You are confused, there is nothing in the link supporting your fringe ideas.
Gravitational collapse
- Thread starter arfa brane
- Start date
You are confused, there is nothing in the link supporting your fringe ideas.
One point about gravitational collapse, it isn't all about black holes.
Since a cloud of gas and dust can collapse to a star (or to a body like Jupiter, which modern astronomy claims would have been another star if it had been able to accrete more matter), hence, any planet or other stable structure is the result of gravitational collapse of a larger volume of matter.
Clearly neither the sun or any of the planets is homogenous or pressureless. Do you need to integrate over a density gradient, or doesn't it matter, and why not?
Since a cloud of gas and dust can collapse to a star (or to a body like Jupiter, which modern astronomy claims would have been another star if it had been able to accrete more matter), hence, any planet or other stable structure is the result of gravitational collapse of a larger volume of matter.
Clearly neither the sun or any of the planets is homogenous or pressureless. Do you need to integrate over a density gradient, or doesn't it matter, and why not?
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.
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.
Complete nonsense.
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:
Even the vacuum has an energy density, which is the same thing as pressure.
--http://en.wikipedia.org/wiki/Cosmological_constant
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.
http://www.eftaylor.com/download.html#general_relativity
Still my favorite science book.
Yes, thanks brucep that does look like a good link. Personally, I find Baez's approach a bit confusing at times (perhaps he needs to be more specific about what he means when using terms like "the metric", although this is usually taken to mean the metric tensor).
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.
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.
Markus Hanke
Registered Senior Member
The second one of these equations basically gives a relation for how much the volume of a unit ball around some point in space-time changes; that on the other hand is exactly the geometric definition of the Ricci tensor, if the original volume is taken to be infinitesimal. That gives you the connection between the two equations. See also :
http://math.ucr.edu/home/baez/einstein/node10.html
I think you have not read Baez's link completely. Here i have quoted Baez's conclusion on cosmology from his link.
Baez said:
Obviously you did not read the next section where he explains that the simplified model is incorrect.
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.
In his final equation, Newtonian component( as per the simplified model) is still there. It only differs by 'lambda/R' factor.
...meaning that the cosmological model is NOT "Newtonian", contrary to your fringe ideas.
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.
Note that the reference to geodesic and field equations is about their Newtonian versions, not their (fully) relativistic ones.
--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.
Good Post.
This is not a fringe idea.
In cosmology many equations are equivalent with Newtonian Equations. For example Friedmann Equations are equivalent to Newtonian Interpretaion.
MOND theory which is considered as alternative to Lambda-CDM model, is based on Newtonian Equation.
More-ever in cosmology space itself becomes Euclidean, which is a significant transition from 4D to 3D.
Sure it is, like most (if not all) of your posts. I already pointed out to you the error in your attempt at reading the Baez page(s).
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?
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?
Take a look at a graphic depiction of the Earth's gravity map, and then consider how the Earth's rotation would complicate things. Form a distance the gravitational field of most large objects can be treated as uniform, while close or at their surface their gravity maps will be unique. And trying to map or model the spacetime curvature, in any real sense becomes more difficult. The sort of thing that requires supercomputers and time...
Gravity seems uniform from our perspective on earth, for two reasons. First the variations require sensitive equipement to measure and second we are within the Earth's inertial frame of reference.