Galaxies Lecture 10

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

A question that arose when Hubble began classifying elliptical galaxies by their eccentricity is whether the observed “flattening” is the result of the galaxy’s rotation. We can characterize this by comparing the energy of rotation to the energy of gravitation:

{\displaystyle {\begin{aligned}E_{rot}&=E_{grav}\\{\frac {1}{2}}Iw^{2}&={GM^{2} \over R}\\{\frac {1}{2}}MR^{2}{v_{c}^{2} \over R^{2}}&={\frac {1}{2}}M\sigma _{v}^{2}\\{v_{c} \over \sigma _{v}}&=1\\\end{aligned}}\,\!}

Experimentally, we have recently measured ${\displaystyle {{v_{c} \over \sigma _{v}}<1}}$, so we know that elliptical galaxies are not rotationally supported. This calculation was just for stars. If we were to do this for gas, we would have to include the effects of pressure. However, most galaxies are not pressure supported, so this result should hold. Spiral galaxies are rotationally supported, with a ratio of about 10.\

Now we will talk about two kinds of systems: those with power law density distributions and disks.

Systems with Power Law Density Distributions

We’ll say that the density is given by:

${\displaystyle \rho (r)=\rho _{0}\left({r_{0} \over r}\right)^{\alpha }\,\!}$

We would like to derive expressions for the surface density ${\displaystyle \Sigma (R)}$, ${\displaystyle v_{c}}$, and ${\displaystyle \Phi (R)}$. ${\displaystyle R}$ is the distance from the center along the surface that we see. If ${\displaystyle x}$ is the vertical distance along the line of sight to the center of a galaxy, we can say:

{\displaystyle {\begin{aligned}dx\cos \phi &=rd\phi \\{x \over R}=\tan \phi \\R=r\cos \phi \\x=r\sin \phi dx=r{d\phi \over \cos \phi }\\\end{aligned}}\,\!}

Now ${\displaystyle r=(x^{2}+R^{2})^{\frac {1}{2}}}$, so ${\displaystyle \phi (x)=\phi _{0}r_{0}^{\alpha }(x^{2}+R^{2})^{-{\alpha \over 2}}}$. We then have:

{\displaystyle {\begin{aligned}\Sigma (R)&=2\int _{0}^{\infty }{\rho (x)dx}\\&=2\rho _{0}r_{0}^{\alpha }\int _{0}^{\pi \over 2}{{R^{(1-\alpha )} \over \cos ^{(}1-\alpha )\phi }{d\phi \over \cos \phi }}\\\end{aligned}}\,\!}

This yields:

${\displaystyle {\Sigma (R)={2\rho _{0}r_{0}^{\alpha } \over R^{\alpha -1}}\int _{0}^{\pi \over 2}{\cos ^{(\alpha -2)}\phi d\phi }}\,\!}$

For the special case that ${\displaystyle \alpha =2}$, we have:

${\displaystyle \Sigma (R)={2\rho r_{0}^{\alpha } \over R^{\alpha -1}}\int _{0}^{{\pi \over 2}d\phi }\,\!}$
${\displaystyle {\Sigma (R)={\pi r_{0}^{2}\rho _{0} \over R}}\,\!}$

This is called a singular isothermal sphere. In the more general case:

{\displaystyle {\begin{aligned}\int _{0}^{\pi \over 2}{\cos ^{h}\phi d\phi }&={{{\sqrt {\pi }} \over 2}\Gamma \left({n+2 \over 2}\right) \over \Gamma \left({n \over 2}+1\right)}\\\Sigma (R)&={\rho _{0}r_{0}^{\alpha } \over R^{\alpha -1}}{(-{\frac {1}{2}})!\left({\alpha -3 \over 2}\right)! \over \left({\alpha -2 \over 2}\right)!}\\M(R)&={4\pi \over 3-\alpha }\rho _{0}r_{0}^{\alpha }r^{3-\alpha }\\v_{c}^{2}(R)&={4\pi \over (3-\alpha )}G\rho _{0}r_{0}^{\alpha }r^{(}2-\alpha )\\\end{aligned}}\,\!}

Note that for a singular isothermal sphere, ${\displaystyle v_{c}(R)}$ is constant.

{\displaystyle {\begin{aligned}v_{e}^{2}({r^{\prime }})2\int _{0}^{\infty }{{GM({r^{\prime }}) \over {r^{\prime }}^{2}}d{r^{\prime }}}&={8\pi G\rho _{0}r_{0}^{\alpha } \over (3-\alpha )(\alpha -2)}r^{2-\alpha }\\&={2v_{c}^{2} \over \alpha =2}\\\end{aligned}}\,\!}

This requires ${\displaystyle \alpha >2}$. Thus, we have:

${\displaystyle \left({v_{e} \over v_{c}}\right)^{2}={2 \over \alpha -2}\,\!}$

Disks

For disks, we need to write a Poisson Equation in cylindrical coordinates.

