The total rest energy of all the matter in the observable universe
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Last updated Dec 1, 2012
Jordan 
#1
Dec 1, 2012
What does the LambdaCDM model say about the initial massenergy 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 massenergy 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 massenergy 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 LambdaCDM, 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 LambdaCDM 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 massenergy of the universe at any moment (this includes dark matter and dark energy). The universe has a constant amount of total massenergy (EC) that is equal to the total observable massenergy at this moment (which is known as the "age of the universe"). It is true that the universe has a constant amount of massenergy, 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 
#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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