Dynamical processes affecting satellite galaxies

Galaxies that form in dark matter halos come in two varieties: those which live at the centre of the dark matter halo are called central galaxies and those which go around the centrals are called satellite galaxies. Given that satellite galaxies are orbiting the dark matter halo in which they reside they are susceptible to some dynamical effects. We will discuss these next.

Dynamical friction

Satellite galaxies fall in to the dark matter halo of a central galaxy along with their dark matter subhalos. The orbit of the satellite subhalo can be computed by using the potential of the main halo. But the halo itself is made up of dark matter particles which themselves get affected by the presence of the subhalo. This causes some rearrangement of the particles within the halo, which is localized to a region close to the subhalo. This induces extra forces on the subhalo than you would obtain if you just use a smooth potential. These were calculated by Chandrasekhar in 1943.

Imagine a big point mass particle of mass M, moving in a sea of particles with mass m, with \(M\gg m\). Let us go to the frame in which the bigger particle is at rest. The sea of particles is approaching the particle with bigger mass with a velocity \(V_0\). The particles which are at an impact parameter \(b\) get deflected by an angle which is approximately given by

\begin{equation} \alpha = \frac{2GM}{bV_0^2} \end{equation}

Due to momentum conservation, the bigger mass \(M\) will move such that

\begin{equation} M v_M = -m V_0(1-\cos(\alpha)) \sim -m V_0 2\sin^2 (\alpha/2) \sim m V_0 \alpha^2/2 \end{equation}

The particles with impact parameters between \(b\pm db/2\) will approach the bigger particle at a rate \(2\pi b db V_0 n\). The rate of the change of velocity of the bigger mass \(M\) will be given by

\begin{eqnarray} \frac{d\vec{v}_M}{dt} &\sim& -\int_{b_{\rm min}}^{b_{\rm max}} 2\pi b db V_0 n \frac{m}{2M} V_0 \left(\frac{4GM}{bV_0^2}\right)^2\\ &\sim& -16\pi \ln\Lambda \frac{\rho}{MV_0^2}G^2M^2 \end{eqnarray}

The Coulomb logarithm in this case is given by

\begin{equation} \Lambda= \frac{b_{\rm max}V_0^2}{GM}\,. \end{equation}

This simple calculation is a factor 4 off from the correct formula but has the correct behaviour with the relevant parameters (the issues are using the simple deflection formula for \(\alpha\) and the integration limits, see Binney and Tremaine for full derivation). Continuing still with our approximations for getting some insights,

\begin{eqnarray} \frac{d\vec{v}_M}{dt} &\sim& \ln\Lambda \frac{\rho}{M_{\rm sh}}\frac{G^2M_{\rm sh}^2}{V_0^2} \\ &\sim& \ln\Lambda \frac{GM_{\rm halo}}{R_{\rm halo}^3}\frac{GM_{\rm sh}}{V_0^2} \end{eqnarray}

This shows that if subhalos enter the halo with the same velocities \(V_0^2 \propto GM_{\rm halo}/R_{\rm halo}\), the subhalos which are more massive, will experience more dynamical friction. This means the higher mass subhalos which host more luminous galaxies will sink quickly towards the center.

The correct equation for the deceleration force on the particle \(M\) turns out to be

\begin{equation} M\frac{d\vec{v}_M}{dt} = - 4\pi G^2 M^2\ln \Lambda \int_0^{v_M} m 4\pi f(v_{\rm m}) v_{\rm m}^2d{v}_{\rm m} \frac{\vec{v}_M}{v_M^3} \,. \end{equation}

where \(b_{\rm max}\) is the maximum impact parameter for the system. The deceleration force is proportional to \(M^2\) and depends upon the density of the sea of particles the bigger mass \(M\) is floating in. In the case that the particle is moving fast, the force falls off as \(v_M^{-2}\). For slow moving particles, the force is proportional to \(v_{\rm M}\). Even subhalos which are on circular orbits will lose energy due to dynamical friction and spiral inwards. This is how satellite galaxies lose their orbital angular momentum and sink to the center.

  • If the subhalos are experiencing dynamical friction, where does the energy dissipate? Remember we just used gravity and our systems should conserve energy, not lose them with time.

Tidal truncation of subhalos

In the above case we treated subhalos as if they were point mass particles, but in reality they are quite extended. When the subhalos move through the bigger halo, they can undergo tidal effects.

The forces that hold the subhalo together are gravitational. If these forces are overwhelmed by the tidal forces, material will start to strip off. For a subhalo of mass \(m\) at a halo centric distance \(R\), the tidal radius will be the solution to the equation,

\begin{equation} \frac{Gm(\lt r_{\rm tide})}{r_{\rm tide}^2} = -\frac{\partial }{\partial R} \left[ \frac{GM}{R^2} \right] r_{\rm tide} \end{equation}

We can solve this equation by assuming mass profiles for \(m(r)\) and \(M(R)\). As expected for two subhalos with the same profile, the one closer in will have a smaller tidal truncation radius.

In addition there is also another resonant radius, where one can show that the orbital time around the main halo at that radius is smaller than the orbital time around the subhalo. Particles beyond this resonant orbit also get stripped. The tidal truncation radius then turns out to be smaller of the two.

For subhalos which have a density profile given by the Navarro Frenk and White profile, the quantity \(\rho r^2\) peaks near the scale radius \(r_{\rm s}\). Thus the halo get tidally stripped when its tidal radius becomes comparable to the scale radius.

The closer you get to the halo, the smaller is the truncation radius. The particles in the outskirts of the subhalo become unbound and therefore become part of the main halo. A subhalo thus loses mass rapidly at its pericentric passage, where the tidal effects can be quite strong. Observationally the tidal truncation radius at the current radius is actually related to the tidal radius at the closest passage ever in the orbit of the subhalo.