${\displaystyle {1 \over R}{\partial \over \partial R}\left(R{d\Phi \over dR}\right)+{\partial ^{2}\Phi \over \partial z^{2}}=4\pi G\rho \,\!}$

Using seperation of variables, we’ll say ${\displaystyle \Phi (R,z)=J(R)Z(z)}$. Then

${\displaystyle {1 \over J(R)}{d \over dR}\left(R{dJ \over dR}\right)=-{1 \over Z(z)}{d^{2}Z \over dz^{2}}\,\!}$

This last looks like a harmonic oscillator equation with constant ${\displaystyle k={\sqrt {{1 \over J(R)}{d \over dR}\left(R{dJ \over dR}\right)}}}$:

{\displaystyle {\begin{aligned}{d^{2}Z \over dz^{2}}&=-k^{2}Z(z)\\Z(z)&=Se^{\pm kz}\\\end{aligned}}\,\!}

We can then work out that:

${\displaystyle \Phi (R,z)=e^{\pm kz}J_{0}(kR)\,\!}$

{\displaystyle {\begin{aligned}\Sigma (R)&=\Sigma _{0}e^{-\left({R \over R_{d}}\right)}\\\Phi (R,z)&=-2\pi G\Sigma _{0}R_{d}^{2}\int _{0}^{\infty }{{J_{0}(kR)e^{-k\left|z\right|} \over (1+(kR_{d})^{2})^{\frac {3}{2}}}dk}\\\Phi (R,0)&=-4\pi G\Sigma _{0}R\left[I_{0}\left({R \over R_{d}}\right)K_{1}\left({R \over R_{d}}\right)-I_{1}\left({R \over R_{d}}\right)K_{0}\left({R \over R_{d}}\right)\right]\\v_{c}^{2}(R)&=R{d\phi \over dR}=4\pi G\Sigma _{0}R_{d}\left({R \over R_{d}}\right)^{2}\left[I_{0}\left({R \over R_{d}}\right)K_{0}\left({R \over R_{d}}\right)-I_{1}\left({R \over R_{d}}\right)K_{1}\left({R \over R_{d}}\right)\right]\\\end{aligned}}\,\!}

Where ${\displaystyle K_{n}}$’s are the modified Bessel Functions.

{\displaystyle {\begin{aligned}I_{\nu }(\left({R \over R_{d}}\right))&=e^{-i\pi \nu \over 2}J_{\nu }(i\left({R \over R_{d}}\right))\\K_{\nu }(\left({R \over R_{d}}\right))&=\lim _{\nu \to 0}{\pi \over 2}{I_{-\nu }(\left({R \over R_{d}}\right))-I_{\nu }(\left({R \over R_{d}}\right)) \over \sin \nu \left({R \over R_{d}}\right)}\\\end{aligned}}\,\!}

Thus we have:

{\displaystyle {\begin{aligned}I_{0}(\left({R \over R_{d}}\right))&=J_{0}(i\left({R \over R_{d}}\right))\\I_{1}(\left({R \over R_{d}}\right))&=-iJ_{1}(i\left({R \over R_{d}}\right))\\I_{-1}(\left({R \over R_{d}}\right))&=iJ_{1}(i\left({R \over R_{d}}\right))\\\end{aligned}}\,\!}

To remind us of what these modified Bessel functions are:

{\displaystyle {\begin{aligned}K_{0}(x)&\approx -\ln(x)\\K_{1}(x)&={\pi \over 2}{iJ_{1}(ix)+iJ_{1}(ix) \over \sin x}\\&=\pi {J_{1}(ix) \over \sin x}\\&J_{\nu }(x)\sum _{k=0}^{\infty }{{(-1)^{k} \over k!(\nu +k)!}\left({\frac {1}{2}}x\right)^{\nu +2k}}\\\end{aligned}}\,\!}

For large x, ${\displaystyle I_{\nu }(x)\to {e^{2} \over {\sqrt {2\pi x}}}}$ and ${\displaystyle K_{\nu }(x)\to {\sqrt {\pi \over 2x}}e^{-x}}$.\For small x, ${\displaystyle I_{\nu }(x)\to {1 \over x!}\left({\frac {1}{2}}x\right)^{\nu }}$, and ${\displaystyle K_{\nu }(x)\to {(\nu -1)! \over 2}\left({\frac {1}{2}}x\right){-\nu }}$.\Thus, ${\displaystyle J_{\nu }(x)={\sqrt {2 \over \pi x}}\cos(z-{\frac {1}{2}}\nu \pi -{\frac {1}{4}}\pi )}$.\Plugging this into the equations we’ve derived, we find that as ${\displaystyle \left({R \over R_{d}}\right)\to 0}$, a near-zero approximation gives us:

${\displaystyle {v_{c}^{2}(R)=4\pi G\Sigma _{0}R_{d}\left(\left({R \over R_{d}}\right)\right)^{2}\left[\ln {2R_{d} \over R}-{\frac {1}{2}}\right]}\,\!}$

We can graph this, but it looks vaguely like a random distribution with a left bound. We have now developed the tools we need to be able to compute ${\displaystyle v_{c}^{2} \over R}$ for the sphere and the disk, and since ${\displaystyle \Phi _{TOT}=\Phi _{1}+\Phi _{2}+...}$, we can compute the total ${\displaystyle v_{c}}$ of most galaxies. We can plot ${\displaystyle v_{c}}$ versus radius and observe that up to ${\displaystyle R_{c}}$, ${\displaystyle v_{c}}$ is dominated by the bulge, between ${\displaystyle R_{c}}$ and ${\displaystyle 2R_{d}}$ we see a superposition of the two influences, and beyond ${\displaystyle 2R_{d}}$, ${\displaystyle v_{c}}$ is dominated by the disk.

${\displaystyle v_{c}^{2}(r)={4\pi \over (3-\alpha )}G\rho _{0}r_{0}^{\alpha }r^{2-\alpha }\,\!}$

However, this predicts that as ${\displaystyle r}$ increases, ${\displaystyle v_{c}\to 0}$. We know this is not the case: in fact, ${\displaystyle v_{c}\to constant}$. This tells us there must be some matter there (which we are not seeing), that is behaving like an isothermal sphere–that is, it must have a ${\displaystyle {1 \over r}}$ density profile.