It is a sufficiently well-known fact that the rotation curves of spiral galaxies are approximately flat in the middle to far regions of the disk. This is in apparent contradiction with the fact that most luminous matter is at the center of the galaxy, and its abuncance dereases rapidly as we move further away from the center. This is resolved by considering there is a massive dark matter halo that surrounds and the vissible parto of the galaxy. My aim is to find analitically and with as few assumptions as possible an expression for the density profile of such a dark matter hallo, under the assumptions that:

  • The rotation curve is in good aproximation, flat
  • Luminous (barionic) matter's density decreasses rapidly as we move away from the center.

For this, I find it necessary to give an expression for the radial distribution of luminous mass in the galaxy, using $$\rho(r)=\rho_0·e^{-r}$$Is this a reasonable assumption?If so,from that point on, how would you find the dark matter halo's density distribution?

Note that I'm not trying to find an experimentally accurate expression such as the Navarro–Frenk–White profile and I am not paying attention to the possible divergence of the total mass of the halo.


1 Answer 1


This is not a trivial task, finding a reasonable DM halo that fits observations is kind of a hot topic today. To give you an idea: the dark matter density profile in the inner few kpc can give you an idea of the nature of dark matter itself (whether it is hot or cold), and that is something we desperately would like to know. But you can give it a try with this!

If $\Phi$ is the gravitational potential (I will assume you have spherical symmetry, otherwise this gets complicated really fast) then the circular velocity at radius $R$ is

$$ v_c^2(R) = R\frac{\partial \Phi}{\partial R} \tag{1} $$

If you only have two components in your model (bar: baryonic matter, DM: dark matter), then

$$ v_c^2(R) = R\frac{\partial (\Phi_{\rm bar} + \Phi_{\rm DM})}{\partial R} = R\frac{\partial \Phi_{\rm bar}}{\partial R} + R\frac{\partial \Phi_{\rm DM}}{\partial R} = v_{c, {\rm bar}}^2 + v_{c, {\rm DM}}^2 \tag{2} $$

The term $v_{c,{\rm bar}}$ you already know because you "know" (or at least assumed) its density profile

$$ \rho_{\rm bar}(r) = \rho_0 e^{-r} \tag{3} $$

with the help of Poisson's equation you can find the potential generated by this profile

$$ \Phi_{\rm bar} = (...) \tag{4} $$

With this you can replace in Eq (2) to get the term $v_{c,{\rm DM}}$:

$$ v^2_{c,{\rm DM}}(R) = v^2_{c}(R) - v^2_{c,{\rm bar}}(R) \tag{5} $$

On the RHS you already know the term $v^2_{c,{\rm bar}}$ from Eq. (4), and you assume a model for $v_c(R)$ that fits your needs. From this you can find $v_{c,{\rm DM}}$ and from there $\Phi_{c,{\rm DM}}$.

Once this is done you can find again $\rho_{\rm DM} \sim \nabla^2 \Phi_{c,{\rm DM}}$


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