# Derivation of Planck Radiation Formula

1. Nov 9, 2014

### Fantasist

I am having some issues with the derivation of the Planck Radiation Formula, as for instance given in http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html . My point is that the assumption of cavity modes implies the formation of standing waves inside the cavity walls. In most situations in nature (e.g. the sun) it is however absolutely inconceivable that here standing waves could develop, simply because there are no walls or other sharp boundaries. So with the fundamental theoretical assumption in the derivation not applicable at all, why is it that the Planck Radiation Formula still describes the solar spectrum to a good approximation?

2. Nov 9, 2014

### dextercioby

The sun is almost the perfect radiator of electromagnetic waves. You see, in the typical example, you're dealing with a black body, in the form of a cavity with a small hole through which the incoming radiation is highly unlikely to escape. In other words, the black hollow box is the prototype of the perfect absorber. There's a connection between the perfect radiator and the perfect absorber: their distributions of the radiation behave in the same way. See http://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation

Last edited: Nov 9, 2014
3. Nov 10, 2014

### Fantasist

Hi,

Thanks for your reply. It doesn't quite answer my question though, which was regarding the model assumptions made to derive the Planck spectrum (i,e. standing cavity wave modes), and the fact that these assumptions are not applicable e.g. for the sun. Normally, if I try to model a physical problem theoretically, I have to justify the model assumptions by them being a more or less close description of reality, but how can the assumption of standing cavity wave modes even only remotely describe the situation on the sun (or other natural radiation sources for that matter).

4. Nov 11, 2014

### Jano L.

Mathematically, the reflective walls and standing waves are not vital part of the derivation of the spectral function - it can be derived in the same way even if the walls are removed from the consideration. The vital parts of the usual derivation are Poynting expressions for EM energy and probability distribution for the possible states of the field.

The ideally reflective walls are mentioned because they are theoretically needed to prepare macroscopic region of space containing radiation in exact thermodynamic equilibrium with matter inside. It is an idealization - such walls cannot be made and some radiation always passes through. Measurements of the spectral curve of thermal radiation were originally made with radiation from inside of real metallic cavity, with reflectivity lower than 100 %.

So in reality we expect all radiation to be different from the equilibrium radiation (which is an abstraction).

Just as the radiation of a real cavity, the radiation of the Sun differs from the equilibrium radiation - it has preferred direction of propagation and its spectrum has absorption and emission lines which are not present in equilibrium radiation.

Still, the solar radiation spectrum is quite close to the Planck curve. One possible interpretation of this is that even without the reflective walls, the mutual interaction of the energetic charged particles in the dense part of the Sun is frequent and chaotic enough to produce radiation close to equilibrium radiation.

5. Nov 11, 2014

### Fantasist

Is there any generally accessible reference for this?

6. Nov 11, 2014

### Jano L.

I do not know. It seems to be trivial - take any derivation of the spectral curve using the Fourier series of the EM field with boundary condition appropriate for perfectly reflective walls and notice that similar expansion into Fourier series is possible for any field, without any boundary conditions. This is because mathematically, the Fourier expansion of a function on a finite interval works for any function, not just a standing wave with zero amplitude at the ends. The rest of the derivation is the same.

7. Nov 14, 2014

### vanhees71

It doesn't matter how you shape your cavity to evaluate the black-body spectrum in the thermodynamic limit. A black-body spectrum always occurs if you look at radiation that comes from a source of electromagnetic radiation in thermal equilibrium at a given temperature. The sun is (up to the Fraunhofer absorption lines and other deviations from the Planck spectrum due to the fact that the em. radiation must go through the material around the sun) opaque for photons, i.e., before the photons come out of the sun they have scattered many times within the plasma the sun consists of and thus are in thermal equilibrium. Thus the base spectrum of the em. radiation from the sun is a black-body spectrum, which is modified due to the above mentioned effects.

8. Nov 14, 2014

### Fantasist

A Fourier transformation is hardly applicable for the radiation spectrum of the sun (or other natural sources) as the Fourier theorem implies that all frequencies are locked in phase. In the solar spectrum, photons with different frequencies will in general be completely uncorrelated.however.

Last edited: Nov 14, 2014
9. Nov 15, 2014

### Jano L.

I do not understand you here at all. I mentioned Fourier series, not Fourier transformation.

What do you mean by Fourier theorem?

What do you mean by

In the solar spectrum, photons with different frequencies will in general be completely uncorrelated.

?

The fact the solar radiation does not have comb-like spectral curve does not prevent expansion of its EM field into Fourier series.

10. Nov 15, 2014

### vanhees71

The issue in the derivation from thermal QED is to evaluate the partition sum of the free electromagnetic field,
$$Z=\mathrm{Tr} \exp(-\beta \hat{H}).$$
The most simple way is to use the mode decomposition of the field operator in the totally radiation-gauge fixed formulation, which gives
$$\hat{H}=\sum_{\vec{p},\lambda} \omega(\vec{p}) \hat{N}(\vec{p},\lambda),$$
where $\vec{p}$ runs through the values $\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}$, where $L$ is the length of a large cube within the thermal radiation field. For simplicity I've assumed periodic boundary conditions. The specific form of the boundary conditions is pretty unimportant in the thermodynamic limit, which we have to take at the end of the calculation. $\omega(\vec{p})=|\vec{p}|$ is the energy of a photon of momentum $\vec{p}$ and $\lambda \in \{-1,1 \}$ are the two helicity values (polarizations) of a photon.

