Puchinita5 said:
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?
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.
Puchinita5 said:
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.
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.
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?
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.