# Flat rotation curves and gravitational potential

I have been reading about spiral galaxies rotation curves, and I have a question I would like to clarify.

For example, many of them have flat rotation curves after some characteristic distance $r>r_{c}$ from the center. If this is so, then, when one computes the gravitational potential by solving:

$$v^{2}=r\dfrac{d\phi}{dr},$$

it means that at large distances, where $v(r)$ is almost constant, $\phi(r)$ increases as a logarithm of the distance.

If one keeps putting massive particles in the outer parts of the galaxy; one has two options:

1. The velocity is almost constant in a finite distance range and then, almost at the edge of the galaxy, starts to decrease.
2. The velocity keeps being constant till $r\to+\infty.$

If 1) is true, then why it is necessary dark matter to embbed the whole galaxy in a halo, since at the edge of the galaxy everything becomes keplerian again?

If 2) is true, then, since a classical massive particle can't tunnel the potential barrier created by $\phi(r)\sim \ln(r/r_{c})$, doesn't it mean that there's a radial cutoff $r_{\Lambda}$ for every particle such that it can't move beyond that orbit of radius $r_{\Lambda}$? Does't it mean that the galaxy is a self-bound and finite object?

## 1 Answer

(1) is correct: the velocity remains roughly constant until some finite radius at which point it starts to fall-off (approaching Keplerian). This occurs because the extent of the dark-matter halo extends to large radii than the galaxy---usually a factor of about 2-10 or so (but it is hard to measure) for the 'virial radius' (which measures the extend of the halo) compared to the 'half-light/effective radius' of the galaxy

• You're considering a specific type of halo, for example, NFW or Burkert. They give $v(r)\sim 1/r$ for large values of $r.$ But NFW faces many problems and Burkert is phenomenological, so there's no physics behind. then, 2) doesn't seem that crazy – Ernesto Lopez Fune Nov 11 '16 at 19:25
• @user115376 There aren't observations of rotation curves at large enough radii (100's of kpc) to show the dropoff. (2) doesn't make sense because it requires an infinite amount of mass for all halos (not to mention that particles are moving dynamically instead of quantum mechanically). At 100s of kpc, halos become truncated due to tidal interactions with the local environments, and at only slightly larger scales (Mpc) you start running into other halos and larger scale structure. – DilithiumMatrix Nov 11 '16 at 19:25
• If the total gravitational potential (baryons+dark matter) increases with distances, since a particle can't have an infinite energy, otherwise no structures form, there's a radius above which a massive particle with energy E can't move. Integrating till that cutoff radius, no infinite mass is required. – Ernesto Lopez Fune Nov 11 '16 at 19:29
• @user115376 Sure, but you have to consider particles of all energies, including those approaching zero (i.e. unbound) and beyond... The universe isn't a single, isolated, and perfectly smooth halo/potential. If you'd like more information you should look into cosmological dark matter simulations (e.g. Millennium) are the resulting density structure and profiles. – DilithiumMatrix Nov 11 '16 at 19:37
• Extreme-energy cosmic rays have energies exceeding $5\times10^{19} eV$: this energy is still finite. Charged particle will radiate. The problem in 2) arise when neutral high energy particles: photons+neutrinos. Photons, no problem, the effective potential, only accounts for gravitational lensing. Neutrinos, two cases. a) Stick to Standard Model and consider massless. The same as photons. b) Go to BSM and consider the mass. They interact weakly and non-electromagnetically, doesn't matter if they could reach "infinite energies" at the end you won't see them. This is not a contradiction. – Ernesto Lopez Fune Nov 12 '16 at 9:14