**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**

_{m}

_{n}**– ½ R**

_{ }**g**

_{m}

_{n}**=**

_{ }**T**

_{m}

_{n}**+ L**

_{ }**g**

_{m}

_{n}

_{ }
After
inserting the stress energy tensor equations into the Einstein field equations
one gets:

(dR/dt
) (/R)

^{2}= (8 p)/ 3 (G r) – k / R^{2}
whence:

(d

^{2}R/ dt^{2})/ R = - 4 p G_{m}_{n}**(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

**of 10***Planck energy*^{19 }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 “

**” of his life and thereafter agreed that the best approach was to let L = 0 so that the result conforms with the observations.***the biggest blunder*
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 ‘**’ – where***megaparsec*__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/ R

^{2}so that:**V =****ò**^{¥}^{ }_{o}_{ }**(GM m/ R**^{2}) dR
Then:

**K = mv**

^{2}/ 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 R**

^{2}[ H^{2}/ 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:

**H**

^{2}/ 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 H^{2}/ 8**p**

**G**

This works out – using the current estimated value of H – to about 9.3 x 10

^{–27}kg/ m^{3}
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/ m^{3}^{}

^{}

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/ m

^{3 }. Thus, the cosmic density at t = 0.03 sec after the primordial fireball was just over 100 million times more dense than plutonium!__Problems__:

1) Write out in long form the full sum (all
terms) for the interval

**ds**^{2 }=**g**_{m}_{n}**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.

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