# I Atom as a harmonic oscillator of radition

1. Aug 11, 2017

### PainterGuy

Hi

Q1:
I was reading about ultraviolet catastrophe and it was said that atoms were assumed to be harmonic oscillators of radiation.

I believe that two harmonic oscillators could have the same frequency but different amplitudes so it would mean that two different atoms (i.e. two harmonic oscillators) might be emitting same frequency electromagnetic waves but with different amplitudes. Do I have it right? Please keep your answer simple.

Assuming that what I said above is correct, the electromagnetic wave with greater amplitude would produce more brightness compared to a blue color electromagnetic wave with half the amplitude. The overall emitted blue color electromagnetic wave would have an average amplitude.

Q2:
The following is an excerpt from the Wikipedia article on ultraviolet catastrophe.

The ultraviolet catastrophe results from the equipartition theorem of classical statistical mechanics which states that all harmonic oscillator modes (degrees of freedom) of a system at equilibrium have an average energy of .

An example, from Mason's A History of the Sciences,[2] illustrates multi-mode vibration via a piece of string. As a natural vibrator, the string will oscillate with specific modes (the standing waves of a string in harmonic resonance), dependent on the length of the string. In classical physics, a radiator of energy will act as a natural vibrator. And, since each mode will have the same energy,
most of the energy in a natural vibrator will be in the smaller wavelengths and higher frequencies, where most of the modes are.

According to classical electromagnetism, the number of electromagnetic modes in a 3-dimensional cavity, per unit frequency, is proportional to the square of the frequency. This therefore implies that the radiated power per unit frequency should follow the Rayleigh–Jeans law, and be proportional to frequency squared. Thus, both the power at a given frequency and the total radiated power is unlimited as higher and higher frequencies are considered: this is clearly unphysical as the total radiated power of a cavity is not observed to be infinite, a point that was made independently by Einstein and by Lord Rayleigh and Sir James Jeans in 1905.

Source: https://en.wikipedia.org/wiki/Ultraviolet_catastrophe#Problem

I don't understand the first statement in red. I believe that modes are distinct frequencies. How come most of the modes are in higher frequencies. The equal number of modes could be in low frequencies too. There might even be radio waves. Please have read the highlighted part in this text: http://imageshack.com/a/img924/7230/jFDaZK.jpg. I have attaching the image as an attachment too but not sure it would show up clearly. Even this text says that most of the energy is contained in high frequencies without really explaining the reason, at least as I see it. Please guide me.

I think that the second statement in red means that the number of vibrators (atoms) emitting an electromagnetic wave of certain frequency is proportional to the square of the frequency. Do I have it right?

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2. Aug 11, 2017

### sophiecentaur

Take a string with transverse waves on it. There is a fundamental mode and there is a first overtone that would be a second harmonic - and no more. That would mean there is only one mode in the first octave (1). The third and fourth harmonics are in the next octave (2). 5th, 6th 7th and 8th (4) are in the next octave. 9,10,11,12,13,14,15,16 (8) in the next octave etc etc, until ultraviolet. If all modes had the same energy, you have your catastrophe.
There isn't room on the string for more than a few modes per Octave at the low frequencies.

3. Aug 11, 2017

### PainterGuy

Thank you.

I'm sorry but the answer didn't help me.

An atom when it is radiating an electromagnetic wave wouldn't care much if its radiated electromagnetic wave would fit inside the cavity or not. It was also thought that a blackbody would could emit and absorb all electromagnetic radiation without any reflection. So, my question is still that what factor was dictating the atoms to start with some fundamental frequency which fits in the cavity and then emit all other higher frequencies which are harmonics. Please guide me.

Could someone please comment on Q1 from my first post and the following question which is also from my original post. "I think that the second statement in red means that the number of vibrators (atoms) emitting an electromagnetic wave of certain frequency is proportional to the square of the frequency. Do I have it right?"

Thank you for the help.

4. Aug 11, 2017

### sophiecentaur

How could the radiated frequency be different from the frequency of the cavity? What about the boundary conditions that always need to apply?

