# How dark matter makes ordinary matter far from the center of a disc galaxy rotate faster?

I was reading several questions on this subject, incluiding duplicates, and even received the very useful help of Kyle Oman in this question to understand the behavior and distribution of DM.

But I'm not sure if I actually understand why the galaxy mass "rotate as a whole" instead of "swirling" like the disc of a forming planetary system.

Trying to understand it I imagined a given object in the border of the visible portion of a disc galaxy, the 2nd part of the shell theorem cancels the gravitational effect of the DM further from the center, but the extra inner mass increases the speed of rotation of this object. Then a closer object, with higher rotation speed due to visible matter mass, again by the shell theorem ignore some of the DM affecting the first object. Thus the inner DM mass increases its speed a bit less. And so subsequently with objects closer to the center.

Is this idea correct?

Excuse the plainness of my question and thanks for your help in advance. Best regards!

• – Kyle Oman Aug 4 '15 at 21:49
• @KyleOman thanks again for the suggestion, I readed it before asking, but I was still missing the Differential Rotation issue. – Leopoldo Sanczyk Aug 4 '15 at 22:40

Disk galaxies don't rotate like solid bodies (think frisbee), but rather rotate differentially (think whirlpool). The rotation speed as a function of radius is called a rotation curve, and is often interpreted as a measurement of the mass profile of a galaxy, as:

$$v_c(R) = \sqrt{\frac{GM(<R)}{R}}$$

where $M(<R)$ is the total mass enclosed$^1$ within radius $R$ and $v_c(R)$ is the rotation speed at radius $R$.

One of the observations supporting the existence of dark matter is that galaxies rotate faster in their outskirts than you'd expect from estimating their mass distribution from visible stars, dust, gas, etc. This can also be interpreted as evidence that gravity (general relativity) is incorrect.

$^1$To get an accurate measurement of $M(<R)$ by observing $v_c(R)$, the mass distribution must be spherically symmetric (it generally is not, e.g. disk galaxies are obviously not exactly spherical, though the DM halo dominates the mass and is roughly spherical). That DM is the dominant form of mass at all radii is more true for smaller (dwarf) galaxies, so these are often the focus of debates about the DM distribution. The orbits of the tracer being observed must also be circular (they often aren't, but it is typically possibly to measure the circularity to get a handle on this source of uncertainty). For an extensive discussion see this paper (full disclosure: I am an author on it).

• Can you emphasise the applicability limits of your equation. It is tiresome to keep fending off DM-deniers complaints that using the shell theorem is incorrect for non-symmetric mass distributions (which is true). – ProfRob Aug 4 '15 at 22:17
• @KyelOman Nice equation! Why I didn't see it while surfing Wikipedia? Anyway, I should have thought the point between the 2 extremes again! A fixed disc Milky Way shouldn't have spiral arms and it's not too swirled to 12 billion years of rotation? I should try to improve the wiki article with your explanation. Thanks a lot for your help again! – Leopoldo Sanczyk Aug 4 '15 at 23:07
• @RobJeffries true. Will try to remember to get that tomorrow. – Kyle Oman Aug 5 '15 at 3:29
• @LeopoldoSanczyk that equation is just a = v^2/r for circular motion, re-arranged and with the gravitational acceleration. As for the spiral arms... there's a few questions on here about how spiral arms form. They don't exactly "wind up" over 12 billion years. e.g. physics.stackexchange.com/questions/69949/… (both answers) – Kyle Oman Aug 5 '15 at 3:32
• @RobJeffries and done. – Kyle Oman Aug 5 '15 at 17:18