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On Feb 20, 2012, at 2:29 AM, jfwoodward@juno.com wrote:

Let's start with the rock being swung on a string and then move on to Sciama and other issues.

The rock: I pull on the string as I give it some initial velocity and set it into uniform circular motion.  In order to maintain that motion -- neglecting friction and windage -- I must continue to apply a constant centrepital force on the string that is communicated to the rock.  At any instant I ask: what force(s) are acting on the rock?

Jack: Sure, so what?

Jim: The obvious force in play is that exerted by the string, analyzed microscopically it is electrical,

Jack: Correct.

Jim: but that is not important.  What is important is that this force is the one that figures into Newton's second law, the equation of motion.  

Jack: True, of course. That real electrical force pushes the rock off a timelike geodesic in the real geometrodynamic field most aptly expressed by the GCT local scalar invariants (same in every coincident set of LNIFs) {e^I, S^I^J} from which ds^2, Ruvwl etc. can be computed.

Jim: But Newton's laws of mechanics tell us that that is not the whole story.  The third law tells us that whenever a force acts on a massive object (the string on the rock), there is an equal and opposite reaction force (the rock on the string).  

Jack: This is expressed LOCALLY in classical field theory simply by  Tuv^;v = 0 in a LNIF. Locally 1905 SR works. That's the EEP, so that we also have in a coincident LIF TIJ^,J = 0 for all the matter fields. One can go from a classical field theory to a particle picture if one likes, but this is good enough.

Action-reaction 3rd law is contained in TIJ^,J = 0 from local spatial translation invariance. End of story, no need for cosmology here - no need for some ineffable nonmeasurable meta-theoretical phi = c^2.


Jim: That means that when the string acts to produce the centrepital force, the rock -- because it has inertial mass -- exerts an equal and opposite force on the string.  As Paul notes, if a spring scale is inserted between the string and the rock, it is the inertial reaction force of the rock on the string that causes the scale reading to be non-zero.  This is Newton speaking I note.  I trust you don't find him obscure.

Jack: I still find your writing a bit obscure. So now do we agree

1) "inertial mass" = "gravitational mass" = relativistic mass M = m0(1 - v^2/c^2)^-1/2?

Newton's 2nd law is

2) DP^u/ds = F^u

3) P^u = Mdx^u/ds

4) ds^2 = guv(LNIF)dx^udx^v = nIJ(LIF)e^Ie^J  EEP

Newton's 3rd Law is contained in LOCAL

5) TIJ^,J (Matter) = 0

Yes? No?

Jim: The question is: what is the origin of the reaction force -- an inertial reaction force -- on the string?  This has nothing to do with the Higgs process.  That's a red herring.  Read Wilczek's book.

Jack: I have read Wilczek's book. I find your above remark still unintelligible.

The Higgs process + QCD + nuclear + atomic +solid state physics "explain" the origin of m0 for any material object. So what precisely are you saying here? What mathematical object in the theory are you talking about?

Jim: Paul's answer is that the force is produced by the rock as a consequence of it having mass.

Jack: Right, it comes from m0.

Jim: He chooses not to call it a force acting on the rock that is transmitted through the rock to the string.

Jack: Unintelligible to me. What's the problem? The string pulls the rock off a geodesic with an electrical force. The rock pulls back on the string with an equal and opposite ELECTRICAL FORCE according to the equation

TIJ^,J(LIF) = 0

One must include all relevant LOCAL degrees of freedom of source fermion particles and gauge force mostly VIRTUAL bosons. What we have however is a coarse-grained lumped parameter description with many microscopic details averaged out - as in elasticity theory.

Jim: I assume he affects this position because he wants to explain inertial forces as some local action of the quantum vacuum on the rock that produces a force on the string, but not on the rock.

Jack: I see no need to invoke the quantum vacuum here as any zero order dominant factor.

Jim: My position is that a force -- gravity in fact -- acts on the accelerating rock to produce the inertial reaction force -- a real force,

Jack: Your position then, Jim, I find completely wrong from the git-go. Newton's theory of gravity force is not adequate for these fundamental questions. There is no such thing as a real gravity force in Einstein's theory.

Newton's gravity force F = - Gmm'/r^2 is completely contained in the Levi-Civita-Christoffel torsion-free metric connection {^uvw} in the GEODESIC FORCE-FREE equation

DP^u(test particle)/ds = 0  i.e. Newton's FIRST LAW of MOTION

Newton's 2nd law of motion is

DP^u(test particle)/ds = F^u(electromagnetic-weak-strong)

Newton's 3rd law of motion of action-reaction is

DP^u/ds = 0

for pairs of interacting particles.

