> For the complete documentation index, see [llms.txt](https://slowdiveptg.gitbook.io/notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://slowdiveptg.gitbook.io/notes/galactic-dynamics/week2.md). # 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** ![](https://1509032923-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MPMxe8Bu9WDT3p-DA_8%2Fsync%2F82af2d08496cba0dab7d8434ffc1d9dd9268fc42.eps?generation=1608873191463710\&alt=media) 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 ![](https://1509032923-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MPMxe8Bu9WDT3p-DA_8%2Fsync%2F9c983aa39d1f073c87da5113df3bf01f8af3051c.eps?generation=1608873191757528\&alt=media) 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} $$