Galaxies Lecture 02

Timescale for collisional (star-star) relaxation

Suppose Star A passes Star B with impact parameter $b$. This will deflect the trajectory of Star A in a direction $\perp$ to the original trajectory. We’ll say that a trajectory is significantly altered when $\Delta v_{\perp }\sim v$. Integrating over all possible $b$, and integrating over all pairs of stars, we will get an estimate of the timescale for star-star interactions in a galaxy.

$F_{\perp }={Gm^{2} \over x^{2}+b^{2}}\cos \theta \,\!$

Using $\cos \theta =\left({b \over x^{2}+b^{2}}\right)^{\frac {1}{2}}$, we have:

$F_{\perp }={Gm \over b^{2}}\left(1+{x^{2} \over b^{2}}\right)^{-{\frac {3}{2}}}\,\!$

then using $F={d \over dt}{p}$, and $x\approx vt$, we have:

$F_{\perp }={Gm \over b^{2}}\int _{-\infty }^{\infty }{\left(1+\left({vt \over b}\right)^{2}\right)^{-{\frac {3}{2}}}dt}=\Delta v_{\perp }\,\!$

Substituting $s={vt \over b}$,

${\Delta v_{\perp }={2Gm \over bv}{s \over (1+s^{2})^{\frac {1}{2}}}{\bigg |}_{0}^{\infty }={2Gm \over bv}}\,\!$

This breaks down when $v^{2}\sim {2Gm \over b}$, giving us a minimum interaction distance

$b_{min}={2Gm \over v^{2}}\,\!$

For $m\sim 2\cdot 10^{33}$, and $v\sim 2\cdot 10^{6}$, $b_{min}$ is of order 1 AU.\

Note, by the way, that:

$\Delta v_{\perp }=\overbrace {Gm \over b^{2}} ^{F_{gravity}}\overbrace {2b \over v} ^{interaction \atop timescale}\,\!$

Anyway, we wanted to integrate over all stars. We expect that if we throw a star through a galaxy that its net deflection $\int _{*}{\Delta v_{\perp }}=0$ because stars are probably distributed symmetrically. In order to get a real measure for the interactions going on in a galaxy, we want to calculate $\Delta v_{\perp }^{2}$:

$\sum _{*}{\Delta v_{\perp }^{2}dn}=\left({2Gm \over bv}\right)^{2}{2N_{tot} \over R^{2}}b\ db\,\!$ ${\begin{aligned}\Delta v_{\perp }^{2}&=\left[8\left({Gm \over Rv}\right)^{2}N_{tot}\right]\int _{b_{min}}^{R}{db \over b}\\&=\left[8\left({Gm \over Rv}\right)^{2}N_{tot}\right]\ln {R \over b_{m}in}\end{aligned}}\,\!$

And we define $\Lambda \equiv {R \over b_{min}}$. Using the definition of $b_{min}$, we have:

$v^{2}={GM \over R}={GmN_{tot} \over R}\,\!$

And since $\Delta v_{\perp }^{2}=8v^{2}{\ln \Lambda \over N}$, we have

${\Delta v_{\perp }^{2} \over v^{2}}=8{\ln \Lambda \over N}\,\!$

and ${\Delta v_{\perp }^{2} \over v^{2}}\approx 1$ represents the condition for a galaxy to have lost its “memory” of its initial conditions. Using $\Lambda ={Rv^{2} \over 2Gm}$ and ${Rv^{2} \over m}=N_{tot}$, we have

${\Delta v_{\perp }^{2} \over v^{2}}={12.5\ln N \over N}\,\!$

The time for a star to cross a galaxy is $t_{cross}={2R \over v}$, and the number of crossings required to relax is $n_{relax}={0.1N \over N}$. Thus, the relaxation time is

${\begin{aligned}t_{relax}&=t_{cross}n_{relax}\\&={0.1N \over \ln {N}}t_{cross}\\&={0.1N \over \ln N}{2GM \over v^{3}}\end{aligned}}\,\!$

To evaluate whether various galaxies have relaxed, we’ll make a table:

System	${0.1N \over \ln N}$	v (km/s)	R (pc)	t cross (yrs)	t relax (yrs)	Age
Old Open Cluster: Pleideas (50)	1.28	1	2	$2\cdot 10^{6}$	$1-2\cdot 10^{6}$	$1\cdot 10^{5}$ ?
Globular Cluster ($10^{6}$)	$7.2\cdot 10^{3}$	3	3	$10^{6}$	$7\cdot 10^{9}$	$1\cdot 10^{10}$
Milky Way ($10^{11}$)	$4\cdot 10^{8}$	100	$10^{4}$	$10^{8}$	$4\cdot 10^{16}$	$1\cdot 10^{10}$
Dwarf Galaxy ($10^{7}$)	$6.2\cdot 10^{4}$	15	$5\cdot 10^{2}$	$3\cdot 10^{7}$	$3\cdot 10^{14}$	$1\cdot 10^{10}$
Galaxy Cluster ($10^{3}$)	14	300	$10^{6}$	$3\cdot 10^{9}$	$5\cdot 10^{10}$	$1\cdot 10^{10}$

The takeaway point here is that stars in large galaxies still remember their original trajectories.

An important number to remember is $1{km \over s}=1{pc \over My}$. If a galaxy is relaxed, we may expect a thermalized velocity distribution of stars, but depending on the geometry of a galaxy, this may only apply within velocities along a particular axis.

When galaxies are ripped apart, streams of stars can be torn off into moving groups which can be identified by their common velocities.