<p>This expression comes from considering a volume element of gas inside a star in hydrostatic equilibrium. If the pressure changes, the gas is compressed or expanded, and the volume element moves a small distance <span class="math-container">$dr$</span>, until the pressure is balanced.</p>
<p>To calculate what happens to the volume element, we make two assumptions:</p>
<h3>Assumption #1: The gas is ideal</h3>
<p>This is usually a good approximation in stars where quantum effects can be neglected (i.e. not in stellar remnants).
For an ideal gas of pressure <span class="math-container">$P$</span>, temperature <span class="math-container">$T$</span>, and mass density <span class="math-container">$\rho$</span> (and uniform composition so that the mean molecular weight <span class="math-container">$\mu$</span> is constant), the equation of state is
<span class="math-container">$$
P = \frac{R}{\mu} \rho T,
$$</span>
where <span class="math-container">$R$</span> is the gas constant. Differentiating wrt. <span class="math-container">$r$</span> tells you how much the pressure changes as you move your little volume of gas:
<span class="math-container">$$
\begin{array}{rcl}
\frac{dP}{dr} &amp; = &amp; \frac{R}{\mu}\left( \rho\frac{dT}{dr} + T\frac{d\rho}{dr}\right)\\
              &amp; = &amp; \frac{P}{T}\frac{dT}{dr}  + \frac{P}{\rho}\frac{d\rho}{dr}. \tag{1}
\end{array}
$$</span></p>
<h3>Assumption #2: The gas is adiabatic</h3>
<p>The movement of the gas happens on a &quot;dynamical timescale&quot;, which in stars is much, much smaller (~hours) than the &quot;thermal timescale&quot; (mega-years), and can hence be considered adiabatic. In this case,
<span class="math-container">$$
P \propto \rho^\gamma, \tag{2}
$$</span>
where <span class="math-container">$\gamma$</span> is the adiabatic index. With this relation, you can write Eq. 1 as
<span class="math-container">$$
\begin{array}{rcl}
\frac{dP}{dr} &amp; = &amp; \gamma \rho^{\gamma-1}\frac{d\rho}{dr} \\
              &amp; = &amp; \gamma \frac{P}{\rho} \frac{d\rho}{dr}. \tag{3}
\end{array}
$$</span></p>
<h3>Combine the assumptions</h3>
<p>Combining Eqs. 1 and 3 then gives you
<span class="math-container">$$
\gamma \frac{P}{\rho} \frac{d\rho}{dr} = \frac{P}{T}\frac{dT}{dr}  + \frac{P}{\rho}\frac{d\rho}{dr},
$$</span>
or
<span class="math-container">$$
\left.\frac{dT}{dr}\right|_\mathrm{ad} = (\gamma-1)\frac{T}{P}\frac{P}{\rho}\frac{d\rho}{dr}. \tag{4}
$$</span></p>
<p>Differentiating Eq. 2,
<span class="math-container">$$
\begin{array}{rcl}
\frac{dP}{d\rho} &amp; = &amp; \gamma\rho^{\gamma-1}\\
                 &amp; = &amp; \gamma \frac{P}{\rho}.
\end{array}
$$</span>
so Eq. 4 can be written
<span class="math-container">$$
\left.\frac{dT}{dr}\right|_\mathrm{ad} = \frac{\gamma-1}{\gamma} \frac{T}{P}\frac{dP}{dr} \tag{5}
$$</span>.</p>
<p>This is one version of the adiabatic temperature gradient, but we can also express it in terms of pressure: Since we're physicists — not mathematicians — we have no problems treating <span class="math-container">$dr$</span> as a finite variable, so eliminating that from Eq. 5 gives you
<span class="math-container">$$
\frac{\gamma-1}{\gamma} = \frac{P}{T}\frac{dT}{dP},
$$</span>
or, using @uhoh's fact that <span class="math-container">$dx/x=\ln x$</span>,
<span class="math-container">$$
\boxed{
\frac{\gamma-1}{\gamma} = \left.\frac{d\ln T}{d\ln P}\right|_\mathrm{ad}.
}
$$</span></p>

CC BY-SA 4.0
2022-06-29T19:54:01.340Z
Thank you for the detailed answer, it was super helpful!
Loika
CC BY-SA 4.0
2022-06-28T11:41:53.253Z
2022-06-29T09:31:41.393Z
2022-06-29T09:31:41.393Z
pela
<p>We define the adiabatic temperature gradient as <span class="math-container">$$\Delta_{ad}= \Big(\frac{\partial \log \mathrm{T}}{\partial \log \mathrm{P}}\Big)_{ad}$$</span></p>
<p>The goal of this gradient is to show how the temperature changes under adiabatic compression/expansion. However, I don't understand why we use the logarithms of T and P, and not just simply <span class="math-container">$\frac{\partial\mathrm{T}}{\partial\mathrm{P}}$</span>.</p>
<p>Thank you for any help!</p>

CC BY-SA 4.0
2022-06-27T20:28:39.653Z
Just a guess because my morning coffee hasn't kicked in yet; for things $y$ that vary monotonically over a large range $dy/dx$ gets small when $y$ gets small,  but if you normalize like $(1/y) dy/dx$ you can see trends at different scales equally. Note that $d \log(y)/dx = (1/y) dy/dx$, so I think this definition is just $(x/y)dy/dx$ or something like that.
uhoh
CC BY-SA 4.0
2022-06-28T10:08:50.893Z
Thank you, this could be it I think! Also I guess maybe because in stellar evolution most plots (e.g. the HRD) have their axes logarithmic, so that could be another reason, for convenience?
Loika
<star><stellar-evolution><stellar-astrophysics>
Stellar evolution temperature gradient - why the logarithm?
Loika
# Stellar evolution temperature gradient - why the logarithm?

