# “Super dense” black body radiation

My question is simple: is it possible to create radiation with the same shape of spectrum as black body radiation, isotropic, but with greater density (number of photons with given frequency in unit volume) than black body radiation, which is (energy per unit volume per unit frequency):

$$S_\nu = a \frac{8\pi{h}}{c^3} \frac{\nu^3}{e^{hc/\lambda{k}T}-1}$$

where $a = 1$.

If I am inside a solid sphere with temperature, say 6000 K, the radiation inside is black body, $a = 1$. If only a small part of the cavity has this temperature (and the rest is cooled to near 0 K, or imagine dark sky in space with sun), the radiation still has the correct spectrum, but is not isotropic and density is smaller that black body radiation ($a<1$). Is there any mechanism which would produce higher density, that is $a>1$?

According to Kirchhoff's law of thermal radiation, there is a proportionality between emissivity and absorptivity of a body. Since a black body has got the maximum absorptivity at every wavelength, the emissivity at every wavelength should be maximal as well. In your formula, a = 1 would thus be an upper bound.

However, I recently came across this paper which claims that you could overcome black body radiation at a given frequency by realizing "double negative metamaterials with arbitrary small loss and arbitrary high absolute values of permittivity and permeability (at a given frequency)." Nevertheless, this seems very theoretical to me, and I think, it is safe to say that a black body represents an upper bound for the spectral density of emitted electromagnetic radiation.

• Kirchhoff's law is valid only for bodies in thermodynamic equilibrium. A body that is not in such equilibrium can radiate more (in finite spectral range) than Planck's formula would predict. For example, see en.wikipedia.org/wiki/Radioluminescence – Ján Lalinský Mar 24 '18 at 10:39
• While this is true, the formula in the question uses a specific temperature T. It only makes sense to apply this formula if T can be defined. This is why I quietly assumed thermal equilibrium – lmr Mar 24 '18 at 12:08
• But the formula in the question does not require that the emitting bodies have the same temperature $T$, or any temperature at all. – Ján Lalinský Mar 25 '18 at 0:35
• It is true that the bodies do not need to have the same temperature. But we are not going to compare a black body at 1K with a non-black body at 6000K as in the example presented in the question. The assumption of thermal equilibrium and equal temperatures of the bodies simply makes sense and I still think, this is what was implicitly meant... – lmr Mar 25 '18 at 11:14

Short answer: If that formula refers specifically to electromagnetic radiation, then $a = 1$ exactly.

If you forcibly set up a distribution of photons with $a \neq 1$ in your formula, that system is not in equilibrium, and will evolve to a different temperature (and the same total energy) such that $a=1$. The new state, with $a=1$ and $T^{\prime} \neq T$ will be the maximum entropy distribution for a bath of photons with given energy.

More generally, if we're considering the energy density of all kinds of generalized "radiation" then $a$ counts the number of (approximately) massless degrees of freedom. For the kind of temperatures we typically talk about, photons are the only particles with $m \ll T$ so $a=1$. In the early universe, several particles satisfied this constraint, and $a \gg 1$. In more detail:

1. The Planck blackbody distribution is equivalent to the Stefan-Boltzmann law for radiation.

2. The Stefan-Boltzmann law can be derived by looking at the spectrum of waves in thermal quantum field theory (which is the appropriate model when describing relativistic particles; since $m \ll T$ the particles we're modeling are moving at relativistic speeds).

3. In the thermal QFT framework, $a$ counts the number of (approximately massless) waves/fields/particles in the theory. Certain fields can also have fractional contributions to $a$ if they are not exactly massless, but only approximately so. For a summary of how $a$ evolves through the history of the universe (based on our current understanding) see section 21.3.2 here.

If only a small part of the cavity has this temperature (and the rest is cooled to near 0 K, or imagine dark sky in space with sun), the radiation still has the correct spectrum, but is not isotropic [...]

Further, that is not an equilibrium situation, so the discussion above doesn't apply.

• "If that formula refers specifically to electromagnetic radiation...". Yes, I am interested only in photons (temperatuures of thousands K). – Leos Ondra Mar 27 '18 at 8:02
• Thousands of K is a low enough temperature that photons are the only relevant degrees of freedom (so are neutrinos, but they interact weakly and could be neglected for this discussion). – Siva Mar 28 '18 at 5:38

Well if i look at your argument logically, with sun and dark sky, all you did was putting the sun away from me, aka spreading the rays over a distance (from sun to me) The inversion of that line of thinking would, I'd say, be to insert a lense and create a focal point.

• IMHO, lens cannot create super dense black-body radiation. – Leos Ondra Mar 30 '18 at 21:02
• i compared it with his example of putting the sun in a bigger space to look on, aka "zooming out". – Leviathan Apr 4 '18 at 8:21

This might be simpler than what you're looking for, but note that for sufficiently low frequencies, all blackbody spectra are all identical up to a scale factor, because they obey the Rayleigh-Jeans law. Physically, in this regime every mode has energy $$k_B T$$, so scaling $$T$$ just scales up the spectrum. So if you want to imitate $$a > 1$$ for temperature $$T_1$$, just take an object at high temperature $$T_2 \gg T_1$$ and filter out the high frequencies appropriately!

I don't think what you're asking for can be done with only objects at thermodynamic equilibrium at temperature $$T_1$$, as the other example explains: since $$a$$ is also the fraction of radiation that a blackbody absorbs, it can't be greater than one unless you have an exotic setup.