Inertia is about something, not nothing. Nothing has no inertia. :-)

By "inertia" I mean REST MASS of real particles m0. I don't know what you and Jim mean by the same word.

I do not think that rest masses need Mach's principle. Higgs field and QCD explain them LOCALLY.

So Newton's 2nd Law in curved space-time is

F = DP/ds

P = MV

M = m0(1 - (V/c)^2)^-1/2

D/ds = covariant derivative with respect to proper time ds and the Levi-Civita Chistoffel connection

DP/ds = MdV/ds + VdM/ds

V = Dx/ds = dx/ds

On Jan 12, 2012, at 1:44 PM, Paul Zielinski wrote:

Jack if this is your way of conceding fine. But I hope you can acknowledge at least to yourself that "inertial reaction" is not
about electrical contact forces, it's about inertia. You push on an object, and you push yourself back at the same time.
Doesn't matter whether there is direct contact or not.

SORRY Z BUT YOU STILL DON'T GET IT!

I don't?

I'm afraid you don't.

Action-Reaction is from conservation of linear momentum, there need not be PUSHES at all!

Right, It could be a pull. If you pull an object towards you, you pull yourself towards the object at the same time. Because both you and the object possess inertia.

NO! It need not be a "pull" either. Push or pull - another Laputan difference that does not make a difference!

There need not be any direct contact for this principle to apply. Central forces will do quite nicely -- whether repulsive or attractive.

That's what I told you. However, there are no central forces in gravity. It's purely geodesic no forces at all. No real pushes or pulls. By definition, a push and pull transforms a geodesic to an off-geodesic path in the local curvature field.

Then clearly contact forces have nothing to do with third law action/reaction.

In the EM case, the static EM field has linear momentum (e/c)A at the position of the charge!

kinetic momentum - (e/c)A is a local gauge invariant

so that is a CONTACT interaction.

Conservation of linear momentum comes from space translation symmetry in the dynamical actions of the systems involved via Noether's theorem.

You keep repeating this but what you are saying here has no basis in Noether's theorem, There is nothing in Neother's theorems about conservation of linear momentum, for example, "coming from" translational symmetry of the Lagrangian. It's a logically independent proposition that you are simply taking on to Noether's work.

Another not-even-wrong difference that makes no difference.

How can Noether's theorems explain why the Lagrangians of all Newtonian closed n-body systems have translational symmetry? You haven't answered this question Jack. Because you can't.

What does "explain" mean here?  Noether's theorem connects translation symmetry to conservation of linear momentum.

Conservation of linear momentum in the two-body problem gives equal opposite reaction quite trivially.

P(A,B) = P(A) + P(B)

dP(A,B)/ds = 0 conservation from invariance of the action S(A) + S(B) + S(A,B) under rigid global T3 group.

Therefore,

DP(A)/ds + DP(B)/ds = 0

i.e. F(A) = - F(B)

Elementary physics.

Indeed, in the two-body pure gravity problem there are no PUSHES (INERTIAL REACTION FORCES) because both masses are on force-free timelike geodesics in their combined composite curvature field.

This is a Newtonian system Jack. Why are you mixing things up? Aren't you just blowing smoke?

No Z you fail to understand the physics here. Even in Newton's theory there is no inertial reaction force associated with Newton's "fictitious" gravity force. You still don't get the difference. You are weightless when Newton's "gravity force" acts on you. In contrast, if you carry an electric charge floating in space, you will feel the electric field pushing you off the local timelike geodesic! You never feel Newton's gravity pseudo force. You always feel Maxwell's electric force.

If you really think inertia plays no role in this Newtonian example, specifically with regard to conservation of linear momentum, then I really don't know what to say.

All that matters here is

DP(A)/ds + DP(B)/ds = 0

where

DP/ds = MdV/ds + VdM/ds

And M does appear. Is that what you mean?

I keep telling you (and even Jim sometimes) not to handwave with words when an equation will pin down what you mean.

PUSHES OFF MINKOWSKI GEODESICS BY NEWTON'S GRAVITY FORCE ARE A CHIMERA, A DELUSION, BECAUSE NO-WEIGHT NO-g-force is felt. NEWTON'S GRAVITY FORCE IS NOT LIKE AN ELECTRICAL FORCE IN THAT REGARD.

So the Newtonian gravitational pull of the earth doesn't pull the moon toward it? And the moon doesn't reciprocate by
pulling the earth toward it? And all this is no governed by Newton's third law? And there is no action/reaction in this
example?

That's right. There is no REAL PULL there is only GEODESIC MOTION. Pull means local g-force from Newton's 2nd law with F =/= 0.

You can say there is a PSEUDO-PULL - AS IF A FORCE ACTED.  BUT IT'S NOT A REAL FORCE. A REAL FORCE CAUSES OFF-GEODESIC MOTION RELATIVE TO THE ACTUAL TENSOR CURVATURE FIELD.

There is no g-force in any purely gravitational system.

The above case is different from when you step on a scale on Earth - there you have quantum EM near field intermolecular forces pushing both you and Earth off timelike geodesics of the total local curvature field. Of course the perturbation of Earth's worldline is generally too small to be detected.

Timelike geodesics have nothing to do with this. This is Newtonian physics. You are getting everything mixed up.

NO Z, EVEN IN NEWTON'S THEORY THERE IS NO g-FORCE IN A PURELY GRAVITY INTERACTION. UNLIKE THE EM INTERACTION. YOU FAIL TO UNDERSTAND EINSTEIN'S THEORY AND EVEN NEWTON'S THEORY.

THE PHYSICAL QUESTION IS WILL THE POINTER OF AN ACCELEROMETER MOVE OR NOT!

Mach's Principle is completely irrelevant to both examples above UNLESS YOU HAVE GRAVITY AND/OR EM WAVES emitted - you may need Mach's Principle for the Dirac-Wheeler-Feynman-Cramer transactions as a future de Sitter event horizon total absorber of last resort.

That's another issue. I'm just trying to get some basic Newtonian physics straightened out here.

Z.