Galaxies – How to Interpret Line of Sight Velocity Distributions and Dispersions

dispersiongalaxiesgravitysimulationsvelocity

What I did: I am analysing some simulation snapshots of the Milky Way and it is modelled as a DM halo – bulge – disc system. I produced some mean line of sight velocity (LOSV) and dispersion of mean LOSV plots that I am attaching here. To produce these plots I simply extracted the vx,vy,vz components of the velocities of the stars and i picked z-axis as the line of sight. I overlaid a grid on the galaxy snapshot and at different inclinations(theta=0,45,90 degrees)of the galaxy I plotted the mean vz in the grid cells (vz is the z-component of the velocities). For dispersion, I just plotted the std deviation of the vz in the grid cells.

My question: I want to now understand and interpret my plots. What can I deduce from the plots for the different angles of inclinations? what can we deduce about the orbits the stars have at the bulge and disc? I mainly want to understand why the line of sight velocity distribution looks the way it does on my plots form a physics point of view..that is, due to gravity or mass distribution perhaps…please refer to equations if you want. Any help is much appreciated, thanks so much!

Best Answer

why do the mean losv and dispersion look this way?

The radial velocity profile tells you that the stars in the disk are rotating in a coherent fashion. You can calculate the rotational velocity by correcting for the inclination: $v = {v_{rad}\over \sin(\theta)}$. If you plot the velocity profile as a function of the distance $r$ form the center of the galaxy, you will obtain the famous flat rotation curve often used as proof for the existence of dark matter.

Indeed, from the velocity profile you can infer the density profile as follows. In order for a star to be in a circular orbit, its velocity must satisfy

$${v(r)^2 \over r} = -{d \Phi(r) \over d r}$$

Poisson equation in a spherical potential (the DM halo is approximately spherical and it contributes the most to the potential) is:

$${1 \over r^2} {d \over dr}(r^2 {d\Phi \over dr}) = -4\pi G \rho$$ Substituting the first equation in the second one obtains $$4\pi G \rho = {1 \over r^2} {d \over dr}(r v^2) = {1 \over r^2}(v^2 + 2rv {dv \over dr} )$$

$$\rho(r) = {v(r)^2 \over 4 \pi G r^2} (1 + 2 {r \over v(r)}{dv(r) \over dr})$$

If $v(r)$ is constant (flat profile) then $\rho(r) \propto r^{-2}$, also called singular isothermal sphere profile and this should roughly match the radial density profile of your DM halo.

In the bulge, things are different. Stars are not rotating coherently, each start follows its own orbit. This leads to a low average radial velocity (because the random velocities average to zero) and to a high radial velocity dispersion $\sigma_{rad}$. Assuming that the velocity distribution is isotropic, you can calculate the total velocity dispersion as $\sigma = \sqrt 3 \sigma_{rad}$.

From it, you can estimate the total mass of the bulge using the virial theorem.

$$2K = -U \implies M\sigma^2 = \alpha {G M^2 \over R}$$ $$M = {R \sigma^2 \over G \alpha}$$

Where $R$ can be the radius of the bulge and $\alpha$ is a geometrical factor of the order of $1$, that depends on the specific density profile of the bulge.

Additionally, you could use the $M-\sigma$ relation to estimate the mass of the central black hole, but in your case you don't mention any black hole, so I think this may not be relevant

Related Solutions

Newtonian Gravity – Velocity Distribution in Plummer’s Models and Other Mass Distributions

The Plummer model has a potential of the form

$$ \Phi(r)=-\frac{1}{\sqrt{r^2+1}} $$ (obviously ignoring all constants). Equating the above with the kinetic energy, you get $$ \frac12v_e^2+\Phi(r)=0\to v_e(r)=\sqrt{2}\left(r+1\right)^{-1/4} $$ This velocity is the maximum velocity you can have at a radius $r$, so we must have that $0\leq v\leq v_{e}$. In order to get the velocity, we need to know its distribution function; this can be found in this book. It reduces down to $$ g(v)dv\propto\left(-E\right)^{7/2}v^2dv $$ where $E=-v_e^2+\frac12v^2$. If we let $q=v/v_e$, then the above becomes $$ g(q)=\left(1-q\right)^{7/2}q^2 $$ with $0\leq q\leq1$. It then becomes a matter of using either rejection sampling (which I generally do not recommend) or the Newton root-finding method (which I usually do recommend) to find a fit. Then you just randomize the angles $\theta,\,\phi$ and get your x,y,z velocities:

vel = x*sqrt(2.0)*(1.0+r(i)*r(i))**(-0.25)
theta = acos(rand(-1.0,1.0))
phi = rand(0.0, 2.0*pi)
vx(i) = vel*sin(theta)*cos(phi)
vy(i) = vel*sin(theta)*sin(phi)
vz(i) = vel*cos(theta)

You can find more information in this great pdf book.

General Relativity – Computing and Interpreting Schwarzschild Black-Hole Metric Line Element in Cartesian Coordinates

In addition to @Andrea's answer about interpreting timelike, spacelike, and null paths, there is a computational issue with your method as well.

The metric, $g_{\alpha\beta}$, is not the same everywhere in spacetime. It is a function of radial distance from the center. In your calculation you evaluate the metric at the starting point ($\vec{r}_0$), and then use that value for the whole path.

$$s = \sqrt{\pm g_{\alpha\beta}(\vec{r}_0) \Delta r^\alpha\,\Delta r^\beta}$$ $$\Delta\vec{r} = \vec{r}_1 - \vec{r}_0,$$ where $\vec{r}$ is a four-vector with components $r^\alpha = (t,x,y,z)$. If the displacement is small, then this approximate calculation should be acceptable. The correct calculation would integrate $ds$ over the path.

$$s = \int_{\vec{r}_0}^{\vec{r}_1} ds = \int_{\vec{r}_0}^{\vec{r}_1} d\tau \sqrt{\pm g_{\alpha\beta}[\vec{r}(\tau)] \frac{d r^\alpha}{d\tau}\, \frac{d r^\beta}{d\tau}},$$ where $\tau$ is an affine parameterization of the path. For timelike paths, proper time is a common choice.

The integral accounts for the fact that $g$ takes on a different value at each point along the path.

For a spacelike path, you would use the $+$ under the square root and interpret the $s$ as the proper length separating the points. For a timelike path, you would use the $-$ and $s$ is the proper time taken.

Best Answer

Related Solutions

Newtonian Gravity – Velocity Distribution in Plummer’s Models and Other Mass Distributions

General Relativity – Computing and Interpreting Schwarzschild Black-Hole Metric Line Element in Cartesian Coordinates

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