<blockquote>
<p>Why does climate science divide total insolation by 4?</p>
</blockquote>
<p>For two reasons, each of which halves the effects of insolation:</p>
<ul>
<li>At any point in time, only half of the Earth is illuminated by the Sun. Another name for this factor is &quot;daytime&quot; versus &quot;nighttime&quot;.</li>
<li>At any point in time, the surface area of the Earth's hemisphere that is illuminated by the Sun is about <span class="math-container">$2\pi r^2$</span>, where <span class="math-container">$r$</span> is the radius of the Earth. But the cross section of that hemisphere to solar radiation is the area of a circle of radius <span class="math-container">$r$</span>, which is <span class="math-container">$\pi r^2$</span>, half of the area of that hemisphere.</li>
</ul>
<blockquote>
<p>? 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?</p>
</blockquote>
<p>The Earth rotates fairly quickly, so that <strong>is</strong> 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.</p>
<p>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.</p>
<p>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.</p>

CC BY-SA 4.0
2021-03-05T21:34:47.690Z
He didn't ask that.
User123
CC BY-SA 4.0
2021-03-05T21:35:22.657Z
@User123 The title of the question most certainly asks for exactly that.
David Hammen
CC BY-SA 4.0
2021-03-05T21:56:43.143Z
Ok, it is better now.
User123
CC BY-SA 4.0
2021-03-05T22:18:17.703Z
"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?
Leon Hiebert
CC BY-SA 4.0
2021-03-05T21:34:06.457Z
2021-03-05T21:56:18.217Z
2021-03-05T21:56:18.217Z
David Hammen
<p>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.</p>
<p><strong>Why do we divide by 4?</strong></p>
<p>Imagine you see earth from the perspective of the sun. What you see is a disc. The area of this disc is <span class="math-container">$\pi r^2$</span> and therefore the total energy earth received from the sun is <span class="math-container">$S_0 \pi r^2 (1-\alpha)$</span>, where the <span class="math-container">$(1-\alpha)$</span> takes into account that the solar radiation is <a href="https://en.wikipedia.org/wiki/Albedo" rel="nofollow noreferrer">partly reflected</a>. As stated in the other answers,  the idea is to distribute the energy over the whole surface of earth <span class="math-container">$4\pi r^2$</span>. Thus, energy received per <span class="math-container">$m^2$</span> is <span class="math-container">$\frac{S_0 (1-\alpha)}{4}$</span>.</p>
<p><strong>Why is it reasonable to assume that the energy is distributed evenly over the whole globe at an instant?</strong></p>
<p>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.</p>
<p><strong>A quantitative example</strong>:</p>
<p>Let's compare the time it takes the atmosphere to adjust to an imbalance <span class="math-container">$N$</span> to the time it takes the earth to rotate once. A full rotation takes approximately <span class="math-container">$t_{rot} \approx 60\times60\times24s = 86400s$</span>.</p>
<p>The imbalance problem requires a few explanations up front. Suppose we double <span class="math-container">$CO_2$</span> in the atmosphere. This causes a <a href="https://en.wikipedia.org/wiki/Radiative_forcing" rel="nofollow noreferrer">forcing</a> of <span class="math-container">$F_{2 \times CO_2} \approx 3.7 \frac{W}{m^2}$</span> according to <a href="https://www.ipcc.ch/site/assets/uploads/2018/03/TAR-06.pdf" rel="nofollow noreferrer">IPCC</a>. We now want to know how long it takes the atmosphere to adjust to this forcing.</p>
<p>Let's setup a model: Following <a href="https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2003GL018747" rel="nofollow noreferrer">Gregory et. al.</a> we model the imbalance linearly as</p>
<p><span class="math-container">$N = F_{2 \times CO_2} + \lambda T$</span>,</p>
<p>where <span class="math-container">$T$</span> is the temperature relative to some reference <span class="math-container">$T_0$</span> and <span class="math-container">$\lambda$</span> is the (negative) <a href="https://en.wikipedia.org/wiki/Climate_change_feedback" rel="nofollow noreferrer">feedback parameter</a> with units <span class="math-container">$\frac{W}{m^2 K}$</span>. 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. <a href="https://link.springer.com/article/10.1007/s00382-019-04686-4" rel="nofollow noreferrer">here</a>)</p>
<p><span class="math-container">$N = C\frac{\text{d}T}{\text{d}t}$</span>,</p>
<p>where <span class="math-container">$t$</span> is time and <span class="math-container">$C$</span> is the heat capacity of the atmosphere. Combining the two equations above we find</p>
<p><span class="math-container">$C\frac{\text{d}T}{\text{d}t} = F_{2 \times CO_2} + \lambda T$</span>.</p>
<p>The solution to the differential equation above is</p>
<p><span class="math-container">$T(t) = \frac{-F_{2\times CO_2}}{\lambda} \left(1 - e^{\frac{\lambda}{C}t} \right)$</span>.</p>
<p>We can see the adjustment process takes an infinite time, however two thirds of the process are done within the <span class="math-container">$e$</span>-folding time <span class="math-container">$\tau$</span> (when the term in the exponent is <span class="math-container">$-1$</span>). The exponent is <span class="math-container">$-1$</span> if <span class="math-container">$t = \tau = \frac{-C}{\lambda}$</span>.
Note that if <span class="math-container">$t$</span> is approaching infinity we have <span class="math-container">$T(\infty) = \frac{-F_{2\times CO_2}}{\lambda}$</span> (This is called Equilibrium climate sensitivity in the case of doubling <span class="math-container">$CO_2$</span>).</p>
<p>What's left to do is estimating <span class="math-container">$C$</span> and <span class="math-container">$\lambda$</span>. We can estimate the heat capacity of the atmosphere (just a column) to be</p>
<p><span class="math-container">$C = c_p \frac{p_s}{g} = \frac{1005 \frac{J}{K kg} 10^5 Pa }{9.81 \frac{m}{s^2}} = 1.02 \times 10^7 \frac{J}{m^2 K}$</span>.</p>
<p><a href="https://www.ipcc.ch/site/assets/uploads/2018/02/WG1AR5_TS_FINAL.pdf" rel="nofollow noreferrer">IPCC tells us</a> that <span class="math-container">$T(\infty)$</span> is likely to be between <span class="math-container">$1.5°C - 4.5°C$</span>. Let's set <span class="math-container">$T(\infty) = 3°C$</span> and calculate</p>
<p><span class="math-container">$\lambda = -\frac{3.7 \frac{W}{m^2}}{3 K} = -1.23 \frac{W}{m^2 K}$</span>. Finally we find</p>
<p><span class="math-container">$\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} $</span>.</p>
<p><strong>Caution</strong>
In reality the time scales differ a lot more. The oceans heat capacity is much higher than the atmospheres. However, the calculations above should convince you that even if all forcing changes atmosphere only time scales differ by at least <span class="math-container">$\mathcal{O}(2)$</span>.</p>

