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kurious
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In the following discussion we will use an equation which is a possible candidate for quantum gravity to demonstrate that the size of the universe at the time of the Big Bang could have been much greater than is currently thought by the majority of physicists.
Using mvr = nh/2pi from Bohr’s theory of the hydrogen atom:
v = nh / 2pi m r
acceleration = (v1 – v2) / t = (n1/m1r1 – n2/m2r2) ( h / 2pi ) / t
acceleration = q^2/ 8 pi^2E0 h t x [ 1/n1 – 1/n2 ] using r = 4 pi x E 0 n^2h^2/ me^2
acceleration = 10^12 ( 1 – 1/ n2)
assuming that for the minimum velocity n1 = 1 and t = 10 ^ - 8 seconds which is what t is for electrons in most atoms.
At the surface of the sun acceleration = 10^2 m/s or thereabouts.
This means n2 must be about 1.0000000001 in the equation above.
If we imagine the universe reached the size of the Sun, then bearing in mind its mass is 10^22 times greater than the mass of the Sun and assuming 1/n2 is proportional to mass then 1/n2 at the surface of the universe was about 1 x 10^22 for a quark, let’s say.
So acceleration = 10^12 x 10^22 = 10^34 m/s^2.
A quark sits at the surface of the universe.
Let’s see what happens if we assume Newtonian physics or at least some of it applies to quarks.We assume also that the the quark consists of a sphere of electric charge and that the charges on this sphere repel one another and that the force of repulsion exactly matches the force of gravity trying to compress the quark at 10^8 metres:
Force = quark mass x acceleration ( we’ll use the mass of an up quark )
Force = 10^ -28 x 10^ 34 = 10^ 6 Newtons.
Now the force of repulsion is given in classical physics by:
kq^2/ r^2
So kq^2 / r ^ 2 = 10^6
10^9x 10^ -38 / r^ 2 = 10^ 6
r = 10^ -18 metres.
There are about ten quarks per cubic metre in the universe, so since
total number of quarks in a particular direction is about 2 (cubed root of ten)
per metre nowadays, there can be about 10^26 quarks in a particular direction
(10^26 = maximum size of universe) nowadays and in the past.If each quark was about 10^ -18 metres in diameter at the start of the universe then the size of the universe was 10^26 x 10^ -18 metres = 10^ 8 metres.
Just what we said it was at the beginning of this exercise!
Using mvr = nh/2pi from Bohr’s theory of the hydrogen atom:
v = nh / 2pi m r
acceleration = (v1 – v2) / t = (n1/m1r1 – n2/m2r2) ( h / 2pi ) / t
acceleration = q^2/ 8 pi^2E0 h t x [ 1/n1 – 1/n2 ] using r = 4 pi x E 0 n^2h^2/ me^2
acceleration = 10^12 ( 1 – 1/ n2)
assuming that for the minimum velocity n1 = 1 and t = 10 ^ - 8 seconds which is what t is for electrons in most atoms.
At the surface of the sun acceleration = 10^2 m/s or thereabouts.
This means n2 must be about 1.0000000001 in the equation above.
If we imagine the universe reached the size of the Sun, then bearing in mind its mass is 10^22 times greater than the mass of the Sun and assuming 1/n2 is proportional to mass then 1/n2 at the surface of the universe was about 1 x 10^22 for a quark, let’s say.
So acceleration = 10^12 x 10^22 = 10^34 m/s^2.
A quark sits at the surface of the universe.
Let’s see what happens if we assume Newtonian physics or at least some of it applies to quarks.We assume also that the the quark consists of a sphere of electric charge and that the charges on this sphere repel one another and that the force of repulsion exactly matches the force of gravity trying to compress the quark at 10^8 metres:
Force = quark mass x acceleration ( we’ll use the mass of an up quark )
Force = 10^ -28 x 10^ 34 = 10^ 6 Newtons.
Now the force of repulsion is given in classical physics by:
kq^2/ r^2
So kq^2 / r ^ 2 = 10^6
10^9x 10^ -38 / r^ 2 = 10^ 6
r = 10^ -18 metres.
There are about ten quarks per cubic metre in the universe, so since
total number of quarks in a particular direction is about 2 (cubed root of ten)
per metre nowadays, there can be about 10^26 quarks in a particular direction
(10^26 = maximum size of universe) nowadays and in the past.If each quark was about 10^ -18 metres in diameter at the start of the universe then the size of the universe was 10^26 x 10^ -18 metres = 10^ 8 metres.
Just what we said it was at the beginning of this exercise!
