# The Hydrostatic Equation $\rho_{r} \frac{d^2r}{dt^2} = -\frac{dP_{r}}{dr} - G \frac{M_{r}\rho_{r}}{r^2}$ or: $\frac{dp}{dz} = -g\rho$ Where $dz$ is the change in depth within a fluid. >[!Success]- Proving the Hydrostatic Equation > >Sure, the partial derivative might seem intimidating at first, but it's actually not that bad! > >When thinking about hydrostatics, we have to think in 'shells'. Before heading it, it's important that we provide ourselves with a method to visualise this - such as in this diagram below! > > > >The term '**hydrostatic equilibrium**' refers to the equilibrium between the pressure from the air above this imaginary slab and of the upward pressure caused by the air below the slab pushing against it. If we let the density of the air be a constant '$\rho , we can get functions for the mass $m$ and weight $W$ of this slab of air respectively: >$m = \rho \delta z$ >$W = g\rho \delta z$ >As atmospheric pressure decreases from $z$ to $\delta z$, we know that $\delta p$ must be negative. The gravitational force at $z + \delta z$ must counteract this change in the pressure along the slab, so we get the equation: >$\frac{\delta p}{\delta z} = -g\rho$ >Which is our hydrostatic equation. With time, it'll make sense - but you have to understand the relative pressure gradients and how they scale with height! # The Equation in Stars A star must be in hydrostatic equilibrium. A star applies an enormous inward pressure to the core of the star/ The pressure is sufficiently strong to ignite nuclear fusion which exerts a pressure back outward and prevents the star from collapsing further. So that means we can find an expression for the **change in the pressure** within the star, as the *Explanation taken from the BAAO R2 2021-2022 Q3 (a)! I couldn't even begin with this when I tried deriving it myself in the BPhO, but I'm pretty sure you bunch will be more competent than me!* First, let's consider a box of height $dr$ nestled some radius $r$ within a star. ![[hydrostaticboxmodel.png]] *The box model, with some captions to help guide you!* For the star to be stable, the ***difference in the outward fusion pressure between r and r+dr*** must balance the **difference in the gravitational force between r and r+dr.** So, let's first create an expression for the force difference between the shells: $dF = \frac{GMm}{r^2}$ Where in this case, $m$ is the mass between the two 'spheres' with radius $r$ and $r+dr$. So, since $dr$ is small, we can approximate this difference to: $dF = \frac{GM(4\pi r^2\rho)dr}{r^2}$ Where $\rho$ is the density of the star. (Questions will usually take this as constant so DON'T WORRY! If they don't, then write $\rho$ in terms of $r$ and integrate across the surface area of whatever object's in the question to find the mass at that point. $M = \rho V = \int \rho \, dA$!) Now, back to hydrostatic equilibrium - remember that the pressure on any object is always equal to the Force divided by the surface area ($P = \frac{F}{A}$). Since the surface area this pressure gradient acts on is the surface area of the spherical shell at the base, and the force that creates this radiation pressure acts opposite the direction of the gravitational force, we can write: $dP 4\pi r^2 = -\frac{GM(4\pi r^2\rho)dr}{r^2}$ Which simplifies to: $\frac{dP}{dr} = -\rho\frac{GM}{r^2}$ Or $\frac{dP}{dr} = -\rho g$ Simple, eh? This lets us find the internal pressures at certain levels of a star. Without this outward force of fusion being checked, the star would blow itself apart, or implode, if you will. This happens in stars higher than 250 solar masses, which is theorised to be the maximum limit for a stable star as the Universe is now. This is known as the **eddington limit.**