Chapter 6: The Entropic Principle

Thanks for the further clarifications on the relationship of gravity and entropy. I’m still not entirely sure how to reconcile all of that with what I’ve read elsewhere. But I appreciate that you have much more insight into this topic. So I will keep trying to sort it all out.

As for the relationship of the sun and the earth as mediated via light, I’m wondering if you can comment on the observation that the light arriving from the sun is via fewer, higher energy photons from a specific spatial direction, while the light departing the earth is via more lower energy photons in all directions. I understand this to mean that the arriving photons have fewer degrees of freedom. Is that correct? If so, why can that not be connected to the relative entropy of those photons?

Thanks.

I will be happy to help you make this reconciliation. From a personal standpoint, I am also curious of your sources if they claim that the entropy of “gravitational degrees of freedom” is important in cosmological structure formation - this is well known to be false.

First of, photons are quantum objects, so their entropy is given by the von Neuman entropy instead of the usual Boltzmann S= k ln Ω (where Ω is the number of microstate). The entropy of a photon is therefore related to how mixed the quantum state of the photon is. Therefore, the proper way to answer this question is to compare the quantum states of incoming photons from the Sun with the quantum states of emitted photons from Earth. A shortcut might be made due to the photons being blackbodies.

I did not do this exercise, but I don’t need to to know what you will get. Because I believe in the 2nd law, I am sure that in the end you will get that all re-radiated photons from the Earth will have greater entropy than all the received photons from the Sun. Edit: actually, I forgot about the entropy stored in the Earth itself, so one cannot use the 2nd law as a shortcut here. I edited the required calculation at the end of this post.

Regardless, my point is that comparing entropies is not useful to assess the ability for systems on Earth to use energy from the Sun. It is hard to explain this without a concrete example in mind, so try computing the amount of energy that is necessarily wasted due to the 2nd law from a heat pump with the Sun as the high temperature reservoir and the Earth as the low temperature reservoir.

Now do the problem again with a “faint sun” of temperature only 500K. I hope you will then be convinced that the huge difference in temperature between the Earth and the Sun plays a major role in the ability for life to harvest useful energy from the Sun. It’s not all in the entropy difference, and that without a reference to the temperatures, this exercise of “comparing entropies” is really physically meaningless!

Edit: @AndyWalsh, here is the calculation for whether arriving photons from the Sun have less entropy than emitted photons from the Earth

  1. Imagine a point on the surface of the Sun. The photons leaving this point are blackbody, therefore the entropy is given by (4/3)U/Ts, where U is the energy of the photon gas, and Ts is the effective temperature of the Sun (~5777K)
  2. Photons arrive on Earth from a particular direction. In particular, from the point of view of the Sun it is confined to a solid angle 4Re²/AU², where Re is the radius of the Earth and AU the astronomical unit. Therefore the flux of energy being sent to Earth is: F= σTs⁴ x (4Re²/AU²)/(4π).
  3. Remember, that was flux from a single point on the surface of the Sun. We integrate over the Earth-facing surface of the sun to get the total flux to be F= σTs⁴ x (4Re²/AU²)/(4π) x (2πRs²), where Rs is the radius of the Sun.
  4. From 1. the flux of entropy received on Earth is therefore: F_S(Sun) = (4/3)σTs³ x (4Re²/AU²)/(4π) x (2πRs²). This quantity has units of “rate of entropy change”, Entropy/(Time).

Now, what is the rate of entropy change due to the Earth’s radiating thermal blackbody radiation?

  1. The luminosity of the Earth is Le = 4πRe²σTe⁴. Te is the effective temperature of the Earth (~300K).
  2. Since this is also blackbody, then the rate of entropy change is F_S(Earth) = -(16/3)πRe²σTe³. The negative sign is there because this is entropy loss from the Earth.

Is the entropy received from the Sun larger or smaller than entropy loss from the Earth’s infrared radiation?

  1. The net rate of change of entropy is given by R_S = F_S(Sun) + F_S(Earth)
  2. R_S = (4/3)σTs³ x (4Re²/AU²)/(4π) x (2πRs²) - (16/3)πRe²σTe³
  3. Collecting like terms, R_S = (16/3)Re²σπ x [Ts³2Rs²/(4πAU²) - Te³]
  4. The entropy emitted by the Earth is therefore larger than that received from the Sun if Te³>Ts³2Rs²/(4πAU²)
  5. The temperature term signifies that larger temperature photons have larger entropies, while the 2Rs²/(4πAU²) term signifies the entropy penalty due to the photons coming from a particular source (a small angular area)
  6. 2Rs²/(4πAU²) is ~10⁻⁶, which means that Ts³2Rs²/(4πAU²)~664000 cubic Kelvin
  7. Te³~27 million cubic Kelvin
  8. Therefore, Te³>Ts³2Rs²/(4πAU²) and the entropy emitted by the Earth is larger than that received by the Sun

This was a fun exercise, but as I talked about in the body of this post, comparing “entropies” this way really does not have any physical meaning with regards to converting the Sun’s heat to usable energy without a reference to the large temperature difference between the Earth and the Sun.

