From Nature:

We divide by $4$ since the solar energy is spread over the surface of the planetary sphere. The Earth intercepts a circular area of incoming sunlight, and this area is spread over a sphere with the same radius as the circle (area of circle / area of sphere of same radius = $0.25$).

It would seem by dividing by 4 it results in the sun's energy being evenly distributed over the entire globe at the same time as if the earth was flat, and weakening the actual energy required to create a climate. I don't see any logical reason to do this as we know this doesn't physically happen.

What is the explanation? Why do you need to average out the sun's energy over the whole planet at the same time when that's not what actually happens?

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    $\begingroup$ 4 is the ratio between the surface of incident flux, which is $A_{in} = \pi r^2$, and the surface of outgoing flux, which is $A_{out} = 4 \pi r^2$. It's as simple as that. The other complications in the other answers, go into detail of how to find the temperature from the flux. $\endgroup$ – AtmosphericPrisonEscape Mar 8 at 12:20
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    $\begingroup$ You need to compare apples to apples. As was explained earlier, the factor 4 comes into play for an averaged model of Earth, that is represented either by a point (0-D) or a 1-D profile. An averaged model does not have pole-caps or a hot equator, where ice would melt. If you want all those details, do a 3-D model, then you won't worry about this factor of 4. Nonetheless 0-D and 1-D models are useful to showcase general atmospheric physics, especially the GHG, and give correct average temperatures. $\endgroup$ – AtmosphericPrisonEscape Mar 8 at 18:10
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    $\begingroup$ (cont.) Your example car exists somewhere as a point on a 3-D sphere where the insolation is larger than the average, and additionally has a significant GHG effect from the atmosphere + window. Furthermore you need a certain photon intensity, not a flux, to raise the temperature. To require a flux for a temperature is just ill-defined in a radiatively diffusive atmosphere. Beyond that, remember that in the diffusive (GHG) atmosphere the local photon intensity increases as $(1+\tau)^{1/4}$, where $\tau$ is the optical depth. $\endgroup$ – AtmosphericPrisonEscape Mar 8 at 18:14
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    $\begingroup$ You use 0-1-D models because they are toy models (simplified models) that showcase the important physics. Try and understand a full 3-D model from scratch, you won't have any luck there. Holder's inequality is a mathematical inequality, and not a physical model. I don't see how you say "i want no math" but then you refuse to take a look into a 1-D model and critisize them. The S-B law gives the power of an emitting body into vacuum, WITHOUT ghg. Again, the latter is given by $T^4=T^4_0 (1+\tau)$. /wiki/Illustrative_model_of_greenhouse_effect_on_climate_change $\endgroup$ – AtmosphericPrisonEscape Mar 9 at 9:28
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    $\begingroup$ sigh "I can still apply S-B to anything I want on earth however.". Wrong. The local radiative intensity $J$ goes as $J\sim J_0(1+\tau)$, hence increasing at optical depth > 1. The Equilibrium temperature is then found by $J-\sigma T^4/\pi=0$. There are no problems in 'this'. 'This' is very well understood to be a working, simplified, illustrative model in the scientific community, and better ones exist. I'm out. $\endgroup$ – AtmosphericPrisonEscape Mar 10 at 10:17

Why does climate science divide total insolation by 4?

For two reasons, each of which halves the effects of insolation:

  • At any point in time, only half of the Earth is illuminated by the Sun. Another name for this factor is "daytime" versus "nighttime".
  • At any point in time, the surface area of the Earth's hemisphere that is illuminated by the Sun is about $2\pi r^2$, where $r$ is the radius of the Earth. But the cross section of that hemisphere to solar radiation is the area of a circle of radius $r$, which is $\pi r^2$, half of the area of that hemisphere.

? Why do you need to average out the sun's energy over the whole planet at the same time when that's not what actually happens?

The Earth rotates fairly quickly, so that is what happens. If the Earth was tidally locked with the Sun, there would be one side of the Earth that would be very hot and one side of the Earth that would be very cold. That is not what happens.

If the Earth rotated once per month (which is what our Moon does do), the sunlit side would eventually come close to being in thermal equilibrium with the incoming sunlight while the dark side would cool to much colder temperatures than are experienced anywhere on the face of the Earth. That also is not what happens. Even the sunlit Sahara does not get as hot as do equatorial regions of the Moon, and even South Pole does not get as cold as do portions of the Moon.

The periods of daytime heating by sunlight and periods of nighttime cooling by the lack of sunlight are both far too brief to bring those portions of the Earth into thermal equilibrium with the sunlight at daytime or with the 2.7K cosmic microwave background radiation at nighttime.

