# Galaxies Lecture 20

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### Galactic Density Profiles

We now ask the question: what is the form of ${\displaystyle g_{z}}$ for intermediate z (300 pc to 30-50 kpc)? We know that:

${\displaystyle \nabla ^{2}\phi =-{\vec {\nabla }}\cdot {\vec {g}}=4\pi G\rho _{TOT}\,\!}$

Then using Stokes’ Law:

${\displaystyle \int _{V}{{\vec {\nabla }}\cdot {\vec {g}}dV}=\int _{S}{{\vec {g}}dA}=4\pi G\int {\rho _{TOT}dA}\,\!}$

Because the mass at intermediate z is dominated by stars (dark matter is still spherically symmetric around our system), we can treat all the mass as being concentrated in an infinite sheet at the midplane. Thus, this problem becomes identical to the electrostatics problem of the flux from an infinite charged sheet, and we can invoke Gauss’s Law and the symmetry of the problem to say that for a box drawn to include a section of the midplane, with faces parallel to the sheet, we have that:

${\displaystyle {\vec {g}}dA=2g_{z}A\,\!}$

where ${\displaystyle A}$ is the area of a face parallel to the sheet. Thus, we have

${\displaystyle {{\vec {g}}=-2\pi G\Sigma _{*}{\hat {z}}}\,\!}$

Where ${\displaystyle \Sigma _{*}}$ is the surface density of stars.

For large z,

{\displaystyle {\begin{aligned}\nabla _{r}^{2}\Phi &=4\pi G\rho \\{1 \over r^{2}}{dr^{2} \over dr}{d\phi \over dr}&=4\pi G\rho _{TOT}\\-g_{r}&={Gm \over r}\\\end{aligned}}\,\!}

Thus we have an overall picture of ${\displaystyle g_{z}}$ vs. z which increases linearly for small z, is flat for a long time, and then decreases as ${\displaystyle 1 \over r}$ for large z. In the flat regime we can calculate:

${\displaystyle g_{z}=2\pi G\Sigma _{*}={2\pi (6.67\cdot 10^{-8})\cdot 70\cdot 2\cdot 10^{33} \over \left(3\cdot 10^{18}\right)^{2}}=6.2\cdot 10^{-9}\,\!}$

Also, ${\displaystyle g_{z}\equiv {\ddot {z}}}$, so we can say ${\displaystyle {\ddot {z}}=4\pi G\rho _{0}z}$, which has solution ${\displaystyle z=Ae^{i\omega t}}$ for ${\displaystyle \omega ^{2}=4\pi G\rho _{0}}$. Thus, for a star close to the midplane (like our sun):

${\displaystyle \omega ={\sqrt {4\pi (7\cdot 10^{-8})\cdot 0.1\cdot 2\cdot 10^{33} \over \left(3\cdot 10^{18}\right)^{3}}}=2\cdot 10^{-15}\,\!}$

We can work out that the period of this oscillation is about 83 million years. It has been suggested that this period for midplane crossings corresponds to the period for mass extinctions on earth by comet/asteroid collisions.

For a midrange star, ${\displaystyle {\ddot {z}}=2\pi G\Sigma _{*}}$, so ${\displaystyle z=\pi G\Sigma _{*}t^{2}}$. This says that the period depends on where the star starts. We’ll say 1 kpc, which gives us the solution:

${\displaystyle {\frac {1}{4}}period=\left({z \over \pi G\Sigma _{*}}\right)^{\frac {1}{2}}=1.06\cdot 10^{9}years\,\!}$

So we’ve worked out something of the motion of an individual star, but now we’d like to work out the distribution of stars in z for a population. We’ll assume that we have an infinite plane-${\displaystyle \|}$ sheet.

{\displaystyle {\begin{aligned}{d \over dz}\left[{1 \over \rho }{d(\rho {\bar {v}}_{z}^{2}) \over dz}\right]&=-4\pi G\rho _{TOT}\\{d \over dz}\left[{1 \over \rho _{*}}{d(\rho _{*}{\bar {v}}_{z}^{2}) \over dz}\right]&=-4\pi G\rho _{*}\\\end{aligned}}\,\!}

We’ll now change variables ${\displaystyle \rho \to \Lambda }$, and ${\displaystyle z\to \zeta }$:

{\displaystyle {\begin{aligned}z&=\zeta \left({{\bar {v}}_{z}^{2} \over 4\pi G\rho _{0}}\right)^{\frac {1}{2}}\\\rho &=\rho _{0}\Lambda (\zeta )\\\end{aligned}}\,\!}

and differentiating, we have:

{\displaystyle {\begin{aligned}dz&=d\zeta \left({{\bar {v}}_{z}^{2} \over 4\pi G\rho _{0}}\right)^{\frac {1}{2}}\\{d \over d\zeta }{1 \over \Lambda (\zeta )}{d\Lambda (\zeta ) \over d\zeta }&=-\Lambda (\zeta )\\\end{aligned}}\,\!}

And this has the ultimate solution:

${\displaystyle {\Lambda (\zeta )=sech^{2}\left({\zeta \over {\sqrt {2}}}\right)}\,\!}$

where, you recall, ${\displaystyle sech(u)\equiv {2 \over e^{u}+e^{-u}}}$. From this messy equation we can infer the units of the following:

${\displaystyle {v_{z}^{2} \over 4\pi G\rho _{0}}\to cm^{2}\,\!}$

So we’ll define

${\displaystyle h_{*}\equiv \left({v_{z}^{2} \over 2\pi G\rho _{0}}\right)^{\frac {1}{2}}\,\!}$

as our scale height, giving us that

${\displaystyle {\rho =\rho _{0}sech^{2}\left({z \over h_{*}}\right)}\,\!}$

To find what z is when ${\displaystyle \rho _{*}={\frac {1}{2}}\rho _{0}}$, we’ll define ${\displaystyle u\equiv {z \over h_{*}}}$ so that:

${\displaystyle {\frac {1}{2}}\rho _{0}=\rho _{0}\left({2 \over e^{u}+e^{-u}}\right)^{2}\,\!}$

We find that this has solution ${\displaystyle u=0.8814}$, so:

${\displaystyle z_{\frac {1}{2}}=0.8814\left({{\bar {v}}_{z}^{2} \over 2\pi G\rho _{0}}\right)^{\frac {1}{2}}\,\!}$

From Binney & Merrifield, we have that in the solar vicinity ${\displaystyle v_{z}=18{km \over s}}$, and ${\displaystyle \rho _{0}=0.1{M_{\odot } \over pc^{3}}}$, so we have for a thin disk:

${\displaystyle z_{\frac {1}{2}}=304pc\,\!}$

For the thick disk, ${\displaystyle v_{z}=39{km \over s}}$, so ${\displaystyle z_{\frac {1}{2}}=1.4kpc}$. From observations, we find that ${\displaystyle z_{\frac {1}{2}}}$ does not change with radius. Note that for small heights, ${\displaystyle sech^{2}\left({z \over h_{*}}\right)}$ is close to ${\displaystyle e^{-\left({z \over h_{*}}\right)^{2}}}$. Similarly, in the asymptotic limit of ${\displaystyle z\to \infty }$, we find that

${\displaystyle \Sigma _{*}\left({z \over h_{*}}\right)=\left({2\rho _{0}v_{z}^{2} \over \pi G}\right)^{\frac {1}{2}}=69M_{\odot }pc^{-2}\,\!}$

which agrees well with observation (${\displaystyle 71\pm 6M_{\odot }pc^{-2}}$).