4
$\begingroup$

I am reading "The Essential Cosmic Perspective" by Jeffrey O. Bennett, Megan O. Donahue, Nicholas Schneider, Mark Voit. In Chapter 14, it is stated that an evidence of the presence of dark matter in our galaxy is that the rotational curve does not match that obtained from calculation. In the calculation, the following formula is used $$M_r=\frac{rv^2}{G}$$ where $M_r$ is the mass enclosed inside a radius $r$ from the galactic center.

My understanding is that the above equation is a result of the shell theorem. But we know that shell theorem applies to spherically symmetric mass distribution only, and most galaxies are disks. So why can we still do this?

$\endgroup$
  • $\begingroup$ Can you substantiate your claim (that galaxy rotation curves are calculated from what you call the shell theorem)? For near-spherical object, the error one makes with this assumption is very small (at most a few %). $\endgroup$ – Walter Feb 5 '15 at 22:33
  • $\begingroup$ question edited with source provided $\endgroup$ – velut luna Feb 6 '15 at 1:54
  • $\begingroup$ @Kyson My answer stands as is, regardless of the source. The shell theorem only applies to spherically symmetric mass distributions. This is an approximation to give the gist of the argument. $\endgroup$ – Rob Jeffries Feb 6 '15 at 7:47
  • $\begingroup$ The authors of that book should have mentioned the caveat, that it only introduces a small error, and that in the scientific literature (of which that book is no part) such simplifications are not made (unless the errors introduced are far smaller than other sources of error). If they did not, then this is just another example of poor practice in presenting science to the public. $\endgroup$ – Walter Feb 6 '15 at 9:26
5
$\begingroup$

We can't. That is an over simplification only used in elementary treatments simply to provide the the flavour of the argument. If you see it done somewhere in the refereed literature, it is probably incorrect. Of course it may be true that the mass is almost spherically symmetric, especially if it is dominated by a spherically symmetric dark matter component, but that the visible light is not. Or, it may be true that at large distances from even an asymmetric mass distribution, a Keplerian potential is a good approximation for the orbits of distant objects (e.g. satellite galaxies to the Milky Way). A measured galaxy rotation curve makes no such assumption, it is merely a measurement of rotation speed as a function of radius. It is only the interpretation and modelling that needs to deal with the mass distribution.

The real situation is much more complex. See for example http://ned.ipac.caltech.edu/level5/March01/Battaner/revision.html

However, even if one assumed all the mass was concentrated into a disk-like shape, the only way you can get flat rotation curves is to assume that the mass in the disk does not "follow the light" - that the mass-to-luminosity ratio increases vastly with radius - which is essentially still saying that you have "dark matter", just in a disk.

$\endgroup$
  • $\begingroup$ question edited with source information provided $\endgroup$ – velut luna Feb 6 '15 at 1:55
1
$\begingroup$
  1. Your claim that sphericity is assumed is incorrect when referred to hard scientific evidence.
  2. Even if that was assumed, the resulting error in the rotation speed is small for spheroidal galaxies and smallish for flat galaxies, provided they are centrally concentrated (which their stellar distribution is).

Scientists are, of course, well aware of the errors emanating from this (and other) assumptions (in particular if the situation is as simple as here) and care is taken (if not by every individual scientist, then certainly by the concensus reached by the scientific community) that these errors have no bearing on the scientific conclusions. Your quote from a undergraduate-level textbook proves nothing in this respect.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.