The Reason Stars Don’t Collapse so Easily

We are bounded in a nutshell of Infinite Space: Worksheet # 9, Problem #1: The Reason Stars Don’t Collapse so Easily

1. Consider a cylindrical differential patch of fluid located a distance r from the center of a star/planet. This patch’s height is dr and the top/bottom of the cylinder has area dA. The fluid is in hydrostatic equilibrium, so this differential patch is at rest.

(a) What is the gravitational force on the cylinder if the fluid has density \(\rho(r)\) in terms of dA, dr, and the local gravitational acceleration, \(g(r)\) ?

(b) The cylinder also feels pressure force from the fluid. If the fluid pressure is \(P(r)\), what is the force associated with the pressure in terms of P, dr, and dA?

(c) Since the cylinder is at rest, the two forces have to balance each other. Write down this force balance and take \(dr \to 0\) to obtain the equation of hydrostatic equilibrium, \[\frac{dP}{dr} = - \rho (r) g(r) \]

a. This problem focusses on the concept of hydrostatic equilibrium, the point at which the pressure exerted by the mass contained in the object balances the collapse potential due to gravity. In order to understand this equation, we must first establish what the gravitational and pressure forces are.

For gravitational forces, we can imagine an infinitesimally small cylinder at a distance r from the center of our star/planet/ object with gravity, and describe the force of gravity on it as described in Newton’s equations: \[F_g = -\frac{MMG}{r^2},\] Newton’s second law of motion also describes to us the definition of a force:  \[F= ma,\] and with knowing that what we call g is the acceleration due to gravity: \[g(r) = a,\]  we can simplify the equation down to: \[ g(r) = \frac{GM}{r^2} .\] Next, we have the following factors we need to incorporate: \[ \rho (r) ~~~ dA~~~dr,\] and if we are defining the mass of this cylinder, then from density equations we know the mass is simply the volume times density, which in the case of a cylinder is the height times the area times the density: \[ dM = \rho (r) ~dA ~dr,\] and if we define this as our mass as determined by the changes in radius, we can create another \(F=ma\) equation: \[ F_{g(r)} = - \rho (r) ~dA~dr~ g(r).\]

b. Now, we will find the pressure exerted by the matter in the object, for which we have the following variables: \[ P(r) ~~~ dr ~~~ dA ,\] and from the definition of pressure we know that pressure is just \[ P = \frac{F}{A},\] so we can rewrite this expression of force and area the following way: \[ F = A~P.\] We can also deduce how the pressure varies from one radius to another, seeing how it changes across the length of the cylinder we are placing in this pressure gradient:  \[\Delta P = (P(r) - P (r + dr) ) ,\] so now it is simply a matter of placing this into our definition of force, which is: \[ F = da~\Delta P\]

c. Finally, we reconcile these two equation to find the general expression of how pressure and gravity equalize each other. Stating our definitions of forces again: \[ F_{g(r)} = - \rho (r) ~dA~dr~ g(r)\] \[ F = da~\Delta P,\] we can now equalize them and start simplifying:  \[- \rho (r) ~dA~dr~ g(r) = da~\Delta P\] \[- \rho (r) ~dr~ g(r) = \Delta P,\] now giving the full definition of \(\Delta P\): \[- \rho (r) ~dr~ g(r) = P(r) - P (r + dr) \] \[- \rho (r) g(r) = \frac{P(r) - P (r + dr)}{dr} ,\]  and the problem establishes we should take \[ dr \to 0,\] which is another way of describing a limit (which is the base for of a derivation: \[- \rho (r) g(r) = \lim_{dr\to 0} \frac{P(r) - P (r + dr)}{dr} ,\] and ultimately, we get our result, the equation for Hydrostatic Equilibrium: \[- \rho (r) g(r) = \frac{dP}{dr}.\]