# How can it be that the sun emits more than a black body?

As far as I know, a black body is an ideal emitter. So how can it be that a non-ideal emitter emits more radiation than a black body?

This happens only in a very limited area at around 500nm, but it still happens: it looks like at the maximum it is around 15% above black body.

This seems impossibile for my understand of a black body. Especially because just for that there is the emissivity value ε, or better ε(λ)

Wikipedia Emissivity: Quantitatively, emissivity is the ratio of the thermal radiation from a surface to the radiation from an ideal black surface at the same temperature. The ratio varies from 0 to 1

Means 0 ≤ ε(λ) ≤ 1

What is the right interpretation? What is the sun doing there, seems like ε(500nm)=1.15?

• OK - in future I'd recommend writing out any abbreviations to avoid confusion. Aug 9 '14 at 15:39
• Dave, the alternative is that you will learn absolutely standard acronyms like AFAIK, FYI, TTYL, BTW, USA, ETC. ;-) Aug 9 '14 at 15:42
• The primary source of your confusion is that image, which comes from wikipedia. The 5250 C black body curve is unreferenced, and it is wrong. There's no knowing where the author of that image obtained that erroneous value. Aug 10 '14 at 17:06
• The plot is mislabelled. The temperature is incorrect. e.g. commons.wikimedia.org/wiki/File:Solar_spectrum_en.svg Oct 17 '21 at 15:43

The total radiative power emitted by the Sun is equivalent to the total radiative power emitted by an ideal black body with a temperature of 5778 K and a surface area equal to that of the Sun. This 5778 K is the Sun's effective temperature. The spectrum of the Sun is very close to that of a 5778 K black body, but there are deviations. Some are due to absorption and emission, but others result from three key items:

• There is no such thing as black body. The concept of a black body is an idealization based on some simplifying assumptions. The Sun doesn't exactly satisfy those simplifying assumptions.

• That effective temperature of 5778 K is based on total radiative power, the area under the curve of the Planck distribution. If the spectrum of sunlight falls short of the 5778 K black body spectrum some wavelengths it must necessarily rise above the 5778 K black body spectrum at others.

• The primary reason the Sun fails to satisfy the assumptions that underly the Planck distribution is that we are seeing light from multiple temperature sources. The rest of this answer goes into this in detail

The Sun is not a solid body. It doesn't have a surface from which the radiation originates. The radiation we see from the Sun comes primarily from the Sun's photosphere, a roughly 500 kilometer thick layer near the top of the Sun. The chromosphere, transition region, and corona are above the photosphere. While these higher layers do make solar radiation deviate from the ideal black body curve, the primary source is the photosphere itself.

The amount of light that is transmitted into empty space is a sharply increasing function of distance from the center. However, it is not a delta distribution. The light that does get through from those deeper layers has a higher temperature than the layers above it. The bulk of the radiation we see from the Sun comes from a ~500 km thick layer called the photosphere. The top of the photosphere has a temperature of about 4400 K and has a pressure of about 86.8 pascals. The bottom has a temperature of about 6000 K and a pressure of about 12500 pascals.

What we see is a blend of the radiation from throughout the photosphere. Some of the light comes from the top of the photosphere, some from the middle, some from the bottom, roughly weighted by pressure. The total spectrum looks close to that of a 5778 K black body, but the contribution from the bottommost part of the photosphere tilts the spectrum away from the ideal a bit, making the a tiny bit heavy for shorter wavelength radiation.

• I'll take that one paragraph out. Aug 9 '14 at 19:35
• Nice answer. Or at least it seems reasonable to someone like me who knows nothing about solar physics. Two questions, purely out of curiosity: 1) Is the layered temperature structure actually measured directly, or is it inferred from other data e.g. magnetohydrodynamic modelling? 2) What is the length scale over which the gas appears locally in equilibrium? (i.e. how thick are the layers, approximately) Aug 9 '14 at 20:29
• @MarkMitchison I think this is a good answer. The solar photosphere is physically about 100-200 km thick. ie. most of the light we see originates from this relatively thin layer. The temperature structure is deduced from physical models but can be confirmed with temperature diagnostics that don't just use Wien's law (e.g. line ratios). The T gradient in the photosphere is about -5 K/km. Nov 14 '14 at 21:53
• @RobJeffries Great, thanks very much for the extra info. Nov 21 '14 at 10:57
• How close are black holes to black bodies? Mar 20 '19 at 18:27

The Planck function that describes the blackbody emission is a function of temperature and wavelength: $$B_\lambda(T)=\frac{2hc^2}{\lambda^5}\cdot\frac1{e^{hc/\lambda k_BT}-1}$$ Due to the temperature dependence, blackbodies at different temperatures have different emissions. The graph below, from Wikipedia shows the drastic changes for blackbodies of temperatures 3000 K, 4000 K, and 5000 K (also shown is the Rayleigh-Jeans regime where $B_\lambda(T)\propto\lambda^{-4}$ which blows up to infinity at low wavelengths).

The spectral class of the sun is G2V. The G2 signifies that the surface temperature of the sun is about 5800 K, not 5250 C (about 5520 K) in your diagram. Thus, the emissions observed from the sun should be larger than that of the modeled blackbody you show.

Plotted below is the Planck function for a 5520 K emitter, a 5777 K emitter and a 5800 emitter. Both the 5777 K and 5800 K blackbodies have a peak that is about 30% larger than the 5520 K blackbody (left axis is W/sr/m$^3$, bottom axis is $\mu$m).

Don't forget temperature differences across the surface, light from hotter depth that eventually finds its way out, other light-producing phenomena (spectral emission from excited electrons recombining with atoms then ionizing again, photons created in scattering, decay, annihilation, etc processes), etc. A star is a complex system with a lot happening all the time. It's safe to assume that blackbody radiation (while the primary source) isn't the only source of radiation.

It appears that your image wants to fit the curve to the larger wavelengths, rather than the peak--this is where your confusion lies. If you fit the peaks, then surely $\varepsilon\leq1$ is satisfied (so long as you note the above comment from Jim, that the sun really isn't a pure blackbody).

• Sorry, minus one. The OP's diagram doesn't show 5250 K. It shows 5250 Celsius degrees. Check it out. Aug 9 '14 at 15:57
• @LubošMotl: True, but 5250 C is 5520 K, still about 300 K off from what the surface temperature of the sun. Aug 9 '14 at 15:59
• Don't forget temperature differences across the surface, light from hotter depth that eventually finds its way out, other light-producing phenomena (spectral emission from excited electrons recombining with atoms then ionizing again, photons created in scattering, decay, annihilation, etc processes), etc. A star is a complex system with a lot happening all the time. It's safe to assume that blackbody radiation (while the primary source) isn't the only source of radiation.
– Jim
Aug 9 '14 at 16:17
• @stefans: the lower temp one looks to be at least a closer fit for the higher wavelength, rather than the peak. I would expect 5800 K to give a better result with regards to $\epsilon$. Aug 9 '14 at 17:04
• An even better value is 5778 K (or 5777 K, depending on who you read), the Sun's "effective temperature". The effective temperature of a star is the temperature of a black body with the same surface area as the star and that emits the same total radiated power as does the star in question. Aug 9 '14 at 17:08

The black body emission is the one of a hot body. It's radiation comes from it's temperature. The sun is energy producing, and this energy emits. Parts of it emit because of the pure surface temperature, but there are other processes, which convert energy.