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Massimo Capaccioli » 16.Galaxy dynamics - Part III

Solutions of the CBE

In the general case, the CBE has non less than 7 independent variables (\textbf{x}, \textbf{v}, t) and cannot be solved. One way to reduce the complexity of the problem is to consider moments of the CBE (e.g. the continuity equation, the Jeans equations, the Virial Theorem), which are defined in the 3D space and can be connected to observations. They allow us to evaluate the general properties of the systems under exam and the properties of the CBE solutions, but they do not provide the full physics of the stellar systems, which is stored in the exact CBE solutions. Exact CBE solutions can be obtained for simple systems.

In this lecture we are interested in deriving some of these solutions as an exercise to understand how observed quantities can be a reflection of the internal dynamics of equilibrium systems. We start by deriving an important theorem of the galaxy dynamics: the Jeans Theorem.

First of all we recall that a function of the phase-space coordinates I(\textbf{x},\textbf{v}): R^6 \rightarrow R,  is an integral of motion if and only if  dI[x(t), v(t)]/dt=0,  which means that:

I[x(t),v(t)]=I(x_0, v_0).

Integral of motions are important to define orbits. In fact, if at any given time we know 5 integral of motion in the 6-D space, each of them defines a 5-D hyper-surface and their intersection defines a 1-D trajectory, the orbit.

Jeans Theorem

Jeans Theorem: Any steady-state solution of the CBE depends on the phase-space coordinates only through integrals of motion in the galactic potential, and any function of the integrals yields a steady-state solution of the CBE. Suppose f is a steady-state solution of the CBE:

\frac{df}{dt}=0 \rightarrow \frac{\partial f}{\partial t} + \vec{\nabla} f \cdot \frac{\partial \textbf{x}}{\partial t} + \frac{\partial f}{\partial \textbf{v}} \cdot \frac{\partial \textbf{v}}{\partial t}=0 ,

otherwise written as:

\textbf{v} \cdot \vec{\rm \nabla} f - \vec{\rm \nabla} \Phi \cdot \frac{\partial f}{\partial \textbf{v}}  = 0,

which is means that f is an integral of motion.


Solutions of the CBE

Conversely, if I_1 \dots I_n, are n integrals of motion and f is any function of n variables then:

\displaystyle{\frac{d}{dt} f [I_1(\textbf{x},\texbf{v}),\dots, I_n(\textbf{x},\textbf{v})]=\sum_{m=1}^n \frac{\partial f}{\partial I_m} \frac{\partial I_m}{\partial t}=0},

because by definition of integral of motion:


Thus f is a solution of the CBE.

We will be mainly interested in the second proposition of the Jeans Theorem in the sense that any distribution function, which is written in terms of integral of motion is assured to be a solution of the CBE. There is a strong version of the Jeans Theorem (that we do not prove here), which states that the distribution function of steady-state galaxy in which almost all orbits are regular with non-resonant frequencies may be presumed to be a function only of three independent isolating integrals.

Applications to spherical systems

Spherical systems admit 4 integrals of motion: the energy and angular momentum. A function f(E, \vec{ L}) of these four integrals is a solution of the CBE. If the system is also spherically symmetric, then f depends only on the magnitude of L; that is: f=f(E,L). We remark that, if the potential is provided by the tracer’s mass, then self-consistency holds, and:

\displaystyle{\nabla^2\Phi =- 4 \pi G \rho= -4 \pi G \int f d^3v} ,

or, using spherical symmetry,

\displaystyle{ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d \Phi}{dr} \right) = -4 \pi G \int f (\frac{1}{2} v^2 + \Phi ) d^3 \vec{v}}.



Applications to spherical systems

For the following calculations we will define a relative potential: \Psi =- \Phi +\Phi_0 , where  \Phi_0  is some constant value, and the relative energy of a star:

\epsilon = -E + \Phi_0= \Psi - \frac{1}{2} v^2 .

In particular \Phi_0 is chosen in order to have f>0  for \epsilon>0, and f=0  for \epsilon \leq 0.

