# Stellar evolution temperature gradient - why the logarithm?

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!

• 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
Jun 27, 2022 at 20:28
• 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? Jun 28, 2022 at 10:08

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

• Thank you for the detailed answer, it was super helpful! Jun 29, 2022 at 19:54