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I am putting together a toy program which shows how stars move around in the galaxy.

To run the simulation I need to know strength of the Milky Way's gravitational field at any location in it. I'm looking for a model (e.g. a collection of uniformly dense planes/rods) rather than a database of potentials.

Where can I get such a model?

I could simply construct an infinite plane of uniform density, but is that good enough? This is only a toy so I'm looking for something which preserves integrity of the overall shape and statistics of the galaxy, rather than worrying about the specific location of any particular star.

Qmechanic
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spraff
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    Relavent link: http://www.astro.cornell.edu/academics/courses/astro201/rotation_flat.htm It states that mass is proportional to distance. – John Alexiou Jan 26 '15 at 14:49

2 Answers2

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Note first that there are three different sources of gravitational potential: the disk, the bulge, and the dark halo.

There are a few different models of the gravitational field of the disk, two of the more common potentials are:

  • Kuzmin model:
    $$\Phi(r,z)=-\frac{GM}{\sqrt{r^2+(a+|z|)^2}}$$
  • Miyamoto-Nagai model:
    $$\Phi(r,z)=-\frac{GM}{\sqrt{r^2+(a+\sqrt{z^2+b^2})^2}}$$ where $a$ and $b$ are scale lengths.

For the bulge, you can use spherically symmetric potentials such as

  • Plummer model:
    $$\Phi(r)=−\frac{GM}{\sqrt{r^2+a^2}}$$
  • Jaffe model: $$\Phi(r)=\frac{GM}{a}\ln\left(\frac{r}{r+a}\right)$$ where $a$ also is a scale length and not necessarily the same as those for the disk.

The dark halo takes a spherical form, $$ \Phi(r)=\frac12V_h^2\ln\left(r^2+a^2\right) $$ where $V_h$ is the radial velocity of the galaxy at far distances ($\sim200$ km/s) and $a$ another scale length that isn't necessarily the same as above.

See also

Kyle Kanos
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  • Excellent resources, thank you. Quick sanity check -- since the positions of stars are usually quoted in heliocentric coordinates, will I have to rotate them all such that the sun is lifted up from the central plane in order to be in the same coordinate system as is used in your second resource? – spraff Jan 26 '15 at 15:04
  • Both sources use galacto-centric coordinates. In that case, our sun is $r\sim8.5 kpc$ and $z\sim100$ pc. – Kyle Kanos Jan 26 '15 at 15:09
  • I've just read this in more detail. Can you please confirm I've got this right? The potential will always be directed in the north-south axis so I should take $\frac{\partial}{\partial z}$ to get the acceleration, yes? Also, what are the units of G? Thanks a lot. – spraff Jan 26 '15 at 15:56
  • Using $F=-\nabla\Phi$ will get you the acceleration (remember that some of these are multi-dimensional and all are non-Cartesian). $G$ has units of $\rm N\cdot(m/kg)^2$. – Kyle Kanos Jan 26 '15 at 15:59
  • Problem is the kinematics are calculated in rectangular coordinates. How do I translate this? Also, G has those units normally but these equations are in kpc and solar masses so won't the constant change? – spraff Jan 26 '15 at 16:01
  • You can always transform the coordinates. The link for $G$ contains the value with other unit systems, but I presume you'd be able to convert it just as well. – Kyle Kanos Jan 26 '15 at 16:03
  • I can transform into cylindrical coordinates no problem, but when I apply $\nabla\Psi$ how do I interpret this equation? What do those hats mean? Is the result in cartesian coordinates? If not how do I interpret addition of terms? – spraff Jan 26 '15 at 17:52
  • Those are unit vectors in the radial, azimuthal and axial directions. You have to convert those into the appropriate Cartesian unit vectors (done in the previous link). – Kyle Kanos Jan 26 '15 at 17:56
  • I'm having trouble with the concept of a non-Cartesian vector. Vectors are defined by linearity, aren't they? – spraff Jan 28 '15 at 17:02
  • If by "linearity" you mean that they are linearly independent, this is generally true. If it makes you more comfortable, you can always use $r^2=x^2+y^2$ for cylindrical potentials ($\Phi(r,z)$) and $r^2=x^2+y^2+z^2$ for spherical potentials ($\Phi(r)$). – Kyle Kanos Jan 28 '15 at 17:06
  • I mean $\Phi(a,b)+\Phi(p,q)\neq\Phi(a+p,b+q)$, right? And because of this fact I can't simply transform $\nabla\Phi(r,z)$ into Euclidian coordinates to get the Euclidian vector which corresponds to $\nabla\Phi(x,y,z)$ can I? – spraff Jan 29 '15 at 12:32
  • No, you can do the vector transformation $\nabla\Phi(r,z)$ to $\nabla\Phi(x,y,z)$. It might be complicated (haven't done it), but it's absolutely possible. My suggestion was to to $\Phi(r,z)\to\Phi(x,y,z)$ and then take the gradient to get your force. – Kyle Kanos Jan 29 '15 at 13:28
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Deriving the galactic mass from rotation has the following chart (on the right) for the enclosed mass as a function of radius

pic

John Alexiou
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