# Galaxies Lecture 03

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### Coalescence of Satellite Galaxies

There is increasing evidence that galaxies form via the coalesence of satellite galaxies. We will discuss dynamical friction as a mechanism for combination. As an illustration, let’s consider a point of mass ${\displaystyle M}$ falling into a galaxy.

Let us say this galaxy is a cluster of stars of mass ${\displaystyle m\ll M}$ arranged on a flat grid. As our point mass travels through this grid, we expect the stars to fall toward it, ultimately gathering along the path behind the point mass. This will create an overdensity of stars directly behind the point mass that will serve to increase the force of gravity on the mass exiting the system relative to the force of gravity on the mass as it fell inward. If this drag is enough to keep the point mass bound to the system, we expect this process to proceed until there is an equipartition of energy between the point mass and the stars of the system.

If this point mass has a gas cloud associated with it, we’d expect this to further increase the drag of the point mass through the system as a result of shocks.

Let us formulate this mathematically. As this point mass interacts with a star at distance ${\displaystyle {\bar {r}}={\bar {x}}_{m}-{\bar {x}}_{M}}$, we will see a force between them:

${\displaystyle {mM \over m+M}{\bar {r}}=-{GMm \over r^{2}}{\hat {e}}_{r}\,\!}$

We expect the changes in the velocites of the star and the point mass that result from this interaction to be:

{\displaystyle {\begin{aligned}\Delta v_{m}-\Delta v_{M}=\Delta V&={\bar {r}}\\M\Delta v_{m}-v\Delta V_{M}&=0\\\Delta V-m&={m \over M+m}\Delta V\\\end{aligned}}\,\!}

Consider that the point mass and star pass each other at a distance (impact parameter) ${\displaystyle b}$. Since ${\displaystyle m\ll M}$, we will for the moment consider the point mass to be stationary, and consider the angle ${\displaystyle \theta }$ by which the path of the star deviates from a straight line. This situation is described by the Rutherford scattering equation in terms of ${\displaystyle \Psi _{0}}$, which is the angle to the "knee" of the curve describing the path of the star. ${\displaystyle \Psi }$ in general is the parameterized angle from the scatterer to the scatteree as a function of time:

${\displaystyle {d^{2}u \over d\Psi ^{2}}+u={G(M+m) \over L^{2}}={F({1 \over u}) \over L^{2}}\,\!}$

where ${\displaystyle u\equiv {1 \over r}}$, and ${\displaystyle L\equiv bv_{0}}$, where ${\displaystyle v_{0}}$ is the initial velocity of the star. This equation has the general solution of a rosette (a precessing elliptic curve). For our problem, this solution looks like:

{\displaystyle {\begin{aligned}u&={1 \over r}=C\cos(\Psi -\Psi _{0})+{G(M+m) \over L^{2}}\\{d \over dt}(r)&=Cr^{2}\Psi \sin(\Psi -\Psi _{0})\\&=Cbv_{0}\sin(\Psi -\Psi _{0})\end{aligned}}\,\!}

Using that ${\displaystyle \theta =2\Psi _{0}-\pi }$, ${\displaystyle L=r^{2}{\dot {\Psi }}}$, and the boundary condition that ${\displaystyle {d \over dt}{r}=-v_{0}}$, we have:

{\displaystyle {\begin{aligned}1=cb\sin(-\Psi )\\{1 \over r}=C\cos(\Psi _{0})+{G(m+M) \over b^{2}v_{0}^{2}}\\{\tan \Psi _{0}={-bv_{0}^{2} \over G(m+M)}}\\\end{aligned}}\,\!}

We may then determine what this means in terms of parallel and perpendicular velocities:

{\displaystyle {\begin{aligned}v_{\perp }&=v_{0}\sin \theta =v_{0}\left|\sin 2\Psi \right|=v_{0}{\left|\tan \Psi _{0}\right| \over 1+\tan ^{2}\Psi _{0}}={v_{0}^{3}b \over G(m+M)}\left(1+{b^{2}v_{0}^{4} \over G^{2}(m+M)}\right)^{-1}\\v_{\|}&=v_{0}-v_{0}\cos \theta =v_{0}\left|1+\cos 2\Psi _{0}\right|=2v_{0}\left[1+{b^{2}v^{4} \over G^{2}(m+M)^{2}}\right]^{-1}\\\end{aligned}}\,\!}