  • $\begingroup$ He didn't ask that. $\endgroup$ – User123 Mar 5 at 21:34
  • $\begingroup$ @User123 The title of the question most certainly asks for exactly that. $\endgroup$ – David Hammen Mar 5 at 21:35
  • $\begingroup$ Ok, it is better now. $\endgroup$ – User123 Mar 5 at 21:56
  • $\begingroup$ "The periods of daytime heating by sunlight and periods of nighttime cooling by the lack of sunlight are both far too brief to bring those portions of the Earth into thermal equilibrium with the sunlight at daytime" You're saying because the earth spins so fast, it doesn't experience the full brunt of the sun on one spot. But what would the blackbody temperature be of this diluted sunlight after you divide by 4? $\endgroup$ – Leon Hiebert Mar 5 at 22:18

My answer is mainly here to give some quantitative reasoning, but for the sake of completeness I'll also answer what my predecessors already answered.

Why do we divide by 4?

Imagine you see earth from the perspective of the sun. What you see is a disc. The area of this disc is $\pi r^2$ and therefore the total energy earth received from the sun is $S_0 \pi r^2 (1-\alpha)$, where the $(1-\alpha)$ takes into account that the solar radiation is partly reflected. As stated in the other answers, the idea is to distribute the energy over the whole surface of earth $4\pi r^2$. Thus, energy received per $m^2$ is $\frac{S_0 (1-\alpha)}{4}$.

Why is it reasonable to assume that the energy is distributed evenly over the whole globe at an instant?

The short answer is: If we use this assumption we are usually dealing with timescales much longer than a full rotation of the earth. Thus, daily variations are negligible.

A quantitative example:

Let's compare the time it takes the atmosphere to adjust to an imbalance $N$ to the time it takes the earth to rotate once. A full rotation takes approximately $t_{rot} \approx 60\times60\times24s = 86400s$.

The imbalance problem requires a few explanations up front. Suppose we double $CO_2$ in the atmosphere. This causes a forcing of $F_{2 \times CO_2} \approx 3.7 \frac{W}{m^2}$ according to IPCC. We now want to know how long it takes the atmosphere to adjust to this forcing.

Let's setup a model: Following Gregory et. al. we model the imbalance linearly as

$N = F_{2 \times CO_2} + \lambda T$,

where $T$ is the temperature relative to some reference $T_0$ and $\lambda$ is the (negative) feedback parameter with units $\frac{W}{m^2 K}$. Thus an increase in temperature reduces the imbalance. Additionally we assume that the imbalance will lead to an increase in temperature over time (as done e.g. here)

$N = C\frac{\text{d}T}{\text{d}t}$,

where $t$ is time and $C$ is the heat capacity of the atmosphere. Combining the two equations above we find

$C\frac{\text{d}T}{\text{d}t} = F_{2 \times CO_2} + \lambda T$.

The solution to the differential equation above is

$T(t) = \frac{-F_{2\times CO_2}}{\lambda} \left(1 - e^{\frac{\lambda}{C}t} \right)$.

We can see the adjustment process takes an infinite time, however two thirds of the process are done within the $e$-folding time $\tau$ (when the term in the exponent is $-1$). The exponent is $-1$ if $t = \tau = \frac{-C}{\lambda}$. Note that if $t$ is approaching infinity we have $T(\infty) = \frac{-F_{2\times CO_2}}{\lambda}$ (This is called Equilibrium climate sensitivity in the case of doubling $CO_2$).

What's left to do is estimating $C$ and $\lambda$. We can estimate the heat capacity of the atmosphere (just a column) to be

$C = c_p \frac{p_s}{g} = \frac{1005 \frac{J}{K kg} 10^5 Pa }{9.81 \frac{m}{s^2}} = 1.02 \times 10^7 \frac{J}{m^2 K}$.

IPCC tells us that $T(\infty)$ is likely to be between $1.5°C - 4.5°C$. Let's set $T(\infty) = 3°C$ and calculate

$\lambda = -\frac{3.7 \frac{W}{m^2}}{3 K} = -1.23 \frac{W}{m^2 K}$. Finally we find

$\tau = - \frac{1.02 \times 10^7 \frac{J}{m^2 K}}{-1.23 \frac{W}{m^2 K}} \approx 8.7 \times 10^6 s \approx 100\times t_{rot} $.

Caution In reality the time scales differ a lot more. The oceans heat capacity is much higher than the atmospheres. However, the calculations above should convince you that even if all forcing changes atmosphere only time scales differ by at least $\mathcal{O}(2)$.

