5.1. Collisionless vs. collisional dynamics¶
Galaxies consist of large numbers \(N\) of stars and most probably even larger numbers of dark matter particles. For a typical galaxy like the Milky Way, the number of stars is \(N \approx 10^{11}\). One might think that the strong, gravitational interactions between individual stars are important drivers of the evolution of galaxies. In this case we would say that the dynamics is collisional (because collisions, even if they are not direct physical collisions, but simply near misses, are important). However, as we will see in this section, this is not the case and close encounters are not important. Galaxies are collisionless. More generally, we refer to effects as being “collisional” if they result from the fact that stellar systems are made up of a finite number of stars or gas clouds (or dark matter if the dark-matter mass is high).
The first thing of note is that the gravitational force drops off as \(1/r^2\) and is therefore a long-range force (a short-range force would be one where the force drops off exponentially or similarly fast). To a first approximation the density in galaxies is roughly constant and, as we demonstrated in Chapter 2.1, in a uniform medium each shell surrounding a given point contributes the same amount of gravitational force. Encounters between individual stars must be well within 1 pc to have a large effect and there are many more shells beyond 1 pc than within it. Therefore, the gravitational force on a star in a galaxy is dominated by the sum of the forces of distant stars rather than by the force from the nearest stars. Because the sum is over the contributions of \(\approx 10^{11}\) stars, this gravitational force is smooth in space and time and stars follow smooth orbits.
In more detail, we can compute the relaxation time as the time over which the combined effect of many close encounters has changed a star’s velocity by \(100\,\%\). To estimate the relaxation time of galactic systems, we will approximate them as consisting of \(N\) equal-mass bodies with mass \(m\) (for a total mass \(M = Nm\)) that are uniformly distributed over a sphere with radius \(R\). To first approximation, the effect of a two-body encounter is to deflect the orbits of the participating bodies and we can compute this deflection using the impulse approximation. In this approximation, we work in the coordinate frame centered on the first body and approximate the trajectory of the second body for the purpose of computing the deflection as following a straight line at constant speed \(v\) as shown in Figure 5.2.
[6]:
from galpy.potential import KeplerPotential
from galpy.orbit import Orbit
kp= KeplerPotential(normalize=1.)
r_init= 5.
v_init= 20.*kp.vcirc(r_init)
phi_init= numpy.pi-0.2
o= Orbit([r_init,v_init*numpy.cos(phi_init),
-v_init*numpy.sin(phi_init),phi_init])
o.integrate(numpy.linspace(0.,1.2,101),kp)
# Location of first mass
figure(figsize=(5,3.375))
o.plot(yrange=[-0.1,1.25],gcf=True)
plot([0.],[0.],'ko',ms=10.)
# Impule approx. trajectory
axhline(o.y(),ls='--')
# Impact parameter
plot([0.,0.],[0.,o.y()])
text(0.2,0.7,r'$b$',fontsize=20.,
horizontalalignment='left',
verticalalignment='center')
# theta line and label
force_arrow_t= 0.15
plot([0.,o.x(force_arrow_t)],[0.,o.y(force_arrow_t)])
text(-0.4,0.3,r'$\theta$',fontsize=20.,
horizontalalignment='center',
verticalalignment='center')
# Force arrow
gca().annotate("",
xy=(o.x(force_arrow_t),o.y(force_arrow_t)-0.4),
xytext=(o.x(force_arrow_t),o.y(force_arrow_t)),
arrowprops=dict(arrowstyle="->"))
text(o.x(force_arrow_t)-0.1,o.y(force_arrow_t)-0.2,r'$F_\perp$',
fontsize=20.,
horizontalalignment='right',
verticalalignment='center')
# x annotation
text(o.x(2.5*force_arrow_t),o.y()+0.05,
r'$x$',
fontsize=20.,
horizontalalignment='center',
verticalalignment='bottom')
gca()._frameon= False
gca().xaxis.set_visible(False)
gca().yaxis.set_visible(False);
Figure 5.2: The geometry of a two-body encounter.
