The total rest energy of all the matt...

The total rest energy of all the matter in the observable universe

Posted in the Science Forum

Essam Maroun

Jordan

#1 Dec 1, 2012
What does the Lambda-CDM model say about the initial mass-energy of the universe and what does Maroun say?

We were hoping to find the quantity of the initial mass of the cosmos through "Einstein's radius of the universe":

r =(constant#r)/( d ^ 1/2 )…(1)

(constant#r)= c /( 4 pi G )^ 1/2

r: Einstein's radius of the universe

d: Density of the space of the universe

c: The speed of light constant

G: The Newtonian gravitational constant

pi = 3.14159

and by favoring a sphere universe, the volume of the universe ( V ) is given by the formula:

V = 4/3 pi r^3 …(2)

Finally, we get the final result by applying the formula:

m = V d …(3)

m: The mass-energy of the universe

Cosmology substitutes Planck length as the initial radius of the universe ( r0 ) in eq #1, remember that r0 = 10^-35 m.

As a result d0 = 10^97 kg/m^3

And V0 = 10^-105 m^3

This makes m0 = 10^-8 kg, or a consequence equals Planck mass

So the outcome of m0: The initial mass of the universe from Einstein's radius of the universe is wrong and makes no sense because the quantity 10^-8 kg is too big for the subatomic scale universe, but we get correct results by applying the current radius of the universe ( r2 ) which approximately equals around 10^25 m.

Substituting r2 in eq #1 gives a final mass of m2 = 10^52 kg.

m2: The estimated total mass-energy of the observable universe

Anyway, I did the clean work in physics and cosmology, and I found a formula that applies at the subatomic and the large scale at the same time, a formula that sets m0 and m2 in one theory that can be called the theory of everything.

In contrast with the Lambda-CDM, my theory works everywhere and at any moment. Maroun's equations share the same results with the modern theories on the large scale universe, but we know that the Lambda-CDM fails in the subatomic scale.

I found out that:

m2 = 10^52 kg.

m0 = 10^-38 kg.

This is by using my discovery of the general formula of the universe.

m = constant ( t ^ 3 )^ 1/2

constant = c^3 / G *(1/t2)^1/2

t: Time which is determined through the interval [ t0, t2 ]

t2: The instant age of the universe at this moment according to the big bang theory

t0: Planck time constant

Remember that r0 = c t0

- What is the amount of energy released in the Big Bang?

I value Dr. Gerald Schroeder opinion, and I keep in mind that the Jews are the first masters of the Bible after God: when a Jewish master explains science, you will understand that cosmology does not contradict the Bible.

I hope this was good introduction !

Maroun complete picture of the universe & the equations of all times (a solution for the initial state of the universe).

Maroun general formula of the cosmos unifies the subatomic universe with the large scale. Through my equations, where I consider the law of conservation of energy, you can calculate the quantity of the total observable mass-energy of the universe at any moment (this includes dark matter and dark energy). The universe has a constant amount of total mass-energy (EC) that is equal to the total observable mass-energy at this moment (which is known as the "age of the universe"). It is true that the universe has a constant amount of mass-energy, but virtually, the physical quantities of the observable universe are changing and here I introduce a set of equations to estimate them.

Sincerely,

Essam Maroun
FREE SERVANT
#2 Dec 1, 2012
If the universe is a sphere, then all points are equally distant from a center. If the universe is a geometric figure generated by the revolution of a semicircle about its diameter then the earth could be at its center after all. All the universe could be moving from the earth. Just sayin......

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