## Stars

A star is a ball of gas, so why doesn’t it collapse under its own weight. Yup! The outward pressure of the hot stuff pushing outwards! But how much hot stuff does it take to reach a **Hydrostatic Equilibrium**?

A stars density and pressure increases towards its centre and for it to be stable, that outward pressure must balance the inward pull of gravity.

For a sphere,

*∫ _{0}^{R }4πr^{3}(dP/dr) = [P(r) 4πr^{3}]_{0}^{R} – 3∫_{0}^{R} P(r)4πr^{2} dr*

Complicated looking I know, but pretty much summed up by saying that the average pressure is,

<P> = -U/3V

i.e. The average pressure is one third of the gravitational potential energy density.

In most stars, the dominant particles are much slower than c (the speed of light) and are therefore non-relativistic. Remembering (as you do) that the velocity of a particle is **v** = (v_{x},v_{y},v_{z}) and momentum is **p** = (p_{x},p_{y},p_{z}); for a box with sides of length *L*, that contains *N* particles,

*p _{x} = (N/L^{3})<p_{x}v_{x}>*

And if the particle motions are isotropic (they very likely are as they are random),

*<p _{x}v_{x}> = <p∙v> */3

So for a non-relativistic gas, **p∙v** =mv^{2} , ρ =-U/3v and E_{tot} = -U/2

However for a relativistic gas, **p∙v** =pc , ρ =-U and E_{tot} = 0

∴ A relativistic gas (as would be found in a white dwarf or very luminous star) is gravitationally barely bound and unstable. The addition of even a small amount of energy would make the star gravitationally unbound. The **Free Fall Collapse Time** would be

*t*_{ff} = √(3π/32*Gρ*) (a function of *ρ* only!)

Stellar Origin

For a cloud with mass M, radius R, total number of particles N, average particle mass m̅, and temperature T; the gravitational potential is,

*U = -C*

*(GMNm̅/R)
*

The constant *C* depends on the gas distribution within the cloud, but will be close to 1 for most purposes. The total kinetic energy is,

*K = (3/2)NkT (k* is the usual Boltzmann constant)

∴ The cloud will collapse when |U| > K

The **Jeans Mass** represents the mass required with R for collapse,

*M _{J} ≈ (kT/(Gm̅))R
*

And the **Jeans Density** is,

ρ_{J} ≈ 1/(πM)^{2}(kT/Gm̅)^{3}

Observations show that stars from in clusters which suggests a multi-step process in which a large cloud collapses and fragments into protostars.

The free fall to a radius of 10^{11}m (at which point the collapse would be stabilised by the heating of the protostar) is ≈ 20,000 years.