# rotation curve in the innermost part of galaxies

I read in Sofue's articles (1999) that he used a method to investigate the bulge of galaxies to obtain rotation curve of this part of galaxies. So my question is how could Vera Rubin et al. obtained the rotation curve in their article (1978), especially the steep increase in the begenning of the curve since in Sofue's article he seems to say it was rather difficult, at that time, to investigate galactic bulges ?

• Could you be a bit more specific about the source? Which article by Sofue (there are 10 from 1999), and where in the article is this said? – Peter Erwin Apr 6 '17 at 15:12
• For example in this article arXiv:astro-ph/9910004 (second sentence of the introduction). So my question is what was the Rubin's method for the inner part ? – ketherok Apr 7 '17 at 6:37

So they are talking about different scales. Rubin is finding the rotation curve on kpc scales. And Sofue is talking about how hard it is to get the rotation curve in the central less than a kiloparsec. In the center you cannot use the same techniques as Rubin did.

Rotation curves of galaxies are most often measured by the gas speed. The speed of the stars can be used, but owing to dynamical heating over time, the stellar rotation speed does not indicate the true rotation speed of the galaxy as well as the gas speed.

When looking for the speed of something, astronomers will look for the redshift or blueshift of the stars or the gas which they are observing. So, they need to identify a gas, which will be present in all galaxies, which has a well known and easy to observe "line" in the spectra which will allow a velocity determination by seeing how red or blueshifted it is compared to that emission line in a laboratory.

Two such lines are H$\alpha$ and molecular CO (the one mentioned in that paper you've linked).

The H$\alpha$ line is an optical line at 656.28 nm. The CO lines are radio waves at 2.6 and 1.3 mm.

It looks like the Rubin+1978 paper took spectra of the H$\alpha$ lines, and from Figure 2, it looks as if they've collected spectra from various radii from the center. If you do it to many different points inside the galaxy at different distances from the nucleus, you can see how fast they are moving relative to each other.

That steep increase at the beginning of the curve usually occurs for around the extent of the bulge, a few kiloparsecs (as in this image from Rubin+1978).

It looks like they had a resolution of about a kpc, which would not tell you much about the center, except that the rotation curve is very high, very quickly (you assume the very center is 0 rotation):

Sofue+1999 seems to be talking about even more central than that, like the "central few hundred pc region" which will require a new technique.

A large limitation, as mentioned in the Sofue+1999 paper, is that HI is often obscured in the very centers (<0.5 kpc) of these galaxies by dust. Galactic dust, in general extincts shorter wavelength light much more effectively than long wavelength light. In this way, looking at the galaxy in longer wavelengths allows you to see through the dust. The HI line is much more easily erased by galactic extinction than the CO line at a longer wavelength. So CO makes it easier to look into the galactic central hundred parsecs than H$\alpha$. You can see this in the galactic extinction curve:

The CO line is in the radio band (mm range), a type of astronomy that was just picking up speed in the 1960s and 1970s. Having been predicted in the 1800s by Maxwell’s equations, scientists were unable to observe any radio signals from the Galaxy until a serendipitous observation of Sagittarius A* by Karl Jansky at Bell Labs in the 1930s.

However, radio astronomy comes with a problem of resolution, the resolving power of a telescope is basically:

$\Theta = 1.22 * \frac{\lambda}{D}$

Where lambda is the wavelength of the observation and D is the diameter of the telescope. And, for reference, the distances to two galaxies from that rotation curve above are:

NGC4594 = 9.45 Mpc
NGC7217 = 15 Mpc

At the distances of these galaxies, we need to be looking in the right wavelength to penetrate, and we need enough resolution to get the rotation speed at maybe 10 parsec scales. This resolution, at 15 Mpc, is maybe 0.1 arcseconds.

$3600 * arctan(\frac{10}{15e6}) * \frac{180}{\pi} = 0.1375 arcseconds$

To get that resolution observing at the 2.6 mm you need for the CO line, you need a:

$1.22 * 3600 * \frac{180}{\pi} * \frac{2.5e-3}{0.1375} = 4581.48$

4.5 km telescope… Hmmmmm. A bit of a problem considering the unfortunate tale of the 300 foot telescope.

Then, computers to the rescue. Interferometry is an idea whereby one can use two telescopes to increase their resolving power. When combining their signals, they have an effective diameter of their separation. Of course, they are not equivalent to a single dish telescope of that diameter, because they collect far less total starlight.

The technique was successfully utilized scientifically as early as the 1920s, to resolve the star Betelgeuse. But it was difficult to use this technique in it’s modern form, combining tens or hundreds of telescopes at once without the advent of computers which could handle the calculations in the 1960s and 1970s.

And now, the final piece. Interferometry grows much more difficult the smaller your wavelength. Interferometry is only possible if amplitude and phase of the incoming wave is recorded and recombined precisely. So an optical interferopeter needs: more precisely polished mirrors, more precisely measured separations, stable signal information transport from the dishes (fiberoptics), and more computing power to run more fourier transforms. While radio interferometry was performed in the 1920s, it took all the way until the 1990s to do optical interferometry, and it's still far from routine (see the Large Binocular Telescope).

So, to summarize:
1) To get a rotation curve of a galaxy, we need a molecular line that we can get a redshift or blueshift from. HI (optical) and CO (mm) are good choices.

2) The very centers of galaxies are obscured by dust, which destroys optical signals, but allows mm to pass, so CO is the molecule of choice for rotation curves of the nuclei of these galaxies, as H$\alpha$ signal is poor in the centers.

3) However, for a mm telescope to be able to spatially resolve small parts of the galaxies at the large distances of 10+ MPC in this wavelength, we need a telescope a few kilometers wide.

4) We can solve this with interferometry, but only recently has the technology to do interferometry at this wavelength been accomplished.

The first interferometer observing at this wavelength, the SubMillimeter Array (SMA; eight, 6-meter telescopes), opened to guest astronomers for the first time in 1990. The SMA was still being expanded at the time of the paper you linked.

These days, the Atacama Large Millimeter Array (ALMA) is just now nearing completion, and is focusing on the same wavelength range only with huge resolution and surface area (66 telescopes from 7-12 meters). They are collecting CO rotation curves to such precision that they claim to be able to estimate the central black hole masses of galaxies to 5%.

Stuff:

• Thank you for your answer but if Rubin could measure H$\alpha$ lines in the galactic bulge, then why Sofue is saying "However, the distribution of dark-matter in the inner disk and bulge are not thoroughly investigated yet because of the difficulty in observing rotation curves of the innermost part of galaxies." ? – ketherok Apr 13 '17 at 22:39
• I've added a couple of things to my answer, hope that helps. – ohrkzt Apr 14 '17 at 7:30
• Thank you for this comprehensive answer. So do you mean that,in Rubin's curves, the steep increase is extrapolated towards zero or that if we zoom on the data we can see that it stops before reaching zero? – ketherok Apr 16 '17 at 21:29
• The Rubin method (optical, H$\alpha$ line) could not look into very center (~<1 kpc) because of dust. mm observations (CO line) can see into the center, but had poor resolution until the 1990s, which is why Sofue says it's hard. Yes, they assume the center to have a rotation speed of 0 km/s. – ohrkzt Apr 17 '17 at 1:34
• Sure thing. Feel free to mark it as answered if you think this might help some other people. – ohrkzt Apr 17 '17 at 15:22