Week2

Determine the trajectory

Methods: Lagrangian/Hamilton Dynamics

Lagrangian

$$\mathcal{L}=K-V=\mathcal{L}(t,q,\dot{q})$$

where $K$ is the kinetic energy, $V$ is the potential energy, and $q$ corresponds to a set of general coordinates

Euler-Lagrange equation

$$\frac{\text{d}}{\text{d}t}\left(\frac{\partial \mathcal{L}}{\partial\dot q}\right)-\frac{\partial \mathcal{L}}{\partial q}=0$$

General momentum

$$p=\frac{\partial \mathcal{L}}{\partial\dot q}$$

Hamiltonian

$$H=\sum p_i\dot q_i-\mathcal{L}=H(t,q,p)$$

and

$$\text{d}H=-\frac{\partial \mathcal{L}}{\partial t}\text{d} t-\frac{\partial \mathcal{L}}{\partial\dot q}\text{d}q+\dot q\text{d}p$$

Hamilton's canonical equations

$$\dot p=\frac{\partial \mathcal{L}}{\partial q}=-\frac{\partial H}{\partial q}\\ \dot q=\frac{\partial H}{\partial p}$$

Solve the two-body trajectory

We know that the trajectory is in a plane - two parameters (general coordinates) - $r$ and $\theta$

(Specific) Lagrangian

$$\mathcal{L}=\frac12\left(\dot r^2+r^2\dot\theta^2\right)+\frac{Gm}{r}$$

General momenta

$$p_r=\dot r,\quad p_\theta=r^2\dot\theta$$

Hamiltonian

$$H=\sum p_i\dot q_i-\mathcal{L}=\frac12\left(p_r^2+\frac{p_\theta^2}{r^2}\right)-\frac{Gm}{r}$$

Here the Hamiltonian is the (conserved) energy $E$ of the system

Hamilton's canonical equations

$$\left\{\begin{align*} \dot r&=p_r\\ \dot \theta&=\frac{p_\theta}{r^2}\\ \dot{p}_r&=\frac{p_\theta^2}{r^3}-\frac{Gm}{r^2}\\ \dot p_\theta&=0 \end{align*}\right. \Rightarrow p_\theta=L,\quad p_r=\pm\sqrt{2\left(E+\frac{Gm}{r}\right)-\frac{L^2}{r^2}}$$

$$\Rightarrow\frac{\text{d}r}{\text{d}\theta}=\frac{p_r}{p_\theta/r^2}=\frac{\pm\sqrt{2\left(E+\frac{Gm}{r}\right)-\frac{L^2}{r^2}}}{L/r^2}$$

In fact

$$\frac{L}{r}-\frac{Gm}{L}=\sqrt{2E+\frac{G^2m^2}{L^2}}\cos(\theta+\theta_0)\Rightarrow r=\frac{L^2/Gm}{1+\sqrt{1+\frac{2EL^2}{G^2m^2}}\cos\theta}$$

We let

$$p=\frac{L^2}{Gm},\quad e=\sqrt{1+\frac{2EL^2}{G^2m^2}}$$

$$\Rightarrow r=\frac{p}{1+e\cos\theta}$$

Typical Polar equations of conic sections!

1. $E<0\Rightarrow e<1$ - ellipse

2. $E=0\Rightarrow e=1$ - parabola

   $$E=0\Rightarrow\frac{v^2}{2}=\frac{Gm}{r}\Rightarrow v=\sqrt{\frac{2Gm}{r}}$$

   For parabola orbits, it is easy for us to calculate the velocity!

   If

   $$\left|\frac{2EL^2}{G^2m^2}\right|\ll1$$

   the orbit looks quite like a parabola so we can also estimate the velocity in this way

   • Example 1 - stars around a SMBH

     $v=100\text{ km/s}$, $r=1\text{ pc}$, $m=10^8 M_{\odot}$

     $$\left|\frac{2EL^2}{G^2m^2}\right|\sim10^{-17}$$

     quite parabolic!

   • Example 2 - two galaxies in a cluster

     $v=500\text{ km/s}$, $r=100\text{ kpc}$, $m=10^{11} M_{\odot}$

     $$\frac{2EL^2}{G^2m^2}\sim10^3$$

     hyperbolic!

3. $E>0\Rightarrow e>1$ - hyperbola

   Scattering

   

   The (specific) energy and angular momenta are straightforward

   $$E=\frac{v_\infty^2}{2}=\frac{v_f^2}{2},\quad L=v_\infty b$$

   And it is easy to tell that throughout the scattering,

   $$\delta \theta=2\theta_{cri}-\pi$$

   where

   $$\theta_{cri}=\pi-\arccos(a/c)=\arccos(-1/e)$$

   In this way,

   $$\delta\theta=2\arccos(-1/e)-\pi=2\arcsin(1/e),\quad e\gg1\Rightarrow\delta\theta\ll1$$

   In fact, when $e\gg1$, we have

   $$\delta\theta\sim1/e=\left(1+\frac{2EL^2}{G^2m^2}\right)^{-1/2}\sim\sqrt{\frac{G^2m^2}{2EL^2}}=\frac{Gm}{{v_\infty^2} b}=\frac{r_{inf}}{b}$$

   $r_{inf}$ is a typical impact parameter with which $\Delta \theta$ becomes significant

   $$e\sim\sqrt{2}\Rightarrow\frac{2EL^2}{G^2m^2}\sim1\Rightarrow b\sim\frac{Gm}{v^2_\infty}\equiv r_{inf}$$

