# Conclusive Proof that Einstein's Field Equations are Wrong

It is quite astonishing that the negative pressure in Einstein's equations has gone unnoticed for such a long time. Or, is it that it has been completely ignored? The errors involved are:

1. The pressure is negative implying implosion, and not explosion as the inflationary scenario would have it;

2. The pressure does not contribute to the mass of a gravitating body; on the contrary, in a state of hydrostatic equilibrium it acts to oppose the action of gravity;

3. In stellar formation, energy and matter are radiated away: the binding energy is related to the mass defect by Einstein's formula. Consequently, a star will end up with a mass M that is smaller than the initial mass M_0 of the gas cloud, and not as Einstein's field equations predict that it will be larger than the initial gas cloud!

For a perfect fluid, the four Einstein field equations can be summed to read:

-a**/a=(4 pi/3) G(rho+3p/c^2),

where the asterisks denote differentiation with respect to time, a is a dimensionless scale factor in the FRW metric, rho is the mass density and p is the hydrostatic pressure. The right hand side is proportional to the trace of the energy-stress tensor of a perfect fluid.

In the last blog we have shown that a**/a=-G rho. If the mass M is constant rather than the density a**/a=-GM/r^3, which is tidal acceleration. Introducing this into the above equation results in

M=M_0(1+3p/rho c^2),

where M_0=(4 pi/3)rho r^3, is the initial mass in the gas cloud. The above equation would have us believe, therefore, that the final mass M is greater than the initial mass M_0 because the pressure contributes to the mass of a gravitating body!

Rather, the condition for hydrostatic equilibrium is

dp/dr=-GM_0/r^2 rho.

To understand this equation, consider a spherical cloud, inside of which is a thin shell. The shell experiences a pull from all the mass M_0 inside the inner radius of the shell. The density rho is that at the center of the thin shell, and it may be different at different parts of the cloud, or be uniform. If the gravitational force acted alone, it would cause the shell to collapse like a balloon being deflated. The acceleration would be proportional to G rho, the square of the Newtonian free fall time. However, inside the inner boundary of the shell is a force due to the pressure exerted by the gas particles which presents its collapse. The above equation expresses the condition of hydrostatic equilibrium where the acceleration of the shell vanishes.

The quantity on the left is the pressure gradient, and it can be approximated by the slope of the p vs. r curve, -p/r. The slope must be negative since the pressure must decrease as the radius of the gas cloud increases. Introducing this into the equation of hydrostatic equilibrium and rearranging lead to

p/rho c^2=GM_0/rc^2,

the ratio of the "classical" Schwarzschild radius, R_s=GM_0/c^2, to the radius r. Now, introducing this into the second term of the expression for the mass leads to the conclusion that the mass of a star will increase due to the pressure that resists gravitational attraction:

M=M_0(1+3M_0/rc^2).

However, it is well known that star formation involves a mass defect---and not a mass increase---due to the radiation of light, neutrinos, etc.

Any scenario based on Einstein's equations has to be condemned for the same reason. A case in point is the so-called inflationary scenario which uses as it due ex machina a negative pressure which creates an expansion of the early universe.

Finally, on the left hand side of the Einstein field equations are the Ricci tensor components. Yet, we know that it is the tidal force that brings particles together. This is the fundamental object of any theory of gravity and not their accelerations. For this one needs the full Riemann tensor, and not the Ricci tensor which is its contraction.