CC BY-SA 4.0
2021-03-06T18:07:40.990Z
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.
Leon Hiebert
CC BY-SA 4.0
2021-03-06T19:22:32.937Z
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.
Joscha Fregin
CC BY-SA 4.0
2021-03-06T21:05:50.307Z
-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,
Leon Hiebert
CC BY-SA 4.0
2021-03-06T22:35:42.810Z
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.
Joscha Fregin
CC BY-SA 4.0
2021-03-06T23:54:45.623Z
See here https://www.researchgate.net/figure/Global-heat-flows-Kiehl-and-Trenberth-1997_fig2_242180386  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.
Leon Hiebert
CC BY-SA 4.0
2021-03-07T00:11:49.500Z
"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." Could you please expand on this? Thanks
Leon Hiebert
CC BY-SA 4.0
2021-03-07T08:48:28.060Z
I did expand using the example. Adaption to radiation imbalances takes much longer than a day. Therefore, intraday variability has a very small or no effect on the adaption process. Take the sea in spring and summer: By the end of the summer it is nice and warm. By beginning of spring there maybe a warm day, but the sea is still cold. Warming takes time and a single day has almost no effect. The sea does not warm significantly (for us to swim in) within the timescale of days. It's rather weeks or months.
Joscha Fregin
CC BY-SA 4.0
2021-03-07T14:06:39.920Z
I See what you mean now. Ok 168 reach ground, but you can't just use that to determine temperature, because some is absorbed by atmosphere + we got some energy flux. But of course you can (even though it is not meaningful) solve $168 = \sigma T^4$ and get some temperature.
Joscha Fregin
CC BY-SA 4.0
2021-03-07T21:27:13.137Z
" But of course you can (even though it is not meaningful) solve 168=σT4 and get some temperature. "  If it's not meaningful, then why is it useful to determine heat/m2?  Can you  use SB to solve watts per m2 of radiation for the inside of a car with a temperature of 51C? Conversely, the same back knowing W/m2?  I think it's actually 500 W/m2 at the higher latitudes and more like 900 at the eq. This is not 168, therefore 340 at TOA must be incorrect. Because total energy was divided by 4.  Sure, on paper you can do what ever you like with math but there's contradiction between what's observed.
Leon Hiebert
CC BY-SA 4.0
2021-03-07T21:37:38.693Z
It is not meaningful because you only consider a fraction of the energy that actually reaches earth (and is not reflected). You can't just separate atmosphere from surface in this framework. I suggest you do some reading. Probably start with reading your linked article properly.
Joscha Fregin
CC BY-SA 4.0
2021-03-07T21:44:57.267Z
With all due respect, I'm not interested in the rest because it doesn't answer my question.  https://www.omnicalculator.com/physics/stefan-boltzmann-law If only a fraction of the energy reaching earth is 168, then that's -39.   This isn't what is observed.  626W/m2 is needed to heat 1m2 inside a car to 51C.  There's a disconnect.
Leon Hiebert
CC BY-SA 4.0
2021-03-07T21:53:41.007Z
The inside of a car is hardly comparable to the atmosphere of earth.
Joscha Fregin
CC BY-SA 4.0
2021-03-07T22:01:25.730Z
I would never use that comparison. Of course it isn't. I am saying if only 168 reached the earth, then a car would only be -39 on a sunny summer day. As per SB, it must be 626, because it reaches 51C. We're basing that on a meter sq.
Leon Hiebert
CC BY-SA 4.0
2021-03-07T22:12:11.977Z
Well, you just did use that comparison again. I'll drop out of this conversation here, because I think we're drifting too far away from the initial question. One last suggestion though: read wikipedias article on greenhouses (the thing you grow plants in).
Joscha Fregin
CC BY-SA 4.0
2021-03-07T22:16:41.403Z
I'm sorry to see you go because this question never gets answered. But you do understand where I'm coming from I hope.  340 at TOA is less than the 626 needed to heat 1m2 of the inside of a car and has nothing to do with extra GHG heat.  Until this contradiction is adequately answered, it will always be a sticking point. Enjoy the rest of your day.
Leon Hiebert
CC BY-SA 4.0
2021-03-07T22:20:56.630Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/120553/discussion-between-j-fregin-and-leon-hiebert).
Joscha Fregin
CC BY-SA 4.0
2021-03-06T17:30:03.520Z
2021-03-06T17:48:24.627Z
2021-03-06T17:48:24.627Z
Joscha Fregin
<p>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.</p>
<p>Imagine that we roast the chicken. We can approximate <a href="https://en.wikipedia.org/wiki/Spherical_cow" rel="nofollow noreferrer">the chicken as a sphere</a>. 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.</p>
<p>Now the calculations: How much is the instantenous heat?
<span class="math-container">$$(1-a)\Omega\pi r^2$$</span>
But this heat is divided evenly to the whole hemisphere, so:
<span class="math-container">$$\frac{(1-a)\Omega\pi r^2}{2\pi r^2}=\frac{(1-a)\Omega}{2}$$</span></p>
<p>But we said the the average heat is just a half of that, so we get wanted:
<span class="math-container">$$\frac{(1-a)\Omega}{4}$$</span></p>

