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This question is inspired from The Sun is giving us a low entropy, not energy.

I accept the claim that sunlight must be a low entropy source of energy, given that it maintains the entropy gradient necessary for life (along with other sustained departures from thermodynamic equilibrium, like weather). We can also show that sunlight is low entropy by comparing its spectrum to a blackbody. Sunlight transmits an energy flux of ~1000W/m² to Earth's surface. Were sunlight a blackbody, then its temperature would be ~360K according to Stefan-Boltzmann. But its actual temperature is closer to 5000K. So sunlight must be far from a blackbody spectrum. (I'm actually not wholly convinced by this last argument. I present it here so that others can point out its flaws.)

Here's my confusion: photons that are emitted in the Sun's core take something like a 100,000 years to reach its surface due to the numerous collision along the way. This ought to thermalize the photons to a high-entropy blackbody radiation. So why is sunlight suddenly low-entropy when it reaches Earth's surface?

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    $\begingroup$ Hint: consider a sun-powered heat engine on Earth vs in the Sun's photosphere. $\endgroup$
    – PM 2Ring
    Jan 8 at 2:48
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    $\begingroup$ according to you argument, if it was possible to compute the blackbody temperature based on the energy flux at earth, then you would also be able to do the same on mars, or jupiter, or pluto... and you would get a different answer for each planet - which is obviously not correct $\endgroup$
    – pygri
    Jan 8 at 13:16
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    $\begingroup$ Btw, the 100,000 years photon bouncing is bullshit. Within the sun's core, a few high energy gamma rays are generated in each fusion reaction (on the scale of mega electron volts), but a huge number of low energy photons are radiated from the sun's surface (on the scale of a few electron volts). Thus, each core gamma ray causes literally millions of photons to be emitted from the surface. And that happens by atoms absorbing photons and subsequently emitting more lower energy photons than what was absorbed. $\endgroup$ Jan 8 at 13:59
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    $\begingroup$ Radiation from the Sun is mostly thermalized, and its spectrum is similar shape to a blackbody spectrum at 5800 K, but much weaker on Earth, due to radiation intensity falling off as inverse square of distance. This weakening with distance has nothing to do with radiation temperature or entropy of sunlight being high or low, it is a geometrical effect (because the Sun radiates in all directions roughly equally). How do you define entropy of sunlight? What makes it "low entropy", low compared to what? $\endgroup$ Jan 8 at 15:59
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    $\begingroup$ I dare say that photons from the sun don’t have low entropy. Photons from the sun here, but not there, have low entropy. It’s the existence of a strong gradient against a place with significantly lower energy that results in low entropy. $\endgroup$ Jan 8 at 19:49

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Solar energy at Earth is low-entropy because all of it comes from a region of the sky with a diameter of half a degree of arc. The rest of the sky (with a few negligible exceptions) is illuminated only by the cosmic microwave background, whose temperature is at somewhere under three kelvin. Some of us are more likely to interact with Earth's surface or atmosphere than we are to interact with the darkness of the sky; but Earth's surface and atmosphere are also substantially cooler than the Sun.

This low-entropy arrangement allows you to construct a heat engine which takes sunlight on one side as a heat source, and radiates waste heat in other directions.

Within the Sun's photosphere, where (as you point out) the photon spectrum is mostly thermalized and photons are roughly equally likely to travel in any direction, such a heat engine is impossible to build: there is no way to connect to the cold reservoir.

