Integrating from r to infinity ¥ we obtain the gravitational potential energy V=-GMm/r.
Now we imagine a large enough spherical volume of radius R (~10Mpc) so that space can be considered homogeneous and the force acting on a particle m at R from the interior volume acts as if all the mass M=4/3prr³ within the volume is a point mass R distant and the rest of the universe exerts no net force on m (the rest of the universe cancels for r>R -- shells dr with r>R do not give a net force on a particle at R).
Which gives the evolution of the separation of r from the origin or, invoking the cosmological principle, the separation r between any two particles...we move to a comoving coordinate system, a system carried along with the expansion of the universe. Since the expansion is uniform, the relationship between r and a comoving distance x can be written
The coordinate grid expands with time as the universe expands, but local inertial reference frames remain at fixed locations in the comoving system. Here, a(t) is the scale factor that determines how physical separations change with time. If a(t) doubles, the separation of all local inertial frames doubles.
Now rewrite the total energy using r = a(t) x = a x resulting in
Then multiply by 2/ma²x² to get the Friedmann equation
In this equation k must be independent of x like all the other terms (homogeneity) so while U is constant for a given particle m it changes with separation x, Uµx².
The k=-2U/mc²x² is often called the curvature and in the Friedmann universe retains its value throught its evolution.
If U<0 the system is "bound" and the velocity goes to zero at a radius Rmax=-GM(m/U)and the particle would fall back.
If U>0 the system is unbound. For any given velocity v there exists a corresponding critical mass that makes U=0:
Now we can put the cosmological constant Lambda into our energy equation:
The effect of the new term is to make the expansion rate increase exponentially with time. The new term dominates when
The latest results imply L=0.73 and the matter density 0.27 (total is unity) which gives an older Universe than if L were zero, since the expansion was slower in the past, relieving the "Universe is too young" problem.
Then a change of E in time dt is
while the volume changes by
and if dS=0 (reversible) the first law gives the fluid equation
| dr/dt + 3(da/dt)/a (r+p/c²)=0. |
The density evolution is due to volume increase (adot/a times r) and loss of energy due to work done as the volume of the universe is increased (adot/a times p/c²) = the energy converted into gravitational potential.
Now if we only knew how P varies, the equation of state ... P=P(r,T??)
and substituting for rhodot from the field equation we have
so from the Friedmann equation again we see
and yes, that is +3p/c², any pressure actually decreases (decelerates) the expansion!
(The curvature k cancelled out.)
then k has units t-2 (and time is measured in metres).
Since H=H(t) we denote the value of H we observe today Ho and, since Ho>0 we know the universe is expanding. (This is in terms of our Friedmann universe, but the Hubble expansion can be explained in simpler ways and usually is.) But using H² for (adot on a) squared we have
So H=H(t) and H should decrease with time as the expansion of the universe is slowed by gravitational attraction. If the Friedmann model is right and if there is no extra "dark physics".
The redshift used to justify the big bang is a result of the time dependent scale factor adot on a... the time between emission and absorption of a photon is dt=dr/c,
where dl=l-lo, lo is the "lab" wavelength, l is the (redshifted) wavelength, and integrating:
We usually talk in terms of a "redshift parameter" z=Dl/lo (if you don't see a subscript it is there!) then "redshift" z is related to the scale of the universe by
So if the wavelength has doubled, z=1 and the universe is twice as large as when the light was emitted. (The universe was half the present age when the light was emitted.)
for dust becomes dr/dt + 3r(da/dt)/a=0
i.e. d(ra³)/dt=0 (zero times a³ if you wonder where that factor went!)
Integrating, ra³=constant. Hardly surprising, the density goes as one over the volume! But this is useful since as we are only interested in a ratio of adot to a we can rescale a(t) and it is usual to choose a(present)=1. Then the physical and comoving coordinate frames coincide at the present time and we can write
and if we substitute this into the Friedman equation with k=0 we get
which obviously :) has a solution a µ tq... hmmmm.... the LHS would be adot squared goes as t to the minus 2q-2 and the RHS 1/a goes as t to the minus q so setting the powers equal we have 2q-2=-q or q=2/3. (Wow!) So a(t) goes as t2/3 and since (we set now=to) a(t)=[t/to]2/3 and
The universe expands forever but the expansion rate decreases with time: H goes as 2 over 3t
Solving this as before we find now r=ro/a4 instead of /a³. Applying this to the Friedmann equation and guessing a solution :) we get a=a(t)=(t/to½) so
The expansion is slower if radiation dominated (extra deceleration due to pressure) but as before density falls off as t². The density falls as the fourth power of the scale factor because of increasing volume (three of the four) and redshift of the radiation (the other one.) The drop can also be thought of as the work done p dV by the pressure as the univeerse expands.
The total density is rtot = rdust + rrad where rdustµa-3 and rradµa-4.
Usually one or the other will dominate, but if radiation dominates note that the rapid fall with scale will eventually cause the "dust" to become important and after a long enough time dominate. At present dust dominates and so will continue to do so for the rest of time.