<p>@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.</p>

<p>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,
$$
\alpha' G \rho R^2 = v^2 .
$$</p>

<p>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.</p>

<p>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.</p>

CC BY-SA 3.0
2018-04-26T10:59:42.207Z
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](https://astronomy.stackexchange.com/questions/26056/a-rough-analytical-expression-for-the-milky-ways-radial-mass-distribution/26063#comment44970_26056) and check the question again.
uhoh
CC BY-SA 3.0
2018-04-26T11:14:24.907Z
"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 $v = constant$
Kornpob Bhirombhakdi
CC BY-SA 3.0
2018-04-26T11:19:06.807Z
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.
uhoh
CC BY-SA 3.0
2018-04-26T11:22:19.007Z
Yeah. You are right. Really forgot about that. I was thinking distant object.
Kornpob Bhirombhakdi
CC BY-SA 3.0
2018-04-26T11:24:01.880Z
And $g(R)$ has a functional form as defined up there. I just did not write it out.
Kornpob Bhirombhakdi
CC BY-SA 3.0
2018-04-26T11:26:14.293Z
What does the Milky Way's radial mass distribution roughly look like? Is it flat, trapezoidal, triangular, gaussian, parabolic, erf-like? I haven't a clue!
uhoh
CC BY-SA 3.0
2018-04-26T11:29:04.820Z
Given $v = constant$, $\rho \propto R^{-2}$.
Kornpob Bhirombhakdi
CC BY-SA 3.0
2018-04-26T11:37:06.943Z
How close is that to the situation for the Milky Way? I'm looking for a statement that describes our galaxy. If this is what it looks like, then please include a statement to that effect in your answer, because this is what I'm asking, and introducing this as a "given" rather than stating it upfront means it might or might not be true.
uhoh
CC BY-SA 3.0
2018-04-26T12:34:38.613Z
Rotational curve is approximately constant. To get roughly very close to a certain galaxy profile, you need to fine tune those parameters. I do not know exact parametets for MW. So I will leave my answer like that as a general concept.
Kornpob Bhirombhakdi
CC BY-SA 3.0
2018-04-26T12:50:14.220Z
Googling I've just found [Modelling mass distribution of the Milky Way galaxy using Gaia billion-star map](https://arxiv.org/abs/1612.07781) for example. Citing this and possibly including either one or both parts of figure 1 (https://i.sstatic.net/oKYNy.png and https://i.sstatic.net/d7dl9.png) or anything similar and adding the words "roughly flat" would do it. Just a though.
uhoh
CC BY-SA 3.0
2018-04-26T10:11:27.463Z
2018-04-26T10:11:27.463Z
Kornpob Bhirombhakdi
<blockquote>
<p>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?</p>
</blockquote>
<p>The simplest example astronomers work with is the radial density profile of a <a href="https://en.wikipedia.org/wiki/Singular_isothermal_sphere_profile" rel="nofollow noreferrer">Single Isothermal Sphere</a> (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 &quot;temperature&quot;, hence isothermal). The density profile takes the form:</p>
<p><span class="math-container">$$\rho(r) = \frac{v^2}{4\pi Gr^2}$$</span></p>
<p>where <span class="math-container">$v$</span> is the rotational velocity. Note that you may see other formulations which use <span class="math-container">$\sigma_v$</span> rather than <span class="math-container">$v$</span>. In this case they're using the <em>velocity dispersion</em> which is slightly different than the rotational velocity.</p>
<p>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 <a href="https://en.wikipedia.org/wiki/Navarro%E2%80%93Frenk%E2%80%93White_profile" rel="nofollow noreferrer">NFW profile</a> and the <a href="https://en.wikipedia.org/wiki/Einasto_profile" rel="nofollow noreferrer">Einasto profile</a>.</p>
<p>The NFW profile is a two-parameter function, given by</p>
<p><span class="math-container">$$\rho(r) = \frac{\rho_0}{\frac{r}{R_S}\Big(1+\frac{r}{R_S}\Big)^2}$$</span></p>
<p>where <span class="math-container">$\rho_0$</span> and <span class="math-container">$R_S$</span> are two, halo-dependent parameters.</p>
<p>The Einasto Profile is again a two-parameter model given by</p>
<p><span class="math-container">$$\rho(r) \propto \exp(-Ar^\alpha)$$</span></p>
<p>where <span class="math-container">$A$</span> and <span class="math-container">$\alpha$</span> are configurable parameters.</p>
<blockquote>
<p>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?</p>
</blockquote>
<p>The Shell theorem for gravity does <em>not</em> 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.</p>