Thus, we approximate the true trajectory (the blue curve, from left to right) as following the dashed straight line. This line makes a closest approach to the first body (the big dot) with a relative distance given by the impact parameter \(b\). We compute the deflection as the change in the component of the velocity perpendicular to the straight-line trajectory, \(\delta v_y\) in Figure 5.2. Following Newton’s second law (Equation 3.2), this change can be computed as \begin{equation}\label{eq-impulsedv-int} m\,\delta v_y = \int_{-\infty}^\infty \mathrm{d} t\, F_\perp\,, \end{equation} where \(F_\perp\) is the perpendicular component of the force, in this case given by \begin{equation} F_\perp = -{Gm^2\over x^2+b^2}\,\cos \theta\,, \end{equation} where \(\theta\) is the angle between the two lines indicated above. From trigonometry, we have that \(\cos \theta = b/\sqrt{x^2+b^2}\) and from the impulse approximation we have that \(x = vt\), such that the deflection becomes \begin{align} \delta v_y & = -Gm\,b,\int_{-\infty}^\infty \mathrm{d} t\, {1 \over (x^2+b^2)^{3/2}}= -{Gm \over b\,v},\int_{-\infty}^\infty \mathrm{d} \tilde{x}\, {1 \over (\tilde{x}^2+1)^{3/2}} = -{2\,Gm \over b\,v}\,,\label{eq-impulse-dv} \end{align} where we used \(\tilde{x} = x/b\) and that \(\int_{-\infty}^\infty \mathrm{d} \tilde{x}\, {1 / (\tilde{x}^2+1)^{3/2}} = 2\). This is an often-used formula in different contexts and an easy way to remember it is to approximate the integral over force in Equation (5.1) as the force at the impact parameter acting over the time \(\Delta T = 2b/v\) that it takes to cross from \(x=-b\) to \(x=b\) \begin{equation} \int_{-\infty}^\infty \mathrm{d} t\, F_\perp \approx \Delta T \times F_\perp(b) = {2b \over v}\times -{GM \over b^2} = -{2\,Gm \over b\,v}\,. \end{equation}
This approximation works quite well in practice. For instance, for the example shown above, which has a noticeable deviation from the impulse approximation, we can compute the exact \(\delta v\) and the approximation (which we compute using that \(-GM/b = \Phi[b]\)); the difference is \(\approx 1.5\%\):
[7]:
print(
f"dv_exact = {o.vy(1.2):.5f}, "
f"dv_approx = {2*kp(o.y(),0.)/o.vx():.5f}, "
f"difference = {(2*kp(o.y(),0.)/o.vx()-o.vy(1.2))/o.vy(1.2)*100.:.2f}%"
)
dv_exact = -0.22170, dv_approx = -0.22510, difference = 1.54%
Now consider a “subject” star passing through our galaxy of size \(R\). Crossing the galaxy, our “subject” star encounters approximately \(N/(\pi R^2) \times 2\pi\,b\mathrm{d}b = 2N\,b\mathrm{d}b/R^2\) stars in a shell of impact radius \(b\) (this is the surface density of stars \(N/(\pi R^2)\) times the area of the shell \(2\pi\,b\mathrm{d}b\)). If the impacts are at random angles, the velocity change from this shell will exhibit a random walk, leading the mean squared velocity to increase by \begin{equation} \langle \Delta v^2 \rangle(b) \approx (\delta v)^2 \frac{2N}{R^2}\,b\mathrm{d}b = \frac{8 G^2\, m^2\, N}{v^2\,R^2} \frac{\mathrm{d}b}{b}\,. \end{equation} Integrating over all \(b\) gives \begin{equation} \langle \Delta v^2 \rangle\approx 8N\left(\frac{Gm}{Rv}\right)^2\,\ln \Lambda\,, \end{equation} where \(\Lambda = b_{\mathrm{max}}/b_{\mathrm{min}}\), the minimum and maximum impact parameter. The minimum can be set to the impact parameter where the change in velocity from a single kick is equal to the velocity, \(v \approx \delta v = 2Gm/(b_{\mathrm{min}}v)\) or \(b_{\mathrm{min}} = 2Gm/v^2\) (we will give a better justification of this in Chapter 19.4.1). The maximum can be set to the size of the system, \(b_\mathrm{max} = R\). Then \(\Lambda = Rv^2/(2Gm)\).