5. Aug 11, 2017

### PainterGuy

I don't know what to say. I'm sure that you and many other senior members with lots of knowledge are here to help others. I'm of the opinion that when helping someone, you need to keep in mind the level of knowledge of the person who is asking the question. I hope that you won't mind my saying that your answer was not so much different from the ones floating around the internet. Einstein was right when he said that every problem should be made as simple as possible but not simpler. The sad part is that so much mumbo jumbo and extensive mathematics is thrown at the learners that it just scares them. There is no doubt that mathematics really captures the intuitive and conceptual understanding in a succinct manner but still not all of the intuitive and conceptual could be sacrificed for the sake of mumbo jumbo.

I believe that science shouldn't be treated as a rigid set of rules and theorems. I'm sure that everything can be made simple if we completely understand the anatomy of problem. People like you could really understand the anatomy of most problems but it's just that you people need to figure out the way to make the solution more simple so that it could be accessible by everyone, or at least the person who requested you for help.

The worst part is that many forums are moderated in such a way that no one is willing to take a criticism in a positive way, and instead they start crucifying others.

Anyway, I don't know anything about frequency of cavity or boundary conditions, and yes, I also don't know quantum mechanics etc. I still appreciate your effort. I'm sure that I would find answers to my questions some day.

Thank you.

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6. Aug 11, 2017

The explanation that resolved the ultraviolet catastrophe is that the modes of the cavity do in fact exist, but their average occupation number, (i.e. the average number of photons in each mode), is computed by the Bose-Einstein factor to be $\bar{n}_s=\frac{1}{e^{hc/(\lambda kT)}-1}$, instead of the result from the equipartition theorem. Thereby, the Planck blackbody function, (which uses this factor), gives the correct result for the radiated spectrum across the entire spectrum. For shorter wavelengths $\lambda$, such as in the ultraviolet (UV), this $\bar{n}_s$ becomes a small number, and also agrees extremely well with all experimental results.

Last edited: Aug 11, 2017
7. Aug 11, 2017

### jbriggs444

The universe is not obliged to arrange itself to fit this prescription.

8. Aug 11, 2017

### ZapperZ

Staff Emeritus
Wait... you're asking about EM wave in cavity, but you're dismissing the physics involved with it. Do you not want to learn about what a cavity is?

When you have a cavity such as the one described here, it is similar to having a waveguide, i.e. only certain modes of EM wave can exist. This is not something we can make up. It is what we have seen and what Mother Nature (remember her?) has dictated. These mathematics that you ignoring is a reflection of the physics principles that Mother Nature has given us.

And we're making this "as simple as possible but NOT SIMPLER", because unless you're willing to spend effort to know about boundary conditions of EM wave and why only such modes exists, then you might want to attempt to explain to a 2-year old the same thing.

How do I force the solution to be more "simple" when it isn't? What you're saying is that if we simply lower the net, and maybe not care about the lines, then everyone should be able to be a pro-tennis player. What makes you think that someone who doesn't have the skill to be a tennis player should be as good as someone who is a pro and has spend years practicing and becoming good at it?

And the worst part about many forums is the sense of entitlement that many people seem to have, as if they are entitled to understand intimately what essentially takes years of learning. And you really should play by your own rules, because when you received replies that were beyond your comprehension, you turn around and BLAME it on the people who responded, as if was their fault that you don't have the capability to understand what you were given.

Zz.

9. Aug 11, 2017

### PainterGuy

I wasn't blaming or criticizing anyone. That was not my intention in any sense. What I was trying to do was just a request to all knowledgeable persons out there to make it little simple whenever they can so that their energy and effort to help others could be more effective and appreciated in a proper manner. And that was my own opinion which might be wrong. Thank you for your understanding.

10. Aug 12, 2017

### sophiecentaur

That is exactly what Physics is trying to do all the time. The fact is that it is not simple it cannot be described in the sort of terms you want.
When the answers you are getting are too hard for you, you cannot complain. All you can do is study the subject and find what they actually mean. There is no short cut.

11. Aug 12, 2017

### Cutter Ketch

In thinking about atoms with different frequencies, I think you may be starting from a bad place. The Rayleigh-Jeans model and the Planck model of the radiation from a black-body were developed without reference to atoms. The atomic model didn't exist. It was being developed at the same time and using many of the same leaps of understanding that fixed the ultraviolet catastrophe. Instead these questions were addressed in regards to standing waves in a cavity of a blackbody radiator. Physicists were aware that if you took a hollow object like a metal sphere with a small hole in it and heated it you would get a very specific spectrum. This is what they were trying to describe.

