How does quantization solve UV catastrophe in black body radiation? What would happen if there was no Planck constant $h$?

Planck's Law is $$I(\nu,T)=\frac{8\pi\nu^3}{c^2}\cdot\frac{1}{e^{h\nu/kT}−1}$$

This solves the UV catastrophe. For higher frequencies, intensity goes to zero.

It does so because of $e^\nu$ not because there is a $h$ constant. If we remove $h$ parameter from the equation, it still goes to zero for higher frequencies. Right?

So why is "UV catastrophe is solved by quantization" said?

Note: I'm not saying if h = 0. I'm saying if there was no h constant. Like this:

$$I(\nu,T)=\frac{8\pi\nu^3}{c^2}\cdot\frac{1}{e^{\nu/kT}−1}$$

• In classical physics you don't have the exponential term at all. The whole factor $\frac{1}{e^{h\nu/kT} -1 }$ stems from quantum mechanical derivations. – taupunkt Jul 11 '14 at 7:12
• If you're setting $h=1$, you still have $h$, only with a value of 1. However, you then won't be able to fit $I(\nu,T)$ to your experimental value. In Hooke's Law, you also can't say that $F=-x$ and thus setting $k=1$, because (in most cases) your formula won't fit your experimental values. – Timitry Jul 11 '14 at 7:22
• have a look at this answer physics.stackexchange.com/q/125921 . It links to a source that clears up the whole black body quantum /classical. You will see that quantization is in the hypothesis that derives the formula. – anna v Jul 11 '14 at 7:48
• That makes no sense, you cannot just remove a constant from a equation. Physical quantities have units and $\nu/kT$ is not dimentionless as it need to be in order for $\exp(\nu/kT)$ to make sense. – Winther Jul 11 '14 at 12:35

By omitting $h$ from the formula you are just setting $h = 1$, so it's still the quantum mechanics description, just with a different value of $h$.

If you don't include quantum mechanics the expression you get is the Rayleigh-Jeans law:

$$I(\nu,T)=\frac{2\nu^2kT}{c^2}$$

The two laws agree for small $\nu$ because for small $\nu$:

$$e^{h\nu/kT} \approx \frac{h\nu}{kT}$$

and in this limit Plank's law reduces to:

$$I(\nu,T) \approx \frac{8\pi\nu^2kT}{hc^2}$$

which is the same as the Rayleigh-Jeans law give or take a multiplicative constant.

• No. I'm not saying if h = 0. I'm saying if there was no h. I will edit my question. – user50322 Jul 11 '14 at 7:06
• @user50322: I've updated my answer to respond to your comment – John Rennie Jul 11 '14 at 7:23
• I still don't understand how can anyone see quantization concept from this formula. If not h parameter, which parameter makes this formula representing quantization? If energy is related with frequency, then uv catastrophe is solved. Then we don't need quantization. But they say quantization solved the problem. I can't see any quantization at this formula. This is really hard. – user50322 Jul 11 '14 at 8:10
• How would a formula have to look like for you to consider it as quantized? Planck used concepts of quantum mechanics, here that energy of a wave comes in packages of $h\nu$, to derive the formula. Without the concept of quantization you don't get the formula, that's why we say it's a formula of quantum physics. – taupunkt Jul 11 '14 at 8:19

The difference between getting a classical formula for a cavity radiator or a quantum mechanical formula lies in the fact that the photon energies are not continuous, that is what quantization means. Planck assumed quantization to get the formula.

Published in 1900, it originally described the proportionality constant between the energy (E) of a charged atomic oscillator in the wall of a black body, and the frequency (ν) of its associated electromagnetic wave. Its relevance is now integral to the field of quantum mechanics, describing the relationship between energy and frequency, commonly known as the Planck relation:

So h is the quantization constant with the appropriate units to get the relation above, i.e. that a quantized harmonic oscillator emits energy in packets given by the formula. This assumption fitted the data and tied well with the photoelectric effect.

The number is not arbitrary, it depends on the system of units, and it can be 1. In high energy physics theories often c=h=1 is assumed and when one wants to get real numbers one has to untangle the units. Setting h to 1 does not negate quantization, it just changes the units.

• The last sentence is pretty key. – Kyle Kanos Aug 31 '14 at 2:13

It doesn't make any sense to just remove $h$, because the units are then incorrect ($h$ has units of action). The question as stated is unanswerable.

Putting aside the ridiculous attempt of user50322 for normalizing Physics Units System to make h=1. He still makes a point when he keeps nagging about Plank Law not being a discrete math equation.

(8π(ν^3)/c^2) / (Exp(hν/kT)-1) is a continuous function with a singularity when ν=0. It does not indicate just by itself the quantum nature of light. It may be said that it is the result of a probability distribution that makes unlikely that very high frequency photons be emitted by a hot black body.

The explanation user50322 is looking for is that : For a large ν, a hν energy amount is difficult to reach for the oscillators (atoms) of a black body at any given high Temperature T. In other words, the random disturbance commotion originated by the Temperature in the walls of the black body has a low probability of energizing any of its oscillators at at-least hν if ν is large, and if the oscillator is short of hv, no photon will be emitted (there the quantization of the damn thing), so just a few photons are emitted at a large frequency ν in comparison with a not so large frequency.

The black body walls will have very few regions where by chance the commotion accumulates energy that surpasses hν.

• What's wrong with a units system where $h=1$? – Kyle Kanos Aug 31 '14 at 2:13
• Nothing is wrong with making h=1 and then adjusting the other units to it, I´m sorry I wrote that it was ridiculous. Maybe I was bothered by the digression from the theme in question. In fact h/2Pi, ħ, has been normalized to 1 for convenience in some contexts. – user58245 Aug 31 '14 at 3:24