We can now define and compute the relaxation time. The relaxation time is defined as the typical time it takes for individual encounters to change a star’s velocity (squared) by 100%; because the specific process that we are considering concerns interactions between two bodies, this process is known as two-body relaxation. First, we compute the number of crossings necessary, from the equation for \(\langle \Delta v^2 \rangle\) per crossing: \begin{equation} n_\mathrm{relax} \approx v^2 / \langle \Delta v^2 \rangle = v^4 R^2 / (8N\,G^2\,m^2\,\ln \Lambda)\,. \end{equation} We can simplify this expression by considering the circular velocity \(v_c\) at \(R\) as a typical velocity \(v\). In Chapter 2.3, we showed that \(v_c^2 = GM(<r)/r\) such that in this case \(v^2 = v_c^2=GNm/R\) or \(v^4 R^2 / (G^2\,m^2) = N^2\). We can also simplify the expression for \(\Lambda = Rv^2/(2Gm)\) for this typical velocity: \(\Lambda = N/2 \approx N\). Therefore, for a typical velocity we get \begin{equation} n_\mathrm{relax} \approx \frac{N}{8\,\ln N}\,. \end{equation} Each crossing takes approximately a dynamical time \(t_\mathrm{cross} = 2\pi R/v\) and the relaxation time is therefore \begin{equation}\label{eq-spherequil-trelax} t_\mathrm{relax} = n_\mathrm{relax} \times t_\mathrm{cross} \approx \frac{N}{8\,\ln N}\,\frac{2\pi R}{v} = \frac{N}{8\,\ln N}\,t_\mathrm{dyn}\,. \end{equation}
For galaxies, \(N \approx 10^{11}\) and \(t_\mathrm{cross} \approx 100\) Myr. Therefore,
[8]:
trelax= 10.**11./8./numpy.log(10.**11)*100*u.Myr
print(trelax.to(10**10*u.Myr))
4.935164567082407 1e+10 Myr
or \(t_\mathrm{relax} \approx 5\times 10^{10}\) Myr, much much much larger than the age of the Universe (\(\approx 10^4\) Myr). There’s enough room in this argument for the effect of simplifying assumptions that any galaxy has a two-body relaxation time much larger than the age of the Universe. Therefore, we can safely say that galaxies are collisionless.
The situation in disk galaxies is slightly more complex than the above argument makes it appear. Even when only considering two-body relaxation, disks have shorter relaxation times than elliptical galaxies for the same \(N\). This is because the stars in disks are (a) packed closer together for the same size \(R\) (as they stay close to the mid-plane) and (b) move at smaller relative velocities with respect to each other, causing velocity kicks by Equation (5.3) to be larger. In fact, the two-body relaxation time in a razor-thin disk is \(t_\mathrm{relax} \approx t_\mathrm{dyn}\). But even after accounting for the thickness of a disk galaxy and its (small) internal velocity dispersion, one can show that the two-body relaxation time is longer than the age of the Universe for disk galaxies. Collisional effects, however, can also occur due to collective effects, for example, the tendency of disks to form spiral structure. Spiral structure can appear without any outside influence, caused simply by the finite number of stars and gas clouds that form a disk galaxy; spiral structure influences the orbits of stars in disks, which are then affected by a collisional effect. Similarly, stars in disks get deflected on their orbits by giant molecular clouds, which is also a direct collisional effect that changes the distribution function of stars. Most of these effects act on time scales larger than the dynamical time, thus, over a few dynamical times we can still mostly consider disks to be collisionless. We will come back to this question in later chapters.
The two-body relaxation argument fully breaks down for regions of high stellar density, such as within tens of parsecs of the centers of galaxies or in the cores of globular clusters. As we saw in Chapter 1.1, globular clusters have about \(N = 10^5\) stars in a region of a few pc. Their typical velocity is therefore \(v \approx 5\,\mathrm{km\,s}^{-1}\) and their crossing time \(t_\mathrm{cross} \approx 1\) Myr. Therefore,
[9]:
trelax= 10.**5./8./numpy.log(10.**5)*1*u.Myr
print(trelax.to(u.Gyr))
1.0857362047581296 Gyr
or \begin{equation} t_{\mathrm{relax,GC}} \approx \frac{10^5}{40\,\ln 10}\,1\,\mathrm{Myr} \approx 1\,\mathrm{Gyr}\,. \end{equation} Two-body relaxation therefore plays a large role in the evolution of globular clusters. However, in these dense environments, it is still the case that \(t_\mathrm{relax} \gg t_\mathrm{dyn}\) and over a few dynamical times, we can still approximate the dynamics as being collisionless. In the context of equilibrium models, this means that we can apply the tools from collisionless modeling to the dynamics of galactic nuclei and globular clusters to investigate their instantaneous equilibrium state, and relaxation slowly changes this (collisionless) equilibrium state through the slow effect of two-body interactions.
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