Lecture 3: Gravitational Potential and N-Body Equations I.

Physics 141/241

Conservative Force Fields

In a one-dimensional system it is always possible to define a potential energy corresponding to any given f(x); let

(1) MATH

where $x_{0}$ is an arbitrary position at which $U=0$. Different choices of $x_{0}$ produce potential energies differing by an additive constant; this constant has no influence on the dynamics of the system.

In a space of $n>1$ dimensions the analogous path integral,


may depend on the exact route taken from point MATH to $\overrightarrow{x}$; if it does, a unique potential energy cannot be defined. One condition for this integral to be path-independent is that the integral of the force f(x) around all closed paths vanishes. An equivalent condition is that there is some function U(x) such that

(3) MATH $.\ \ $

Force fields obeying these conditions are conservative. The gravitational field of a stationary point mass is the simplest example of a conservative field; the energy released in moving from MATH to MATH with $r_{2}<r_{1}$ is exactly equal to that consumed in moving back from MATH to MATH.

Gravitational Potential

In simulation projects it's natural to work with the path integral of the acceleration rather than the force; this integral is the potential energy per unit mass or gravitational potential, MATH, and the potential energy of a test mass m is just MATH. For an arbitrary mass density MATH, the potential is

(4) MATH $,$

where MATH is the gravitational constant and the integral is taken over all space. Poisson's equation provides another way to express the relationship between density and potential:

(5) MATH MATH $.$

Note that this relationship is linear; if MATH generates MATH and MATH generates MATH then MATH MATH generates MATH. Gauss's theorem relates the mass within some volume V to the gradient of the field on its surface:

(6) MATH $,$

where the infinitesimal vector MATH is an element of surface area with an outward-pointing normal vector.

Spherical Potentials

Consider a spherical shell of mass $M$ :

(a) the acceleration inside the shell vanishes

(b) and the acceleration outside the shell is $\ -GM/r^{2}$.

From these results, it follows that the potential of an arbitrary spherical mass distribution is

(7) MATH

where the enclosed mass is

(8) MATH

Elementary examples

A point of mass $M$ :

(9) MATH

This is known as a Keplerian potential since orbits in this potential obey Kepler's three laws. If G=1 is set, the velocity of a circular orbit at radius $r$ is MATH

A uniform sphere of mass $M$ and radius $R$:

(10a) MATH $r\leq R,$

(10b) MATH $r>R,$

where $\rho =$ $M/(4\pi R^{3}/3)$ is the mass density. Outside the sphere the potential is Keplerian, while inside it has the form of a parabola; both the potential and its derivative are continuous at the surface of the sphere.

Galactic Potential, Scales and Units

Galactic potential is the collective self-consisten field of all stars within the galaxy. It is determined by the distribution function MATH which accounts for the mechanical state of the galaxy.

Sun is located in our Milky Way which has $\sim 10^{11}$ visible stars and about $\sim 10^{10}$ solar masses of gas MATH with MATH. In comparison, MATH. Gas has little effect on main features of galactic dynamics.

Most of the stars in the galaxy travel on nearly circular orbits in a thin disk whose radius is of the order of $10$ $kiloparcecs$ and thickness of the order of $1$ $kpc$ with MATH. Typical circular speed of stars is of the order of $200$ $km/s$ and the time required to complete a galactic orbit at $10kpc$ is about $3\cdot 10^{8}yr$. The dispersion in velocities is about $40km/s$. Age of the galaxy MATH. Typical disk star (like our Sun) has completed over 30 revolutions. Galaxy is in steady state. Why and how is explained by the steady state solution of the distribution function MATH from the collisionless Boltzmann equation.

Scales and units

Consider first a practical choice on units for galaxy collisions: MATHThe gravitational constant is MATH in cgs units. Typical galaxy size is measured in tens of kiloparsecs, large scale structure of the universe is measured in Megaparsecs. Let us determine the value of $G$ in the new galactic units:

(11) MATH


This value of $x$ is an inconvenient value for $G$ to carry in simulations. Let us try galactic units which set $x=1$. One possible illustration is MATH In these units

(12) MATH



In these units we can use $G=1$ in the simulations. With this choice two units are arbitrary, the third unit is fixed. There is great flexibility to interpret simulation results in physical units at the very end of the runs with dimensionless numbers.

$\QTR{bf}{Pointers}$ $\QTR{bf}{to}$ astronomy units $\QTR{bf}{and}$ Kepler orbits.

MATH $\QTR{bf}{of}$ Kepler orbits (1 earth year) $\QTR{bf}{and}$ one Saturn year