# How galactic density falls using Oort constants

I am given the observations that Oort constants $$A$$ and $$B$$ are, respectively: $$14.5 \text{km s}^{-1} \text{kpc}^{-1}$$ and $$-12 \text{km s}^{-1} \text{kpc}^{-1}$$. From these, I am supposed to conclude that the galaxy density falls off as $$\sim R^{-2}$$. (Source of problem: Paper 2 Question 7 of 2016, Cambridge Astrophysics Tripos - see page 17.)

I don't see how I can pull this off, would appreciate any help.

In the earlier part of the question, I proved that at the solar location, it is true that

$$\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}+2\left(A^{2}-B^{2}\right)$$

which might help somehow, but I don't see how.

• Just fyi $14.5 \text{km s}^{-1} \text{kpc}^{-1}$ can also be generated by only doing the superscripts with MathJax: 14.5 km s$^{-1}$ kpc$^{-1}$ as 14.5 km s$^{-1}$ kpc$^{-1}$ and without using to MathJax at all: 14.5 km s<sup>-1</sup> kpc<sup>-1</sup> works nicely in posts but doesn't work in comments. There's nothing wrong with the way you did it, but I sometimes find these alternate ways helpful.
– uhoh
May 9, 2021 at 8:45
• I am not sure German lecture notes are helpful for you - but the first 8 pages contain some other expressions for $A$ and $B$. May 9, 2021 at 22:03
• My main challenge is how to relate a radial velocity profile $V(R)$ to a radial density profile. I guess you know that $\omega_0 =A+B$ and $\left(\frac{dV}{dR}\right)_0=-(A+B)$. May 9, 2021 at 22:15
• Can you assure us that this is not part of an assessment? May 9, 2021 at 22:27
• @ProfRob, it is part of a 2016 assessment (see link in question), now I am doing problems for practicing (but I am not graded or assessed on this particular problem). I'll have an exam on this topic (but I doubt they'll ask the same question as in 2016, so yes, you can be assured, this isn't assessment. :) ) May 9, 2021 at 22:45

From the definitions of the Oort constants we know that $$A + B = -\frac{dv}{dr}$$ at the solar radius.

But $$A + B$$ is small, implying that $$dv/dr \simeq 0^{\dagger}$$ and we are dealing with a flat(ish) rotation curve.

Assuming a spherically symmetric mass distribution with a density that is proportional to $$r^\alpha$$, we can write the centripetal acceleration at a radius $$R$$ $$\frac{v^2}{R} = \int^R_0 \frac{4\pi G r^2 \rho(r)}{R^2} \ dr\ .$$ Then, if $$v$$ is constant, we can just deal in proportionalities: $$R^{-1} \propto R^{-2}\int^R_0 r^2 r^\alpha \ dr\ ,$$ $$R^{-1} \propto R^{-2} R^{3+\alpha}\ ,$$ and so $$\alpha =-2$$.

Another variant is to tell you that $$A=0$$ in some galaxy. This means $$dv/dr = v_0/r_0$$, which means $$v \propto r$$ (i.e. solid body rotation). This in turn means $$\alpha =0$$.

$$\dagger$$ by small, what we mean is that the velocity gradient is small compared with $$v_0/R_0$$ where $$v_0$$ is the rotation speed where the Oort constants are defined. Since $$A-B = v_0/R_0$$ this means that $$|A+B| \ll |A-B|$$ which in this case is 2.5 vs 26.5 km/s per kpc.

• First I was unsure what we mean by "small", but then read the "Flat rotation curve" of this: en.wikipedia.org/wiki/Oort_constants, and now I think get it: $|A+B|<<|A|, |A+B|<<|B|$. Thank you! May 10, 2021 at 9:51
• @zabop yes, I should add something along those lines for a full explanation. It is basically that $dv/dr \ll V/R = A-B$. May 10, 2021 at 10:10
• Ohh alright, that's even better. (I am allowed to mark bounty in 11 hours, feel free to add anything if you feel like it, I like the post as it is now). May 10, 2021 at 10:18