Please forgive they way I have presented these equations. I seem to be having trouble entering equations onto the site so I have used a fortran style.
As this part of the website is here to air ATM theories I want to put forward a purely geometric quantum theory of gravity whose value matches that Newtonian gravity within the confines of the solar system where we know Newtonian gravity works well (except for the pioneer anomaly).
To do this I will only use simple geometry but a paradigm shift in thinking is required. I will contend that the presence of matter (therefore energy) generates a super-luminal quantum field around itself and a mass-less particle considered at rest, will generate the largest quantum field.
Mass will have a retarding effect on the generation of an objects quantum field and the base of any measurement will be taken to be that of the quantum field of a mass-less particle considered at rest.
I assert that the radius of a quantum field generated by a mass-less particle in one second will be equal to ‘c’ or 299 792 458 m giving a volume of V = 4/3*pi*c^3 and here V = Phi = 1.129 x 10^26 m^3 s^-1.
To make this ATM a quantum theory I need a time and space quanta which will be 1/Phi = 8.86 x 10^-27 m^-3 s^-1.
Space will be measured by the radius of the quantum field and time will be measured by the volume of the quantum field. This simple contention will separate time and space during measurement but space-time will remain one entity.
As I said earlier mass will retard or distort the quantum field from its true value. I will use the volume of the distortion of the Earth’s quantum field caused by the Earth’s mass to gain a space-time constant ‘STC’. I will use it to calculate the quantum field distortion caused by any mass and give an example of the Sun’s quantum gravity field.
At the Earth surface and according to Newton, gravity has a distortion of g = 9.81 m s-2 so the volume of this distortion caused by the Earth will be Vde = (4/3*pi*(Re + g)^3) – (4/3*pi*Re^3) = 5.01 x 10^15 m^3 where Re = the radius of the Earth.
Now I will simply divide the volume of Earth’s distortion by the mass of the Earth to gain the space-time constant, so STC = Vde/Me = 8.386 x 10^-10 where ME is the mass of the Earth.
Now let’s see if this geometric quantum method can match Newtonian gravity for the gravity field of the Sun.
As stated earlier the base of any measurement will be the flow of a quantum field generated by a mass-less particle considered at rest. Here space is ST = ((3*t*Phi)/(4*pi))^1/3 where ‘t’ is the time input in seconds.
The Sun’s quantum field will be retarded by a small amount because of its mass. The volume of the distortion of the Sun’s quantum field will be Vds = Ms x STC = 1.67 x 10^21 m3 where Ms = Mass of the Sun. Now I need to convert this value to time using the space-time quanta.
The time distortion for the Sun will be tds = (1/Phi)*Vds = 1.48 x 10^-5 s. This means the Sun’s quantum field is retarded by tdS and the flow of the Sun’s quantum field itself will be ts = t – tds.
Using this method the measurement of space and time are completely interchangeable. It will take a mass-less particle considered at rest t = ((4*pi*Rs^3)/(3*Phi)) = 12.49 s, where Rs = radius of the Sun, for the quantum field to reach the surface of the Sun from its centre of mass so ST = (3*t*Phi/4*pi)^1/3 = 6.956 x 10^8 m.
The retardation of the Sun’s own quantum field will be STS = ((3*ts*Phi)/(4*pi)) = 6.955 997 256 79 x 10^8 m giving a quantum distortion value for the Sun’s gravity of qgs = ST – STs = 274.321 m s^-2 if we check this with Newtonian gravity Ngs = G*Ms / Rs^2 = 274.321 m s^-2 and as you can see they are in agreement.
Combining the geometric equations and checking the value of the Sun’s gravity at 1 AU we have ST = (3*(((4/(3*phi))*pi*AU^3)*phi)/(4*pi))^(1/3) = 1.495 978 706 91 x 10^11 and
STs = ((3*(((4/(3*phi))*pi*AU^3)-TdisSun)*phi)/(4*pi))^(1/3) = 1.495 978 706 909 94 x 10^11
Giving a value for the distortion of quantum gravity at 1 AU from the Sun of qgs = ST -STs =5.93 x 10^-3 m s^-2
The value for Newtonian gravity will be Ngs = GMs/AU^2 = 5.93 x 10^-3 m s^-2.
Newtonian gravity and quantum gravity differ by only 9.1 x 10^-9 m s^-2 over this distance.
This method of calculating quantum gravity remains very close to Newtonian gravity for any planet or moon over any distance.
If anyone would like to test this for their self, I will be happy to provide the equations in a form that can be directly entered into the free Console calculator which is accurate to 80sf and can be downloaded at http://ccalc.shanebweb.com/
This paradigm shift in thinking about time and space will take some people by surprise but the mathematics work and work well. I hope you will try it yourself and that it gives everyone food for thought.