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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 of the stellar number density of the Milky Way. Note that your browser needs to support WebGL.

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    $\begingroup$ Yes, the density depends on absolute distance from the plane. $\endgroup$ – Rob Jeffries May 21 '15 at 21:19
  • $\begingroup$ Supplementary question: should it be |Z - Z0| instead of |Z + Z0|? Z0 is the Z-coord of the Sun. $\endgroup$ – John O May 21 '15 at 23:55
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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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