CC BY-SA 4.0
2021-03-05T22:09:37.367Z
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?
Leon Hiebert
CC BY-SA 4.0
2021-03-05T22:13:14.777Z
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.
User123
CC BY-SA 4.0
2021-03-05T22:37:58.057Z
Are you suggesting the "4" is dependent on rotation speed?
Leon Hiebert
CC BY-SA 4.0
2021-03-06T01:32:08.940Z
@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!
uhoh
CC BY-SA 4.0
2021-03-06T10:00:22.287Z
@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 $P/2$.)
User123
CC BY-SA 4.0
2021-03-06T10:02:01.907Z
Radiant heat, not conducted heat; not sure temperature is the correct unit. Watts per square metre? A quarter of a Fahrenheit temperature won't be a quarter as hot btw.
Ken Fabian
CC BY-SA 4.0
2021-03-05T21:44:06.360Z
2021-03-05T21:44:06.360Z
User123
<p>From <a href="https://www.nature.com/scitable/knowledge/library/the-global-climate-system-74649049/" rel="nofollow noreferrer">Nature</a>:</p>
<blockquote>
<p>We divide by <span class="math-container">$4$</span> 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 = <span class="math-container">$0.25$</span>).</p>
</blockquote>
<hr />
<p>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.</p>
<p>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?</p>