  • $\begingroup$ Thank you for taking the time to write that out. However, just being a layman, I won't be able to work out all the math. I will say what I know though. After all of those calculations they determine only 168W / M^2 hits the surface. Can you tell me what that works out to in temperature? Thanks. $\endgroup$ – Leon Hiebert Mar 6 at 18:07
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    $\begingroup$ I'm not sure I understand your question. Surface receives around $E = 1362(1-0.29)/4 W/m^2= 242 W/m^2$. Using Stefan-Boltzmann law $E = \sigma T^4$ we find a temperature of around -18°C. This is approximately earths temperature without considering the greenhouse effect. $\endgroup$ – J. Fregin Mar 6 at 19:22
  • $\begingroup$ -18 would be the earth's temperature in the middle of the atmosphere is it not? That's what I understand. The amount of sunlight hitting 1m^2 is 168W as per Trenberth et al. That's the number you get after dividing by 4. I was hoping you could tell me what that works out to in temperature. eta: or is 280W the quotient. Pardon my memory I will have to check. 2nd edit: right 340 at the atmosphere, 168 or 163 at the surface depending on what diagram you're looking at, $\endgroup$ – Leon Hiebert Mar 6 at 21:05
  • $\begingroup$ You must confuse some things. Check the link you provided. The numbers in there are similar to those in my previous comment. $340 W/m^2$ at top of atmosphere is reasonable, 168 at surface is not. You are right, somewhere in the middle of the atmosphere we do have -18°C. My comment assumes no greenhouse effect. In that scenario surface temperature is indeed around -18°C. The equations provided in your linked article will yield the same result. $\endgroup$ – J. Fregin Mar 6 at 22:35
  • $\begingroup$ See here researchgate.net/figure/… Yes you get 340 because that's the quotient of 13668 TSI as we know. But then you end up with 168 at the surface. How hot is that per m^2? I don't know what it is but something seems off when it can reach 50C in a car in a couple of hours but only 168W is coming in. Is there a way to calculate 168W/M^2? Thanks. $\endgroup$ – Leon Hiebert Mar 6 at 23:54

Climate science studies the processes on the Earth over the course of many centuries. The rotation of the Earth is so small compared to many centuries.

Imagine that we roast the chicken. We can approximate the chicken as a sphere. The sphere (or chicken) is continuously revolving, thus the source of heat is changing the hemisphere that is being roasted. At the instant of time, only one hemisphere is being roasted. But imagine that the chicken spins at 1000 revolutions per second. (Of course, roasting chicken on a drill is not recommended.) You see, that the effects cancel out, and the total heat is equal to half of the instantenous heat.

Now the calculations: How much is the instantenous heat? $$(1-a)\Omega\pi r^2$$ But this heat is divided evenly to the whole hemisphere, so: $$\frac{(1-a)\Omega\pi r^2}{2\pi r^2}=\frac{(1-a)\Omega}{2}$$

But we said the the average heat is just a half of that, so we get wanted: $$\frac{(1-a)\Omega}{4}$$

  • $\begingroup$ Imagine roasting 200lb pig on a spit over a fire. It takes 24hrs on a slow rotation to cook the pork evenly. The temperature is 160F. If I use your equation, and four sources of 40F surround the pig instead of a fire, the pork will still be raw. Averaging temperature serves no purpose. Further, (1−a)Ω4 suggests all sunlight is reaching the earth at the same time. This is not the case. Why not (1−a)Ω2? $\endgroup$ – Leon Hiebert Mar 5 at 22:09
  • $\begingroup$ Imagine this: $dt$ of the time on one side, $dt$ on the opposite. So half of the time $2dt$ is on one side, same on another. Now add these small amounts of time. $\endgroup$ – User123 Mar 5 at 22:13
  • $\begingroup$ Are you suggesting the "4" is dependent on rotation speed? $\endgroup$ – Leon Hiebert Mar 5 at 22:37
  • $\begingroup$ @LeonHiebert roasting is a nonlinear process; it involves a threshold. Heat diffusion is basically a linear process, though blackbody radiation is of course very nonlinear, being $T^4$. Your question turns out to be quite interesting! $\endgroup$ – uhoh Mar 6 at 1:32
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    $\begingroup$ @LeonHiebert No, fast rotation is needed only to heat a sphere uniformly. If the rotation is slower, one hemisphere of a sphere gets warmer and the other one colder. (And opposite after $P/2$.) $\endgroup$ – User123 Mar 6 at 10:00

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