<p>First, let's acknowledge this fact: <span class="math-container">$$c_p=R+c_v \tag{1}$$</span>,
where <span class="math-container">$R$</span> is the specific gas constant.</p>
<p>This means that <span class="math-container">$x=R/c_p$</span>.
Rearranging the equation, we can see that <span class="math-container">$$P+I=\int{c_p\left(\frac{P}{P_{00}}\right)^{R/c_p}\Theta dM}\tag{2}$$</span>
Notice that <span class="math-container">$\left(\frac{P}{P_{00}}\right)^{R/c_p}$</span> is the <a href="https://en.wikipedia.org/wiki/Exner_function" rel="nofollow noreferrer">Exner function</a>. By extension,<span class="math-container">$$P+I=\int{c_p T dM}\tag{3}$$</span> If we rearrange (1) we can get</p>
<p><span class="math-container">$$P+I=\int{c_v T+RT dM}\tag{4}$$</span>.
You observed correctly that <span class="math-container">$I=\int{c_v T}dM$</span>. This leaves <span class="math-container">$$P=\int{RT dM}\tag{5}$$</span>.</p>
<p>Using the ideal gas law,<span class="math-container">$$P=\int{P\alpha dM}$$</span>, where <span class="math-container">$\alpha$</span> is specific volume. This makes (4) look like <span class="math-container">$$P+I=\int{c_v T+P\alpha dM}\tag{4}$$</span> which is just a restatement of the first law of thermodynamics: <span class="math-container">$\delta q = c_v dT + p d\alpha$</span></p>

CC BY-SA 4.0
2021-02-10T01:18:19.473Z
@gansub The original problem has $x=(c_p-c_v)/c_p$ which is actually $x=R/c_p$ in disguise.
BarocliniCplusplus
CC BY-SA 4.0
2021-02-11T07:16:46.680Z
Can you tell me what does standard pressure p00 means? Is it a constant?
 Hou
CC BY-SA 4.0
2021-02-11T07:40:24.967Z
Typically, $p_{00}$ is 1000 hPa. But if you are thinking of applying to a different planet, then I would recommend a different value.
BarocliniCplusplus
CC BY-SA 4.0
2021-02-14T11:03:27.053Z
Thanks, but I still have one more question, where is the gravitational energy? In your equations it seems that it doesn’t represent the gravitational energy, it’s just the work of the pressure.  What does total potential energy means? Does it includes the gravitational energy, or it is just the heat?
 Hou
CC BY-SA 4.0
2021-02-14T15:38:46.900Z
@Hou The work of the pressure is about equivalent to the gravitational energy exchange. Consider the [derivation of the dry adiabatic lapse rate](https://en.wikipedia.org/wiki/Lapse_rate#Dry_adiabatic_lapse_rate). If I recall right, it is part of the [Lorenz Energy Cycle papers](https://en.wikipedia.org/wiki/Lorenz_energy_cycle), so total potential energy is the amount of energy that can sustain the mean flow as well as generate the kinetic energy of eddies.
BarocliniCplusplus
CC BY-SA 4.0
2021-02-14T15:56:35.337Z
For example, if you want to consider a dry adiabatic atmosphere, $RT=R \times (T_0-g*z/c_p)$. But keep in mind, the actual atmosphere is neither adiabatic nor dry.
BarocliniCplusplus
CC BY-SA 4.0
2021-02-14T16:10:15.887Z
@BarocliniCplusplus So in a word, the total potential energy (P +I) here is the part of energy in the atmosphere that can be converted into kinetic energy, and by its definition, it is actually the enthalpy($c_v T + R T = c_p T = h$), right?
 Hou
CC BY-SA 4.0
2021-02-14T18:06:21.880Z
@Hou If you want to think of it that way, sure (as long as $h$ is enthalpy, not height). Also, keep in mind, that this is an integrated quantity. That is $\int{c_p TdM}=\int\int\int{\rho c_p TdV}=\int\int\int{\rho h dV}$. This is the 'total' part of the 'total potential energy.'
BarocliniCplusplus
CC BY-SA 4.0
2021-02-15T03:00:50.280Z
@BarocliniCplusplus So the $h$ here is actually the specific enthalpy, right?
 Hou
CC BY-SA 4.0
2021-02-15T03:09:05.613Z
That is what you indicated as the symbol for enthalpy..
BarocliniCplusplus
CC BY-SA 4.0
2021-02-16T06:38:40.330Z
@BarocliniCplusplus Well...I think there is a misunderstanding, I am totally aware that $h$ is defined by myself as the symbol for enthalpy....And I am thinking that $h$ here should be the specific enthalpy since $c_p$ is the specific heat, is that correct?
 Hou
CC BY-SA 4.0
2021-02-09T17:57:00.620Z
2021-02-09T17:57:00.620Z
BarocliniCplusplus
<p>In one of Lorenz’s paper (I mean the meteorologists Edward Lorenz), he stated that the total potential energy of the whole atmosphere <span class="math-container">$P + I$</span> (means the sum of potential energy and internal energy) is</p>
<p><span class="math-container">\begin{equation}
P + I = c_p p_{00}^{-x} \int p^x \Theta dM
\end{equation}</span></p>
<p>where <span class="math-container">$p$</span> is pressure, <span class="math-container">$p_{00}$</span> is a standard pressure, and <span class="math-container">$x$</span> is the ratio <span class="math-container">$(c_p - c_v)/c_p$</span>, <span class="math-container">$c_v$</span> and <span class="math-container">$c_p$</span> are the specific heats of air,<span class="math-container">$\Theta$</span> is the temperature and M is the mass.</p>
<p>Based on my knowledge, the internal energy should be a simple form that
<span class="math-container">\begin{equation}
I = \int c_v T dM
\end{equation}</span>
and the potential energy is actually the gravitational potential energy, which is
<span class="math-container">\begin{equation}
P = \int g h dM
\end{equation}</span></p>
<p>where <span class="math-container">$g$</span> is the gravitational acceleration and <span class="math-container">$h$</span> is the height. So my question is how did Lorenz derived his form of the total potential energy of the atmosphere? The original paper is <em>Energy and numerical weather prediction</em><a href="https://www.tandfonline.com/doi/pdf/10.3402/tellusa.v12i4.9420" rel="nofollow noreferrer">Lorenz 1960</a>, and this equation was the first equation in the paper.</p>

