Monday, May 11, 2009

Galaxy rotation curves

Ok, so finally I think we can look at rotation curves. We'll make the simplifying assumption that the objects we are interested in are in a perfectly circular orbit about the center of the galaxy, an assumption which doesn't really change anything so it's ok (another larger point about physics: quite often [in fact, almost always], we take a complicated problem and approximate it into something smaller that we can solve [often called the "spherical cow" approach - we would approximate a cow to be a sphere and go from there]. The question then often becomes "how good was the approximation?" In this case, there is no real difference between circular and elliptical orbits, so the approximation is fine and the conclusions are valid).

We know the equation of circular motion, F=mv2/r. And by hypothesis, the only force acting on the object in orbit is the force of gravity, F=G*m1*m2/r2. In this case, m1 is the mass of the galaxy, and m2 is the mass of the object. We equate the forces, so mv2/r = G*m1*m2/r2. Now, the mass from the circular motion equation is just the mass of the object in orbit, so m2 will cancel. All that remains is to solve for the velocity, since that's what we measure using the Doppler effect and red shift.

m2*v2/r = G*m1*m2/r2

First, divide both sides by m2

v2/r = G*m1/r2

Next, multiply through by r

v2 = G*m1/r

Now, take the square root of both sides

v = Sqrt(G*m1/r)

And that's it. We have derived that the velocity of an object in orbit about a galaxy should be proportional to the square root of the mass of the galaxy divided by the orbital radius. There is one more thing we should be aware of, which is that I haven't made any assumptions yet about the size of the mass of the galaxy. A galaxy is a very large thing, and what happens if you're inside part of it? For example, the Sun and the Earth are somewhere inside the Milky Way galaxy. We do orbit the Milky Way center, but part of the Milky Way is outside our orbit. The answer is that in this case, m1 refers the total mass inside the orbit. It doesn't matter how spatially extended it is in space, as long as the object in which we're interested is outside of the galaxy, the equations are fine. And since I'm particularly interested in the mass of the bright part of the galaxy, it's easy to know when we're outside that part, so everything holds.

Now, let's look at a plot. If all the mass were in the bright part of the galaxy, then outside the bright part (say a radius of 100, just to make the plot look right), from the last equation, we would expect the velocity to fall like 1/Sqrt(r) (G and m1 would be constant). That would look like this:

Instead, we measure a flat line, like this:

Therefore, by deduction, we know that either Newtonian gravity is wrong (a possibility, I'll admit), or that there is more mass than we thought, mass that is not contained in the bright part of the galaxy. In fact, we know the distribution of that mass, as it has to increase like 1/Sqrt(r) or else the the velocity would not be flat.

This is what some of the actual data looks like:

These are measurements of galaxy rotation curves (Begeman, Broeils and Sanders, Mon. Not. R. astr. Soc., 1991, 249, 523). I apologize for the image quality, but velocity is on the y-axis and radius is on the x-axis, and all the black points are actual measurements. You can see at small radius the velocity increases. This is where the bright part of the galaxy is, and as the radius increases, we are containing more mass in the orbit. At larger radius, we would expect to see the velocity drop. But instead, it stays constant. The dashed and dotted lines are the the components of the mass, the bright part and the dark part. This data provides evidence that there is matter that we are not seeing, that is not interacting with light, but is dark matter.


  1. I think I would like to know what the consequences are of discovering or measuring dark matter. Also, does what you are doing have any relation whatsoever to things like the Hubble telescope or general space travel that people seem to be doing more and more of? Might your discoveries, for instance, give us an idea of the future of the universe as we know it?
    xox MOM

  2. I think your Mom would be interested know that Vera Rubin, the astronomer who pioneered the study of galactic rotation curves, took on that project because she was a pregnant post-doc and needed a short term (ha, ha) project to keep herself occupied while away from work. I know this because she lectured at the Hayden Planetarium a few years ago.

  3. What would be the effect of the spiral arms spiralling themselves (faster nearer the centre) on their rotation speed around the galaxy centre?

  4. Dear Anonymous,
    I'm not sure what you mean. The matter contained within the arms would be included in the gravitational force felt by stars at larger radii, so that's already taken into account. To first order, the orbital velocity of an object is entirely determined by the amount of matter within the area swept out by the orbit. You can therefore approximate each location on the spiral arms as feeling the gravitational force of everything inside that location and nothing outside.

  5. Hi Hugh, nice piece of text.

    Take a look at a disk like galaxy. Assume the mass density uniformly distributed but dependent on r: k(r). Assume the thickness of
    the galaxy to be H. The amount of mass inside a sphere of radius r is the integral over r of k(r)*H*2pi*r.
    If this density k(r) is constant this amount of mass would increase like r^2: K*pi*H*r^2.
    This would result in a speed of: sqrt(G*K*pi*H*r^2 / r) = sqrt(G*K*pi*H*r).
    This is not what we see in galaxies.
    Is it possible that the mass density is not constant but steadily decreasing with r, like: k(r) = K / r ?
    If that is the case the mass inside a sphere of radius r is: Integral ( K/r * H*2pi*r *dr) = 2pi*H*K*r.

    Hey, that could be something...

    Speed = sqrt (2pi*H*K*r / r) = sqrt (2pi*H*K) = constant, not dependent on r.
    (Of course the speed is only constant at larger values of r. That is because near the center of the galaxy the mass distribution is


    What could explain that the mass density is proportional to 1/r ?
    Suppose a uniform band of mass that is located near the center of the galaxy.
    That means: The mass of a small ring = A*H*2pi*R*dr with A the mass density at this position R and dr the horizontal thickness of the

    ring (if the galaxy is supposed to lie on a horizontal plane).
    Suppose for a reason that over a large number of years the radius of the ring grows from R to r.
    This means that the original amount of mass in the original small ring is smeared out over a ring with a larger radius.
    The total amount of mass in the larger ring is equal to the total amount in the smaller ring.
    The new density of the expanded ring is thus: A*R/r which is proportional to 1/r.
    If the density of rings of matter had always be nearly the same at a fixed radius R, the mass distribution over radius r is really

    proportional to 1/r if continuously new ring elements of matter keep growing their radius.

    How is it possible that the size of a mass ring could grow in time?
    That is possible by assuming that every rotation of a piece of mass around the center causes exchange of angular momentum to and from

    the inner rings because of tidal effects (see: This is also the case with the moon

    that revolves around Earth. The radius of the moon orbit increases a few centimeters each year.

    If we watch a spherical rotating galaxy for a large number of years it will be flattened and stretched in horizontal direction

    because of the tidal acceleration effects. Around the outer layers of the galaxy the mass distribution is nearly uniform. This means

    that the mass density initially does not depend very much on r at large distance from the center. Because of the tidal acceleration

    the radiusses of rings of mass is growing resulting in a continuous passing of mass rings of equal density seen from a fixed radius

    The described spherical galaxy results in a flat disc of matter uniformly distributed over the ring elements resulting in a mass

    density proportional to 1/r at larger values of r.

    When the source was an elliptically formed rotating galaxy rotating around an axis perpendicular to the long axis of the ellipse,

    i.e. the sides of the ellipse are visually rotating. Because of the same tidal accelleration effects the long axis of the ellipse

    stretches and the angular velocity decreases. This results in a spiral galaxy.

    Regards, Rene