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I've searched around the site and looked at older posts but none of the answers make sense to me.

For a bit of background: I'm a high school student and don't have a great understanding of the more advanced physics topics like statistical mechanics. Similarly the only formula's I am aware of in this area are Planck's Law, Wiens Law and the basic wave equations.

I understand the concept of black body radiation and how the classical theory leads into the UV Catastrophe. I also understand the implications of the quantisation of energy and the basics of the quantum model of light. Furthermore, I'm also aware of the photoelectric effect.

However the one thing that I'm struggling with is how exactly Planck's assumption, that energy is quantised, solved the UV catastrophe. Why is that when we assume energy is quantised, we can now predict the blackbody curve perfectly? Logically, what changes in our understanding of blackbody radiation?

I'd appreciate a more intuitive answer but do not mind mathematical equations.

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I'll aim for an intuitive explanation of how quantization can solve the ultraviolet catastrophe and let others (or online resources) provide accurate formulas.

The "ultraviolet catastrophe" says that a black body would radiate energy not just around a given average wavelength (color), but also in all wavelengths, especially very short ("ultraviolet") wavelengths, and because these wavelengths are very energetic and there is no limit to how short you can make them (if you think $\lambda$ is a very short wavelength, $\lambda/2$ is an even shorter wavelength and the black body also radiates in it), the total amount of energy that would be produced is infinite.

How does quantization solves this? Well, basically it says that you can't radiate less than a "quantum" of radiation in a particular wavelength. According to Plank's quantization formula $E=hc/\lambda$, a very short wavelength has a very large quantum. So if the would-be-radiation predicted by the black-body forumla is large but not as large as this single quantum - absolutely no radiation would be emitted in this wavelength by the given process.

In our case, in a black body in thermal equilibrum only a tiny minority of atoms have very large energy, and only those could possibly radiate some of their energy as one or more quanta of very short wavelengths. All other atoms will radiate none of their energy in those very short wavelengths. All-in-all, the amount of energy radiated in very short wavelength will be very small, approaching zero as the wavelength gets shorter (giving the familiar bell-like shape of the black body radiation curve) instead of infinite.

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how exactly Planck's assumption, that energy is quantised, solved the UV catastrophe.

The term "UV catastrophe" means that high frequency radiation carries so much energy that its total energy comes out infinite. It usually refers to the flaw of the result of the Rayleigh-Jeans calculation, or any similar calculation where classical statistical physics is applied to systems of infinite number of degrees of freedom, such as radiation or mechanical continuum. Planck did not care about this UV catastrophe, the term did not even exist back then, it came about later, probably in Ehrenfests' book on statistical mechanics.

Planck's motivation was to explain (theoretically derive) the blackbody radiation spectral function that was already known from experiments.

Planck found out in different ways that if EM radiation energy changes only via discrete energy steps $\hbar\omega$, this allows him to derive a blackbody radiation spectral function that agrees with measurements. In this limited sense, quantization prevents UV catastrophe. However, the modern formulation of this calculation (canonical quantization) shows there is still a contribution to total EM energy which has infinite energy in high frequencies. So, strictly speaking, canonical quantization allows us to arrive at the blackbody radiation spectral function, but it did not resolve the inherent problem of infinite energy in these calculations.

Both the Rayleigh-Jeans derivation and the quantum-theoretical derivation are based on the Poynting formula for EM energy, and both give increasing functions with infinite total energy.

The RJ derivation assumes radiation energy is given by the Poynting formula, and every Fourier mode of the radiation has the same statistics as classical oscillator in thermal bath, and thus obeys equipartition of energy (every oscillator has energy $k_BT$). This gives spectral function

$$ f_{RJ}(\omega) = c_2 k_BT \omega^2, $$ which is correct only for low frequencies $\omega$, but then it fails to agree with measurements for high frequencies. Mathematically, it suffers from the UV catastrophe (infinite energy in high frequencies). In short, if every oscillator has average energy $k_BT$, then because number of oscillators per $d\omega$ is $c_2\omega^2 d\omega$, we get increasing function of frequency and infinite total energy.

The modern way to derive the Planck function is based on the same Poynting formula, but rejects equipartition of energy for oscillators. Instead, it assumes the oscillators have discrete energies $E_n = \hbar\omega(n + \frac{1}{2})$. This means oscillators of different frequency have different average energies, and equipartition does not hold. Their statistics is assumed to obey the Boltzmann probability distribution, and this produces the spectral function $$ f_{qt}(\omega) = c_1 \hbar \omega^3 + c_2 k_B T \omega^2 \frac{\frac{\hbar \omega}{k_B T}}{e^{\frac{\hbar \omega}{k_B T}} - 1}. $$

There are two terms here. First, there is a new term, proportional to frequency cubed. It is due to zero point energy $\hbar\omega/2$ which the oscillator has even in the ground state $n=0$. This means that we have infinite energy in high frequencies, similarly to the UV catastrophe in the RJ result, only now the infinite energy term does not depend on temperature. Second, the other term is the RJ function multiplied by a "correction factor" that hammers it down for high frequencies.

When we say the quantum result agrees with the experiment, we mean this second term. We conveniently ignore the first term. Sometimes we say this term is because we used the wrong Hamiltonian, sometimes we say the term and the EM energy is there but it is constant and does not contribute to radiation energy.

