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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!

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    $\begingroup$ 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. $\endgroup$
    – uhoh
    Jun 27, 2022 at 20:28
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    $\begingroup$ 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? $\endgroup$
    – Loika
    Jun 28, 2022 at 10:08

1 Answer 1

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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 $dx/x=\ln x$, $$ \boxed{ \frac{\gamma-1}{\gamma} = \left.\frac{d\ln T}{d\ln P}\right|_\mathrm{ad}. } $$

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    $\begingroup$ Thank you for the detailed answer, it was super helpful! $\endgroup$
    – Loika
    Jun 29, 2022 at 19:54

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