(Asked by: username_0 on 2022-06-27)

We define the adiabatic temperature gradient as $$\Delta\_{ad}= \Big(\frac{\partial \log \mathrm{T}}{\partial \log \mathrm{P}}\Big)\_{ad}$$

The goal of this gradient is to show how the temperature changes under adiabatic compression/expansion. However, I don't understand why we use the logarithms of T and P, and not just simply $\frac{\partial\mathrm{T}}{\partial\mathrm{P}}$.

Thank you for any help!

**Tags:** star, stellar-evolution, stellar-astrophysics
> **Comments:**
> **username_1**: Just a guess because my morning coffee hasn't kicked in yet; for things $y$ that vary monotonically over a large range $dy/dx$ gets small when $y$ gets small,  but if you normalize like $(1/y) dy/dx$ you can see trends at different scales equally. Note that $d \log(y)/dx = (1/y) dy/dx$, so I think this definition is just $(x/y)dy/dx$ or something like that.  
> **username_0**: Thank you, this could be it I think! Also I guess maybe because in stellar evolution most plots (e.g. the HRD) have their axes logarithmic, so that could be another reason, for convenience?  

---

## 1 answers

### Answer by username_2 on 2022-06-28. Score: 4 | Accepted answer

This expression comes from considering a volume element of gas inside a star in hydrostatic equilibrium. If the pressure changes, the gas is compressed or expanded, and the volume element moves a small distance $dr$, until the pressure is balanced.

To calculate what happens to the volume element, we make two assumptions:

### Assumption #1: The gas is ideal

This is usually a good approximation in stars where quantum effects can be neglected (i.e. not in stellar remnants).
For an ideal gas of pressure $P$, temperature $T$, and mass density $\rho$ (and uniform composition so that the mean molecular weight $\mu$ is constant), the equation of state is
$$
P = \frac{R}{\mu} \rho T,
$$
where $R$ is the gas constant. Differentiating wrt. $r$ tells you how much the pressure changes as you move your little volume of gas:
$$
\begin{array}{rcl}
\frac{dP}{dr} & = & \frac{R}{\mu}\left( \rho\frac{dT}{dr} + T\frac{d\rho}{dr}\right)\\
& = & \frac{P}{T}\frac{dT}{dr} + \frac{P}{\rho}\frac{d\rho}{dr}. \tag{1}
\end{array}
$$

### Assumption #2: The gas is adiabatic

The movement of the gas happens on a "dynamical timescale", which in stars is much, much smaller (~hours) than the "thermal timescale" (mega-years), and can hence be considered adiabatic. In this case,
$$
P \propto \rho^\gamma, \tag{2}
$$
where $\gamma$ is the adiabatic index. With this relation, you can write Eq. 1 as
$$
\begin{array}{rcl}
\frac{dP}{dr} & = & \gamma \rho^{\gamma-1}\frac{d\rho}{dr} \\
& = & \gamma \frac{P}{\rho} \frac{d\rho}{dr}. \tag{3}
\end{array}
$$

### Combine the assumptions

Combining Eqs. 1 and 3 then gives you
$$
\gamma \frac{P}{\rho} \frac{d\rho}{dr} = \frac{P}{T}\frac{dT}{dr} + \frac{P}{\rho}\frac{d\rho}{dr},
$$
or
$$
\left.\frac{dT}{dr}\right|\_\mathrm{ad} = (\gamma-1)\frac{T}{P}\frac{P}{\rho}\frac{d\rho}{dr}. \tag{4}
$$

Differentiating Eq. 2,
$$
\begin{array}{rcl}
\frac{dP}{d\rho} & = & \gamma\rho^{\gamma-1}\\
& = & \gamma \frac{P}{\rho}.
\end{array}
$$
so Eq. 4 can be written
$$
\left.\frac{dT}{dr}\right|\_\mathrm{ad} = \frac{\gamma-1}{\gamma} \frac{T}{P}\frac{dP}{dr} \tag{5}
$$.

This is one version of the adiabatic temperature gradient, but we can also express it in terms of pressure: Since we're physicists — not mathematicians — we have no problems treating $dr$ as a finite variable, so eliminating that from Eq. 5 gives you
$$
\frac{\gamma-1}{\gamma} = \frac{P}{T}\frac{dT}{dP},
$$
or, using @uhoh's fact that $dx/x=\ln x$,
$$
\boxed{
\frac{\gamma-1}{\gamma} = \left.\frac{d\ln T}{d\ln P}\right|\_\mathrm{ad}.
}
$$

> **Comments:**
> **username_0**: Thank you for the detailed answer, it was super helpful!  

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