Question

I'm trying to understand Plancks formula. I started off with: E= hv/(e^(hv⁄((kT)))-1) but I've found other formulas too. I'd like to use this formula or some other to plot my own radiation curve so I'd like to know what data would work (and it's meaning) so that I can do this. Thanks for any help.

Answer 1

http://phet.colorado.edu/en/simulation/blackbody-spectrum

Answer 2

Thanks Carl I'll have a look at that. Ultimately I'm trying to understand the maths too. I thought I'd be able to put the figures in to get E then dividing this by hv should always give an integer result - well thats what the book I was reading said Planck found. But I couldn't even get the curve right. Thanks for taking an interest again!

Answer 3

I've got the idea of the physics - but I'm a bit puzzled as to how the formula generates the curve - I'll see if anyone in the maths section can throw any light on it. TTFN

Answer 4

It's not E that is primeval, and you derive the number of photons by dividing by hv, it's the number of photons that is primeval, and you derive the energy by multiplying the number of photons by hv. For a given E, you can certainly ask yourself whether that precise E could be the result of a set of photons of one frequency -- you divide by hv and see if you get an integer. If you do, great, it can, and that number is how many photons of frequency v you need. If it can't -- then either that energy is impossible (if you are resrticted to photons of a fixed v) or you need photons of some other frequency to mix in.

I don't actually know how Planck first derived the blackbody curve, since he did it without quantum mechanics. That's an interesting question, actually.

The modern derivation is fairly complex. You will need several physics concepts, such as the normal modes of a standing wave in a cavity, the density of states, the quantum restriction on the amplitude of the modes (that's where Planck's constant and the E = n hv formula comes in) as well as the Boltzmann probability distribution of states and the relationship between thermodynamics energy and the Zustandsumme or partition function. It's usually only worked out thoroughly in graduate statistical mechanics classes, although you often refer to it in undergraduate classes. You'll find a careful derivation in any standard graduate statistical mechanics textbook, for example McQuarrie, "Stastistical Mechanics."

I wish I could think of a quick way to at illustrate the basic ideas, but it's tough. Here's the best I can do:

(1) We're trying to describe the distribution of energy among electromagnetic "normal modes" in a cavity with hot walls (which will also work for empty space near a hot object -- that's just an infinite cavity with one (maybe curved) wall). The importance of the wall is that it serves to easily convert radiation at one wavelength to radiation at another. That is, two green photons can collide with the wall, be absorbed, and then have a blue and a red photon (totally up to the same energy) re-emitted. This ia realistic and very general scenario for the interaction of radiation and matter.

(2) We begin by asking ourselves what kind of electromagnetic standing waves fit into the cavity. Clearly there are restrictions on their wavelength: an even number of wavelengths has to fit into the cavity. These are the "normal modes," just like the normal modes of a vibrating string, consisting of the fundamental, first overtone, and so forth. This gives us a "density of states," meaning how many modes we have within a given frequency range, e.g. between 10 GHz and 11 GHz, or equivalently between a wavelength of 500 nm and 501 nm.

(3) Now we have to decide how the energy is distributed among all these modes. There are good general arguments (and common sense says) the probability of a given mode having energy in it should be indendend of the frequency. So the mode at 501.234567 nm should be just as likely to have 1.5 pJ of energy in it as would be the mode at 501.234568 nm.

(4) This leads to the conclusion that the total amount of radiation we see at a given wavelength should be proportional to the density of states at that wavelength: the higher the density of states (the more normal modes are crammed into each nm), the more energy we should see. And, indeed, if you work this out mathematically, at low frequencies it works out just as you'd expect. You get a curve (of energy at a given frequency) that rises as the density of states rises (roughly as frequency square). This is the Rayleigh-Jeans law, which can be derived classically, and works fine at low frequency.

(5) But the problem is that the density of states rises without limit as frequency rises, which means there is no upper limit to the Rayleigh-Jeans law: the energy emitted at a given frequency should rise and rise with frequency until it reaches infinity, or at least an absurdly huge number. This is the "ultraviolet catastrophe" that Planck was trying to solve.

(6) Quantum mechanics resolves this problem, because it turns out that the probability that a given mode has energy in it is NOT the same, because we can only add energy to a given mode in "units" (quanta) of hv. What that means is that for higher and higher frequencies, we must allocate energy in larger and larger chunks. Why does that matter? Because there are far fewer ways to allocate a given amount of energy in large chunks than in small chunks. So if the energy is disitributed in all the ways it can be, without favoring one way over another, what will happen is that the ways of which there are the most -- which is when the energy is distributed in small chunks, at lower wavelengths -- will overwhelmingly dominate. Those in which the energy is distributed in very high frequency modes will be exceedingly rare. Hence, the probability of seeing energy at higher at higher frequencies drops off and eventually reaches zero, as you'd expect.