Why is that when we assume energy is quantised, we can now predict the blackbody curve perfectly?

It's not exactly that way. We have to ignore the zero point energy and then we get the curve that agrees with experiments.

The standard answer to "why" is that EM radiation actually has discrete energy states, and then Boltzmannian statistics implies the rest.

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The complete derivation is slightly long and mathematical in nature so I'll just mention some key points.

Before Planck came into the picture, physicists wanted to study light-matter interaction. To do so, they experimentally realised a nearly ideal blackbody (using what experimentalists call a 'Jean Cube') and obtained the graph of spectral radiance vs wavelength. The experimentalists were happy, but the theorists needed a framework to understand the graphs obtained.

Instead of using spectral radiance, they found it more useful to start off with energy density function $\rho (f)$. Now, to find the energy density function, the approach the physicists took was:

  1. Count the number of waves that can fit inside Jean's cube. Because radiation inside Jean's cube gets reflected back and forth, it was reasonable to assume the waves inside Jean's cube were stationary waves.
  2. Find the average energy of these waves in thermal equilibrium. This is where statistical mechanics comes to help us. The physicists assumed the entities to be the standing waves here and not particles.
  3. Using the results from 1) and 2), we can calculate the energy between the frequency range $f$ and $f+df$. This effectively gives us the expression of the energy density function $\rho (f)$.

Nice. Step 1 is calculated through a few arguments but nothing to worry about here. What makes the difference is step 2.

Before Planck, everyone assumed energy to be continuous. Therefore, we calculate the average energy of the system (using statistical mechanics) as: $$ \bar{E}=\frac{\int_0^{\infty} E e^{-\frac{E}{k T}} d E}{\int_0^{\infty} e^{-\frac{E}{k T}} d E} $$ It can be easily shown that this becomes $\bar{E} = kT $.

Using the fact: total energy within the interval $df$ equals $($total number of waves in $df$ interval$)$ times $($average energy per wave$)$ times $($volume of jeans cube$)^{-1}$ and moving from a frequency domain to wavelength domain, we end up obtaining: $$\rho(\lambda)d\lambda = \frac{8 \pi kT} {\lambda ^{4}} d\lambda$$

(don't worry about how we exactly got this right now and all what's needed is just the result we obtain)

One can convert the energy density function to spectral radiance, and then it turns out that our theoretical model shows $\frac{1}{\lambda ^{4}}$ behaviour and this is why we obtain the 'Ultraviolet catastrophe'. Clearly, for shorter wavelengths - around the ultraviolet range - the spectral radiance seems to diverge.

This was a big issue during those times. This meant that either our theoretical models were wrong or incomplete. Max Planck claimed that the theoretical models were incomplete in the sense that for long wavelengths, our theoretical graphs tend to match with experimental graphs (so we aren't completely wrong). Because for long wavelengths we get some overlap, it is fair to assume $\bar{E} \approx kT$ is correct in this range. Max Planck successfully solved this puzzle by assuming energy to be discrete. Nobody knows what thought process made Planck make the following assumption (but it worked somehow): $$E=n\varepsilon; n \in N$$ (where $N$ is the set of natural numbers)

Because the energy is now discrete, this means, going back to step 2, our calculations become: $$ \bar{E}=\frac{\int_0^{\infty} E e^{-\frac{E}{k T}} d E}{\int_0^{\infty} e^{-\frac{E}{k T}} d E} \Rightarrow \bar{E}=\frac{\sum_{n=0}^{\infty} n \varepsilon e^{-\frac{n \varepsilon}{k T}}}{\sum_{n=0}^{\infty} e^{-\frac{n \varepsilon}{k T}}} $$

The latter can be calculated using techniques from differentiation and geometric progression to obtain: $$ \bar{E}=\frac{\varepsilon}{e^{\frac{\varepsilon}{k T}}-1} $$

Okay, we got that by assuming energy is discrete. That's fine but what's important are the following observations:

  1. From experimental graphs, as frequency goes to zero, $\bar{E}$ tends to $kT$ and when frequency tends to infinity, $\bar{E}$ tends to $0$.
  2. From the above equation $\bar{E}=\frac{\varepsilon}{e^{\frac{\varepsilon}{k T}}-1}$, we find that as $\varepsilon$ goes to zero, $\bar{E}$ tends to $kT$ and when $\varepsilon$ goes to infinity, $\bar{E}$ tends to $0$.

This means that $\varepsilon$ and frequency $f$ are somehow related, so we postulate the following relation: $$ \varepsilon = hf$$ where $h$ is just some constant.

Okay, we use $\bar{E}=\frac{\varepsilon}{e^{\frac{\varepsilon}{k T}}-1}$ and proceed what we did earlier and we obtain: $$ \rho(\lambda) d \lambda=\frac{8 \pi h c}{\lambda^5} \frac{d \lambda}{\left[e^{\frac{h c}{\lambda k T}}-1\right]} $$

The behaviour of the energy density function is now very much similar to the behaviour seen in the experimental data. All that's left to do is calibrate the constant $h$ so that the theoretical graph matches the experimental data exactly. Doing so gave $h$ a value that we today associate with Planck's constant.

Supplementing this along with Nadav's answer to this question effectively should explain everything as to why quantisation solves the ultraviolet catastrophe.

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