- #1
PainterGuy
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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?
Thank you for your help.
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?
Thank you for your help.