Think of it this way. The highest possible frequency I can observe is where all of the observed energy is allocated to one extremely high frequency photon. How likely is that? Not very! Compare to having the energy distributed among a million photons of much lower frequency, where there are bazillions of ways to distribute those photons. Much more likely! So that's what we see.

You may be horrified to discover by this last bit that the blackbody distribution is a statistical, and not mechanical, result. That is, a given black body at a given instant does not HAVE to follow the blackbody distribution law -- it violates no law of physics if it does not -- but it's very, very unlikely that it won't. Nevertheless, this is the case. It may be comfort to know that quite a lot of other laws -- such as the ideal gas law, Ohm's Law, and the like -- actually fall into the same category. A fair amount of what we observes is only statistically true, but we consider it to be true absolutely because the statistics are such that deviations are unbelievably rare. It's like the fact that in four coin tosses you may easily NOT see two heads and two tails, because half heads is not an absolutely law, but only a statistical law. But in four trillion coin tosses, the deviations from 0.5 x 10^12 heads would be percentages so small as to be difficult to write down.

You may also notice the Second Law of Thermodynamics sneaking in there -- that clearly although I have not used the term the idea of maximizing the entropy of the system of photons is relevant. And indeed that is so. You can tell the Second Law is important because temperature enters the final equations -- and any formula that contains temperature is necessarily statistical, and necessarily involves the Second Law.

There's a lot of key physics hidden in that curve.

Answer 5

Thanks Carl for a really brilliant answer. I've actually been able to generate my own curve from Plancks formula using an excel spreadsheet. Along the way I did actually uncover the fact the the formula I originally had was for a single photon and that the 'correct' version came in two forms - one using frequency and one using wavelength. Now here's something I don't understand in (1) You've written:-
" That is, two green photons can collide with the wall, be absorbed, and then have a blue and a red photon (totally up to the same energy) re-emitted."
OK - but in looking at the photoelectric effect a photon of sufficient frequency (energy) causes an electron to be emitted rather than another photon. I'd love to know what makes the difference. Perhaps I should post another question on this? Anyway thanks for your help again.

Answer 6

Well, if the photons are below the cutoff frequency, then of course they won't eject anything, they'll just be reflected, scattered or absorbed and turn to heat. The cutoff frequency for most metals is in the UV, so this is why metals are shiny - most of the visible light you throw at them comes right back at you. But even above the cutoff frequency, not every photon will eject an electron. Indeed, I expect only a small fraction will.

Answer 7

Yep - now I get it! I'd been a bit confused by the fact that photons mediate the force between electrons. Now I see that a photon with insuffient energy to eject an electron will just be reflected. Actually my understanding is that the energy is absorbed by an electron that jumps to a higher level and it's when it jumps back down again that a photon is emitted? I find all this fascinating and I have lots more questions but I think I need to study a bit harder so my questions make sense. Thanks for helping Carl - I'm a fan and I've kept your original answer in my notes it was so useful!

Answer 8

Well, now if you want to talk about the photon as a mediator for the electromagnetic force, that gets still more complicated -- now you're off in quantum field theory. As I understand it, a particle of spin 1, like the photon, necessarily has three possible polarization, one longitudinal and two transverse. It is the longitudinal photon that mediates the electromagnetic force, and the two transverse photons that constitute the EM radiation we observe. In other words, the photon as a force carrier and photon as a component of EM radiation are two different beasts. (This is not to say the transverse photons do not carry momentum and cannot cause forces to arise -- they can. But they are transverse to the direction of the EM radiation. Forces that are parallel to the vector between source and object affected -- like the Coulomb force -- must be carried by something else.)

Generally, yes, we would describe the scattering of a photon by an electron in a bound state (e.g. in an atom or metal) as absorption and emission. But the absorption does not always have to precede the emission -- the other way around, where the photon is first emitted, and then absorbed, is also possible. You may object that this means energy is not conserved during the process, and you would be correct -- but that's one of the weirdnesses relativity brings to atomic physics. Energy does not need to be conserved in the middle of complex processes.

Answer 9

Thanks Carl for another brilliant answer. I think I can safely say that you've well and truely answered my question. You're right about my delving (at a very amateur level) into Quantum Theory. I've read "The Quantum Universe" by Brian Cox and Geoff Foreshaw. I'm just sort of padding out what prompted the development of Quantum Theory in the first place. There's certainly one part of the book were I didn't understand what they were getting at but that can wait for another time. Thanks again for your time, patience and well expressed answers.