For large ${\displaystyle b}$, ${\displaystyle b^{2}\gg {G^{2}(m+M)^{2} \over v_{0}^{4}}}$, so

${\displaystyle v_{\perp }={2Gm \over bv_{0}}\,\!}$

Thus we have recovered our result from last lecture for large impact parameters. We now have a result for a single encounter, so we need to examine the rate of encounters:

${\displaystyle t={1 \over n\sigma v}\,\!}$

This is the inverse of our rate, and ${\displaystyle \sigma =2\pi b\ db}$. ${\displaystyle v=v_{0}}$, so we just need a figure for ${\displaystyle n}$. We will add a little formality here by introducing a density function ${\displaystyle f(x,y,z,v_{x},v_{y},v_{z})}$ so that

${\displaystyle {\bar {n}}=f({\bar {v}}_{m})d^{3}{\bar {v}}_{m}\,\!}$

This gives an integral for the change in the velocity of the point mass as a function of time:

${\displaystyle {d \over dt}{v_{M}}{\bigg |}_{{\bar {v}}_{M}}={\bar {v}}_{0}f({\bar {v}}_{m})d^{3}{\bar {v}}_{M}\int _{0}^{b_{max}}{{2mv_{0} \over (M+m)}\left[1+{b^{2}v_{0}^{4} \over G^{2}(M+m)^{2}}\right]^{-1}2\pi bdb}\,\!}$

This is an integral of the form ${\displaystyle \int {{k_{1} \over (1+k_{2}^{2}b^{2})}2\pi bdb}}$, which we can solve, giving us the result:

${\displaystyle {d \over dt}{v_{M}}={\bar {v}}_{m}f({\bar {v}}_{m})d^{3}{\bar {v}}_{m}{2mv_{0} \over (M+m)}{(M+m)^{2}G^{2} \over v_{0}^{4}}{\frac {1}{2}}\left(1+{v^{4}b_{max}^{2} \over G^{2}(M+m)^{2}}\right)\,\!}$

Then defining ${\displaystyle \Lambda ={v_{0}^{2}b_{max} \over G(M+m)}}$ and noting that ${\displaystyle \ln \Lambda \approx {\frac {1}{2}}(1+\Lambda ^{2})}$, we have our final result:

${\displaystyle {{d \over dt}{{\bar {v}}_{m}}=-v_{0}\pi ^{2}\ln \Lambda G^{2}m(M+m){\int _{0}^{v_{M}}{f(v_{M})v_{m}^{2}dv_{m}} \over v_{m}^{3}}{\bar {v}}_{M}}\,\!}$

This is the Chandrasekhar Dynamical Friction Formula. Often, we are only interested in the question of how long it takes a black hole to get to the bottom of a system of stars. For this purpose, we can estimate the above equation as:

${\displaystyle {d \over dt}{v_{M}}\approx 4\pi G^{2}{mM\ln \Lambda n \over v_{0}^{2}}\,\!}$

Let’s examine a couple of cases. If ${\displaystyle v_{m}}$ is small, then ${\displaystyle f(v_{m})\approx f(0)}$. Then our integral becomes:

${\displaystyle {d \over dt}{v_{M}}=-{16\pi ^{2} \over 3}\ln \Lambda G^{2}f(0)m(M+m){\bar {v}}_{M}\,\!}$

This says that the acceleration of a point mass is proportional to the velocity. This is the form of viscous friction (a marble in honey). On the other hand, if we say that ${\displaystyle f({\bar {v}}_{m})}$ is Maxwellian, then:

${\displaystyle {d \over dt}{v_{M}}={-4\pi G^{2}\ln \Lambda (M=M)n_{0}m \over v_{m}^{3}}\left[e^{4x}-{2x \over {\sqrt {\pi }}}e^{-x^{2}}\right]{\bar {v}}_{m}\,\!}$

where ${\displaystyle x\equiv {v_{m} \over {\sqrt {2}}\sigma }}$. Here, ${\displaystyle {{\bar {v}}_{m} \over v_{0}^{3}}\sim {1 \over v_{0}^{2}}}$, which is similar to the frictional force of air on a bullet.