# How to derive the hydrostatic equilibrium equation from the variational principle?

I was reading the book "Advanced Stellar Astrophysics" (William K. Rose, 1998) and I came across a different approach to deriving the hydrostatic equilibrium equation. First he defined the total energy of a static, spherical star, $$E = \int udM - \int \dfrac{GM}{r}dM, \tag{1}\label{1}$$ where $$u$$ is the internal energy per gram. Then he defined the Lagrangian change in the internal energy density, $$\Delta u = u(r + \xi(r)) - u_0(r),$$ where $$\xi(r)$$ is the Lagrangian displacement and $$u_0(r)$$ are the undisturbed values of the internal energy density. The next step is completely strange and doesn't make much sense to me: He makes a Lagrangian variation in \eqref{1} and looks like: $$\Delta E = \int \Delta u dM + \int \dfrac{GM}{r^2}\xi(r)dM \tag{2}\label{2}$$ which he calls, $$\Delta u = - P\nabla\bigg(\dfrac{1}{\rho}\bigg) = \dfrac{P}{\rho^2}\Delta \rho$$ How did he arrive at these last two results? I can't deduce that $$r^2$$ in the denominator of the second term of the equation \eqref{2} and neither the positive sign.

• Hint: Substitute $r \to r + \xi(r)$ and subtract from original equation. Remember $\xi(r) \to 0$ Commented Sep 12, 2020 at 4:28
• Thank you! got here doing this and then expanding Commented Sep 12, 2020 at 23:49

So the claim is that we examine a perturbation$$\delta r$$ which is very very small and for any other quantity we define $$\delta F = F(r + \delta r)-F(r)$$ to first order.
We insert it in every single term in the integral (this is the part which I find suspicious) and have $$\delta E =\int\mathrm dM\left(\delta u - \frac{GM}{r+\delta r} + \frac{GM}{r} \right).$$ To answer your first question, the middle term satisfies \begin{align} - \frac{GM}{r+\delta r} &= -\frac{GM}r \frac1{1+\frac{\delta r}r}\\ &= -\frac{GM}r \left( 1 - \frac{\delta r}r + \left(\frac{\delta r}r\right)^2 -\dots \right) \end{align} by the usual rules of geometric series. To first order we keep the first two terms only.
Second there is the claim that $$\delta u = - p~\delta V,$$a standard claim of thermodynamics which again just sort of doesn't make sense here but that's because I don't see exactly what the argument is. Since $$u$$ is the internal energy per unit of mass then $$V$$ should be the volume per unit of mass, which makes it exactly $$V = 1/\rho$$ where $$\rho$$ is the mass density. Doing a Taylor series $$\delta\frac1\rho =-\frac{\rho'~\delta\rho}{\rho^2}$$ to first order.
I think I can make sense of this but I do not think I can make sense of this inside of that integral sign which I would want to rewrite as $$\int \mathrm dM =\int \rho(r)\mathrm dr$$ and then I would never do a Lagrangian path variation on $$r$$ but on $$\rho$$ maybe. I spent a little time trying to Google ways to do thermodynamics as a sort of calculus of variations problem, and got a lot of black hole thermodynamics papers which figures. I suppose one could try to maximize the entropy or so, maybe minimize a free energy assuming that the core of the star is held at a very constant temperature or so, argue that the density profile which minimizes this should be the quasi-static equilibrium? But again, varying $$r$$ just sounds like you are doing a fancy version of the normal, “balance two forces against each other” argument and this integral over $$\mathrm dM$$ is spurious?
• And it worked! In the end it appeared that in order for the condition for hydrostatic equilibrium to be recovered we must have to, $$\dfrac{dP}{dM(r)} = - \dfrac{GM(r)}{4\pi r^4}.$$ which is precisely the Lagrangian version of hydrostatic equilibrium! Commented Sep 13, 2020 at 16:26