Galaxies Lecture 25

More on Epicyclic Frequencies

We began yesterday by deriving our epicyclic frequency:

${\displaystyle \kappa ^{2}=R{\frac {d}{dR}}{\Omega ^{2}}+4\Omega ^{2}\,\!}$

and is a measure of the commensality of orbits in the rotating frame. That is, it measures the oscillation of non-circular orbits around the circular orbital frequency. For our sun, this looks like:

${\displaystyle \kappa ^{2}=-4B\Omega \,\!}$

where the Oort constants are: ${\displaystyle A=14.5{km \over s\cdot kpc}}$, ${\displaystyle B=-12{km \over s\cdot kpc}}$, and ${\displaystyle \Omega =27{km \over s\cdot kpc}}$. Using that the sun is 8 kpc we can get the velocity of our sun in its orbit. Plugging the Oort constants into our equation for the epicyclic frequency, we get that ${\displaystyle \kappa =38{km \over s\cdot kpc}}$. We also showed last time that for keplarian orbits, ${\displaystyle \kappa =\Omega }$. In this case, our star makes 1 orbit on it’s epicycle per revolution around the galactic center. Drawing this out (and remembering that the epicyclic orbits rotate around with the star as it goes around the galaxy), we find that a circular orbit becomes an ellipse. This is an ${\displaystyle m=1}$ perturbation of an orbit, using the perturbation formula:

${\displaystyle {\Omega _{p}=\Omega -{n\kappa \over m}}\,\!}$

For a solid body, we derived last time that ${\displaystyle \kappa =2\Omega }$. Tracing this out, we get a “bar-like” orbit. This is an ${\displaystyle m=2}$ perturbation. Now if we consider many of these orbits nested inside of one another,

In general, we can characterize a potential by plotting ${\displaystyle \Omega }$ vs. ${\displaystyle R}$. For solid-body rotation, we have ${\displaystyle \Omega }$ constant vs. R. Next, let’s say that ${\displaystyle \Omega ={V \over R}}$, and ${\displaystyle V}$ is constant. Then, we have ${\displaystyle \Omega \propto {1 \over R}}$. Now if we draw the epicyclic perturbations ${\displaystyle \kappa =2\Omega }$ and ${\displaystyle \kappa ={\sqrt {2}}\Omega }$, they parallel ${\displaystyle \Omega }$, and converge to 0. We can then consider rotation curves ${\displaystyle \Omega \pm {\kappa \over 2}}$ (which are resonances in our orbital system), and these rotation curves are flat.

Now if we have a constant perturbation (or pattern) ${\displaystyle \Omega _{p}}$ superimposed on a rotation curve, we inevitably have a point where ${\displaystyle \Omega =\Omega _{p}}$, where we have corotation. This, obviously, creates a resonance because the normal rotation feeds energy into the perturbation. Similarly, we have a resonance (called the Outer Lindblad resonance) at ${\displaystyle \Omega +{\kappa \over 2}}$, where rotation feeds energy in phase with the orbit (and this continues for ${\displaystyle \kappa \over 3}$ and so on, but to a lesser extent.

Let’s take an example: if we have a 2 arm spiral that moves round with pattern speed ${\displaystyle \Omega _{p}}$, then a star at radius where ${\displaystyle \Omega =\Omega _{p}}$, we have energy fed into the pattern. Farther out (or closer in), it may not be that ${\displaystyle \Omega =\Omega _{p}}$, but rather, it matches with a combination of the orbital frequency and the epicyclic frequency. This will feed energy into the epicyclic orbit, reinforcing the pattern.

There are also Inner Lindblad resonances (there can be 2 of them if the curve ${\displaystyle \Omega -{\kappa \over 2}}$ turns over). Further inside, we have ${\displaystyle V}$ going linearly with ${\displaystyle R}$, which is solid-body rotation. In this regime, all perturbed orbits rotate around with one another, giving us a bar. This can last out to the corotation point. Outward of this point, these resonances wind back (the phases of the epicycles changes), giving us spirals.

Let’s try to calculate the winding time of a pattern, using ${\displaystyle \phi _{p}=\phi _{0}}$ at ${\displaystyle t=0}$, where ${\displaystyle \phi _{p}}$ is the angle of the major axis of the pattern. The difference in rotation speeds of different points on the pattern causes a drift:

${\displaystyle \phi _{p}(t)=[\Omega -{\kappa \over 2}-\Omega _{p}]t\,\!}$

Earlier, when we did the winding problem, we had ${\displaystyle \cot \alpha =\left|R{d \over dt}{\phi }\right|}$, where ${\displaystyle \alpha }$ was the pitch angle. Plugging this in for our galaxy, we get ${\displaystyle \alpha =1.4^{\circ }}$ (as opposed to ${\displaystyle .25^{\circ }}$, which we calculated earlier) after ${\displaystyle 10^{10}yrs}$. This is a big improvment. We measure angles from ${\displaystyle 5\to 15^{\circ }}$, and if we allow for ${\displaystyle 3\cdot 10^{9}yrs}$ to pass (instead of the Hubble time), we get winding angles this large. Just judging by the rarity of grand-design spirals, we have reason to believe they might be temporal phenomena, so this estimate is not unreasonable.

The rest of what we didn’t cover today is in Chapter 6 of Binney & Tremaine.