# Ultraviolet catastrophe and quantization

1. Jan 22, 2012

### Puchinita5

1. The problem statement, all variables and given/known data

I'm taking physical chemistry 2 right now and we are discussing the basics of quantum mechanics, and i'm trying to understand black body radiation and the ultraviolet catastrophe.

What I don't understand is how quantization and plancks distribution fix the ultraviolet catastrophe. I have googled a million websites, including answers on this forum, and it all sounds like jibber jabber. So here is what is in my head, someone please explain to me what I am really misunderstanding and try to dumb it down as much as possible!

So, a blackbody is something that absorbs all wavelengths of energy, and also emits all wavelengths of energy? And classical physics says that as you decrease the wavelength, the amount of energy reradiated by the object should get bigger and bigger. so theoretically, if you had an infinitely small wavelength, the object should radiate infinite energy? But the models disagree with this.

Then Planck said that energy is quantized. But how does this stop the energy from going to infinity? Because the way i picture it in my head, is that instead of a solid curved line going up to infinity, it should just be a dotted line instead since energy can only be integer values.
How does "energy packets" mean that instead of going to infinity it slopes back downward?

I am obviously way off, I just really cannot find an explanation dumbed down enough for me to understand.

The answers online i keep finding mention "oscillators" and "modes" and I have no idea what these mean so it doesn't help me at all.

2. Jan 22, 2012

### Mindscrape

Long story short (if you want the long story look of the derivation of Planck's law), having discrete wavelengths (think particle in a box) means discrete energies. Having energy packets limits the number of modes at higher energies, shorter wavelengths, and the ultraviolet catastrophe is solved.

3. Jan 22, 2012

### Puchinita5

I guess what I don't understand is what a "mode" is. Why does having energy packets make a difference? Couldn't you add an infinite amount of energy packets and so still an infinite amount of energy?

ahhh quantum is demoralizing.

4. Jan 22, 2012

### silverdiesel

Plancks hypothesis was simply that E=hv.

This is so common now its hard to think otherwise. In fact, in your question, you stated the answer. What you claim is classical, is actually quantum. Higher energy = higher frequency.

Classically, higher energy = higher amplitude. So, classically, you have a Gaussian distribution of wavelengths, some high, some low, evenly spread about the mean. However, a black-body is a squewed Gaussian, meaning there are more in the lower frequencies, and very few in the higher frequencies. Planck solved this by guessing that the energy of emitted radiation is proportional to the frequency. Then, by conservation of energy, higher frequencies are limited, and lower frequencies will be more common.

He did not realize this had implications of quantization until Einstein used his hypothesis (E=hv) to explain the photoelectric effect, which worked very nicely, but required that light be described as discrete particles with E=nhv, n=1,2,3...

A "mode" is a method of vibration which is stable. The easiest case to think about are two pendulums connected with a spring. They can either swing togther, or fully out of phase. Those are the two modes. Anything in between will randomly swing about until it either settles into one of the two modes, or completely looses all the energy to friction. This is a real world, macroscopic example of quantization.

5. Jan 22, 2012

### vela

Staff Emeritus
Take, for example, a standing wave on a string with two fixed ends.

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html#c3

Because of the boundary conditions, there are only certain allowed frequencies and wavelengths, which are labeled by n=1, n=2, etc. Each value of n corresponds to a mode.

Electromagnetic waves behave the same way, for instance, when they're in a metal cavity. Only certain frequencies will satisfy the boundary conditions. These correspond to the allowed modes.

You can do a calculation to find the number of allowed modes in a range of frequencies from $\nu$ to $\nu+d\nu$, and you'll find that it's given by
$$N(\nu)\,d\nu = \frac{8\pi \nu^2 V}{c^3}$$where V is the volume of the cavity.

In the classical picture, you can vary the energy by changing the amplitude of the standing wave. Since the amplitude can take on any continuous value, the energy of each mode can take on any continuous value from 0 to infinity. The probability that a mode will have energy E is given by the Boltzmann distribution
$$p(E) = \frac{1}{kT}e^{-E/kT}$$so the average energy of each mode, classically, is
$$\langle E \rangle = \int_0^\infty Ep(E)\,dE = kT$$The energy per unit volume belonging to the modes with frequencies between $\nu$ to $\nu+d\nu$ is simply the number of modes, $N(\nu)\,d\nu$, times the average energy of each mode, kT, divided by the volume, V, which yields the Raleigh-Jeans formula for blackbody radiation:
$$\rho(\nu)\,d\nu = \frac{8\pi \nu^2}{c^3}kT\,d\nu$$As you can see, as the frequency $\nu$ increases, the energy density $\rho(\nu)$ increases without bound. This is because the number of modes increases with frequency while the average energy per mode is constant. This is the ultraviolet catastrophe.

What Planck said was that a mode with frequency $\nu$ could only have an energy equal to a multiple of $h\nu$. The probability of a mode having an energy $E_n = nh\nu$ is still given by the Boltzmann distribution, but this time the average energy of the mode works out to be
$$\langle E \rangle = \frac{\displaystyle\sum_{n=0}^\infty E_n p(E_n)}{\displaystyle\sum_{n=0}^\infty p(E_n)} = \frac{h\nu}{e^{h\nu/kT}-1}$$A consequence of Planck's idea is that the average energy of a mode is no longer independent of frequency. It goes to 0 at high frequencies. The energy density is now given by
$$\rho(\nu)\,d\nu = \frac{8\pi \nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT}-1}\,d\nu$$So while the number of modes increases with frequency, the average energy decreases. At high frequencies, the exponential in the denominator dominates, and the energy density to go to 0 in agreement with the experimental observation.

6. Aug 6, 2012

### vijayan_t

Actually the Intensity is decreased while the wavelength is decreased.
see this explanation.

Last edited by a moderator: Sep 25, 2014