Galaxies Lecture 02

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Timescale for collisional (star-star) relaxation

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

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

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

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

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

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

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

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

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

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

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

Note, by the way, that:

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

Anyway, we wanted to integrate over all stars. We expect that if we throw a star through a galaxy that its net deflection ${\displaystyle \int _{*}{\Delta v_{\perp }}=0}$ 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 ${\displaystyle \Delta v_{\perp }^{2}}$:

${\displaystyle \sum _{*}{\Delta v_{\perp }^{2}dn}=\left({2Gm \over bv}\right)^{2}{2N_{tot} \over R^{2}}b\ db\,\!}$
{\displaystyle {\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 ${\displaystyle \Lambda \equiv {R \over b_{min}}}$. Using the definition of ${\displaystyle b_{min}}$, we have:

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

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

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

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

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

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

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

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

An important number to remember is ${\displaystyle 1{km \over s}=1{pc \over My}}$. 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.