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Why do astronomers and cosmologists have to invoke the idea of dark matter to explain the motion of stars within galaxies? We need a little physics and a little math to grasp the idea

Stars in galaxies rotate strangely bout their centers. Stars farther out from the center rotate with nearly the same period as stars closer in, as Vera Rubin and Fritz Zwicky noticed. That means they are moving far faster along their orbits than the stars closer in. This is inconsistent with their motion being set by the total mass of all the stars interior to them. That total mass can be estimated from the count of stars at various distances closer than a star of interest. Of course, there’s also a massive black hole at the center of a galaxy, but that makes the problem worse. It looks like most of the mass of the galaxy is close to the center. Stars farther out should have a lower speed and an even longer period of rotation.

The Solar System is an exact analog. The more distant the planet is from the Sun, the longer its orbital period. The period of Mars is 1.88 times that of the Earth. The period of Jupiter is 12 times longer. Newtonian gravity (and even relativistic theory, at the relatively small star speeds) can simply calculate the orbital period of a small object circling a huge central mass. The period increases as the radius of the orbit to the 3/2 power (derived at the end). If that were to happen, ~~the nice spiral arms of galaxies would smear out to little rims within a few revolutions of the galaxies.~~ [No! 22 July 22: See the simulations of galaxy rotations: the outer arms stay rather intact while the inner parts wind up. Mean galaxy rotation times are roughly a billion years, so most nearby galaxies have rotated a lot. The spiral structure does get preserved.] If gravity works, then there must be much mass that’s invisible, and astronomers now call it dark matter – it interacts with stars and other visible matter only through gravity.

Here’s a simplified model of how dark matter affects the orbits of stars. Let’s make a spherical galaxy filled uniformly with dark matter, and take that as the dominant mass (which astronomers do, for a number of reasons). To calculate the orbital speed of a small object such as a star at radius r from the center, we may balance the force of gravity pulling it toward the center against the virtual centrifugal force pushing it out.

It can be shown nicely through calculus that the gravitational pull of a spherical body is identical to that of a single point of the same mass, located at the center. (The math is a bit lengthy; I provide a URL at the end for the calculation.) This is true even if the density of mass varies by radial distance, as long as the distribution is uniform cross all angles. So, let the density of mass of dark matter per unit volume be denoted as ρ (“rho”). The total mass, M, within a sphere of radius r is ρ multiplied by the volume of the sphere, which is . Note that there is more dark matter farther out than the orbit of the star. Calculus also tells us readily that it has no net effect, no pull!

The pull of that mass gravitationally, F_g, is given by the classic equation

Here, m is the mass of our circling object or star, and G is the universal gravitational constant. Let’s express M in terms of the radius and the density of dark matter:

Here, I gathered all the constants into a single constant, k.

The star is moving with a speed v (I call it speed because velocity is a vector, a speed in a specified direction. That direction is constantly changing as the star orbits, but, for a circular orbit we’re considering, the speed stays the same). The centrifugal force, F_c, is then given by another classic equation,

The two forces balance for a stable orbit, giving us an equation that will relate speed v to distance r:

Cancelling out the star mass m on both sides (yes, the mass of the star does not matter! It’s very small compared to all else), we get

So, the speed is proportional to the radius. The orbital distance to be traversed is also proportional to r, so the orbital period is the same for stars at all radii!

There are obvious simplifications, such as using a spherical mass of dark matter rather than a flattened prolate spheroid that looks closer to the shape of the visible galaxy. The latter would give a different pattern of orbital period versus radius, but still one that lets stars orbit faster than around a central mass or the mass of all their companion stars at smaller radii. There’s something to this dark matter hypothesis, and the shape of the swirling galaxy is just the star!

22 July 24: On the way to making an astronomy demo:

Actually, the observed tangential velocities of stars in galaxies tends to be almost constant with radius – see the saved file Lecture 24_ Spiral Galaxies.pdf in ~/websites_all/[s-t-s]/stem_demos. To explain this, we have to invoke a falloff of mass density with radius:

To get cumulative M(r) proportional to r, write

I had thought about a simple demo if v were to increase with r, as I thought originally. Then I could have a student swing a weight on a string in a circle. With a helper to check that the student kept the period the same, that student would feel a centrifugal force proportional to r. Sigh. With v = constant, the centrifugal force declines as 1/r. It’s not that much different from the simple solar system model with . I’d be asking the student to feel the falloff of ½ compared to ¼ between the galaxy model and the solar system model with a doubling of radius.

Whether or not I decided to use centrifugal force to demonstrate the dark matter effect, I’d need to tell students how the tangential velocities are measured using the Doppler effect. That would involve a spectroscope and then telling the students that the Franuhofer lines, which are not readily apparent, shift with line-of-sight velocity, and not by much. I could also use animations of galaxy rotations.

Note about a spherical body acting as a body of the same mass concentrated at the center: the math is a bit lengthy, involving analytical geometry and calculus. It’s presented at https://en.wikipedia.org/wiki/Shell_theorem. Have fun, if you wish!

Second note, about the orbital period of a small mass around a big mass being proportional to the 3/2 power of the orbital radius: We can do the same force balance as in the main text above, but now the central mass M is fixed, independent of the radius r.:

This gives us the equation for the speed v in terms of orbital radius r:

So, v decreases with radius! The orbital period T is the distance around the orbit, 2πr, divided by the speed, v, which is :

A few things we note: First, take the example of Mars, which is 1.524 times farther from the Sun than is the Earth, on average. (That’s comparing their major axes; both orbits are eccentric, or somewhat elliptical.) Its orbital period should be greater than that of Earth by a factor . That makes it about 365*1.8 = 687 Earth days. That’s just as is observed, within the precision quoted here.

The rotation of Mars about its own axis is another matter. When teams of scientists at NASA and JPL kept watch on the Mars rovers, they often set their watches to Mars days, which are 24 hours and 37 minutes long. They got progressively out of sync with the Earth days, ending up sleeping when Earthlings were awake and vice versa. They never saw the Earth and Mars years start on the same Earth day – that is, when Mars ends up right behind the Earth on a January first of some distant Earth year. The ratio of the orbital periods is not a simple ratio of integers. To get them within one day of another takes many centuries. I haven’t worked out the exact math.

Meast. – units; eqpt. – 42 ff.

Eqpt. in all sci. – get used to it – bring out some; pp. 50 ff.