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  • $\begingroup$ Not right. If we remove the sun and install many such lights that are good for plants, then plants are fine. Right? $\endgroup$
    – stuffu
    Jan 8 at 4:07
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    $\begingroup$ I don't think that's right. I think photosynthesis requires a cold reservoir, usually maintained by the water volume in the plant, and by thermal contact with the surrounding air, water, and soil. If you completely surround a plant with hot lamps, so that no cold reservoir is available, the plant will come into thermal equilibrium with the lamps. (You can experiment with this by putting plants into your kitchen oven. I recommend lightly dressing the plants with salt, pepper, and a little olive oil before they go in.) $\endgroup$
    – rob
    Jan 8 at 4:54
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    $\begingroup$ Thank you for the answer. I think the crux of my misunderstanding is that while sunlight in isolation may be at high entropy (though not at maximum entropy, since it's highly directional), when considering the sunlight + earth + rest of space system as a whole, it represents a significant departure from equilibrium and a source of low-entropy energy. $\endgroup$ Jan 8 at 12:15
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    $\begingroup$ @JánLalinský I waffled about that, then wrote it anyway. Since the maximum efficiency of a heat engine goes like the ratio of the temperatures of the hot versus cold thermal reservoirs, a heat engine with a Carnot efficiency of 1% would have to connect thermal reservoirs in different solar layers separated by hundreds or thousands of kilometers. Make the CMB available as a cold reservoir, and that constraint disappears completely. $\endgroup$
    – rob
    Jan 8 at 15:35
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    $\begingroup$ @AlessandroPower Yes, I agree with your self-assessment. If Earth were another sun, then Sol's light wouldn't be so useful for powering things. It's because the Earth is so relatively dead in comparison. It doesn't matter if the top of your waterfall is a billion km below sea level. If the bottom of your waterfall is a billion and one meters below sea level you can still get energy from it. $\endgroup$
    – DKNguyen
    Jan 8 at 16:14
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Sunlight is low entropy because the sun is very hot. Entropy is essentially a measure of how spread out energy is. If you consider two systems with the same amount of thermal energy, then the one where that energy is more concentrated (low entropy) will be hotter.

Sunlight transmits an energy flux of ~1000W/m² to Earth's surface. Were sunlight a blackbody, then its temperature would be ~360K according to Stefan-Boltzmann. But its actual temperature is closer to 5000K. So sunlight must be far from a blackbody spectrum

Sunlight is fairly close to black body radiation. The relevant flux is at the surface of the sun, not the surface of the earth.

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    $\begingroup$ "Sunlight is fairly close to black body radiation. The relevant flux is at the surface of the sun, not the surface of the earth." Thank you, this confirms that my "argument" was indeed flawed. $\endgroup$ Jan 8 at 11:55
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I think your confusion might be partly about the relationship between temperature and entropy when it comes to radiation.

Roughly speaking, the biosphere absorbs Solar radiation at $\sim 6000\,\mathrm{K}$ and emits waste heat into space at $\sim 300\,\mathrm{K}$. To a very good approximation, over long enough time scales all of the absorbed radiation is eventually emitted again.

Now if you have 1 joule of thermal radiation at 6000K, it has entropy $\frac{4}{3} \frac{Q}{T} = 4/3\cdot 1/6000 = 1/4500\,\mathrm{J\,K^{-1}}$, much lower than if you had 1 joule of thermal radiation at 300K, which has $4/3\cdot 1/300 = 1/225 \,\mathrm{J\,K^{-1}}$.

So indeed, the high temperature Solar radiation is a low entropy source of energy, in comparison to the waste heat radiated away from the Earth at ~300K.