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As far as sunlight and life on earth Denton’s new book Children of Light presents some interesting information.

Well, @PdotdQ, you definitely win the “above and beyond the call of duty” award. I appreciate you putting in all of that effort to educate me (and hopefully some other folks in the forum). Clearly my descriptions were incomplete–which was always likely, given the nature of the project, but it’s helpful to have more details on exactly what is missing. At the video chat on Saturday, I will be sure to encourage folks to check out this thread so they can benefit from your work as well.

It would be interesting to play around with the different components to see under what conditions the Sun/Earth relationship flips. In particular, I wonder how big of an angular area the Sun would have to occupy to reverse things. Maybe it will tell us something about why the Sun is the same apparent size as the moon. (I am very skeptical that it actually would and am content for the Sun/Moon size ratio to be a coincidence, but it would be interesting if there were more of an explanation.)

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Can you elaborate?

You meant I won the “biggest procrastinator” award :wink:

This is an interesting line of thought. Indeed, it will flip when the ratio of the temperatures to the third power, (Te/Ts)³, becomes smaller than θ²/4π, where θ is the angular size of the Sun in the sky.

Ah, yes. I know that gambit well. The math I don’t have to do is often more fun than the math I do.

In that spirit, if I’m doing all the math right* and using all the correct values for the constants**, I get a critical value for \theta of \sim 0.06. At that point, \left (\frac{T_e}{T_s}\right )^3 = \frac{\theta^2}{8\pi}. To a first order, that’s 6x the actual value. So probably nothing to do with the size of the moon, but it still answers how much bigger/closer the Sun would have to be to change that entropy relationship.

*Going by your formula in #8 above ( T_e^3 > \frac{T_s^3 2 R_s^2}{4 \pi AU^2} ) I feel like there is an extra factor of 2 that I’m not certain I’m accounting for. My understanding is that \theta \approx \frac{R_s}{AU} because AU \gg R_s. Edit: Mathing while tired is hazardous; Diameter = 2 * Radius; that’s where the extra factor of 2 goes. \theta \approx \frac{2R_s}{AU}.

** For reference, I’m using T_e = 300K, T_s = 5777K, R_s = 695,700km and AU = 149.6 gigameters.

Edit: Corrected some calculation errors.

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Hmm, I didn’t realize that this forum supports latex.

Interesting! I might assign this to my future students.

I don’t guarantee that my formulae are correct to every factors of 2π’s, but I also defined θ as the full angular size of the Sun ~(2Rs/AU) instead of ~Rs/AU. Edit: yup, I dropped a 1/2, it should be θ²/8π.

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Note that after updating for the correct value of \theta and using 8\pi instead of 4\pi, it comes out to roughly 6x the actual value, not 10x.

Yup, radius is not diameter, so I was correct that I wasn’t properly handling a factor of 2. I have edited the post accordingly.

Just a reminder that I’ll be on Facebook Live tomorrow night (11/17) at 7:30pm EST to chat all things Faith across the Multiverse. This will be the final video chat for this year, although I welcome ongoing discussion here in the forum over the holidays.

This should be the link for the video stream when it starts:

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How much bigger would it need to be?

To a first approximation, the answer is almost certainly “astronomically larger.”

I think I see a way to hand-wave some dimensional analysis and get an answer in the correct units by applying Boltzmann’s constant and this equation from @PdotdQ’s work above: \frac{4 \sigma T_s^3}{3} \cdot \frac{4R_e^2}{4 \pi AU^2} \cdot 2 \pi R_s^2 and stating the problem more precisely to involve the flux of entropy from the sun for 1 second. But I’m not entirely clear on where \sigma comes from to calculate an actual numerical result.

@physicists, I don’t understand. Help me?

Fair warning - I’m not claiming any of this makes any physical sense. Just that I might be able to fudge a silly answer to my own silly question.

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I don’t know what you are trying to calculate, but:

σ is the Stefan-Boltzmann constant. The first term in that equation is just

\frac{4}{3} \frac{\rm{Energy \; Flux \; of \; blackbody \; radiation}}{\rm{Temperature}}

This is because the entropy of blackbody photon gas is (4/3)(Energy/Temperature). Note that the energy flux of blackbody radiation is F=\sigma T^4.

