6. Expanding Universe – The Big Bang
It is generally little known or appreciated that when Einstein’s field equations are generalized to take into account the effects on the radius of the universe, the expansion of the universe naturally results.
Einstein's gravitational equations (with cosmological term, L, for the sake of generality) are
R mn – ½ R g mn = T mn + L g mn
After inserting the stress energy tensor equations into the Einstein field equations one gets:
(dR/dt ) (/R)2 = (8 p)/ 3 (G r) – k / R2
(d2R/ dt2 )/ R = - 4 p G mn (p + r/3) + L/ 3
After setting the cosmological constant L = o and eliminating r, one obtains as soln. for R (radius of universe as power law function).
R= (9/ 2GM)1/3 t 2/3
One can deduce from this that at the Planck energy of 1019 GeV, the symmetries of gauge theory were still united in a single force. This is at a time of 10-44 seconds of cosmic age.
This also represents the closest approach of physics to the cosmic singularity (t = 0) but still defines the ‘Big Bang’ since the explosion is already underway and forces are still unified.
Note also that as t increases, R increases, thereby disclosing the expansion of the universe. Originally, Einstein had set out a value for the cosmological constant, which some sources had cited as:
L » - 10 . 3 x 10 -34 sec -2
Which disclosed a retarding force that for any given instant t, allowed the universe to remain static, i.e. rate of instantaneous expansion canceled by retardation. Einstein later admitted this as “the biggest blunder” of his life and thereafter agreed that the best approach was to let L = 0 so that the result conforms with the observations.
It is interesting that we can use basic physics concepts to do with the conservation of energy, along with cosmic expansion as embodied in the Hubble law, i,.e. v = H R
Showing that the velocity v of recession for an object increases with distance R, to find the density near the time of the Big Bang.
The age of the universe (in seconds) related to the Hubble constant by:
t = 1/ H
Currently, we estimate H » 70 km/ sec/Mpc, where Mpc denotes ‘megaparsec’ – where 3.26 light years is equal one parsec)
Since the recession is isotropic (in all directions) we can assume an expanding sphere, for which the total energy must be:
E(total) = K + V
Where K and V respectively denote the kinetic and potential energies (In the latter case, we visualize the work done in displacing a reference mass, say m, away from a central attracting object – doing work against the gravitational force,:
F = . GM m/ R2 so that: V = ò ¥ o (GM m/ R2 ) dR
K = mv2 / 2 and V = - GM m/ R
where R is the radius and G is the Newtonian gravitational constant, G = 6.67 x 10 –11 N- m 2/ kg 2
Let M = r (4 p R 3/ 3)
Where the bracketed quantity denotes the sphere volume and r is the density. Using the above expression for M (total mass) and elementary algebra , we can rewrite the equation for the total energy of the expanding universe as:
E(total) = K + V =
m (HR)2 / 2 - G m r (4 p R 2/ 3)
where the recessional velocity v = HR has been substituted in the first term. We can factor common quantities (to both terms) out and obtain:
K + V = m R2 [ H2/ 2 - G r (4 p / 3)]
Note that the condition for minimal “escape velocity” for a distant object will be attained when E(total) = 0, or the bracketed term is zero., which implies:
H2/ 2 = G r (4 p / 3)
This is exactly the equation that can be used to determine the critical density r c , of the cosmos – beyond which we may expect it to expand forever. Thus:
r c = 3 H2 / 8 p G
This works out – using the current estimated value of H – to about 9.3 x 10 –27 kg/ m3
We can even go beyond this, generalizing the same physics, to obtain an estimate of the cosmic density in the very earliest instants after the Big Bang. We can do so by recognizing that all the above key quantities (H, R, r) are in reality functions of the time t elapsed after the Big Bang. Thus, we can replace R with R(t), H with H(t) and r with r(t). H(t) is not a big worry since we already saw that t = 1/H , so we can re-arrange the earlier equation for critical density replacing H = H(t) with 1/ t, and obtain a simple expression to solve for r(t).
r(t) = 3/ 8 p G t 2
Where various values for t can be substituted into the equation to obtain r(t).- the cosmic density at that instant. For example, say we want to know the density at a time of 0.03 seconds after the Big Bang. Then, substitute t = 0.03 sec, and the value of G (assuming it has not changed with time, and is truly constant!)
r(t). = 1.98 x 10 12 kg/ m3
This is a truly astounding density that fully comports with our expectation that the Big Bang was initiated in an extremely high density state. By way of comparison, plutonium has a density of 19, 200 kg/ m3 . Thus, the cosmic density at t = 0.03 sec after the primordial fireball was just over 100 million times more dense than plutonium!
1) Write out in long form the full sum (all terms) for the interval ds2 = g mn dx m dx n
Be sure to include all terms for the sum that are applicable to standard form.
2 (a) Using the appropriate relations, estimate the density of the universe at a time 0.01 second after the Big Bang.
b) Repeat your computation if the Hubble constant is found to be H = 100 km/ sec/Mpc.
Post a Comment