<atmosphere><atmosphere-modelling><atmospheric-chemistry><atmospheric-radiation><atmospheric-circulation>
How to derive the total potential energy of the atmosphere?
 Hou
# How to derive the total potential energy of the atmosphere?

(Asked by: username_0 on 2021-02-09)

In one of Lorenz’s paper (I mean the meteorologists Edward Lorenz), he stated that the total potential energy of the whole atmosphere $P + I$ (means the sum of potential energy and internal energy) is

\begin{equation}
P + I = c\_p p\_{00}^{-x} \int p^x \Theta dM
\end{equation}

where $p$ is pressure, $p\_{00}$ is a standard pressure, and $x$ is the ratio $(c\_p - c\_v)/c\_p$, $c\_v$ and $c\_p$ are the specific heats of air,$\Theta$ is the temperature and M is the mass.

Based on my knowledge, the internal energy should be a simple form that
\begin{equation}
I = \int c\_v T dM
\end{equation}
and the potential energy is actually the gravitational potential energy, which is
\begin{equation}
P = \int g h dM
\end{equation}

where $g$ is the gravitational acceleration and $h$ is the height. So my question is how did Lorenz derived his form of the total potential energy of the atmosphere? The original paper is *Energy and numerical weather prediction*[Lorenz 1960](https://www.tandfonline.com/doi/pdf/10.3402/tellusa.v12i4.9420), and this equation was the first equation in the paper.

**Tags:** atmosphere, atmosphere-modelling, atmospheric-chemistry, atmospheric-radiation, atmospheric-circulation


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## 1 answers

### Answer by username_1 on 2021-02-09. Score: 3 | Accepted answer

First, let's acknowledge this fact: $$c\_p=R+c\_v \tag{1}$$,
where $R$ is the specific gas constant.

This means that $x=R/c\_p$.
Rearranging the equation, we can see that $$P+I=\int{c\_p\left(\frac{P}{P\_{00}}\right)^{R/c\_p}\Theta dM}\tag{2}$$
Notice that $\left(\frac{P}{P\_{00}}\right)^{R/c\_p}$ is the [Exner function](https://en.wikipedia.org/wiki/Exner_function). By extension,$$P+I=\int{c\_p T dM}\tag{3}$$ If we rearrange (1) we can get

$$P+I=\int{c\_v T+RT dM}\tag{4}$$.
You observed correctly that $I=\int{c\_v T}dM$. This leaves $$P=\int{RT dM}\tag{5}$$.

Using the ideal gas law,$$P=\int{P\alpha dM}$$, where $\alpha$ is specific volume. This makes (4) look like $$P+I=\int{c\_v T+P\alpha dM}\tag{4}$$ which is just a restatement of the first law of thermodynamics: $\delta q = c\_v dT + p d\alpha$

> **Comments:**
> **username_1**: @gansub The original problem has $x=(c_p-c_v)/c_p$ which is actually $x=R/c_p$ in disguise.  
> **username_0**: Can you tell me what does standard pressure p00 means? Is it a constant?  
> **username_1**: Typically, $p_{00}$ is 1000 hPa. But if you are thinking of applying to a different planet, then I would recommend a different value.  
> **username_0**: Thanks, but I still have one more question, where is the gravitational energy? In your equations it seems that it doesn’t represent the gravitational energy, it’s just the work of the pressure.  What does total potential energy means? Does it includes the gravitational energy, or it is just the heat?  
> **username_1**: @Hou The work of the pressure is about equivalent to the gravitational energy exchange. Consider the [derivation of the dry adiabatic lapse rate](https://en.wikipedia.org/wiki/Lapse_rate#Dry_adiabatic_lapse_rate). If I recall right, it is part of the [Lorenz Energy Cycle papers](https://en.wikipedia.org/wiki/Lorenz_energy_cycle), so total potential energy is the amount of energy that can sustain the mean flow as well as generate the kinetic energy of eddies.  
> **username_1**: For example, if you want to consider a dry adiabatic atmosphere, $RT=R \times (T_0-g*z/c_p)$. But keep in mind, the actual atmosphere is neither adiabatic nor dry.  
> **username_0**: @username_1 So in a word, the total potential energy (P +I) here is the part of energy in the atmosphere that can be converted into kinetic energy, and by its definition, it is actually the enthalpy($c_v T + R T = c_p T = h$), right?  
> **username_1**: @Hou If you want to think of it that way, sure (as long as $h$ is enthalpy, not height). Also, keep in mind, that this is an integrated quantity. That is $\int{c_p TdM}=\int\int\int{\rho c_p TdV}=\int\int\int{\rho h dV}$. This is the 'total' part of the 'total potential energy.'  
> **username_0**: @username_1 So the $h$ here is actually the specific enthalpy, right?  
> **username_1**: That is what you indicated as the symbol for enthalpy..  
> **username_0**: @username_1 Well...I think there is a misunderstanding, I am totally aware that $h$ is defined by myself as the symbol for enthalpy....And I am thinking that $h$ here should be the specific enthalpy since $c_p$ is the specific heat, is that correct?  

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