<p>From the definitions of the Oort constants we know that
<span class="math-container">$$A + B = -\frac{dv}{dr} $$</span>
at the solar radius.</p>
<p>But <span class="math-container">$A + B$</span> is small, implying that <span class="math-container">$dv/dr \simeq 0^{\dagger}$</span> and we are dealing with a flat(ish) rotation curve.</p>
<p>Assuming a spherically symmetric mass distribution with a density that is proportional to <span class="math-container">$r^\alpha$</span>, we can write the centripetal acceleration at a radius <span class="math-container">$R$</span>
<span class="math-container">$$\frac{v^2}{R} = \int^R_0 \frac{4\pi G r^2 \rho(r)}{R^2} \ dr\ . $$</span>
Then, if <span class="math-container">$v$</span> is constant, we can just deal in proportionalities:
<span class="math-container">$$ R^{-1} \propto R^{-2}\int^R_0 r^2 r^\alpha \ dr\ , $$</span>
<span class="math-container">$$ R^{-1} \propto R^{-2} R^{3+\alpha}\ ,$$</span>
and so <span class="math-container">$\alpha =-2$</span>.</p>
<p>Another variant is to tell you that <span class="math-container">$A=0$</span> in some galaxy. This means <span class="math-container">$dv/dr = v_0/r_0$</span>, which means <span class="math-container">$v \propto r$</span> (i.e. solid body rotation). This in turn means <span class="math-container">$\alpha =0$</span>.</p>
<p><span class="math-container">$\dagger$</span> by small, what we mean is that the velocity gradient is small compared with <span class="math-container">$v_0/R_0$</span> where <span class="math-container">$v_0$</span> is the rotation speed where the Oort constants are defined. Since <span class="math-container">$A-B = v_0/R_0$</span> this means that
<span class="math-container">$$ |A+B| \ll |A-B| $$</span>
which in this case is 2.5 vs 26.5 km/s per kpc.</p>

CC BY-SA 4.0
2021-05-10T09:51:58.337Z
First I was unsure what we mean by "small", but then read the "Flat rotation curve" of this: https://en.wikipedia.org/wiki/Oort_constants, and now I think get it: $|A+B|<<|A|, |A+B|<<|B|$. Thank you!
zabop
CC BY-SA 4.0
2021-05-10T10:10:49.763Z
@zabop yes, I should add something along those lines for a full explanation. It is basically that $dv/dr \ll V/R = A-B$.
ProfRob
CC BY-SA 4.0
2021-05-10T10:18:16.337Z
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).
zabop
CC BY-SA 4.0
2021-05-09T23:14:46.583Z
2021-05-10T10:14:30.423Z
2021-05-10T10:14:30.423Z
ProfRob
<p>I am given the observations that <a href="https://en.wikipedia.org/wiki/Oort_constants" rel="nofollow noreferrer">Oort constants</a> <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are, respectively: <span class="math-container">$14.5 \text{km s}^{-1} \text{kpc}^{-1}$</span> and <span class="math-container">$-12 \text{km s}^{-1} \text{kpc}^{-1}$</span>. From these, I am supposed to conclude that the galaxy density falls off as <span class="math-container">$\sim R^{-2}$</span>. (Source of problem: <a href="https://www.ast.cam.ac.uk/sites/default/files/Papers_1-4_2016.pdf" rel="nofollow noreferrer">Paper 2 Question 7 of 2016, Cambridge Astrophysics Tripos</a> - see page 17.)</p>
<p>I don't see how I can pull this off, would appreciate any help.</p>
<hr />
<p>In the earlier part of the question, I proved that at the solar location, it is true that</p>
<p><span class="math-container">$$\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}+2\left(A^{2}-B^{2}\right)$$</span></p>
<p>which might help somehow, but I don't see how.</p>

CC BY-SA 4.0
2021-05-09T08:45:01.647Z
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
CC BY-SA 4.0
2021-05-09T22:03:44.427Z
I am not sure [German lecture notes](https://www-astro.physik.tu-berlin.de/sites/www-astro.physik.tu-berlin.de/files/Vorlesung/EAA-2018-WS/EA_A20190111.pdf) are helpful for you - but the first 8 pages contain some other expressions for $A$ and $B$.
B--rian
CC BY-SA 4.0
2021-05-09T22:15:52.083Z
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 $\left(\frac{dV}{dR}\right)_0=-(A+B)$.
B--rian
CC BY-SA 4.0
2021-05-09T22:27:57.480Z
Can you assure us that this is not part of an assessment?
ProfRob
CC BY-SA 4.0
2021-05-09T22:45:21.493Z
@B--rian, thank you, looking into those relations again (once I knew them, now not that confident).
zabop
CC BY-SA 4.0
2021-05-09T22:45:25.023Z
@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. :) )
zabop
<milky-way><rotation><mathematics>
How galactic density falls using Oort constants
zabop
# How galactic density falls using Oort constants

(Asked by: username_0 on 2021-05-07)

I am given the observations that [Oort constants](https://en.wikipedia.org/wiki/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](https://www.ast.cam.ac.uk/sites/default/files/Papers_1-4_2016.pdf) - 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.

**Tags:** milky-way, rotation, mathematics
> **Comments:**
> **username_1**: 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.  
> **username_2**: I am not sure [German lecture notes](https://www-astro.physik.tu-berlin.de/sites/www-astro.physik.tu-berlin.de/files/Vorlesung/EAA-2018-WS/EA_A20190111.pdf) are helpful for you - but the first 8 pages contain some other expressions for $A$ and $B$.  
> **username_2**: 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 $\left(\frac{dV}{dR}\right)_0=-(A+B)$.  
> **username_3**: Can you assure us that this is not part of an assessment?  
> **username_0**: @B--rian, thank you, looking into those relations again (once I knew them, now not that confident).  
> **username_0**: @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. :) )  

---

## 1 answers

### Answer by username_3 on 2021-05-09. Score: 3 | Accepted answer

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.

> **Comments:**
> **username_0**: First I was unsure what we mean by "small", but then read the "Flat rotation curve" of this: https://en.wikipedia.org/wiki/Oort_constants, and now I think get it: $|A+B|<<|A|, |A+B|<<|B|$. Thank you!  
> **username_3**: @zabop yes, I should add something along those lines for a full explanation. It is basically that $dv/dr \ll V/R = A-B$.  
> **username_0**: 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).  

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