To paraphrase Richard Feynman: What I cannot construct on the spot, I do not understand.
On Jan 15, 2012, at 3:10 PM, This email address is being protected from spambots. You need JavaScript enabled to view it. wrote:
No Jack, M is not a constant.  

Jim you did not read my text correctly. Please read again. I said in your theory M is NOT a constant. :-)
That's  a simplifying assumption that is often and incorrectly made.
That's exactly what I have been saying!
I've been through this with several serious physicists.  Getting rid of Mach effects is not so simple.  Take a look at the excerpt from Rindler that I circulated a while back.
Hmmm I don't recall that. Maybe Sirag has it?
Also, the vdm/dt term in the 2nd law is routinely screwed up by even competent physicists.  I've been through all that several times too.
If phi and c are locally measured invariants and phi = c^2, then Mach effects follow.  

I do not understand this last sentence at all.
1) phi and c are NOT locally measured frame invariants in my opinion.
Please provide a detailed math proof of that. I am not even sure what you mean by "phi" and "c" here. When I calculate "c" in the LNIF from ds = 0 it's NOT a scalar invariant at all.
PROOF:
ds^2 = g00(cdt)^2 + g0idx^i(cdt) + gijdx^idx^j
= (cdt)^2[g00 + g0iV^i/c + gijV^iV^j/c^2]
= (cdt)^2[(1 + phi/c^2) + A.V/c + gijV^iV^j/c^2]
ds = 0 for light
 gijV^iV^j = c'^2
(1 + phi/c^2) + A.c'/c + c'^2/c^2 = 0
a quadratic equation for the locally measured speed of light in vacuum c' in an accelerating non-inertial frame.
where c' is the LOCALLY MEASURED SPEED OF LIGHT IN THE LNIF. OBVIOUSLY IT'S NOT FRAME INVARIANT.
INDEED C' HAS TWO DISTINCT ROOTS IN GENERAL THAT MAY NOT EVEN BE REAL NUMBERS IN GENERAL?
2) phi =/= c^2 in my opinion except possibly at a horizon where your phi, A theory as a quasi-EM theory is not correct in my opinion.
No hand waving.  No funny physics.
I have not been hand waving. I have given detailed calculations CONSISTENTLY & REPEATEDLY TO NO AVAIL. Please practise here what you preach. ;-)