Obviously, the relative potential of an isolated system satisfies the Poisson’s equation of the form:

\nabla^2 \Psi = -4 \pi G\rho,

with boundary conditions \Psi \rightarrow \Phi_0  as  |x|\rightarrow \infty.

Isothermal sphere

The equation of state of an isothermal body of gas is p=K \rho, where K is a constant.

The equation of hydrostatic support of an isothermal gas is:

\frac{dp}{dr}=\frac{k_B T}{m} \frac{d \rho}{dr} = - \rho \frac{GM(r)}{r^2} ~~~~~~(1)

where k_B is the Boltzmann constant, {p} and {T} are the pressure and temperature of the gas, m is the mass of the particle, and M(r) is the mass of the system within r. By multiplying the equation above by (r^2m/\rho k_B T) and differentiating, we obtain:

\displaystyle{ \frac{d}{dr}\left( r^2\frac{d \ln \rho}{dr} \right) = - \frac{Gm}{k_B T} 4\pi r^2 \rho },

where we have used  (dM/dr)=4\pi r^2 \rho.

Some application of Jeans Equations

Let us assume now that the stellar system has a distribution function of the form:

\displaystyle{ f(\epsilon)=\frac{\rho_1}{(2\pi\sigma^2)^{3/2}} e^{\epsilon/\sigma^2}=\frac{\rho_1}{(2\pi\sigma^2)^{3/2}} \exp\left(\frac{\Psi-1/2v^2}{\sigma^2} \right)}. \hspace{2cm}(2)

Integrating over all velocities, we derive the number density of the system: \rho=\rho_1 e^ {\Psi/\sigma^2}, \hspace{2cm}(3)

thus, the Poisson’s Equation becomes:

\displaystyle{ \frac{d}{dr}\left( r^2 \frac{d \ln \rho}{dr}\right)=- \frac{4 \pi G }{\sigma^2} r^2 \rho}. \hspace{2cm}(4)

By comparing equation (1) with eq. (4) we see that, if we define: \sigma^2=\frac{k_B T}{m}, the two equations are identical. This means that the structure of an isothermal self-gravitating sphere of gas is identical with the structure of collisionless system of stars whose density in phase space  f(\epsilon)  is given by equation (2).

Some application of Jeans Equations

In particular we can find out which number density distribution is given for a isothermal sphere by solving equation (4).  If we assume the solutions of the form of a power law \rho=C r^{-b}, then the left side of the equation:

\displaystyle{ \frac{d}{dr}\left( r^2 \frac{d \ln \left(C r^{-b}\right)}{dr}\right)= -b },

and the right side:

\displaystyle{- \frac{4 \pi G }{\sigma^2} r^2 \rho = - \frac{4\pi G}{\sigma^2} C r^{2-b}} . They are equal if:  b=2 and  C=(\sigma^2/2\pi G), which implies:

\displaystyle{\rho(r)=\frac{\sigma^2}{2\pi G r^2}}. \hspace{2cm}(5)

This is the density of the singular isothermal sphere because the density is infinite at r=0. It is possible to derive other solutions, which do not have this kind of problem (and are more physical). A well know solution is the so called King’s law which has the form: \displaystyle{ \rho(r) \propto \frac{\rho_0}{ 1+(r/r_0)^2}}, where  \rho_0  is the central density and: \displaystyle{ r_0=\sqrt{\frac{9 \sigma^2}{4\pi G \rho_0}}},  is the King radius.

