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Isothermal v NFW halos

  1. Jan 9, 2006 #1
    Hi all,

    I have currently been reading up on dark matter halos as I’m considering doing a small project on them for my degree. I plan to use an (approximated) analytical equation for the density at r of an isothermal halo in terms of the halos central density and core radius.

    density(r) = density(0) * (a^2)/(a^2+r^2)
    where a = core radius.

    How would I be able to fix either the core radius or central density of in order to model the Milky Way?

    I would love to compare this result with a model of the milky way that has an NFW profile.

    Regarding the NFW halos, I’m having trouble with the virial radius. The virial radius is needed in order to calculate the scale radius, which is used in the equation for the density at r. I understand the virial radius is the radius of a sphere that encompasses around 200 times the critical density. But how does the virial radius compare to the physical radius of a halo. For instance, I've read the virial radius for the Milky Way was 428 kpc or so. Much larger than the radius of the milky way, which is around 15 kpc. Do I need the physical radius of the milky way in order to model it as an NFW halo?

    Any help will be very much appreciated.
    Last edited: Jan 9, 2006
  2. jcsd
  3. Jan 9, 2006 #2


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    Should that not be
    [tex]\rho(r) = \frac{\rho_0 a^3}{(a + r)^2r}[/tex] ?

  4. Jan 9, 2006 #3

    Interesting you should mention that. I have seen several different equations for this isothermal model. The equation I quoted was taken from arXiv:astro-ph/0403064 v1 2 Mar 2004 (page 33).
  5. Jan 9, 2006 #4


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    I took mine from Disk formation in an NFW halo
    Yours looks simpler.

  6. Jan 12, 2006 #5
    think I understand

    Right, a bit more reading into it and I think I understand now. The virial radius, is the physical radius of the halo, which for the Milky Way is around 350 kpc or so, and the radius of the visible disc is something like 15kpc.

    As for obtaining a core radius for that isothermal model, the differential equation describing an isothermal model must be solved numerically, and then the approximation is fitted to the actual solution. For a given mass within a given volume, one can then obtain the central density
  7. Jan 13, 2006 #6


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    These are all fitting formulae (not physically motivated, for the most part), so their form can vary from paper to paper. However, there are some standards. The singular isothermal sphere should always take the simple form:


    where there will be some set of normalizing constants in the above equation. Sometimes it's expressed in terms of the associated rotation velocity and other times just some scaled density, but as long as they're both 1/r2, it's really the same thing.

    When one moves to the non-singular isothermal spheres (meaning the density doesn't approach infinity as r -> 0), the standards are less straightforward. The available formulae can be characterized (and are chosen) based on their limiting behavior. The formula the OP gives approaches the singular isothermal sphere at large radii and a constant density at small radii. This is a decent fit to some galaxies, but a poor fit to most clusters (for which similar formulae are often used). Another example of a non-singular isothermal sphere is:

    [tex]\rho = \frac{Me^{-r/r_t}}{2\pi^{3/2}\gamma r_t (r^2 + r_c^2)}[/tex]

    There are several normalizing constants and scale radii, but the important thing is that it has roughly constant density at the center, approaches 1/r2 at moderate radii, and has an exponential truncation at the outskirts.

    The point behind the isothermal sphere is that it produces flat rotation curves at moderate radii. It's meant to fit to the observational data that suggest a roughly flat rotation curve for spiral galaxies. We can only measure rotation curves out to moderate distance from the galaxy's center (the galaxy becomes too dim at the outskirts), so we don't know how far out the 1/r2 dependence goes. It must change form at some point because the total integrated mass diverges for a halo that is 1/r2 all the way to infinity. The curve the OP gives ignores this problem, presumably because the outer profile doesn't matter for the purposes of the reference paper.

    The profile Garth gives is quite different, both qualitatively and quantitatively. It is the standard NFW profile. This profile was made to fit simulations rather than observations, so if its parameters can be chosen to successfully fit real galaxy data, then it means theory matches observation. Again, look at the limiting behavior. At large radii, it approaches a 1/r3 profile, while at small radii it goes as 1/r. At first glance, it would seem to never have a 1/r2 dependence and thus be inconsistent with observations of a flat rotation curve. However, there must be some transition zone between the two dependences, so it turns out that at the radii to which we can measure rotation curves, the NFW profile is approximately 1/r2.

    There has been a lot of research on how well the NFW profile fits real data and the general consensus seems to be that it's better than isothermal models, but not dramatically so. The biggest discrepancy is near the centers of the halos, where the simulations appear to have a sharper cusp than the data. This was already discussed in another thread:

    Cuspy Halo Problems

    As simulation and observations improve, we'll be able to get a better handle on just how well they match. The fits do seem to be getting better with time, but I would still be cautious to say anything definitive.
    Last edited: Jan 13, 2006
  8. Jan 13, 2006 #7


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    Yes, thank you ST, the OP formula

    [tex]\rho(r) = \frac{\rho_0 a^2}{(a^2 + r^2)}[/tex]

    is the standard density function of the required dark matter halo to fit the rotation curve of a spiral such as the MIlky Way. It delivers a rigid-body rotaiton for small r and a flat rotation curve for large r. Of course it also has to be truncated at some large r.

    My formula quoted

    [tex]\rho(r) = \frac{\rho_0 a^3}{(a + r)^2r}[/tex]

    is the NFW formula that gives a better fit to the DM in galaxy clusters. Which to use depends on at how far out you want to study the halo.

    I was a little confused because S.P.P. mentioned NFW, I thought the focus was to be beyond just the Milky Way.

    Last edited: Jan 13, 2006
  9. Jan 14, 2006 #8


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    If you apply the rules of fluid dynamics, a twisting moment is introduced. This enormously complicates the calculations. About this time, a triplet of bench players from the turbulence family takes over playing the outfield.
    Last edited: Jan 14, 2006
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