See Feynman's Lectures Vol 1 for complications in 3-particle ---> N-particle systems - and one must include the EM field (e/c)A vector potentials, i.e.

P^u(charge) = MdX^u(charge)/ds + (e/c)A^u(EM field)

TO AVOID PARADOXES WITH NON-CENTRAL LORENTZ FORCES ON TWO CHARGES MOVING AT RIGHT ANGLES TO EACH OTHER, FOR EXAMPLE where the Lorentz magnetic forces do not cancel and are in different directions!

Also there is the issue of retardation and advanced time delays, though for the string-rock problem that can be ignored to good approximation I think.

Jim: though it gets called fictitious because being gravitational, the passive gravitational mass and inertial mass cancel in the equation of motion for this force (I know, you will find this obscure; but it isn't; I wrote out the math for this several emails back) -- and that inertial reaction force on the rock is communicated through the rock to the string producing the equal and opposite force required by Newton's third law.

Note that if you deny the existence of the inertial reaction force acting on the string as it exerts a centrepital force on the rock, you are violating Newton's third law.

Now, as to whether Paul's view of inertia as something that's not really a force, though it results in a force versus my claim that the force is gravity.  It is a matter of simple calculation -- Sciama's calculation in fact -- to show that in the vector approximation to GR, the inertial reaction force is due to gravity.  If GR is right (it is), then in our universe at the present epoch, inertial reaction forces are due to gravity.  You can't have it more than one way here.  It's what the theory says.

Why is my last assertion true?  Because we now know as a matter of fact that the universe is spatially flat.  That means that "critical cosmic matter density" obtains.  And that in turn means that phi = c^2.  This is not my interpretation.  This is standard GR cosmology.  It's the reason why Steven Weinberg years ago talked about a three pound universe, and Laurence Krauss now talks about a universe from "nothing".  The gravitational energy just balances the non gravitational energy in a critical cosmic matter density universe; and that means phi = c^2.  To see this, multiply through by the mass M of the universe M phi = Mc^2.

Armed with the knowledge that phi = c^2 for us here and now, we can put this knowledge into Sciama's vector approximation calculation of the gravitational action of the universe on an accelerating object (due to the action of a non-gravitational force pushing it off it's local geodesic).  What we find is:

       E(grav) = - grad phi - (1/c)dA/dt

where A is the gravimagnetic vector potential.  The vector potential has mass current sources, and as Sciama notes, in the instantaneous frame of the accelerating object, all of the matter in the universe appears to be moving past the object with velocity - v.  So when we integrate the mass currents rho v, we can take the v out of the integral and just compute the matter density rho contribution to the integral.  Amazingly enough, that integral just turns out to be phi (times 1/c), so we get:

       E(grav) = - (phi/c^2) dv/dt = - (phi/c^2) a.

Grad phi is zero as there are no local mass concentrations of importance by assumption.  We now multiply this calculated value of the gravelectric field strength seen by the accelerating object by the mass of the object to get the force on the object due to the gravitational action of the universe and take into account our factual knowledge that phi/c^2 = 1 (from the WMAP results) to get:

       F(grav) = E(grav) m = - ma.

This is the inertial reaction force required by Newton's third law.  It is due to the action of gravity, not the quantum vacuum or the Higgs process, or whatever.  You may not like this.  But this is what standard GR says.

I'm going to leave most of the questions about Sciama and the speed of light to another email tomorrow to let this sink in.  But I will make one comment.  You may be wondering: Why aren't there effects like this in electrodynamics?  After all, it's a vector theory too, and almost identical to vector approximation gravity.  The answer is simple.  When you integrate the sources of the EM vector potential -- electric charge currents -- over the universe, you get zero.  Why?  Because the universe, at cosmic scale, is charge neutral.  So rho on average is zero and the vector potential vanishes.  In the case of gravity this isn't so; and the vector potential does not vanish.

If you want to object that Sciama's calculation is just an approximation, be careful.  Inertial reaction forces are not small effects that might be artifacts of the approximation used.  They are as large as the forces that excite them.  That is, very large and not an approximation artifact.  After all, we're talking about the action of the whole blasted universe.

Tomorrow I'll deal with the stuff about the speed of light in GR and why phi, like c, is a locally measured invariant.  Unless, that is, we're still stuck on what an inertial reaction force is and what produces them.  :-)