Relaxation

When a star moves in a cluster, the direction of the velocity will change under the impact of all other stars. We would like to estimate the timescale in which the orbit/initial velocity of a star significantly change ($\Delta\theta$ in velocity $\sim1$)

• When the star moves in the cluster

  $$\left\langle\vec\theta_f^2\right\rangle=\left\langle\left(\vec\theta_0+\sum_{i}\delta\vec\theta_i\right)^2\right\rangle= \left\langle\vec\theta_0^2\right\rangle +2\left\langle\vec\theta_0\sum_{i}\delta\vec\theta_i\right\rangle +\left\langle\sum_{i}\left(\delta\vec\theta_i\right)^2\right\rangle +\left\langle\sum_{i\neq j}\delta\vec\theta_i\delta\vec\theta_i\right\rangle$$

  where $\langle\rangle$ stands for taking average

  Obviously, the second and the fourth terms vanish as each $\delta\vec\theta_i$ is random and independent, so we are extremely interested in the third term

  $$\left\langle\sum_{i}\left(\delta\vec\theta_i\right)^2\right\rangle=N\left\langle\left(\delta\vec\theta\right)^2\right\rangle\sim N\left(\frac{r_{inf}}{b}\right)^2\sim1$$

  where $N$ is the number of stars that have impact on the star we consider

• A more realistic model

  

  The number of stars lying within the ring $b\sim b+\text{d}b$ can be estimated with the surface/column density (number within an unit area) $\sigma$

  $$\delta N=2\pi b\text{d}b\cdot \sigma\sim2\pi b\text{d}b\cdot\frac{N_*}{\pi R_*^2}$$

  where $N*$ is the total star number in the cluster and $R*$ is the cluster's radius

  So every time the star crosses the cluster

  $$\begin{align*} \left\langle\Delta\left(\theta^2\right)\right\rangle&=\int\delta N\left\langle(\delta\theta)^2\right\rangle\\ &\sim\frac{2\pi N_*}{\pi R_*^2}\int_{b_{min}}^{R_*}\left(\frac{r_{inf}}{b}\right)^2 b\text{d}b \end{align*}$$

  but note that the approximation

  $$\delta\theta\sim\frac{r_{inf}}{b}$$

  is valid only when $b$ is large, say, $b>r_{inf}$, so we have to rewrite the integral

  $$\begin{align*} \left\langle\Delta\left(\theta^2\right)\right\rangle &\sim\frac{2N_*r_{inf}^2}{R_*^2}\left[\int_{r_{inf}}^{R_*}\frac{\text{d}b}{b}+\int_{b_{min}}^{r_{inf}}\varphi(b){\text{d}b}\right]\\ &=\frac{2N_*r_{inf}^2}{R_*^2}\left[\ln\frac{R_*}{r_{inf}}+\int_{b_{min}}^{r_{inf}}\varphi(b){\text{d}b}\right] \end{align*}$$

  It's not easy to estimate the second term. For some systems it is negligible, usually when $b{min}$ is large ($\sim R{\odot}$) and is comparable to $r_{inf}$, which is true for stellar clusters. Anyway, we simply drop that term...

  $$\left\langle\Delta\left(\theta^2\right)\right\rangle \sim\frac{2N_*r_{inf}^2}{R_*^2}\ln\frac{R_*}{r_{inf}}\equiv \frac{2N_*r_{inf}^2}{R_*^2}\ln\Lambda$$

  The typical crossing number for that angle to become significant is

  $$N_{cross}=\frac{1}{\left\langle\Delta\left(\theta^2\right)\right\rangle}$$

  And the relaxation timescale

  $$\begin{align*} T_{relax}&=t_{cross}N_{cross}\sim\frac{2R_*}{v}\cdot \frac{R_*^2}{2N_*r_{inf}^2\ln\Lambda}\\ &=\frac{R_*^3}{N_*v\ln\Lambda}\frac{v^4}{G^2m^2}\\ &\sim\frac{R_*^3}{N_*v\ln\Lambda}\frac{G^2m^2N_*^2}{G^2m^2R_*^2}\quad\left(v^2\sim\frac{GM_*}{R_*}=\frac{GmN_*}{R_*}\right)\\ &=\frac{N_*}{\ln\Lambda}\cdot\frac{R_*}{v}\\ &=\frac{N_*}{\ln N_*}\cdot\frac{R_*}{v} \quad\left(\Lambda=\frac{R_*}{r_{inf}}=\frac{R_*v^2}{Gm}\sim\frac{GmN_*R_*}{GmR_*}=N_*\right) \end{align*}$$

  $T{relax}\ll T{Hubble}\Rightarrow$ Collisional - the cluster has changed a lot since it was formed

  • Star cluster

    $$N_*\sim10^5,\ T_{cross}\sim\frac{R_*}{v}\sim\frac{1\text{ pc}}{10 \text{ km/s}}\sim10^5\text{ yr}$$

  $$T_{relax}\sim10^{8-9}\text{ yr}\ll 10^{10}\text{ yr}\sim Age$$

  $T{relax}\gg T{Hubble}\Rightarrow$ Collisionless - the cluster remains what it looked like before

  • Galaxy

    $$N_*\sim10^{11},\ T_{cross}\sim\frac{R_*}{v}\sim\frac{10\text{ kpc}}{100 \text{ km/s}}\sim10^8\text{ yr}$$

    $$T_{relax}\sim10^{17-18}\text{ yr}\gg T_{Hubble}$$