<p>I quote from B. J. Finlayson-Pitts &amp; J. N. Pitts, Jr., &quot;Atmospheric Chemistry: Fundamentals and Experimental Techniques&quot;, John Wiley &amp; Sons (1986), page 8:
&quot;The decreasing temperature with increasing altitude in the troposphere is due to the strong heating effect at the earth's surface from the adsorption of visible and near ultraviolet radiation.&quot;
Therefore, I'd opt for point 2 as main contribution.</p>

CC BY-SA 4.0
2024-09-01T11:07:43.727Z
2024-09-01T11:07:43.727Z
Riccardo
<p>Both is right and needed to explain the gradient simultaneously. It's the adiabatic expansion and cooling of the gas over a warm surface in the presence of gravity which creates the temperature gradient.</p>
<p>Earth's surface is solid material, and indeed does absorb the sunlight much better than the (mostly) transparent air itself. Gas molecules takes on the surface temperature when they 'collide' with the surface. Doing so they also cool the surface as each gas molecule leaves with the velocity equivalent to the temperature of the surface and thus transports away some of the energy, leaving the surface slightly cooler (it also works reversed, thus the transfering or actually warmer, if the surface happens to be cooler than the gas molecule was before). For an ideal gas this relation between velocity <span class="math-container">$v$</span> and temperature <span class="math-container">$T$</span> comes from the equipartition of energy to all (three) degrees of freedom of the gas with molecular mass <span class="math-container">$m$</span>: <span class="math-container">$$ \frac{1}{2}mv^2 = \frac{3}{2}k_BT$$</span>
<span class="math-container">$$ \Longrightarrow T = \sqrt{\frac{3k_BT}{m}}$$</span></p>
<p>Now we only have warm gas over a warm surface. Now it needs another process to distribute this energy. In a planetary atmosphere this works by convection, that hot air is (slightly) less dense than cooler air as we require a pressure equilibrium. Again, from an ideal gas: <span class="math-container">$pV = Nk_BT$</span>. Thus for two elements of air at different temperatures <span class="math-container">$T_1$</span> and <span class="math-container">$T_2 &gt; T_1$</span> we get in the equilibrium of pressure <span class="math-container">$p_1 = p_2$</span>: <span class="math-container">$$ \frac{Nk_BT_1}{V_1} = \frac{Nk_BT_2}{V_2}$$</span> or simplified <span class="math-container">$$\frac{T_1}{V_1} = \frac{T_2}{V_2}$$</span>. And thus with <span class="math-container">$T_2 &gt; T_1$</span> we see that <span class="math-container">$V_1 &gt; V_2$</span> is needed to fulfill the equation - which means the density of the hot air is less than that of cold air. That in turn means that the hot air will rise - and now the adiabatic expansion kicks in, and cools this hot air, creating the adabatic gradient we see (on average) in the atmosphere (as the rising gas volume expands, and in the absence of any direct heating we have adiabatic expansion).</p>
<p>One may now ask: what about <a href="https://en.wikipedia.org/wiki/Heat_transfer" rel="nofollow noreferrer">heat conduction or diffusion</a>? Sure, they can happen, and do happen. But they are <em>less efficient</em> in energy transport than the convection we see, thus do not play an important role in normal atmospheres. If we had a far-out planet with low energy input on the planet's surface, the temperature gradient in the atmosphere might be low enough (lower than the <a href="https://en.wikipedia.org/wiki/Lapse_rate" rel="nofollow noreferrer">adiabatic gradient</a>) that it could be established only via heat diffusion.</p>

CC BY-SA 4.0
2024-10-01T15:21:22.437Z
2024-10-01T15:21:22.437Z
planetmaker
<p>I've come across two different explanations when researching this topic in different areas:</p>
<ol>
<li>Adiabatic lapse rate: temperature falls naturally due to pressure decrease.</li>
<li>Distance from the surface: since air is mostly transparent, the majority of sunlight is absorbed by the Earth's surface, which then conducts heat to the atmosphere.</li>
</ol>
<p>What are the relative contributions of these (or other) factors to the empirical temperature profile of the troposphere?</p>

<atmosphere><temperature><troposphere>
Why does air temperature decrease with height in the troposphere?
Henry
# Why does air temperature decrease with height in the troposphere?