The hole is very black because any light entering has a very small statistical probability of rattling around and coming back out. So logically this is also a nearly perfect blackbody radiator.

So with this device in mind they added up the EM modes that could exist in the cavity. We know that the transverse EM field has to go to zero at the metallic surface. In other words the standing of waves in the cavity have to have nodes at the surfaces. This is where the analogy to a rope comes in. You have standing waves with nodes at the ends just like on a rope. (With some additional counting because the cavity isn't 1 dimensional)

Now we get to one of your main questions: why are the modes more dense at higher frequency. Well, picture the fundamental and first harmonic modes on a rope. Their wavelength and frequency differ by a factor of two. Now picture the mode with 100 antinodes and the mode with 101 antinodes. Their wavelength and frequency differ by only 1%. You can see how the number of modes get closer and closer together in frequency as the number grows.

I hope that helps

12. Aug 20, 2017

### PainterGuy

Yes, you are right about the atomic model. I was trying to get a conceptual hold of blackbody radiation phenomenon and the ultimate catastrophe from historical perspective. I believe that at that time plum pudding model was being used. They were thinking in terms of standing waves and oscillators.

So, a hollow blackbody radiator wasn't just a hypothetical object. A hollow metallic sphere would also emit radiation from its outer surface but I believe that their analysis wasn't concerned about that because only the inner hollow structure approximated a blackbody.

The following is an excerpt from a book. "A good approximation to a blackbody is a hollow box with a small aperture in one wall. Light that enters the aperture will eventually be absorbed by the walls of the box, so the box is a nearly perfect absorber. Conversely, when we heat the box, the light that emanates from the aperture is nearly ideal blackbody radiation with a continuous spectrum."

Why do the EM standing waves need to have nodes at the inside surface of metallic sphere? The following is one of the webpages.

"Electromagnetic standing waves in a cavity at equilibrium with its surroundings cannot take just any path. They must satisfy the wave equation in three dimensions. The solution to the wave equation must give zero amplitude at the walls, since a non-zero value would dissipate energy and violate our supposition of equilibrium."

There should be some reason that the surface only allows those EM standing waves which can have nodes at the surface and others, like radio waves having wavelength in km, which couldn't fit the nodes are not permitted.

In a metallic hollow sphere, EM standing waves don't need to pass the center of sphere, or do they? I mean that there could be some standing waves along the chords of sphere and not along the diameter. Please have a look here if what I'm saying is not clear.

Further, any EM standing wave could be oriented along so many planes like unpolarized light waves. So, is this how each standing wave is oriented along so many planes in a hollow metallic sphere blackbody radiator?

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13. Aug 20, 2017

@PainterGuy For the way the modes get counted, this posting might be helpful: https://www.physicsforums.com/threads/boltzmann-vs-maxwell-distribution.918232/ (See post #4 in particular.)There is also a factor of 2 for the two directions of polarization. The number of modes in $k$ space, and the number of modes in momentum space $p$ is very much a quantum mechanical and statistical physics calculation. That such a thing will work really isn't intuitively obvious IMO, but that is the accepted way of computing them, and it works to give the correct result. The geometry that is used in the derivation is usually rectangular, e.g. a cube, but the result for the number of modes per energy interval is proportional to the volume, regardless of the geometry. In this posting, the problem involved the number of momentum states for the particles of a gas, but somewhat surprisingly, the mode counting is the same for the photon modes in a cavity.

14. Aug 21, 2017

### sophiecentaur

Because they are Standing Waves. That's a definition. You can argue about how valid the main assumption is but that was modified and dealt with years ago by Planck.
You have skipped half way into the reasoning that was used by Rayleigh / Jeans and should really start at the beginning if you want to have a valid opinion.

15. Aug 21, 2017

### f95toli

That is not correct. An atom can only radiate if there is a mode of the right frequency available. The problem here is that you can't think of the atom and the cavity as separate systems; what happens will depend on the combined properties of both the atom and the cavity and it ends up being quite complicated. There is a whole area of quantum physics that deals with these type of systems (cavity Quantum Electrodynamics),

Note also that the atom is MORE likely to radiate (the lifetime of decaying state is reduced) if there is a cavity mode available than if it was in free space; this is the so-called Purcell effect.

Edit: Note, however, that you will only get the Purcell effect if the coupling between the atom and the cavity is strong.

Last edited: Aug 21, 2017