# I Galaxy rotation curves

1. Aug 16, 2017

### redtree

I apologize for the simple question, but I have not been able to find the answer.

For the inner portion of a galaxy rotation curve (where the outer portion is the part invariant to distance and the inner part is where rotational velocity increases with radius), how much is simply due to Newton's shell theorem, i.e., additional mass enclosed by the radius?

2. Aug 16, 2017

### DrSteve

The rotation curve in the inner part of the galaxy is dictated solely by visible, baryonic matter that obeys Kepler's and Newton's law

3. Aug 16, 2017

### redtree

For a circular orbit obeying Newtonian physics, rotational velocity ($\vec{v}_{r}$), given a constant effective point mass:

\begin{split}

\vec{v}_{r} &= \sqrt{\frac{G m}{\vec{r}}}

\end{split}

Such that rotational velocity decreases as the square root of radius.

However, for the inner part of the galaxy rotation curve, rotational velocity increases as a function of radius, implying that mass is a function of radius, i.e., $m(\vec{r})$, for this portion of the curve, which is what I meant in referencing Newton's shell theorem (see https://en.wikipedia.org/wiki/Shell_theorem). Given that increasing rotational velocity as a function of distance distinguishes the inner portion of Galaxy rotation curve, Newton's shell theorem would seem the dominant effect for the inner portion of the galaxy rotation curve. Am I wrong in saying that?

Last edited: Aug 16, 2017
4. Aug 16, 2017

### Bandersnatch

Yes, it's all correct (if put a bit weirdly). The bulge is an approximately spherical distribution of matter with more or less constant density, so you can apply shell theorem to justify using the enclosed mass only:
$v_{r} = \sqrt{\frac{G V(r)\rho}{r}}$
$v_{r} = \sqrt{\frac{G 4 \pi r^3 \rho}{3 r}}$
$v_{r} = r \sqrt{\frac{G 4 \pi \rho}{3}}$
which results in a raising velocity curve.

It's still Newton's laws, though. I wouldn't call it 'an effect'.

5. Aug 16, 2017

### Bandersnatch

$\frac{r^3}{r}=r^2$
and
square root of r squared is r.

6. Aug 16, 2017

### redtree

Sorry. Dumb mistake; got it.

7. Aug 17, 2017

### DrSteve

Then outside the bulge (the aforementioned region of more or less constant density) it should quickly start to fall, but it doesn't.

8. Aug 19, 2017

### redtree

If $\vec{r}$ denotes the radius of the sphere within which the mass of constant density exists, at what radius (in terms of $\vec{r}$) does rotational velocity become constant?

9. Aug 19, 2017

### redtree

Is there a simple geometric equation for the "universal rotation curve" in terms of $\vec{r}$, $M$ and $G$? The best I can find is from arXiv:astro-ph/0506370, where $U(\vec{r})$ denotes the gravitational potential, and $G_0$, $M_0$ and $\vec{r}_0$ denote coupling constants:

\begin{split}
U(\vec{r})&=\frac{G_0 M}{\vec{r}}\left[1+ \sqrt{\frac{M_0}{M}}\left(1- e^{-\frac{\vec{r}}{\vec{r}_0}} \right) \right]
\end{split}

Last edited: Aug 19, 2017
10. Aug 19, 2017

### Bandersnatch

For constant density there is no such radius - orbital velocity increases without bound (well, until you hit the Schwartzschild radius and an event horizon forms).

You get constant velocity if the density is a function of radius such that $\rho \propto 1/r^2$.

11. Aug 19, 2017

### redtree

I am assuming a finite radius for the mass density.

12. Aug 19, 2017

### Bandersnatch

If the sphere of uniform density has a finite radius $R$, then the velocity increases with distance from the centre $r$ as in post #4 as long as distance from the centre is not larger than $R$. Once $r>R$, all the available mass is contained inside, so the velocity curve becomes Keplerian (since as per the shell theorem all that mass can be treated as central point mass).

You can only get a flat rotation curve inside a sphere of non-uniform density, proportional to $1/r^2$.

13. Aug 19, 2017

### redtree

Yes. You are assuming Dark Matter. Perhaps I wasn't clear. I was asking for an equation where $M$ represents the visible matter.

14. Aug 19, 2017

### snorkack

Are there any graphs showing the total density, visible density and inferrable dark density, as a function of radius?

15. Aug 22, 2017

### redtree

I don't know of any. Do you?

16. Aug 22, 2017

### Staff: Mentor

17. Aug 23, 2017

### Bandersnatch

I don't know of any single equation. Usually it's done by adding contributions from bulge+disc+halo.

In my earlier responses I was referring to how it is impossible to have flat rotation curve where the only mass concentration is in the form of a sphere of constant density - which is what your post #8 and #11 seemed to imply. It doesn't matter if you're counting visible or dark matter.

That's an attempt at modified gravity. By it's nature it does not involve simple Newton's laws. I think it's beyond the scope of this thread.

I agree with @fresh_42 - there are plenty of decomposed velocity curves, which after all are derived from densities.

18. Aug 24, 2017

### snorkack

Neither of them is quite what I was asking for.
Yes, the rotation curve is the source of density data of total mass. And that velocity curve is NOT derived from densities - vice versa, because the total density is unobservable.
Note how the line of "curve derived from visible disc" lacks its error bars - and it is the one derived from densities.
That´s a bare prediction. No error bars, no visible matter density for comparison.
And demonstrate how a "quick Google search" has very strong limits unless you know exactly what term to search for, and where.

19. Aug 24, 2017

### Bandersnatch

No, I don't mean the observed velocity curve - you're right, velocities and distances are the only direct observables.

But then you need to figure out how to build a density model, so as to match the curve. This is done separately for bulge, disc, and halo. So when you see a graph such as this one:

(source: https://arxiv.org/pdf/0811.0859.pdf)
The component curves that you're trying to match to the observed one are derived from whatever density model you might have chosen for that component. This is what I meant.
(in that paper they show their density model in Fig.2)

20. Aug 24, 2017

### snorkack

But you should not need any density model to derive the density from observed rotation curve!
The exercise should be simply:
1) For every r, v2*r should be proportional to the total mass enclosed by r
2) Differentiating the observed v2*r with respect to r gives the mass of each shell from r to r+dr
3) Divide it by volume (4πr2*dr) gives the density
Probably the nonspherical shape of disc galaxies introduces some complications, though...