Week1

Measure the universe

From Kepler to Newton

Kepler III

P2a3P^2\propto a^3

where $a$ is the semi-major axis

Acceleration of circular motion

f=v2rf=\frac{v^2}{r}

By assuming circular orbits of planets, we have

f=v2r1rP2rr3=r2f= v^2r^{-1}\propto \frac{r}{P^2}\propto\frac{r}{r^3}=r^{-2}

which is consistent with Newton II

F=Gm1m2r2F=\frac{Gm_1m_2}{r^2}

Various astrophysical objects

Using the tool

v2r=GMr2M=v2rG\frac{v^2}{r}=\frac{GM}{r^2}\Rightarrow M=\frac{v^2r}{G}

we can measure

The sun

For the first time we are able to measure the solar mass

P=2πa3GMM=4π2a3GP2=4π2×(1.5×108 km)3G×(1 yr)2=2×1033 gP=2\pi\sqrt{\frac{a^3}{GM_{\odot}}}\Rightarrow M_{\odot}=\frac{4\pi^2a^3}{GP^2}=\frac{4\pi^2\times(1.5\times10^8\text{ km})^3}{G\times(1\text{ yr})^2}=2\times10^{33}\text{ g}

Star cluster

R1 pc, σ20 km/sM105M, ρ105 M/pc3R\sim1\text{ pc},\ \sigma\sim 20\text{ km/s}\Rightarrow M\sim10^5M_{\odot},\ \rho\sim10^5\ M_{\odot}\text{/pc}^{-3}

Dynamic mass $\sim$ Visible mass

Galaxy

R10 kpc, σ200 km/sM1011M, ρ101 M/pc3R\sim10\text{ kpc},\ \sigma\sim 200\text{ km/s}\Rightarrow M\sim10^{11}M_{\odot},\ \rho\sim10^{-1}\ M_{\odot}\text{/pc}^{-3}

Discovery of dark matter - Dynamic mass $>$ Visible mass

SMBH (Sgr A*)

R0.01 pc, σ500 km/sM2×106M, ρ2×1012 M/pc3R\sim0.01\text{ pc},\ \sigma\sim 500\text{ km/s}\Rightarrow M\sim2\times10^{6}M_{\odot},\ \rho\sim2\times10^{12}\ M_{\odot}\text{/pc}^{-3}

In such high mass density, the average distance between two stars (solar mass) is

D121/3×104 pc20 AUD\sim\frac{1}{2^{1/3}}\times10^{-4}\text{ pc}\sim20\text{ AU}

Cosmology

Typical acceleration

Estimation using hubble timescale and speed of light

av2rc210 Gyrc107cm/s/yra\sim\frac{v^2}{r}\sim\frac{c^2}{10\text{ Gyr}\cdot c}\sim10^{-7}\text{cm/s/yr}

Two-body problem

Assuming two point mass $m_1$ and $m_2$ rotating around the common CoM

m1r¨1=Gm1m2r2r13(r2r1)m2r¨2=Gm1m2r2r13(r2r1)m_1\ddot{\vec{r}}_1=\frac{Gm_1m_2}{|\vec r_2-\vec r_1|^3}\left(\vec r_2-\vec r_1\right)\\ m_2\ddot{\vec{r}}_2=-\frac{Gm_1m_2}{|\vec r_2-\vec r_1|^3}\left(\vec r_2-\vec r_1\right)

Let $\vec r=\vec r_2-\vec r_1$, we have

m1r¨1=Gm1m2r3r,m2r¨2=Gm1m2r3rm_1\ddot{\vec{r}}_1=\frac{Gm_1m_2}{r^3}\vec r,\quad m_2\ddot{\vec{r}}_2=-\frac{Gm_1m_2}{r^3}\vec r

Trajectory

In an effective one-body problem, we consider only the evolution of $\vec r$

r¨=G(m1+m2)r3rGmr3r\ddot{\vec r}=-\frac{G(m_1+m_2)}{r^3}\vec r\equiv-\frac{Gm}{r^3}\vec r
  • Test mass moves around $m_1+m_2$

Energy

E=12m1r˙12+12m2r˙22Gm1m2r=12[m1(m2m1+m2)2+m2(m1m1+m2)2]r˙2Gm1m2r=12m1m2m1+m2r˙2Gm1m2r=m1m2m1+m2[12r˙2G(m1+m2)r]\begin{align*} E&=\frac{1}{2}m_1\dot{\vec r}_1^2+\frac{1}{2}m_2\dot{\vec r}_2^2-\frac{Gm_1m_2}{r}\\ &=\frac{1}{2}\left[m_1\left(\frac{m_2}{m_1+m_2}\right)^2+m_2\left(\frac{m_1}{m_1+m_2}\right)^2\right]\dot{\vec r}^2-\frac{Gm_1m_2}{r}\\ &=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}\dot{\vec r}^2-\frac{Gm_1m_2}{r}\\ &=\frac{m_1m_2}{m_1+m_2}\left[\frac{1}{2}\dot{\vec r}^2-\frac{G(m_1+m_2)}{r}\right] \end{align*}

where

μ=m1m2m1+m2\mu=\frac{m_1m_2}{m_1+m_2}

is the reduced mass, and

12r˙2G(m1+m2)r\frac{1}{2}\dot{\vec r}^2-\frac{G(m_1+m_2)}{r}

is known as the specific energy in the effective one-body system

Angular momentum

J=m1r1×r˙1+m2r2×r˙2=[m1(m2m1+m2)2+m2(m1m1+m2)2]r×r˙=μr×r˙\begin{align*} \vec J&=m_1\vec r_1\times\dot{\vec r}_1+m_2\vec r_2\times\dot{\vec r}_2\\ &=\left[m_1\left(\frac{m_2}{m_1+m_2}\right)^2+m_2\left(\frac{m_1}{m_1+m_2}\right)^2\right]\vec r\times\dot{\vec r}\\ &=\mu\vec r\times\dot{\vec r} \end{align*}

where

r×r˙\vec r\times\dot{\vec r}

is the specific angular momentum

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