3
$\begingroup$

I found the image below in Space.com's article This 3D Color Map of 1.7 Billion Stars in the Milky Way Is the Best Ever Made, although it is not the map mentioned in the title.

The caption for this image reads:

This radial velocity image shows the movement of 7 billion stars. The colors run from blue (stars moving at 50 km/s toward us) to red (stars moving 50 km/s away from us). The white color shows when, on average, the stars are not moving in the line of sight with respect to us. Stars lagging behind as they orbit the center of the Milky Way appear to be traveling away from us, and those speeding up appear to be traveling toward us. Credit: ESA/Gaia/DPAC

If you imagine a band along the galactic equator the dominant velocity shows two positive and two negative "peaks", with a zero crossing in the direction of the galactic center.

Purely for fun I wanted to see if I could reproduce this behavior with a simple calculation based on a 2D calculation assuming circular motion and a radial density distribution $\rho(r)$ which I could then use to figure out a rotational velocity distribution $v(r)$, bun I swiftly realized that I have no idea what the density profile would look like.

  1. For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane?

  2. For spherically symmetric distributions, Newton's Shell theorem allows one to treat all mass inside a sphere defined by an orbit's radius as if it were at the center, and to ignore all mass in the shell outside of that radius. Is there anything like an analog to this for a radial distribution within a plane?

apparent velocity of the Milky Way seen from Earth (Gaia)

$\endgroup$
  • $\begingroup$ It works the other way around. You get the density distribution by looking at the velocity data. $\endgroup$ – Rob Jeffries Apr 26 '18 at 6:31
  • 1
    $\begingroup$ @RobJeffries I'm not trying to be an astronomer, but instead just trying to do a mathematical exercise to gain some better insight, and so I can work in any direction I like. I want to work a simple problem in the direction indicated. $\endgroup$ – uhoh Apr 26 '18 at 6:34
  • 1
    $\begingroup$ You might want to check out this question which doesn't specifically address your question here, but may give you a good launching point. $\endgroup$ – zephyr Apr 26 '18 at 13:04
  • $\begingroup$ @zephyr that's a helpful and also quite an interesting item to read, thank you for the suggestion! $\endgroup$ – uhoh Apr 26 '18 at 13:09
1
$\begingroup$

For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane?

The simplest example astronomer's work with is the radial density profile of a Single Isothermal Sphere (SIS). It is called this because it is spherically symmetric (and thus applicable to a 2D plane for your purposes) and all objects orbit with the same velocity (and thUs have the same "temperature", hence isothermal). The density profile takes the form:

$$\rho(r) = \frac{v^2}{4\pi Gr^2}$$

where $v$ is the rotational velocity. Note that you may see other formulations which use $\sigma_v$ rather than $v$. In this case they're using the velocity dispersion which is slightly different than the rotational velocity.

Other, more realistic density profiles have been found by running simulations of the Universe and matching functional equations to the density profiles of the resulting galaxies. Such popular results are the NFW profile and the Einasto profile.

The NFW profile is a two-parameter function, given by

$$\rho(r) = \frac{\rho_0}{\frac{r}{R_S}\Big(1+\frac{r}{R_S}\Big)^2}$$

where $\rho_0$ and $R_S$ and two, halo-dependent parameters.

The Einasto Profile is again a two-parameter model given by

$$\rho(r) \propto \exp(-Ar^\alpha)$$

where $A$ and $\alpha$ are configurable parameters.

For spherically symmetric distributions, Newton's Shell theorem allows one to treat all mass inside a sphere defined by an orbit's radius as if it were at the center, and to ignore all mass in the shell outside of that radius. Is there anything like an analog to this for a radial distribution within a plane?

The Shell theorem for gravity does not extend to a 2D ring. However, I will say that when talking about the orbits of stars in galaxies, the mass of stars outside a star's orbit are generally considered negligible. The primary reason for this is that it is Dark Matter which comprises most of the mass of a galaxy and contributes the most to defining a star's orbit in a galaxy. The Dark Matter halo is often assumed to be spherically symmetric, in which case Newton's Shell theorem does apply and the mass you're concerned with in determining a star's orbit, is the mass of the Dark Matter halo interior to the star's orbit.

$\endgroup$
  • $\begingroup$ Thanks, I am curious about the Shell Theorem; does it really apply to a 2D cylindrically symmetric distribution in our 3D universe as opposed to a 2D (theoretical) universe? I haven't found a source for this yet. $\endgroup$ – uhoh Apr 26 '18 at 13:36
  • $\begingroup$ -1 Pending proof that the shell theorem applies to a disc (or even non-spherical ellipsoidal distributions). I believe that Newton's "ring theorem" would apply in 2D if gravity diminished as $r^{-1}$. $\endgroup$ – Rob Jeffries Apr 26 '18 at 16:21
  • $\begingroup$ @RobJeffries Now that I look at it, I think you're right. I'll edit my answer. $\endgroup$ – zephyr Apr 26 '18 at 16:28
  • $\begingroup$ Ok, but is most of the mass, say interior to the Sun's orbit dark matter? If not, then you have a complicated non-spherical distribution of mass and the shell theorem does not apply. Isn't this why stellar orbits are not closed? $\endgroup$ – Rob Jeffries Apr 27 '18 at 6:37
0
$\begingroup$

@Rob Jeffries mentioned that "You get the density distribution by looking at the velocity data." I also believe this is what you are looking for, so I will give some calculation details.

Assume spherical symmetry, and circular motion, the gravity equates to the circular motion as $$ \alpha \frac{GMm}{R^2} = \frac{mv^2}{R} $$ where $G$ is gravitational constant, $M$ is the enclosed mass at the radial distance $R$, $v$ is the tangential (not radial) velocity, $m$ is a test mass, and $\alpha$ is a constant for effective potential, which depends on the assumed shape of potential. Then, we can express the mass $M$ as the density profile $\rho$. With the spherical symmetry, the profile $\rho \propto M R^{-3}$. Therefore, $$ \alpha' G \rho R^2 = v^2 . $$

Since observationally we can construct the rotation curve which is $v = f(R)$, the density profile is then a function depending only on $R$: $\rho = g(R)$, i.e., the radial mass distribution.

Some notes include i) the mass $M$ includes dark matter; ii) $v$ is the tangential velocity, not the radial velocity as presented in the figure you mentioned.

$\endgroup$
  • 1
    $\begingroup$ Rob Jeffries is great! But this is an answer to a question I did not ask, and definitely not an answer to the question as asked. See my comment and check the question again. $\endgroup$ – uhoh Apr 26 '18 at 10:59
  • $\begingroup$ "For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane?" A: "Roughly," $\rho = g(R)$ given $v = constant$ $\endgroup$ – Kornpob Bhirombhakdi Apr 26 '18 at 11:14
  • $\begingroup$ Since comments are considered temporary in Stack Exchange, it would be best if you address the question directly in the answer, rather than as additional comments. Also, since I don't have $g(R)$ handy, what I am asking for is an analytical expression that roughly matches the Milky Way's distribution. $\endgroup$ – uhoh Apr 26 '18 at 11:19
  • 1
    $\begingroup$ Yeah. You are right. Really forgot about that. I was thinking distant object. $\endgroup$ – Kornpob Bhirombhakdi Apr 26 '18 at 11:22
  • $\begingroup$ And $g(R)$ has a functional form as defined up there. I just did not write it out. $\endgroup$ – Kornpob Bhirombhakdi Apr 26 '18 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.