Some application of Jeans Equations

In equation (4) we define:  \tilde{\rho}=\rho/\rho_0  and  \tilde{r}=r/r_0,  such that:

\displaystyle{ \frac{d}{d\tilde{r}}\left( \tilde{r}^2 \frac{d \ln\tilde{\rho}}{d\tilde{r}}\right)=- 9 \tilde{r}^2 \tilde{\rho}},

which, using Eq. (3), can be re-written as:

\displaystyle{ \frac{d}{d\tilde{r}}\left( \tilde{r}^2 \frac{d (\Psi/\sigma^2)}{d\tilde{r}}\right)=- 9 \tilde{r}^2 \exp\left[ \frac{\Psi(r)-\Psi(0)}{\sigma^2} \right] }. \hspace{2cm}(6)

The new density law, which can be derived by numerical integration of equation (6), is characterized by the presence of a “core” in the central regions where the density flattens to constant value \rho_0. At large distance (r > 15\, r_0) it goes as the singular isothermal sphere (eq. 5), which in the new coordinates writes: \tilde{\rho}=\frac{2}{9} \tilde{r}^{-2}.


Some application of Jeans Equations

It can be shown that a good approximation for the solution of equation (6) in the central regions (r <2 r_0) is the modified Hubble law:

\displaystyle{ \tilde{\rho}(\tilde{r})=\frac{1}{(1+\tilde{r}^2)^{\frac{3}{2}}}}.

The surface density \Sigma(R) of the isothermal sphere in units of \rho_0 r_0, for R \gg r_0 is:

\displaystyle{ \Sigma(R) = \frac{\sigma^2}{2 G R} = \frac{2}{9} \pi \rho_0 r_0 \frac{r_0}{R}}.

We remark that, if

M(r)= 4 \pi \int r^2 \rho(r) dr

is the mass interior to the spherical radius r, then the circular speed is written as: \displaystyle{ v_c^2=\frac{G M(r)}{r}}, which, by equation(4), can be also re-written as:

\displaystyle{ v_c^2=-\sigma^2 \frac{d \ln \rho}{d \ln r}},

where the term  (d \ln \rho / d \ln r)  tends to -2 at r \gg r_0 for the isothermal sphere. Thus the circular speed at large distances from the center is constant to v_c=\sqrt{2}\sigma.

Distribution Function from the density profile

We finally show how one can derive the distribution function f(\epsilon) of a spherically symmetric star system starting from the density profile \rho(r). By definition we know that the density profile is the integral of the DF over all velocities. If we write f as a function of the energy and we replace the integral over all velocities as the integral over \epsilon using a change of variable, we obtain:

\displaystyle{ \rho(r)= 4 \pi \int_0^\Psi f(\epsilon) \sqrt{2(\Psi-\epsilon)} d\epsilon}. As far as \Psi is a monotonic function of r, we can consider \rho as a function of \Psi and obtain:

\displaystyle{ \frac{1}{\sqrt{8}\pi}\frac{d\rho}{d\Psi}=\int_0^\Psi \frac{f(\epsilon)d\epsilon}{ \sqrt{\Psi-\epsilon}}} .

This is an Abel integral equation of the kind already found previously in this course, which can be solved to obtain f(\epsilon):

\displaystyle{ f(\epsilon)=\frac{1}{\sqrt{8}\pi^2}\frac{d}{d \epsilon} \int_0^\epsilon \frac{d \rho}{d\Psi} \frac{d\Psi}{ \sqrt{\epsilon-\Psi}} }.

Eddington formula


\displaystyle{ f(\epsilon)=\frac{1}{\sqrt{8}\pi^2}\left[ \int_0^\epsilon \frac{d^2 \rho}{d\Psi^2} \frac{d\Psi}{ \sqrt{\epsilon-\Psi}} +\frac{1}{\sqrt{\epsilon}}\left(\frac{d\rho}{d\Psi}\right)_{\Psi=0} \right]},

which is also called Eddington’s formula from its original inventor, sir Arthur Eddington (1916).

This formula states that one can derive the DF (as a function of the energy only) of a spherical (non rotating) systems just by knowing its density distribution. However, there is no guarantee that the solution of this equation will satisfy the physical requirement of being nowhere negative.

One can show that this is true if and only if:

\displaystyle{ \int_0^\epsilon \frac{d \rho}{d\Psi} \frac{d\Psi}{ \sqrt{\epsilon-\Psi}}} ,

is an increasing function of \epsilon.

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