(Asked by: username_0 on 2024-08-07)

I've come across two different explanations when researching this topic in different areas:

1. Adiabatic lapse rate: temperature falls naturally due to pressure decrease.
2. Distance from the surface: since air is mostly transparent, the majority of sunlight is absorbed by the Earth's surface, which then conducts heat to the atmosphere.

What are the relative contributions of these (or other) factors to the empirical temperature profile of the troposphere?

**Tags:** atmosphere, temperature, troposphere


---

## 2 answers

### Answer by username_1 on 2024-09-01. Score: 1

I quote from B. J. Finlayson-Pitts & J. N. Pitts, Jr., "Atmospheric Chemistry: Fundamentals and Experimental Techniques", John Wiley & Sons (1986), page 8:
"The decreasing temperature with increasing altitude in the troposphere is due to the strong heating effect at the earth's surface from the adsorption of visible and near ultraviolet radiation."
Therefore, I'd opt for point 2 as main contribution.


---

### Answer by username_2 on 2024-10-01. Score: 0

Both is right and needed to explain the gradient simultaneously. It's the adiabatic expansion and cooling of the gas over a warm surface in the presence of gravity which creates the temperature gradient.

Earth's surface is solid material, and indeed does absorb the sunlight much better than the (mostly) transparent air itself. Gas molecules takes on the surface temperature when they 'collide' with the surface. Doing so they also cool the surface as each gas molecule leaves with the velocity equivalent to the temperature of the surface and thus transports away some of the energy, leaving the surface slightly cooler (it also works reversed, thus the transfering or actually warmer, if the surface happens to be cooler than the gas molecule was before). For an ideal gas this relation between velocity $v$ and temperature $T$ comes from the equipartition of energy to all (three) degrees of freedom of the gas with molecular mass $m$: $$ \frac{1}{2}mv^2 = \frac{3}{2}k\_BT$$
$$ \Longrightarrow T = \sqrt{\frac{3k\_BT}{m}}$$

Now we only have warm gas over a warm surface. Now it needs another process to distribute this energy. In a planetary atmosphere this works by convection, that hot air is (slightly) less dense than cooler air as we require a pressure equilibrium. Again, from an ideal gas: $pV = Nk\_BT$. Thus for two elements of air at different temperatures $T\_1$ and $T\_2 > T\_1$ we get in the equilibrium of pressure $p\_1 = p\_2$: $$ \frac{Nk\_BT\_1}{V\_1} = \frac{Nk\_BT\_2}{V\_2}$$ or simplified $$\frac{T\_1}{V\_1} = \frac{T\_2}{V\_2}$$. And thus with $T\_2 > T\_1$ we see that $V\_1 > V\_2$ is needed to fulfill the equation - which means the density of the hot air is less than that of cold air. That in turn means that the hot air will rise - and now the adiabatic expansion kicks in, and cools this hot air, creating the adabatic gradient we see (on average) in the atmosphere (as the rising gas volume expands, and in the absence of any direct heating we have adiabatic expansion).

One may now ask: what about [heat conduction or diffusion](https://en.wikipedia.org/wiki/Heat_transfer)? Sure, they can happen, and do happen. But they are *less efficient* in energy transport than the convection we see, thus do not play an important role in normal atmospheres. If we had a far-out planet with low energy input on the planet's surface, the temperature gradient in the atmosphere might be low enough (lower than the [adiabatic gradient](https://en.wikipedia.org/wiki/Lapse_rate)) that it could be established only via heat diffusion.


---