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  • $\begingroup$ "Now if you have 1 joule of thermal radiation at 6000K, it has entropy..." Actually, I think you are reproducing my misunderstanding. Your calculation returns the radiation's change in entropy, not its total entropy. Indeed, the entropy per volume of blackbody radiation scales like T^3. While it's fair to say that sunlight loses less entropy than the Earth gains, that does not imply that sunlight itself is low entropy. I think the confusion between these two ideas is at the heart of my question. $\endgroup$ Jan 8 at 15:04
  • $\begingroup$ @AlessandroPower Using naively the entropy change formula $\Delta S = \Delta Q/T_{Sun} - \Delta Q/T_{Earth}$, Earth does not gain entropy, because in a dynamic thermal equilibrium, energy that comes to Earth at the Sun temperature $5800$K leaves the Earth at the Earth temperature $300$K so it carries away much larger entropy. This is of course a problematic argument too, because the energy exchange via radiation is not like heat transfer over zero temperature difference and the formula need not apply. $\endgroup$ Jan 8 at 16:24
  • $\begingroup$ @AlessandroPower it's not the entropy per volume that's relevant, it's the entropy per joule. This is because to a good approximation, the radiative energy coming in is balanced by the radiative energy going out, whereas this is not true for volume in any sense that I know of. The entropy per joule for thermal radiation is $\frac{4}{3}\frac{Q}{T}$, so that's the relevant formula, not the $T^3$ one. In my answer, I calculated the total energy per joule of both the incoming and outgoing radiation. Per joule, the incoming energy is lower entropy than the outgoing energy. $\endgroup$
    – N. Virgo
    Jan 9 at 9:03
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At the sun's surface, sunlight is similar to black body radiation at the sun's surface temperature. Thus, if you brought a surface next to the sun's surface, it would be able to heat up to that same temperature. And this temperature determines the entropy content of the sunlight.

Now the sunlight leaves the sun and spreads out through space. This has three effects:

  1. The energy flux is spread across a larger and large sphere as the light travels further away. This reduces the sunlight's intensity according to the inverse square law.

  2. The direction of the individual photons become more and more aligned with each other as the image of the sun in the sky becomes smaller and smaller. This increases order in the sunlight.

  3. The spectrum of the sunlight gets misaligned with respect to its intensity. This increases order in the sunlight as well.

The first effect should increase the entropy content of the radiation, but the other effects exactly cancel that. The light's voyage effectively transfers the entropy content from being due to statistical direction into entropy content due to low intensity.

This can easily be shown by focusing sunlight: If you were to build perfect optics that focus the sunlight such that reflected sunlight converges from all around on a single point, an object at that point would be heated to the sun's surface temperature.


Real world applications:

Solar power towers use mirrors to focus sunlight onto a small receiver, producing temperatures as high as the receiver materials allow for. This takes advantage of effect 2.

Flat plate collectors frequently use a special coating that is both optimized to absorb visible light, while simultaneously reflecting thermal infrared light. This takes advantage of effect 3 to decouple the albedo for incoming and outgoing radiation, selectively allowing the visible incoming light to be absorbed while keeping the losses due to outgoing heat radiation low.

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I'm actually not wholly convinced by this last argument. I present it here so that others can point out its flaws.

Earth's surface vs Earth atmosphere

~1000W/m² to Earth's surface

$1\,000 \frac{\mathrm{W}}{m²}$ is indeed a correct order of magnitude for the maximum irradiance on Earth's surface. But some of it has already been absorbed or reflected by the atmosphere. The irradiance you should use is the solar constant, which is $\approx 1\,360 \frac{\mathrm{W}}{m²}$.

This already increases the temperature from $360K$ to $\approx 394K$ in your calculations.

Earth atmosphere vs Sun surface

More importantly, and as mentioned by @Dale: "The relevant flux is at the surface of the sun, not the surface of the earth."

The ratio you need to use is $\frac{\mathrm{Astronomical Unit}}{\mathrm{Sun Radius}} \approx \frac{150\cdot10^9\mathrm{m}}{100\cdot\mathrm{Earth Radius}} \approx 233$.

You need to take inverse-square law into account, as well as Stefan Boltzmann law.

So the correcting factor is $ \approx \sqrt[^4]{{233^2}} \approx \sqrt{233} \approx 15$.

And $394K * 15 \approx 5\,910 K$, which is pretty close to the photosphere's temperature.

If you use a more precise value for Sun's radius (as opposed to simply Earth's radius * 100), you get $5\,770K$.

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Sunlight exhibits low entropy due to its highly ordered nature. It consists of photons moving uniformly in a specific direction, making it a coherent stream of energy. Entropy is a measure of disorder, and sunlight, being organized and structured, has lower entropy compared to more random or disordered forms of energy. This organized characteristic of sunlight contributes to its lower entropy state.

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