> Just sum over the estimated total of 10^80 particles, using mc^2 by first estimating the average mass of those particles for the rest energy, adding their average potential gravitational energy and their average kinetic energy. Why not? AG
You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LC
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AG
> It's claimed the energy is undefined in GR. Regardless, what I am trying to do is estimate what the total energy is, not whether it's conserved for an expanding or contracting universe. AG
On Wednesday, September 4, 2019 at 1:48:15 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AGV = -GMm/r
On Wednesday, September 4, 2019 at 2:37:07 PM UTC-6, Lawrence Crowell wrote:On Wednesday, September 4, 2019 at 1:48:15 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AGV = -GMm/rSure. That's the formula for a particle of mass m at distance r from the center of mass, for mass M. How can you equate that (when integrated from r to infinity) as the potential energy of the FIELD? AG
On Wednesday, September 4, 2019 at 1:48:15 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AGV = -GMm/rRead the following where by using H = 0, zero energy and just Newtoin's laws it is easy to derive the FLRW equations for k = 0 or a flat spatial surface.LC
On Wednesday, September 4, 2019 at 2:37:07 PM UTC-6, Lawrence Crowell wrote:On Wednesday, September 4, 2019 at 1:48:15 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AGV = -GMm/rRead the following where by using H = 0, zero energy and just Newtoin's laws it is easy to derive the FLRW equations for k = 0 or a flat spatial surface.LCBut if the spatial surface is flat, there is no gravity. So how can this be an argument for claiming the total estimated of a universe with gravity is zero? AG
On Friday, September 6, 2019 at 10:31:32 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 2:37:07 PM UTC-6, Lawrence Crowell wrote:On Wednesday, September 4, 2019 at 1:48:15 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AGV = -GMm/rRead the following where by using H = 0, zero energy and just Newtoin's laws it is easy to derive the FLRW equations for k = 0 or a flat spatial surface.LCBut if the spatial surface is flat, there is no gravity. So how can this be an argument for claiming the total estimated of a universe with gravity is zero? AGNot so, for it is embedded in spacetime and there is an extrinsic curvature. You have to research some of this, such as reading Misner, Throne & Wheeler Gravitation Ch 21.LC
On Saturday, September 7, 2019 at 2:05:11 PM UTC-6, Lawrence Crowell wrote:On Friday, September 6, 2019 at 10:31:32 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 2:37:07 PM UTC-6, Lawrence Crowell wrote:On Wednesday, September 4, 2019 at 1:48:15 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AGV = -GMm/rRead the following where by using H = 0, zero energy and just Newtoin's laws it is easy to derive the FLRW equations for k = 0 or a flat spatial surface.LCBut if the spatial surface is flat, there is no gravity. So how can this be an argument for claiming the total estimated of a universe with gravity is zero? AGNot so, for it is embedded in spacetime and there is an extrinsic curvature. You have to research some of this, such as reading Misner, Throne & Wheeler Gravitation Ch 21.LCThanks. I have that book handly and will study your reference. However, on the other issue I raised, I think I am on firm ground that there is no general definition for the potential energy OF a gravitational field; rather just the potential energy of a test particle -- in which case there's something awry wih your additional of gravitation potential energy with rest and kinetic energy. AG
On Sunday, September 8, 2019 at 12:47:28 AM UTC-5, Alan Grayson wrote:
On Saturday, September 7, 2019 at 2:05:11 PM UTC-6, Lawrence Crowell wrote:On Friday, September 6, 2019 at 10:31:32 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 2:37:07 PM UTC-6, Lawrence Crowell wrote:On Wednesday, September 4, 2019 at 1:48:15 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AGV = -GMm/rRead the following where by using H = 0, zero energy and just Newtoin's laws it is easy to derive the FLRW equations for k = 0 or a flat spatial surface.LCBut if the spatial surface is flat, there is no gravity. So how can this be an argument for claiming the total estimated of a universe with gravity is zero? AGNot so, for it is embedded in spacetime and there is an extrinsic curvature. You have to research some of this, such as reading Misner, Throne & Wheeler Gravitation Ch 21.LCThanks. I have that book handly and will study your reference. However, on the other issue I raised, I think I am on firm ground that there is no general definition for the potential energy OF a gravitational field; rather just the potential energy of a test particle -- in which case there's something awry wih your additional of gravitation potential energy with rest and kinetic energy. AGThe definition of energy as some constant of dynamics is difficult in general relativity.LC
On Sunday, September 8, 2019 at 1:28:36 PM UTC-6, Lawrence Crowell wrote:
On Sunday, September 8, 2019 at 12:47:28 AM UTC-5, Alan Grayson wrote:
On Saturday, September 7, 2019 at 2:05:11 PM UTC-6, Lawrence Crowell wrote:On Friday, September 6, 2019 at 10:31:32 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 2:37:07 PM UTC-6, Lawrence Crowell wrote:On Wednesday, September 4, 2019 at 1:48:15 PM UTC-5, Alan Grayson wrote:
On Wednesday, September 4, 2019 at 4:08:58 AM UTC-6, Lawrence Crowell wrote:You also have to include the total gravitational energy or T^{ab} due to local sources and Λg^{ab}.The ADM Hamiltonian constraint is NH = 0 where this Hamiltonian is determined by the traceless transverse part of the extrinsic curvature or Gauss fundamental form. For a general spacetime manifold there is no way to define mass-energy and for most Petrov types the mass-energy is simply no defined. Think of a spherical space with matter throughout. There is no way to construct a Gaussian surface with which to integrate a total mass or energy. Also if that putative surface is embedded in mass-energy then that surface is subject to diffeomorphisms of local curvature. Energy is then not localizable, and in general things that we want invariant are so independent of such diffeomorphisms.LCThe energy of the gravitational field is positive for each particle of average mass. But how does one calculate the negative potential energy for each average mass particle? I can calculate the potential energy of a test particle at some location IN a field, but how can I calculate the total negative potential energy OF the field (for a particle of average mass)? AGV = -GMm/rRead the following where by using H = 0, zero energy and just Newtoin's laws it is easy to derive the FLRW equations for k = 0 or a flat spatial surface.LCBut if the spatial surface is flat, there is no gravity. So how can this be an argument for claiming the total estimated of a universe with gravity is zero? AGNot so, for it is embedded in spacetime and there is an extrinsic curvature. You have to research some of this, such as reading Misner, Throne & Wheeler Gravitation Ch 21.LCThanks. I have that book handly and will study your reference. However, on the other issue I raised, I think I am on firm ground that there is no general definition for the potential energy OF a gravitational field; rather just the potential energy of a test particle -- in which case there's something awry wih your additional of gravitation potential energy with rest and kinetic energy. AGThe definition of energy as some constant of dynamics is difficult in general relativity.LCSince Newtonian gravity doesn't define (negative) potential energy for a gravitational field, and GR doesn't even define (negative) potential energy, do you concede there's no basis for the conclusion that the net estimated energy of the universe is exactly zero? There seems to be nothing negative to add to the positive energies to get zero. AG
Just sum over the estimated total of 10^80 particles, using mc^2 by first estimating the average mass of those particles for the rest energy, adding their average potential gravitational energy and their average kinetic energy. Why not? AG