Have you read the link i provided? This is as per the link and not my conclusion.
You are confused, there is nothing in the link supporting your fringe ideas.
Have you read the link i provided? This is as per the link and not my conclusion.
General relativity is concerned with the mental construct, space-time. Beyond the mental construct, gravity also produces pressure and phase changes of matter, that are dependent on this induced pressure. The iron core of the earth is solid at 8000C, because of the pressure created by earth's gravity. If we left the space-time well of the center of the earth, the iron would become liquid and then vapor at that temperature. This is not GR but material reality.
Gravitational pressure is not the same as space-time effects, since as the pressure increases, the phase transitions of matter get faster and faster. Whereas, the time aspect of space-time slows as gravity gets stronger. The nuclear reactions of a star, which is the fastest phase transition within the star, occur at the bottom of the space-time well where time has slowed the most. Pressure is above and beyond GR.
This can also be inferred quite easily. Space-time is one part space and one part time (space-time). Gravity is an acceleration which is one part space and two parts time; d/t/t. Gravity contains more time (potential) than specified by GR. The extra time potential creates pressure (force/area) allowing time transitions to speed up as space-time slows.
If space-time caused pressure than travel near C via SR would squish us into new phases that speed up; we would eventually undergo fusion and convert to energy as we approach C. But SR velocity (d/t) does not have the extra time potential, as does gravity (d/t/t), so pressure is of no concern.
Sorry, no. In a neutron star for example, pressure contributes significantly to total gravitational energy density. According to Baez, mass density and pressure are comparable in neutron stars.wellwisher said:Pressure is above and beyond GR.
--http://en.wikipedia.org/wiki/Cosmological_constantThe cosmological constant has the same effect as an intrinsic energy density of the vacuum, ρ[sub]vac[/sub] (and an associated pressure)
Baez is an expert gravitational physicist and knows how to use GR physics and mathematics to evaluate the spacetimes and events associated with the GR domain of applicability. That's what it's all about. What you're reading from Baez is a good way to beef up your understanding. This is a nice chapter on spacetime curvature. Easy to read. Chapter 2 of Exploring Black Holes.Sorry, no. In a neutron star for example, pressure contributes significantly to total gravitational energy density. According to Baez, mass density and pressure are comparable in neutron stars.
Even the vacuum has an energy density, which is the same thing as pressure.--http://en.wikipedia.org/wiki/Cosmological_constant
It would be nice to be able to get from Einstein's usual formula $$
R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} + \Lambda\,g_{\mu \nu} = {8 \pi G \over c^4} T_{\mu \nu} $$ to $$
\frac {\ddot V} { V} {\mid _{t = 0}} = -{1\over 2} (\rho + P_x + P_y + P_z) $$. Or the other way. By which I mean, show how these equations are related.
You are confused, there is nothing in the link supporting your fringe ideas.
Baez said:In other words, the equation governing this simplified cosmology is the same as the Newtonian equation for what happens when you throw a ball vertically upwards from the earth! This is a nice example of the unity of physics.
I think you have not read Baez's link completely. Here i have quoted Baez's conclusion on cosmology from his link.
Do you see the "simplified" in the "simplified cosmology" in the quote? Obviously not. As long as you pretend to be doing physics like Farsight, via quote cherry-picking and by not understanding the underlying physics and math, you won't be doing any physics.
It is simplfied because he considered it as homogeneous and isotropic.
Obviously you did not read the next section where he explains that the simplified model is incorrect.
As long as you pretend to be doing physics like Farsight, via quote cherry-picking and by not understanding the underlying physics and math, you won't be doing any physics.
...meaning that the cosmological model is NOT "Newtonian", contrary to your fringe ideas.In his final equation, Newtonian component( as per the simplified model) is still there. It only differs by 'lambda/R' factor.
--http://en.wikipedia.org/wiki/Newtonian_motivations_for_general_relativity.Overview of the Newtonian picture
The geodesic and field equations simply are a restatement of Newton's Law of Gravitation as seen from a local frame of reference co-moving with the mass within the local frame. This picture contains many of the elements of General Relativity, including the concept that particles travel along geodesics in a curved space (spacetime in the relativistic case) and that the curvature is due to the presence of mass density (mass/energy density in the relativistic case). This picture also contains some of the mathematical machinery of General Relativity such as tensors, Christoffel symbols, and Fermi-Walker transport.
Another thing about all this Newtonian/Einsteinian model thingy: mass appears to have various definitions or formulations. Newtonian gravity (where mass is a scalar density) does in fact contain some of the mathematical structures of GR.--http://en.wikipedia.org/wiki/Newtonian_motivations_for_general_relativity.
Note that the reference to geodesic and field equations is about their Newtonian versions, not their (fully) relativistic ones.
...meaning that the cosmological model is NOT "Newtonian", contrary to your fringe ideas.
This is not a fringe idea.
Ok, staying with this notion of pressure being equivalent to energy density.
The planet earth is a gravitationally collapsed 'body', which is stable--its volume is not changing--so the pressures in each 'direction' (here, we need one along the time dimension) must sum to zero. The earth isn't a sphere, but rather a spheroid; it has a bulge at the equator, so gravitational acceleration is greatest at the poles, where the surface is closer to the COM.
So, somewhat non-intuitively, although there is more mass near the equator g is smaller there. This is because (the force of) gravity is the inverse-square of distance between centres of mass.
What's a good way to interpret this in terms of curvature?