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Apparently spectral solar radiation is approximated by a black body at 5800 K. The spectral black body distribution (Planck distribution) is shown below (from Incropera, Fundamentals of Heat and Mass Transfer), with different temperatures including solar radiation at 5800 K.

enter image description here

  1. The heat flux on an Earth surface perpendicular to sun rays (i.e., the solar constant), is roughly 1.36 kilowatts per square meter. Is this simply the integral of the above distribution?
  2. Why is the sun approximated as a black body at 5800 K? Does this mean that the surface of the sun which is emitting radiation has a temperature of 5800 K? That seems kind of low.
  3. Is solar radiation approximated as a black body at 5800 K only on Earth, or is it the same everywhere? Why don't atmospheric effects and scattering change measurements of the solar spectrum on Earth?

EDIT: The solar constant is approximated by considering the Stefan-Boltzmann law (i.e., the integration of the spectral solar emission), the size of the sun, and the distance from the sun to Earth. A good derivation is shown here: https://www.youtube.com/watch?v=DQk04xqvVbU

Qmechanic
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Thermodynamix
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4 Answers4

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  1. Yes - the integral of the spectrum you refer to gives the total power per unit area emitted from the surface of the sun. If you multiply that by a factor of $\left(\frac{\text{Solar Radius}}{1\text{ AU}}\right)^2$ to account for the $1/r^2$ dependence of intensity on distance, then you'll get the solar constant you quote.
  2. Yes. The sun is not at a single uniform temperature - the radiation which reaches Earth is mostly emitted from the photosphere (~6000 K) but the temperature varies dramatically between the different layers of the sun.
    enter image description here

Source

  1. Everywhere. The sun is very nearly an ideal black body. This is a property of the sun, not of a particular vantage point from which you're observing it. Furthermore, atmospheric effects dramatically change measurements of the solar spectrum on Earth.
    The upper atmosphere blocks nearly all of the radiation at higher frequencies than UV, and quite a lot of the IR spectrum is absorbed and scattered by greenhouse gases. Visible light passes through without much trouble (which is a substantial part of why we evolved to be sensitive to those frequencies) but the facts that the sky is blue and that sunsets are beautiful demonstrate that the atmosphere scatters visible light as well.

enter image description here

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Ruslan
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J. Murray
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  • How does intensity have a $1/r^2$ dependence? I think radiant flux has a $1/r^2$ dependence due to the definition of a solid angle: https://en.wikipedia.org/wiki/Radiant_intensity – Thermodynamix Mar 20 '19 at 19:34
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    @Drew I am using the terminology as defined here: https://en.wikipedia.org/wiki/Intensity_(physics) – J. Murray Mar 20 '19 at 19:42
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As a complement to the other answers, I'd like to point out that if you measure the solar spectrum at a very high resolution, you see very large variations in spectral radiance over very short wavelength scales. Here is a figure that I made for my PhD thesis to illustrate this:

High resolution solar spectrum

A high resolution solar spectrum (source here) is plotted in grey and the same spectrum with degraded resolution is plotted in black. The lower panels show close-ups of smaller wavelength regions. This is the solar spectrum as seen from space, i.e. there are no effects from Earth's atmosphere present.

These variations are due to a multitude of absorption lines (normally referred to as Fraunhofer lines) caused by absorption from gases in the colder outer parts of the photosphere (and to some extent in the chromosphere).

Hence, the ~5800 K black body spectrum is not a good description of the solar spectrum at very fine wavelength scales. But it's a reasonable approximation on larger scales.

jkej
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    Out of interest, are the v-shaped all-the-way-to-zero drops at 393½ and 397 nm real or somehow an artifact of the processing? The whole thing looks intuitively like the absolute value of an amplitude that drops below the axis and climbs back up again, but I have trouble imagining a physical effect that would cause it to look like that. – hmakholm left over Monica Mar 20 '19 at 19:26
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    @HenningMakholm These are two of the strongest Fraunhofer lines. They are listed in the Wikipedia article as the H and K lines, also known together as the Calcium II doublet. I don't believe there is any artifact visible in the plot of them. – jkej Mar 20 '19 at 19:44
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    I don't doubt the lines are there; I'm just struggling to understand their shape. Perhaps I should ask that as a new question, though. – hmakholm left over Monica Mar 20 '19 at 19:50
  • @HenningMakholm Ok, I can't say I know exactly why they look like that. I haven't produced the data, so I can't rule out artifacts to 100%, but I would have assumed that they at the very least got the shape of the major Fraunhofer lines correct. – jkej Mar 20 '19 at 20:06
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    Thanks for sharing your work. Good to see some original data. I learned something new today by visiting this seemingly simple question. (As always: when you look really close things become really complicated ;-). ) – Peter - Reinstate Monica Mar 21 '19 at 07:04
  • @HenningMakholm The shape of the atmospheric O2 (Fraunhofer) lines looks "distorted" because there is a pressure gradient in earths atmosphere. So one observes a superposition of many Lorentzian/Voigt lines of different width. Because line width is approximately proportional to pressure it creates a more triangular shaped absorption profile for the atmospheric lines. – Andreas H. Mar 21 '19 at 09:09
  • I apologise if this is too divergent and perhaps should be a separate question, but I saw the close-ups and thought "there are so many lines/dips!". Presumably every single one corresponds to elemental absorption..?

    Then I thought: in school we're taught that the sun's emission spectrum is continuous (ignoring the fraunhofer lines, et al). But is it really? Or is it just the sum of very VERY many emission lines of various elements in the sun. Emissions should be quantised to electron shell transitions, and now I can't see why anything would be truly continuous.