CC BY-SA 3.0
2018-04-26T13:36:47.440Z
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.
uhoh
CC BY-SA 3.0
2018-04-26T16:21:52.667Z
-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 $r^{-1}$.
ProfRob
CC BY-SA 3.0
2018-04-26T16:28:04.263Z
@RobJeffries Now that I look at it, I think you're right. I'll edit my answer.
zephyr
CC BY-SA 3.0
2018-04-27T06:37:53.803Z
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?
ProfRob
CC BY-SA 4.0
2023-11-08T04:56:12.400Z
I was a "little" slow on accepting, *uhoh!*
uhoh
CC BY-SA 4.0
2018-04-26T13:18:19.210Z
2023-11-06T19:03:31.590Z
2023-11-06T19:03:31.590Z
zephyr
<p>I found the image below in Space.com's article <a href="https://www.space.com/40406-gaia-release-color-milky-way-map.html" rel="nofollow noreferrer">This 3D Color Map of 1.7 Billion Stars in the Milky Way Is the Best Ever Made</a>, although <em>it is not the map mentioned in the title</em>.</p>
<p>The caption for this image reads:</p>
<blockquote>
<p>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</p>
</blockquote>
<p>If you imagine a band along the galactic equator the dominant velocity shows two positive and two negative &quot;peaks&quot;, with a zero crossing in the direction of the galactic center.</p>
<p>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 <span class="math-container">$\rho(r)$</span> which I could then use to figure out a rotational velocity distribution <span class="math-container">$v(r)$</span>, bun I swiftly realized that I have no idea what the density profile would look like.</p>
<ol>
<li><p>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?</p>
</li>
<li><p>For spherically symmetric distributions, Newton's <a href="https://en.wikipedia.org/wiki/Shell_theorem" rel="nofollow noreferrer">Shell theorem</a> 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?</p>
</li>
</ol>
<p><a href="https://i.sstatic.net/4Vyep.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4Vyep.jpg" alt="apparent velocity of the Milky Way seen from Earth (Gaia)" /></a></p>

CC BY-SA 3.0
2018-04-26T06:31:33.730Z
It works the other way around. You get the density distribution by looking at the velocity data.
ProfRob
CC BY-SA 3.0
2018-04-26T06:34:50.050Z
@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.
uhoh
CC BY-SA 3.0
2018-04-26T13:04:31.420Z
You might want to check out [this question](https://astronomy.stackexchange.com/questions/16391/what-is-the-shape-along-the-plane-not-up-down-of-stellar-orbits-in-flat-spira/18768#18768) which doesn't specifically address your question here, but may give you a good launching point.
zephyr
CC BY-SA 3.0
2018-04-26T13:09:34.770Z
@zephyr that's a helpful and also quite an interesting item to read, thank you for the suggestion!
uhoh
<galaxy><milky-way><velocity><mathematics>
Is there a rough analytical expression for the Milky Way's radial mass distribution?
uhoh
# Is there a rough analytical expression for the Milky Way's radial mass distribution?

(Asked by: username_0 on 2018-04-26)

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](https://www.space.com/40406-gaia-release-color-milky-way-map.html), 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](https://en.wikipedia.org/wiki/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)](https://i.sstatic.net/4Vyep.jpg)](https://i.sstatic.net/4Vyep.jpg)

**Tags:** galaxy, milky-way, velocity, mathematics

> **Comments:**
> **username_1**: It works the other way around. You get the density distribution by looking at the velocity data.  
> **username_0**: @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.  
> **username_2**: You might want to check out [this question](https://astronomy.stackexchange.com/questions/16391/what-is-the-shape-along-the-plane-not-up-down-of-stellar-orbits-in-flat-spira/18768#18768) which doesn't specifically address your question here, but may give you a good launching point.  
> **username_0**: @username_2 that's a helpful and also quite an interesting item to read, thank you for the suggestion!  

---

## 2 answers

### Answer by username_2 on 2018-04-26. Score: 6 | Accepted answer

> 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 astronomers work with is the radial density profile of a [Single Isothermal Sphere](https://en.wikipedia.org/wiki/Singular_isothermal_sphere_profile) (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](https://en.wikipedia.org/wiki/Navarro%E2%80%93Frenk%E2%80%93White_profile) and the [Einasto profile](https://en.wikipedia.org/wiki/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 $R\_S$ are 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.

> **Comments:**
> **username_0**: 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.  
> **username_1**: -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 $r^{-1}$.  
> **username_2**: @RobJeffries Now that I look at it, I think you're right. I'll edit my answer.  
> **username_1**: 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?  
> **username_0**: I was a "little" slow on accepting, *username_0!*  

---

### Answer by username_3 on 2018-04-26. Score: 1

@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.

> **Comments:**
> **username_0**: 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](https://astronomy.stackexchange.com/questions/26056/a-rough-analytical-expression-for-the-milky-ways-radial-mass-distribution/26063#comment44970_26056) and check the question again.  
> **username_3**: "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 $v = constant$  
> **username_0**: 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.  
> **username_3**: Yeah. You are right. Really forgot about that. I was thinking distant object.  
> **username_3**: And $g(R)$ has a functional form as defined up there. I just did not write it out.  
> **username_0**: What does the Milky Way's radial mass distribution roughly look like? Is it flat, trapezoidal, triangular, gaussian, parabolic, erf-like? I haven't a clue!  
> **username_3**: Given $v = constant$, $\rho \propto R^{-2}$.  
> **username_0**: How close is that to the situation for the Milky Way? I'm looking for a statement that describes our galaxy. If this is what it looks like, then please include a statement to that effect in your answer, because this is what I'm asking, and introducing this as a "given" rather than stating it upfront means it might or might not be true.  
> **username_3**: Rotational curve is approximately constant. To get roughly very close to a certain galaxy profile, you need to fine tune those parameters. I do not know exact parametets for MW. So I will leave my answer like that as a general concept.  
> **username_0**: Googling I've just found [Modelling mass distribution of the Milky Way galaxy using Gaia billion-star map](https://arxiv.org/abs/1612.07781) for example. Citing this and possibly including either one or both parts of figure 1 (https://i.sstatic.net/oKYNy.png and https://i.sstatic.net/d7dl9.png) or anything similar and adding the words "roughly flat" would do it. Just a though.  

---

