This should probably be basic but I've been looking for days and I can't find how to (I'm probably over complicating, but still).

I want to calculate a rotation curve for some spiral galaxies. From what I understand I need to use the brightness profile (which I have) to determine the mass distribution and then use that with Poisson to calculated the $g$. Then I can get the rotation curve.

My problem (for now) is how do I convert the brightness profile to mass-distribution?

the units are $mag/pc^2$ and the $y$-axis scale is 24-->18. it's not a direct relation because the distribution $y$-axis scale goes from small to large (10-->10^2 for example) so I can't just use $100 L/pc^2 = 22.05$ been finding all kind of different equations for different things, just not this.

would appreciate any help.

  • 3
    $\begingroup$ Do you mean you want to predict a rotation curve assuming that mass follows light? $\endgroup$ – Rob Jeffries Mar 8 '15 at 17:44
  • $\begingroup$ yes, exactly what I'm trying to do. $\endgroup$ – mkP Mar 8 '15 at 18:46
  • $\begingroup$ anyone? been checking everyday.... $\endgroup$ – mkP Mar 11 '15 at 17:06

This presentation might be helpful to you.

Based on dust density and mass density in each pixel of an image, the luminosity is calculated as $$L=L_0 \cdot \exp \left( \int \kappa \rho ds\right)$$ where $\kappa$ is a constant and $\rho$ is the density of the interstellar medium (I believe).

A more helpful relation for pixel-mass-to-luminosity is $$\frac{dn}{d \log (m)} \propto m^{- \Gamma}$$ where $$\Gamma= \left\{ \begin{array}{l l} 0.5 & \quad 0.1M_{\odot} \leq m \leq 0.5M_{\odot}\\ 1.15 & \quad 0.5M_{\odot} \leq m \leq 120_{\odot} \end{array} \right.$$

From this, you should be able to generate a rotation curve.

I'm not sure how well this will work for you; another option is the Tully-Fisher relation (for spiral galaxies); the Faber-Jackson relation is used for elliptical galaxies.

Scholarpedia has a helpful article.

  • $\begingroup$ Tully-Fisher won't get you a rotation curve, just an estimate of something like the peak of the rotation curve. $\endgroup$ – Kyle Oman Jun 1 '15 at 16:43

It is probably best to start with a simple/crude estimate which you can then refine as you see fit.

Start by converting your surface brightness profile to units of $L_\odot/{\rm pc}^2$. Very crudely:

$$M - M_\odot = -2.5\log_{10}\frac{L}{{\rm L}_\odot}$$

Note $M_\odot$ is the absolute magnitude of the Sun, not its mass!

Then turn this into a (stellar) mass profile. You'll need a mass-to-light ratio $\Upsilon$ for this. These ratios are a whole sub-discipline unto themselves, so you can refine this endlessly, but a very crude estimate is $\Upsilon\sim1 {\rm M}_\odot/{\rm L}_\odot$ (note that here ${\rm M}_\odot$ is the solar mass - standard notation is messy in this case, sorry). The most useful mass profile for a rotation curve is the cumulative mass profile, which is:

$$M(<R)=2\pi\Upsilon\int_0^R \sigma(R')R'dR'$$

where $\sigma(R)$ is the surface density profile.

Once you have the cumulative mass profile, the rotation curve is trivial to compute:

$$V_{\rm rot} = \sqrt{\frac{GM(<R)}{R}}$$

This assumes that the mass distribution is approximately spherically symmetric (it probably isn't) and that the stars are all on approximately circular orbits (they probably aren't), but it's a start. Hopefully that's enough to get you started, from there it's a matter of reading a big chunk of literature and probably consulting experts to figure out all the details of how to refine your estimate.

Oh and as a bonus I'll save you 5 minutes of unit conversion and give $G$ in useful units:

$$G = 4.031\times 10^{-3} \frac{({\rm km~s^{-1}})^2\,{\rm pc}}{{\rm M}_\odot}$$


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