     – Barney Desmond Mar 22 '19 at 06:36
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    @Barney It is a different question. A very interesting one, but you really should ask it separately. But in short: (1) lines are never infinitely thin because the emitted wave packets have a finite length (“lifetime=uncertainty broadening”): that’s why particle physicists use “decay width” more or less interchangeably with “state lifetime”; (2) at finite temperature and density, this ludicrously small width is increased by several orders of magnitude, mostly by “thermal=Doppler broadening” and “pressure=collisional broadening” in a gas; [cont.] – Alex Shpilkin Mar 22 '19 at 14:29
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    @Barney [cont.] (3) most importantly, in an equilibrium setting (such as that of blackbody radiation), it doesn’t matter whether the absorption at a particular energy is strong or weak as long as absorbing layer is thick enough: it doesn’t matter whether the photon got absorbed on the first or on the billionth encounter as long as it didn’t come out; thus those apparently insignificant tails of the spectral lines end up being as important as the peaks. – Alex Shpilkin Mar 22 '19 at 14:36
  • @Alex Shpilkin: much appreciated, thank you for the answer. It took a little more research and reading, but my mind is thoroughly blown now. The fundamental uncertainties in all of this definitely make it seem worthy of a separate question (pending a search to see whether this has already been asked). – Barney Desmond Mar 24 '19 at 12:29
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  1. The solar constant ($S$) is the energy received per unit area perpendicular to the rays at 1 AU distance (see wikipedia). Thus, mathematically, it represents the total energy emitted by the sun divided by the surface area of a sphere with a radius of $R$=1 AU: $$S=\frac{P_{emitted}}{4\pi R^2}$$ Usually, one can use the Stefan-Boltzman law to compute $P_{emitted}=\sigma T^4\times A_{Sun}$ by the sun which is considered a black body here and where $A_{Sun}$ is the total area of the sun's surface, $\sigma$ the Stefan-Boltzmann constant and $T$ the black body's surface temperature. So: $$S=\sigma T^4\left(\frac{R_{Sun}}{R}\right)^2\approx1380W/m^2$$ By plugging numbers: $R=$1AU$=150\times 10^9m$, $R_{Sun}=6.957\times 10^8m$, $T=5800K$ and $\sigma=5.67\times 10^{-8}W/m^2K$, we get approximately the same Solar constant.

  2. The definition of a black body is a body that absorbs all incoming radiations and that emits light solely due to it's temperature. In astronomy, since stars are really shiny objects and that they are really distant from each other, they can be considered black bodies since we approximate that the light we get from them is almost entirely emitted by their surface and is not reflected light from another star. In other words, they are approximated by black bodies because we can be pretty sure that the light they emit is solely due to their temperature. Approximating the sun as a black body effectively means that it's surface temperature is 5800K. And, indeed, if we look at the most dominant wavelength of the sun spectrum, the maximal intensity comes from green wavelength which corresponds to a black body emitting radiation at a temperature of 5800K. For more info look at the Wien's displacement law.

  3. The sun is considered a black body, not only from the perspective of the Earth but from any perspective. The sun's intensity spectrum is roughly the one of a perfect black body minus absorption by the sun's atmosphere and the Earth's atmosphere depending of where we measure the spectrum. See the sunlight article on wikipedia.

aventurin
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fgoudra
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  • Why do we divide the emitted power from the sun by the surface area of a sphere with radius $R$ = 1 AU? I know that's the distance to the earth, but I don't see how that gives us the heat flux on the earth. Furthermore, the emitted power $P_{emitted}$ is the total power emitted by the sun, and only some of that will hit Earth. – Thermodynamix Mar 20 '19 at 16:52
  • @Drew: The second sentence in that comment is the correct answer to the first one :-) – hmakholm left over Monica Mar 20 '19 at 19:28
  • @HenningMakholm Ah I see now. Thanks. – Thermodynamix Mar 20 '19 at 19:30
  • I am not quite convinced by the sentence "The sun is considered a black body, not only from the perspective of the Earth but from any perspective." The answer by jkej casts serious doubts on it, IMO. – untreated_paramediensis_karnik Mar 20 '19 at 20:29
  • Typo in Stefan-B law. Note that the T of the Sun is defined to be $(L/4\sigma\pi R^2)^{1/4}$. – ProfRob Mar 21 '19 at 07:46
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    @thermomagneticcondensedboson Not really. It's not like there are any absolutes in science - all of our models have a purpose. For mundane purposes, the Sun is close enough to being a blackbody; for things that depend on the imperfections, we often still distinguish the absorption lines etc. by comparison to the ideal blackbody radiation (e.g. stellar spectroscopy). You use the model you need for what you're trying to do, and simplicity trumps precision unless you absolutely need that precision. The blackbody curve is simple; the real spectrum is anything but. – Luaan Mar 21 '19 at 09:35
  • @Luaan I agree with you, though I do not understand why you start your comment by "Not really.". It is fgoudra who brings the Sun as being a black body under any perspective, not me, and this is what I criticized. – untreated_paramediensis_karnik Mar 21 '19 at 15:34
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    Whether something can be modeled as a black body or not does not depend on whether you are actually shining light on it. If I put my mirror into a dark room, it won't turn into a black body by magic, it'll still be a highly reflective object that just happens to be in the dark. What's relevant is, whether light would get absorbed if it were shone on the object. So, please, cut out that part about star light reaching the sun! – cmaster - reinstate monica Mar 21 '19 at 22:47
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A point worth mentioning, is that the temperature of the black body may be a bit surprising. This graph might be useful: enter image description here graph from libretexts

The core of the sun is many millions of degrees. But we never see that temperature. Its surrounded by a convective zone, to the extent that photons can take millions of years to escape from the core, through the radiative and convection zones, to the surface. At the surface, the sun is indeed only a very modest few thousand degrees. (Although specific phenomenas such as flares may be much hotter)

Stilez
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