# What happens at the edge of a galaxy

What happens at the edge (around the optical radius) of a galaxy when it has a flat rotation curve? After some length scale: does the velocity start to decrease or is there a phase-transition-like that keeps the galaxy being finite in size?

• good question : the Milky Way frontiers evolved a lot with the observations and there are a lot of publications about difficulty to define the frontiers, particularly when a fusion with a little galaxy is still in progress. Otherwise, it is a matter of density, as usual, when seen from long distances ... No, there is not a membrane with an inside and an outside – user46925 Sep 12 '16 at 13:01
• thanks for the comments... i read this about the Milky Way, but despites it is the nearest galaxy, i.e. we live in it, measuring its rotation curve it's a nightmare... in another distant galaxy how would it look like? near the edge of the galaxy one could have a keplerian fall of the velocity or some mechanism within the galaxy forbids higher radii orbits ? – user115376 Sep 12 '16 at 13:07
• I forgot to say Milky Way and Andromedia ... Mainly a matter of density. These subjects are controversial because some say that Dark matter is more dense peripheraly. But I don't think that the few velocities measured are the criterium of defining a galaxy frontier. The density falls quickly ; as we can see on images with distant galaxies in the same line of sight, galaxies appear as individual entities. – user46925 Sep 12 '16 at 13:15
• So it means that it is unclear what happens at the edge? – user115376 Sep 12 '16 at 13:26
• Indeed, according to what we were discussing, yesterday I went to a conference with Dr. Asimina Arvanitaki, and she answered the question by saying that there are no compelling data to be sure what happens at the edge, so this is still a mystery. – user115376 Sep 13 '16 at 9:53

Eventually the rotation curves do decrease to smaller velocities. The problem is that this usually happens well outside of the characteristic radii (like the "half-light" radius) after which point it is very difficult to get good measurements.

The velocity as a function of radius is roughly, $$v_{rot} \approx \left(\frac{G M_{enc}(r)}{r}\right)^{1/2}$$
where $M_{enc}$ is the total mass enclosed out to that radius $r$. The distribution of dark matter extends to much larger radii than that of stars, so the mass enclosed tends to increase roughly as fast as the radius does, until you get out to near the "virial radius" (well outside the half-light radius). Eventually the mass enclosed becomes almost constant (because the density gets lower and lower at large radii) and the velocity starts to decrease as $1/r$.

If a star passes the edge with a velocity greater than the local escape velocity $\sqrt{\text{2GM/r}}$ it will leave the galaxy, and if the velocity is smaller it will catch an elliptic orbit. If the velocity is exactly the orbital velocity $\sqrt{\text{GM/r}}$ the orbit will be circular. For $\text{M}$ just use the mass inside the shell (see the shell theorem and the density profiles). So at farther distances where the density sinks the velocities of orbiting objects start to decrease, since all objects travelling with higher velocities would get ejected from the galaxy.

• how can you prove that the gravitational potential near the edge should be the newtonian one? this is the key of the question... – user115376 Sep 12 '16 at 19:41
• the radius of the galaxy is very large compared to its Schwarzschild radius and the orbital velocities are much smaller than the speed of light, so Newton does the job. If you want to take relativity into account divide the orbital velocities by √(1-rs/r), but that will make almost no difference at this distance. – Yukterez Sep 12 '16 at 21:15
• well @Симон Тыран how do you prove it? this is only the case of a Schwarzschild space-time, while in a galaxy there are much masses so Schwarzschild can't be – user115376 Sep 12 '16 at 21:19
• your question was about the edge of the galaxy with mass inside and vacuum outside; then use the outer Schwarzschild solution. if you want the trajectories inside the shell use Newton's shell theorem for practical purposes and the inner Schwarzschild solution if you want to do the extra job de.wikipedia.org/wiki/Schwarzschild-Metrik#Innere_L.C3.B6sung – Yukterez Sep 12 '16 at 21:22
• According to your comment: if the Schwarzschild solution is the key, then, for example, for the Milky Way, whose mass is ‎around 0.8–1.5×10^12 M☉, the associated Schwarzschild radius will be around 2.-3.75×10^12 km. The center of the Milky Way will be dark as a typical black hole, which is not observed. – user115376 Sep 13 '16 at 9:49

Regarding my question, I went to a conference with Dr. Asimina Arvanitaki, and she answered by saying that there are no compelling data to be sure what happens at the edge, so this is still a mystery.