Jack SarfattiOn Feb 1, 2013, at 3:56 PM, Paul Zielinski wrote:

He's saying that inertia is a real Newtonian force, not a fictitious force.

If you don't disagree, then why disagree?

My point here is that in Newtonian terms it isn't an "impressed" force, although it can appear that way in a rotating frame.

I answered: Jim in his book clearly says

"m" = inertia

With that definition "inertia" is NOT a FORCE!

REAL 4-Force on a test particle = Covariant derivative of 4-Momentum of that test particle with respect to that test particle's PROPER TIME

The CONNECTION PART on the RHS contains ALL FICTITIOUS INERTIAL PSEUDO-FORCES from the proper accelerations on the DETECTOR NOT the test particle!

NOW - IN THE VERY SPECIAL CASE OF THE REST FRAME OF THE NOW OFF-GEODESIC TEST PARTICLE a piece of the connection survives and cancels against the Real Force ON THE TEST PARTICLE. IN THIS SPECIAL CASE IT IS THE DETECTOR ITSELF THAT IS ALSO THE CONSTRAINT CAUSING THE REAL FORCE ON THE TEST PARTICLE, AND THE SURVIVING PIECE OF THE CONNECTION IS NOW THE REAL INERTIAL REACTION FORCE BACK ON THE DETECTOR/CONSTRAINT! THUS NEWTON'S THIRD LAW IS OBEYED LOCALLY, BUT THE QUANTUM THEORY EXPLANATION OF THE CAUSE OF BOTH THE REAL FORCE ON THE TEST PARTICLE FROM THE DETECTOR, WHICH NOW IS ALSO THE CONSTRAINT, AND THE INERTIAL REACTION FORCE BACK ON THE CONSTRAINT FROM THE TEST PARTICLE IS ELECTROMAGNETIC-WEAK-STRONG + PAULI EXCLUSION PRINCIPLE LOCAL CONTACT NEAR FIELD FORCES - NOT ANYTHING IN CLASSICAL GR OR MACH'S PRINCIPLE IS NEEDED OR CAN EXPLAIN THE ORIGIN OF "m" - that is a quantum effect! "m" is a free parameter in classical physics.

SEE WHAT LANCZOS WROTE BELOW EQ (5)

(REAL FORCE ON TEST PARTICLE OF INERTIA m)i - (m/g44){^4^4i} = 0 IN THE LNIF REST FRAME OF THE CONSTRAINED TEST PARTICLE

Jack SarfattiPS On Feb 1, 2013, at 4:58 PM, JACK SARFATTI <sarfatti@pacbell.net> wrote:

(REAL FORCE ON TEST PARTICLE OF INERTIA m)i - (m/g44){^4^4i} = 0 IN THE LNIF REST FRAME OF THE CONSTRAINED TEST PARTICLE

NOTE THE REAL HORIZON SINGULARITY

g44 = 0 where the curvature is finite

here, putting in the quantum gravity Planck cutoff

the Unruh temperature on the horizon is hc/LpkB

but far away that temperature redshifts down to hc(LpA^1/2)^1/2kB

i.e. the Planck scale for Hawking blackbody radiation asymptotically redshifts down to the GEOMETRIC MEAN of the Planck scale with scale of the horizon A^1/2 where A is the AREA-ENTROPY of the g44 = 0 horizon.

in cosmology this horizon is observer dependent - the observer always see the above GEOMETRIC MEAN.

In the case of our universe, the COSMOLOGICAL GEOMETRIC MEAN is (10^-33 10^28)^1/2 cm ~ 10^-2 cm.

That Hawking temperature using Stefan-Boltzmann law T^4 gives the observed dark energy density hc/Lp^2A