## R

J L (6.67 X 10~8dyn cm2/g2)(1.67 X 10~24g)2(105cm) _ 6.1 X 1017 cm 0.2 pc

We find the mass by multiplying the density by the volume:

MJ = (4^/3)(1.67 X 10-24 g)(105 cm 3)(6.1 X 1017 cm)3 = 1.5 X 1035 g = 76 M0

We could have obtained the same mass directly from equation (15.5).

As we will see below, not all the mass will end up in the star.

Once a cloud becomes gravitationally bound, it will begin to collapse. We would like to be able to estimate the time for the collapse to take place. We begin by considering a particle a distance r from the center of the cloud. It will accelerate toward the center under the influence of the mass closer to the center than r. The acceleration is given by a(r) = GM(r)/r2

If the acceleration of this particle stayed constant with time, then the free-fall time, the time for it to fall a distance r, would be tff =

Note that the constant (3/2^)1/2 = 0.7, which we can approximate as unity, since we are making an estimate of the time. This gives tff = 1/(Gp )1/2

The free-fall time is independent of the starting radius. Therefore, all matter in a constant density cloud has approximately the same free-fall time. However, as the cloud collapses, the density increases. The collapse proceeds faster. The freefall time for the original cloud is therefore an upper limit to the actual collapse time. However, the result is not very different, since most of the time will be taken up in the early stages of the collapse, when the acceleration is not appreciably different from the one we have calculated. Therefore, we use the free-fall time as a reasonable estimate of the time it will take a cloud to collapse.

There is one important difference between our idealized cloud and a real cloud. A real cloud will probably have a higher density in the center. We can see this as follows. If the cloud is initially of uniform density, all points will have the same inward acceleration. This means that all particles will cover the same inward distance dr (where dr < 0), in some time interval dt. We can see how this changes the density for different volume spheres. If the initial density is p0, then the density of a constant mass collapsing sphere that shrinks from r0 to r is p = Po (ro/ r)3

The change in density dp is found by differentiating to give dp = -3 Po (r0 / r4) dr

The fractional change in density, dp/p, is dp/p =-3 dr/r (15.9)

This means that the smaller the initial sphere we consider, the faster its density will grow.

With a higher density at the center, the freefall time for material near the center will be less than for material near the edge. The material from the edge will lag behind the material closer in. This will enhance the density concentration in the center. The net result is that we end up with a strong concentration in the center. The concentration will eventually become the star, but the material from the outer parts of the cloud will continue to fall in on the star for quite some time.

Example 15.2 Free-fall time

Calculate the free-fall time for the cloud in the above example.

solution

Using equation (15.8) gives f = [(6.67 X 10-8 dyn cm2/g2 )(1.67 X 10-24 g) (105 cm-3)]-1'2

While almost a million years might sound like a long time, it is short compared with the main sequence lifetime of the star that will be formed, or the age of the galaxy.

If the cloud is rotating, then the collapse will be affected by the fact that the cloud's angular momentum must remain constant. The angular momentum L is the product of the moment of inertia I and the angular speed «,

For a uniform sphere, the moment of inertia is

If I0 and m0 are the original moment of inertia and angular speed, and I and m are their values at some later time, conservation of angular momentum tells us that

Using equation (15.11) to eliminate I and I0, we have

(This explains why a figure skater rotates faster as she brings her arms in. The 1/r2 dependence of the angular speed has a dramatic effect.)

To see what effect this has on collapse, we again look at a particle a distance r from the center of a collapsing cloud. The acceleration at that point is still GM(r)/r2. However, the radial acceleration now has two parts: (1) a(r) is associated with the change in magnitude of the radius, and (2) the acceleration associated with the change of direction, r«2. Therefore,

Solving for a(r) gives a(r) = GM(r)/r2 - r«2 (15.14b)

In comparing this with equation (15.6), we see that the acceleration a(r) is less for a rotating cloud than for a non-rotating cloud. The effect of the rotation is to slow down the collapse perpendicular to the axis of rotation.

The effects of rotation will be most significant when the second term on the right-hand side of equation (15.14a) is much greater than the first term, in which case

Multiplying both sides by r2 gives

Noting that v0 = m0 r0 , where v0 is the speed of a particle a distance r0 from the center,

We now solve for r/r0, the amount by which the cloud collapses before the rotation dominates:

For the cloud given in the two previous examples, with an initial rotation speed v0 = 1 km/s, r/r0 = 0.6. This means that, by the time the cloud

ra t

'k Jorb

Jl rai ra2

Fragmentation of a collapsing interstellar cloud. (a) The cloud is initially rotating as shown.As it collapses, the angular momentum J is conserved. (b) As the cloud becomes smaller, its angular speed m must increase to keep the angular momentum fixed.The rotation inhibits collapse perpendicular to the axis of rotation, and the cloud flattens. (c) Unable to collapse any further, the cloud breaks up, with the total angular momentum being divided between the spin and orbital angular momenta of the individual fragments.

reaches half its initial size, the rotation can completely stop the collapse perpendicular to the axis of rotation. Motions parallel to the axis of rotation are not affected by this, so collapse parallel to the axis of rotation can proceed unimpeded, and the cloud will flatten. (We will see that the tendency of rotating objects to form disks will reappear in many astrophysical situations.) Since the collapse is then only in one dimension, it is harder to reach stellar densities. Thus, the effect of rotation is to keep much of the material from becoming a star.

More of the material can end up in stars if the cloud breaks up into a multiple star system. The angular momentum can be taken up in the orbital motion, but individual clumps can continue contracting. This fragmentation process is probably responsible for the high incidence of binary systems. If a cloud shrinks to half its initial size, the average density will go up by a factor of eight. (The density is proportional to 1/volume, and the volume is proportional to r3.) From equation (15.5), we see that the Jeans mass of the denser cloud will be approximately one-third of the original Jeans mass. This means that it is possible for the less massive clumps to be bound and continue their collapse. The fragmentation process (Fig. 15.1) may be repeated until stellar mass objects are reached.

### 15.2 I Problems in star formation

We would like to know the conditions under which stars will form. We would like to know which types of interstellar clouds are most likely to form stars, and which locations within the clouds are the most likely sites of star formation. We would also like to know whether star formation is spontaneous or whether it needs some outside trigger. When we say a trigger is necessary, we mean that the conditions in a cloud are right for star formation, but something is necessary to compress the cloud somewhat to get the process started. Once started, it continues on its own. Sources for triggering star formation that have been suggested are the passage of a supernova remnant shock front, or the compression caused by a stellar wind. (Later in this chapter we will see how expanding HII regions might act as triggers, and in Chapter 17 we will discuss density waves associated with galactic spiral structure. These might also induce star formation.)

Once the collapse to form stars starts, we would like to know how it proceeds, and what fraction of the cloud mass ends up in stars. This is sometimes referred to as the efficiency of star formation. We would also like to know how much of the mass that goes into stars goes into stars of various masses. This is called the initial mass function. By "initial" we mean the distribution of stellar

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