CC BY-SA 4.0
2021-03-08T12:20:19.337Z
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.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-03-08T17:40:58.087Z
But 4 is also total insolation on the whole planet at once. Flat earth. Just because you can mathematically do what you want shouldn't mean that's what actually happens. If the average incident flux is 168 at the surface or 340 at TOA, then that's not enough to melt ice.  SB calculator shows you need 626 W/m2 to raise the temperature of the inside of a car (1m2) to 51C.  That's the average.
Leon Hiebert
CC BY-SA 4.0
2021-03-08T18:10:03.110Z
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.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-03-08T18:14:39.173Z
(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.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-03-08T21:51:32.987Z
"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. " I appreciate your time to educate me but why are you using a 0-d or 1-d profile?  Because of a cross section of the earth in respect to the sun?  Why not use Holder's inequality law instead? It doesn't use 4.
Leon Hiebert
CC BY-SA 4.0
2021-03-08T21:55:01.387Z
" Your example car exists somewhere as a point on a 3-D sphere where the insolation is larger than the average," Well, yes, it's considerably larger than what climate science says it should be. But, again, is that because you use a 1-d flat earth type model? This is perplexing to me.  The S-B law clearly says how much power is required yet when it comes to this, it no longer counts.
Leon Hiebert
CC BY-SA 4.0
2021-03-09T09:28:00.220Z
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
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-03-09T20:50:07.233Z
Yes, I am aware. That's how the Te is determined to be -18 in the middle of the atmosphere.  I can still apply S-B to anything I want on earth however. Concrete, snow, ice, walls, cars etc.  I still don't understand why S-B is ok for the earth but not for a car on earth.  When I feel the sun on my face, I feel the sun, not ghgs and I know it's way more than 168W/m2. That works out to -40.  I think the 1d model and dividing by 4 is causing problems. In fact, don't let the flat earther's know that's how it's done. It would only encourage them.
Leon Hiebert
CC BY-SA 4.0
2021-03-10T10:17:06.743Z
*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.
AtmosphericPrisonEscape
<climate-change><climate><insolation>
Why divide total insolation by $4$?
Leon Hiebert
# Why divide total insolation by $4$?

(Asked by: username_0 on 2021-03-05)

From [Nature](https://www.nature.com/scitable/knowledge/library/the-global-climate-system-74649049/):

> 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?

**Tags:** climate-change, climate, insolation

> **Comments:**
> **username_1**: 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.  
> **username_0**: But 4 is also total insolation on the whole planet at once. Flat earth. Just because you can mathematically do what you want shouldn't mean that's what actually happens. If the average incident flux is 168 at the surface or 340 at TOA, then that's not enough to melt ice.  SB calculator shows you need 626 W/m2 to raise the temperature of the inside of a car (1m2) to 51C.  That's the average.  
> **username_1**: 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.  
> **username_1**: (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.  
> **username_0**: "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. " I appreciate your time to educate me but why are you using a 0-d or 1-d profile?  Because of a cross section of the earth in respect to the sun?  Why not use Holder's inequality law instead? It doesn't use 4.  
> **username_0**: " Your example car exists somewhere as a point on a 3-D sphere where the insolation is larger than the average," Well, yes, it's considerably larger than what climate science says it should be. But, again, is that because you use a 1-d flat earth type model? This is perplexing to me.  The S-B law clearly says how much power is required yet when it comes to this, it no longer counts.  
> **username_1**: 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  
> **username_0**: Yes, I am aware. That's how the Te is determined to be -18 in the middle of the atmosphere.  I can still apply S-B to anything I want on earth however. Concrete, snow, ice, walls, cars etc.  I still don't understand why S-B is ok for the earth but not for a car on earth.  When I feel the sun on my face, I feel the sun, not ghgs and I know it's way more than 168W/m2. That works out to -40.  I think the 1d model and dividing by 4 is causing problems. In fact, don't let the flat earther's know that's how it's done. It would only encourage them.  
> **username_1**: *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.  

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## 3 answers

### Answer by username_2 on 2021-03-05. Score: 9

> 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.

> **Comments:**
> **username_3**: He didn't ask that.  
> **username_2**: @username_3 The title of the question most certainly asks for exactly that.  
> **username_3**: Ok, it is better now.  
> **username_0**: "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?  

