dt 2 = 0, we find that: K = 1 2 Ω (2)

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1 1 1. irial Theorem Last semester, we derived the irial theorem from essentially considering a series of particles which attract each other through gravitation. The result was that d = K + Ω (1) dt where the irial = r dm, Ω is the gravitational potential energy and K is the total kinetic energy. n the case of equilibrium where d dt = 0, we find that: K = 1 Ω () which is very useful when considering stars in hydrostatic equilibrium. Unlike stars, for molecular clouds, there may be a significant external pressure. The virial theorem can be modified to take into account external pressure. n molecular cloud studies, a common way of writing the virial theorem is: 1 d dt = 3 Pd + P r ds + Ω (3) S where 3 Pd is the integral of pressure over the volume of the cloud and is equal to K, P r ds is the integral of the external pressure over the surface of the cloud, and Ω once S again the gravitational potential energy. We define K s = 1 P r ds. f the external pressure is a constant value P S 0, then we can replace: and we can write: K s = 1 P 0 r ds = 3 P 0 (4) S or if we assume a constant pressure 1 d dt = 3 (P P 0 )d + Ω (5) 1 d dt = 3(P P 0) cl + Ω (6)

2 . The Stability of a Cloud (or Clump) Following Spitzer (1978), let s assume that a cloud is in equilibrium, i.e. d /dt = 0. The we rewrite equation 5 so that: assuming a homogenous sphere with sound speed c s 3P 0 = 3 Pd + Ω (7) 4πP 0 Rcl 3 = 3c sm cl 3 GMcl (8) 5 R cl where c s is the sound speed and the ideal gas law gives P = c s ρ. To examine the stability of the cloud to changes in external pressure, we solve the above equation for P 0 and then take the derivative with R P 0 = 1 ( 3c s M cl 3 4π Rcl 3 5 ) GMcl Rcl 4 (9) and dp 0 = 1 ( 9c s M cl dr cl 4π Rcl ) GMcl Rcl 5 (10) Now let s consider what happens if we increase pressure. f the right side of the equation is negative, i.e. R > R crit = 4 GM cl 15 c s (11) then the increase in pressure will cause the cloud to shrink, but the value of P 0 required for equilibrium goes up as well. Eventually, the cloud comes into equilibrium, unless R R crit. At this point, dp 0 /dr cl = 0 and P 0 is at a maximum. ncreasing P 0 will imply that there is no stable equilibrium and the cloud will collapse. For a value of M = 10 4 M, T K = 0 K (and consequently c s = 0.5 km s 1 ), R crit 00 pc. This is larger than most clouds, molecular clouds cannot be supported by thermal pressure. We also note that α = Ω/K = 6, meaning that the cloud would be not in virial equilibrium.

3 3 3. The Effect of the Magnetic Field Magnetic fields may have a stabilizing effect on clouds. Although the clouds are neutral, ionization by cosmic rays, and on the outer layers, U radiation, may create a small number of electrons and ions. These would be coupled to the magnetic field, and in fact, freeze the magnetic field lines to the gas. Thus, the magnetic field pressure could exert a pressure. where 1 d dt = 3 (P P 0 )d + Ω + M B (1) M B = 1 8 π (B B 0)d (13) where B 0 is the external field. M B = 1 8π (B )d 1 6 B Rcl 3 = 1 φ 6π R (14) where b is order unity and B is the mean magnetic field. φ = πrbdr πr B (15) An important value is the ratio of M cl /φ where the graviational potential energy is equal to the magentic energy term, Ω = M B. For a uniform sphere, this occurs when: 3 GMcl 5 R = 1 φ 6 R (16) or when The ratio M cl /φ is known as the mass to flux ratio. f ( ) 1/ M cl 5 φ = 1 18 πg = 1 (17) 1/ πg 1/ M cl φ > 1 πg 1/ (18)

4 4 then gravity dominates over magnetic fields and the cloud is supercritical. Note, that since φ is conserved, the ratio does not change as the cloud collapse. f the ratio is < 1, then the cloud is sub-critical. Note that since N(H ) = M cl, B = φ πrcl πrcl (19) implies that M cl φ = N(H ) B (0) Measurements show that N(H ) B against collapse. 1, suggesting that magnetic fields do not support clouds 4. Turbulent Pressure We now consider the addition of turbulent energy to the kinetic energy term. f we just consider the thermal motions of the gas, the kinetic energy is given by: K = 3 M clc s (1) where c s is the sound speed. Remember that the total 3-D velocity dispersion is We can then define a pressure term: σ = σ x + σ y + σ z = 3c s () P = 3 M cl(c s + σ turb ) (3) where σ turb is the 1-D turbulent dispersion. We can take this from the measured linewidth of the cloud, which is σ tot = c s + σ turb. Then the equation becomes: P tot = P th + P turb = 3 M cl(σ tot) (4)

5 5 For a value of M = 10 4 M, T K = 0 K (and consequently c s = 0.5 km s 1 ), R crit 1 pc. This is similar to the size of most molecular clouds. Plus we know that α = Ω/K is around 1, indicating that it is close to virial equilibrium.