5
$\begingroup$

My review assignment has a question that asks us to use the Navarro–Frenk–White (NFW) profile to find total mass in the galaxy using

$$\rho(R)=\frac{\rho_0}{1+\frac{R}{R_c}}$$ then taking a triple integral of it in spherical coordinates.

I do not have a problem taking the integral or converting it, but I am struggling with what $R_c$ (the critical radius) should be. I must have missed it in class and cannot quite understand it by looking through other online resources. I feel like it would be a constant but I cannot find it online and am wondering if I need to calculate it somehow. Any help with explaining $R_c$ would be greatly appreciated.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ I cleaned your question up a bit... please make sure I didn't change it in a bad way (I think you mean $\rho$ where you wrote $p$, and $\rho_0$ where you wrote $P(not)$) $\endgroup$ – Floris Mar 4 '16 at 0:48
  • 1
    $\begingroup$ The density distribution section of your link talks of a "scaling" radius. Could that be what you need? It also seems to have a slightly different form of the density equation. $\endgroup$ – Floris Mar 4 '16 at 0:49
  • $\begingroup$ Thank you very much for cleaning up my question, my laptop battery died when I was working on this assignment and so I was forced to use my phone. And I think the scaling radius may be the same, however I do not know, I cannot find any mention of scaling radius in my textbook and I cannot seem to find critical radius online. The different form is what makes me think they are different, it may be the virial radius, I am not sure as there seem to be hundreds of forms for this formula and every reasources I've looked at seems to use different notations. $\endgroup$ – user110240 Mar 4 '16 at 11:07
  • $\begingroup$ Yes - I did notice there were various forms. But of course you could just pick a value and reach an answer expressed in terms of the ratio between the limits of integration and the scaling factor. That gives the answer in closed form "up to a term you can make up" - because I am pretty sure that value is something that is not well pinned down. Maybe gets you a B on the question... $\endgroup$ – Floris Mar 4 '16 at 12:47
  • 1
    $\begingroup$ I'm a little confused. First, that's not the NFW profile. Next, the NFW profile is for dark matter, not the total mass of the matter in a galaxy. Lastly,I'm not sure what resources you've found, but this NED-IPAC document on the NFW profile seems to be pretty clear about what $R_c$ (though they've written it as $r_s$) is supposed to mean and do. $\endgroup$ – Kyle Kanos Mar 6 '16 at 20:00
2
$\begingroup$

Well, first of all, that's not the NFW profile, instead you should have:

$$\rho(r) = \frac{\rho_0}{\frac{r}{r_s}(1+\frac{r}{r_s})^2}$$

The radius $r_s$ is usually called the scale radius, and is the place where the logarithmic derivative of the density is $-2$. This isn't especially physically meaningful, but is mathematically convenient. The integral is not particularly straightforward, but the result is well known:

$$M(<r) = 4\pi\rho_0r_s^3\left[\ln\left(\frac{r_s+r}{r_s}\right)-\frac{r}{r_s+r}\right]$$

Note that you need to pick a radius $r$ to integrate out to, because attempting to integrate to infinity will give an infinite mass. One common radius to pick is the "virial" radius, which can be defined in a few ways. One is the radius $r_{200{\rm c}}$ where the average enclosed density is 200 times the critical density of the Universe $\rho_{crit}=\frac{3H^2}{8\pi G}$. This radius is related to the scale radius by the concentration parameter $c$ as:

$$r_{200{\rm c}}=cr_s$$

The concentration parameter can be fixed for a halo of a given mass by using a mass concentration relation (e.g. Ludlow et al. 2014). This extra constraint reduces the NFW profile from a model with two free parameters to one with a single free parameter, which you could choose to call the mass of the halo.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.