The second term takes into account that the Sun spans a certain angular size in the sky as seen from a point on the Earth. The third term takes into account that an entire half of the Earth receives entropy from the Sun.

Join the club! In another thread there was some calculations related to the size of the human genome in bits. The thread referenced a news story that was ambiguous in how it was using ‘bit’. Josh knows that I’m interested in overloaded terms and brought it to my attention. I made a highly amusing (read: dubious) joke involving yet another meaning of ‘bit’ (half of a quarter, as in ‘shave and a haircut, two bits’). Then I doubled down on the (again, dubious) humor by calling back to the discussion here about adjusting the relative size of the Sun, based solely on the fact that entropy also appears in information theory and its bits.

For my sins, I then immediately fell down my own pit trap of wondering whether it would be actually possible to calculate an “actual” answer to my nonsensical question. To wit: how much bigger would the human genome have to be (in information-theory-bits) to have more entropy than the entropy arriving on Earth in one second from the Sun? I have no delusions that the answer will be meaningful in any sense other than the tomfoolery of making units match up.

Thanks!

Seeing as how I went awry previously by trying to do math late at night, I think I will put off the actual calculation for now. But I do appreciate the added info.

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We receive entropy from the Sun? I hate to admit it but I am confused again…and I want to know the answer to your nonsensical question.

Yes, the Sun sends entropy to the Earth through photons, much like the Sun sends energy to the Earth. Entropy is a thermodynamically extensive quantity, like mass, energy, and number of particles. These things can typically be transported from one system to another.

@AndyWalsh I must say I wanted to know the answer too. I don’t know what the answer might mean physically, but if anything it might be a good question to put in a final exam.

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And I will come back to this tomorrow, and show my work. But I need to get some sleep first. Sorry for the suspense.

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OK, let’s do this.

Constants (using SI base units so we can follow them all):

  • \sigma = 5.67 \cdot 10^{-8} \frac{kg \cdot m^2}{s^3 \cdot m^2 \cdot K^4} = 5.67 \cdot 10^{-8} \frac{kg}{s^3 \cdot K^4}
  • T_s = 5777K
  • R_e = 6.38 \cdot 10^6 m
  • R_s = 6.96 \cdot 10^8 m
  • AU = 1.50 \cdot 10^{11} m
  • \kappa = 1.38 \cdot 10^{-23} \frac{kg \cdot m^2}{s^2 \cdot K}

Using the flux of entropy formula \frac{4\sigma T_s^3}{3} \cdot \frac{4R_e^2}{4\pi AU^2} \cdot 2\pi R_s^2 we get \frac{4}{3} \cdot 5.67 \cdot 10^{-8} \frac{kg}{s^3 \cdot K^4} \cdot (5777K)^3 \cdot \frac{4 \cdot (6.38 \cdot 10^6 m)^2}{4\pi \cdot (1.50 \cdot 10^{11} m)^2} \cdot 2\pi \cdot (6.96 \cdot 10^8 m)^2 = 133 \frac{kg \cdot m^2}{s^3 \cdot K}. if we multiply by 1 second to get the entropy of photons arriving over 1 second, we get 133 \frac{kg \cdot m^2}{s^2 \cdot K} or 133 Joules per Kelvin.

Now here’s where we get into territory where I’m less certain if what I’m doing is actually meaningful. If we use the formula for entropy H = \kappa \ln W where W is the number of microstates, then we have a relationship between the thermodynamic entropy and the number of bits needed to specify which of the W microstates the photons are actually in.

So 133 \frac{kg \cdot m^2}{s^2 \cdot K} / 1.38 \cdot 10^{-23} \frac{kg \cdot m^2}{s^2 \cdot K} = 9.6 \cdot 10^{24}. Notice that all the units cancel out, unless we understand \kappa to be in terms of Joules per Kelvin per nat and then the answer is in nats. We can then do a logarithm base conversion from e to 2 to get 1.4 \cdot 10^{25} bits.

And how many bits are in the human genome? If we say it has 3.2 \cdot 10^9 base pairs and use the maximal information content of 2 bits per base pair (an overestimate biologically, but I hope it is obvious that a factor of 2 or 4 isn’t going to matter in the answer we are about to get), then we get 6.4 \cdot 10^9 bits. Which means the human genome would need to be 2.1 \cdot 10^{15} times larger (in terms of bits) to eclipse the entropy of the photons arriving on Earth from the Sun in one second. To try to put a little more perspective on that, if we say there are roughly 10 million eukaryotic species (the upper range of the estimate) and assume that to a first approximation their genomes have the same information content as humans’ (a very crude estimate), we could use the bits in all eukaryotic genomes and still be short by a factor of 100 million.

Now you know, and knowing is half the battle.

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