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### Answer by username_4 on 2021-03-06. Score: 4

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](https://en.wikipedia.org/wiki/Albedo). 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](https://en.wikipedia.org/wiki/Radiative_forcing) of $F\_{2 \times CO\_2} \approx 3.7 \frac{W}{m^2}$ according to [IPCC](https://www.ipcc.ch/site/assets/uploads/2018/03/TAR-06.pdf). 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.](https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2003GL018747) 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](https://en.wikipedia.org/wiki/Climate_change_feedback) 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](https://link.springer.com/article/10.1007/s00382-019-04686-4))

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

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](https://www.ipcc.ch/site/assets/uploads/2018/02/WG1AR5_TS_FINAL.pdf) 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 $\mathcal{O}(2)$.

> **Comments:**
> **username_0**: 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.  
> **username_4**: 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.  
> **username_0**: -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,  
> **username_4**: 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.  
> **username_0**: See here https://www.researchgate.net/figure/Global-heat-flows-Kiehl-and-Trenberth-1997_fig2_242180386  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.  
> **username_0**: "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." Could you please expand on this? Thanks  
> **username_4**: I did expand using the example. Adaption to radiation imbalances takes much longer than a day. Therefore, intraday variability has a very small or no effect on the adaption process. Take the sea in spring and summer: By the end of the summer it is nice and warm. By beginning of spring there maybe a warm day, but the sea is still cold. Warming takes time and a single day has almost no effect. The sea does not warm significantly (for us to swim in) within the timescale of days. It's rather weeks or months.  
> **username_4**: I See what you mean now. Ok 168 reach ground, but you can't just use that to determine temperature, because some is absorbed by atmosphere + we got some energy flux. But of course you can (even though it is not meaningful) solve $168 = \sigma T^4$ and get some temperature.  
> **username_0**: " But of course you can (even though it is not meaningful) solve 168=σT4 and get some temperature. "  If it's not meaningful, then why is it useful to determine heat/m2?  Can you  use SB to solve watts per m2 of radiation for the inside of a car with a temperature of 51C? Conversely, the same back knowing W/m2?  I think it's actually 500 W/m2 at the higher latitudes and more like 900 at the eq. This is not 168, therefore 340 at TOA must be incorrect. Because total energy was divided by 4.  Sure, on paper you can do what ever you like with math but there's contradiction between what's observed.  
> **username_4**: It is not meaningful because you only consider a fraction of the energy that actually reaches earth (and is not reflected). You can't just separate atmosphere from surface in this framework. I suggest you do some reading. Probably start with reading your linked article properly.  
> **username_0**: With all due respect, I'm not interested in the rest because it doesn't answer my question.  https://www.omnicalculator.com/physics/stefan-boltzmann-law If only a fraction of the energy reaching earth is 168, then that's -39.   This isn't what is observed.  626W/m2 is needed to heat 1m2 inside a car to 51C.  There's a disconnect.  
> **username_4**: The inside of a car is hardly comparable to the atmosphere of earth.  
> **username_0**: I would never use that comparison. Of course it isn't. I am saying if only 168 reached the earth, then a car would only be -39 on a sunny summer day. As per SB, it must be 626, because it reaches 51C. We're basing that on a meter sq.  
> **username_4**: Well, you just did use that comparison again. I'll drop out of this conversation here, because I think we're drifting too far away from the initial question. One last suggestion though: read wikipedias article on greenhouses (the thing you grow plants in).  
> **username_0**: I'm sorry to see you go because this question never gets answered. But you do understand where I'm coming from I hope.  340 at TOA is less than the 626 needed to heat 1m2 of the inside of a car and has nothing to do with extra GHG heat.  Until this contradiction is adequately answered, it will always be a sticking point. Enjoy the rest of your day.  
> **username_4**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/120553/discussion-between-j-username_4-and-username_0).  

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### Answer by username_3 on 2021-03-05. Score: 1

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](https://en.wikipedia.org/wiki/Spherical_cow). 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:
$$\frac{(1-a)\Omega}{4}$$

> **Comments:**
> **username_0**: 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?  
> **username_3**: 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.  
> **username_0**: Are you suggesting the "4" is dependent on rotation speed?  
> **username_5**: @username_0 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!  
> **username_3**: @username_0 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 $P/2$.)  
> **username_6**: Radiant heat, not conducted heat; not sure temperature is the correct unit. Watts per square metre? A quarter of a Fahrenheit temperature won't be a quarter as hot btw.  

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