View Single Post
gravity pressure
Old
  (#1 (permalink))
AntonioLao
Raider of the lost time
AntonioLao is just really niceAntonioLao is just really nice
 
AntonioLao's Avatar
 
Status: Offline
Posts: 5,318
Thanks Given: 728
Thanked 124x in 122 Posts
Join Date: Nov 2003
Rep Power: 74
   
gravity pressure - 11-05-2007, 12:44 PM

An implication stemmed from Einstein’s theory of general relativity is that not only mass and energy can create gravitational field, pressure defined as force per unit area also can. When it is negative, it is antigravity. It is formulated from a relation to the dimensional ratio of local acceleration of gravity (g) over the universal constant of gravitation (G), K=g/G. In the MKS system, the dimension of K is kilograms per meter squared (kg/m). The values of K vary since g is a local variable depending on the mass (m). For near earth systems, K is 1.5x10 kg/m.

Therefore, gravity pressure (P) is the difference of normal pressure N/S and Ka, where a is the generalized absolute acceleration and N is the normal force or weight, equation 1: P = KaN/S. If N=mg then P = g(1.5x10¹ºam/S), where S is the surface area to the normal. The absolute magnitude of the generalized absolute acceleration a can be determined independently from equation 2: a·r=c or |a||r|cosq=c, where r is the infinitesimal spacetime metric and c is the speed of light. For positive cosq, the domain of q is [0°,90°) and (270°,360°]. The condition for antigravity is that Ka << N/S or 1.5x10¹ºa << m/S. For an object with m equal 1 kilogram and S is 1 meter squared, the antigravity’s distance |r| is 1.29 billion billion billion meters, which is approximately the radius of the observable universe if light had traveled for 13.7 billions years. If the universe is spherical in shape then positive gravity pressure can be expressed as total energy per unit volume of this radius. Furthermore, the infinitesimal change of radii of the singularity from the big bang must satisfy the uncertainty principle of quantum mechanics: equation 3: Dr·Dp³h/2p, where p is the linear momentum and h is Planck’s constant of action.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
Reply With Quote
The Following User Says Thank You to AntonioLao For This Useful Post:
Lloyd Gillespie (11-06-2007)