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Was reading about dark matter and the distribution of it throughout the galaxy. it said "For example, if rotation curves are flat this means-" what exactly does this mean?

Ray Kay
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  • If you plot the velocity that stars are moving against their distance from the centre of the galaxy then you get the rotation curve. An explanation of how this relates to dark matter can be found here. – lemon Sep 06 '14 at 02:13
  • so would a flat rotational curve be one in which the stars travel at a constant speed regardless of distance from the center of the galaxy? – Ray Kay Sep 06 '14 at 02:26

1 Answers1

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When you look at an image of a galaxy like our own, you can see that most of the visible mass is concentrated in the core and that the density of stars in this core region is approximately constant:

enter image description here

So let's compute the expected rotation curve.

The orbital velocity of a star at a distance $r$ from the centre of the galaxy is $$ v=\sqrt{M(r)/r} $$ where $M$ is the mass of the galaxy inside the orbit $r$, and I've taken $G=1$.

Now let's pick a radius $r$ that is somewhere within the core region. Since the density $\rho$ (mass per unit area) is roughly constant in this region, the mass $M$ inside $r$ will be approximately $$ M(r)=\pi r^2\rho\qquad\mathrm{(core)} $$ And so the orbital velocity in the core region will scale as: $$ v\sim\sqrt{r^2/r}=\sqrt{r}\qquad\mathrm{(core)} $$

Now pick a radius $r$ far outside the core. Since the vast majority of the inner mass $M$ is contained within the core, then as we increase $r$ the mass $M$ will remain roughly constant. The velocity will therefore drop off as $$ v\sim 1/\sqrt{r}\qquad\mathrm{(outer)} $$

Plotting these two limiting cases ($\sqrt{r}$ in blue, $1/\sqrt{r}$ in red, and a 'hybrid' dashed):

enter image description here

we can see the sort of shape the rotation curve should have. However, the actual observed curve is roughly flat as $r$ increases outside of the core region:

enter image description here

This tells us that there is more mass than we can see (i.e. dark matter). Also, from the equation $v=\sqrt{M/r}$, we can see that this is only possible if the mass $M$ increases roughly linearly with $r$.

lemon
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  • With the expected line its velocity goes up than drops. Why would this be expected? Shouldn't the velocity be expected to drop as leave the center? – Ray Kay Sep 06 '14 at 03:41
  • @RayKay I have updated my answer to explain that. – lemon Sep 06 '14 at 12:00
  • Is it possible that inside the core the velocity should increase linearly with radius instead of with the squareroot as $M \sim r^3$ and $v \sim \sqrt{M/r} \sim r$ ? – McLawrence Nov 17 '17 at 16:42
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    @McLawrence My answer had been incorrectly edited to $M\sim r^3$ (now reversed). It should indeed be $M\sim r^2$ since we are treating the galaxy as flat. – lemon Nov 17 '17 at 17:48
  • Where does the $4$ in $4\pi r^2$ come from? Shouldn't it be $\pi r^2$? – Fabian Schn. May 22 '20 at 16:10
  • Well spotted, @FabianSchn. Corrected. – lemon May 25 '20 at 14:51
  • This isn't really correct. You can't say the expected rotation rate is that of a Keplerian orbit that assumes spherical symmetry and then use cylindrical symmetry to get the mass interior to $r$. – ProfRob May 25 '20 at 17:11
  • I don’t understand that chart image… why does’t the curve for the dark matter scenario match the observed? – br3nt Aug 14 '23 at 22:11