<p>It may make more sense to convert the exponential into the product of the individual exponential terms:
$$
\exp(-\frac{R}{L}) \exp(-\frac{Z}{H}) \exp(\frac{Z_{\odot}}{H})
$$
This makes it clearer that the $Z_{\odot}$ term is a constant, which is just part of the overall normalization (similar to the $\exp(R_{\odot}/L)$ term earlier in the original equation).</p>

<p>And, yes, $Z$ is assumed to be always positive, so $|Z|$ would be slightly more correct.</p>

CC BY-SA 3.0
2015-06-05T13:45:09.997Z
2015-06-05T16:24:46.783Z
2015-06-05T16:24:46.783Z
Peter Erwin
<p>The paper is :
<a href="http://www.astro.washington.edu/users/ivezic/Publications/tomographyI.pdf" rel="nofollow">http://www.astro.washington.edu/users/ivezic/Publications/tomographyI.pdf</a></p>

<p>The equation is equation #23 in the paper. It's a model for the density of stars in the Milky Way's disk. It has an exponential dependence on both $R$ and $Z$. $R$ is the distance from the center of the galaxy, and $Z$ is the distance above/below the plane of the disk.</p>

<p>The relation is roughly this:</p>

<p>$$\varrho=\text{constant}\times e^{-\frac{R}{L}-\frac{Z+Z_0}{H}}$$</p>

<p>The problem I'm seeing is the $Z$-dependence of the formula. $R$ and $Z$ here are standard cylindrical coordinates. Thus, $Z$ can be positive or negative. The problem is that the formula blows up when $Z$ is negative. This is unphysical, since the number density must generally decrease going further away from the disk.</p>

<p>Am I missing something? Or should the equation really have an absolute magnitude of $|Z + Z_0|$?</p>

<p>[Added Later:]
I finished making a <a href="http://johnohanley.ca/sky/milky_way_stellar_density/form.html" rel="nofollow" title="3D demo">3D demo</a> of the stellar number density of the Milky Way. Note that your browser needs to support <a href="https://get.webgl.org/" rel="nofollow">WebGL</a>.</p>

CC BY-SA 3.0
2015-05-21T21:19:29.637Z
Yes, the density depends on absolute distance from the plane.
ProfRob
CC BY-SA 3.0
2015-05-21T23:55:34.503Z
Supplementary question: should it be |Z - Z0| instead of |Z + Z0|? Z0 is the Z-coord of the Sun.
John
<milky-way><stellar-dynamics>
Milky Way stellar number density : is the stated equation in this paper incorrect?
John
# Milky Way stellar number density : is the stated equation in this paper incorrect?

(Asked by: username_0 on 2015-05-21)

The paper is :
<http://www.astro.washington.edu/users/ivezic/Publications/tomographyI.pdf>

The equation is equation #23 in the paper. It's a model for the density of stars in the Milky Way's disk. It has an exponential dependence on both $R$ and $Z$. $R$ is the distance from the center of the galaxy, and $Z$ is the distance above/below the plane of the disk.

The relation is roughly this:

$$\varrho=\text{constant}\times e^{-\frac{R}{L}-\frac{Z+Z\_0}{H}}$$

The problem I'm seeing is the $Z$-dependence of the formula. $R$ and $Z$ here are standard cylindrical coordinates. Thus, $Z$ can be positive or negative. The problem is that the formula blows up when $Z$ is negative. This is unphysical, since the number density must generally decrease going further away from the disk.

Am I missing something? Or should the equation really have an absolute magnitude of $|Z + Z\_0|$?

[Added Later:]
I finished making a [3D demo](http://johnohanley.ca/sky/milky_way_stellar_density/form.html "3D demo") of the stellar number density of the Milky Way. Note that your browser needs to support [WebGL](https://get.webgl.org/).

**Tags:** milky-way, stellar-dynamics

> **Comments:**
> **username_1**: Yes, the density depends on absolute distance from the plane.  
> **username_0**: Supplementary question: should it be |Z - Z0| instead of |Z + Z0|? Z0 is the Z-coord of the Sun.  

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## 1 answers

### Answer by username_2 on 2015-06-05. Score: 1

It may make more sense to convert the exponential into the product of the individual exponential terms:
$$
\exp(-\frac{R}{L}) \exp(-\frac{Z}{H}) \exp(\frac{Z\_{\odot}}{H})
$$
This makes it clearer that the $Z\_{\odot}$ term is a constant, which is just part of the overall normalization (similar to the $\exp(R\_{\odot}/L)$ term earlier in the original equation).

And, yes, $Z$ is assumed to be always positive, so $|Z|$ would be slightly more correct.


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