bump for Kazet!

And please read up on probability cause you really need it.

You don't analyze the case L = 0, since that doesn't allow any EM mode. You start with some finite L, which allows wavenumbers (in 1D, generalization to 3D is straightforward) k = 2*pi*n/L with n integer. Taking the limit of L to infinity allows a continuum integral formulation appropiate for thermodynamics, and allows photons will arbitrarily low energies instead of the lower limit corresponding to k = 2*pi/L in the finite L case.