To get the Planck law for the occupation number distribution we evaluate the somewhat more general expression
$$Z[J]=\mathrm{Tr} \exp(-\sum_{\vec{p},\lambda} j(\vec{p},\lambda) \omega(\vec{p}) \hat{N}(\vec{p},\lambda).$$
To evaluate the trace we use the occupation-number eigenstates $|\{N(\vec{p},\lambda) \}$, where $N(\vec{p},\lambda) \in \{0,1,2,\ldots\}$. This gives
$$Z[j]=\prod_{\vec{p},\lambda} \sum_{N(\vec{p},\lambda)=0}^{\infty} \exp[-j(\vec{p},\lambda) \omega(\vec{p}) N(\vec{p},\lambda)] = \prod_{\vec{p},\lambda} \frac{1}{1-\exp[-j(\vec{p},\lambda) \omega(\vec{p})]}.$$
To proceed further we evaluate the logarithm, and convert the sum to an integral by taking the large-volume limit. Then in each volume in momentum space we have $\frac{L^3}{(2 \pi)^3} \mathrm{d}^3 \vec{p}$ states:
$$\ln Z[j]=-V \sum_{\lambda} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \ln \left \{1-\exp[-j(\vec{p},\lambda)\omega(\vec{p})] \right \}.$$
The energy density of the radiation field is then given by
$$\mathrm{d} U=-\mathrm{d}^3 \vec{p} \frac{1}{V} \sum_{\lambda} \left ( \frac{\delta Z}{\delta j(\vec{p},\lambda)} \right )_{j(\vec{p},\lambda)=\beta}.$$
This gives
$$\mathrm{d} U=\frac{1}{4 \pi^3} \mathrm{d}^3 \vec{p} \frac{\omega(\vec{p})}{\exp[\beta \omega(\vec{p})]-1}.$$
Further we have
$$\vec{p}^2=\omega^2 \; \Rightarrow \; \mathrm{d}^3 \vec{p} = \mathrm{d}^2 \Omega \omega^2 \mathrm{d} \omega,$$
and thus finally the Planck radiation formula
$$\frac{\mathrm{d} U}{\mathrm{d} \omega}=\frac{\omega^3}{\pi^2} \frac{1}{\exp(\beta \omega)-1}.$$

11. Nov 15, 2014

### Jilang

12. Nov 15, 2014

### Fantasist

Whether you have a discrete (Fourier series) or continuous (Fourier integral) spectrum. the Fourier representation only makes sense if the Fourier coefficients (including the phase) are constant in time. This is certainly not the case for the solar radiation because of the random nature of the radiation field (atomic emissions at different frequencies are completely uncorrelated both in space and time, and the phase of each photon (wave train) randomly jumps because of atomic collisions).

13. Nov 15, 2014

### Fantasist

Thanks for your elaborate analysis, but I think lastly the derivation is equivalent to slightly more intuitive approaches like in http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/phodens.html#c1
The point is that in either case the derivation is based on the assumptions of oscillator modes set up in an actual macroscopic cavity, when in fact there is no such cavity (of whatever shape or form).

I think the only consistent way to bring in the Rayleigh-Jeans factor is to consider the spectral energy density of a radiating harmonic oscillator, which is also proportional to the square of the frequency ( http://farside.ph.utexas.edu/teaching/qmech/Quantum/node120.html ) Together with the Boltzmann distribution over the energy levels, this would then also yield the Planck spectrum, but now based solely on microscopic properties.

14. Nov 15, 2014

### Fantasist

That as such doesn't explain the Planck spectrum. The latter is not a result of radiation transport but of local conditions.

15. Nov 15, 2014

### Jano L.

No, you have some misconceptions about mathematics of Fourier series. Any function of time continuous on some interval can be expressed as Fourier series there.

Random nature of radiation does not prevent expressing its electric field as a Fourier series. Randomness of thermal radiation means we do not know which definite function is realized. But all possible functions have their corresponding Fourier series.

I am not sure what you mean by this. There is generally always some correlation, since the particles interact with each other. Thermal radiation has non-zero correlation function that is related to its spectral function.

Irrespective of correlations, any EM radiation can be described by EM field. Far from charge singularities, this can always be expressed as Fourier series.

16. Nov 15, 2014

### Jilang

Sure it does, it's explains the thermal equilibrium.

17. Nov 16, 2014

### vanhees71

My "quantization volume" is not meant as a cavity. It's a large volume within the radiation field (large compared to the thermal wavelength of the considered radiation). It's a thought construct. Thermal equilibrium is a state of maximum entropy, i.e., independent of the history of how the system has been prepared, and thus it doesn't matter so much how your volume is shaped and which atoms make up the wall etc. The spectrum only depends on temperature not on the specific way how the em. field has come to equilibrium with the thermally equilibrated matter it interacts with. Thus, there are many ways to derive the spectrum, which are all leading to the same spectrum. E.g., Planck's original derivation started from the interaction of electromagnetic waves with a harmonically bound charges. For Planck there were no photons but only the interaction of the em. field with these oscillators.

18. Nov 16, 2014

### Fantasist

Yes, but that is quite different to Rayleigh's method of considering oscillator modes in a macroscopic cavity (which is practically always used here in educational texts). The result may be the same (yielding the ν2 dependence in the low frequency limit), but the energy density of atomic oscillators will always show this behaviour, even though they may be far from macroscopic equilibrium (only the exponential part of the Planck formula should depend therefore on the existence of a macroscopic equilibrium).