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On Feb 6, 2013, at 10:49 AM, JACK SARFATTI <sarfatti@pacbell.net> wrote:

Using my number uncertain state |U> = x|0> + y|1> instead of your truncated coherent state,
I calculate that these two outcomes have exactly the SAME AMPLITUDE.

This fact is the essence of the refutation..

agreed that is the question.

On Feb 6, 2013, at 10:41 AM, nick herbert <quanta@cruzio.com> wrote:

>>>>In other words, even though the |0>|1>|0>|1> outcome may produce "anti-fringes", it has nowhere near the amplitude to cancel the "fringes" caused by the |1>|0>|1>|0> outcome....since the former outcome describes a right-going photon being reflected (extremely rare due to vanishing 'r') while the latter outcome describes a right-going photon being transmitted (very likely due to 't' approximately equal to 1).<<<<

A very plausible argument
But restore the missing term, Demetrios,
Do the calculation.
Then see if you still believe
that the |1>|0>|1>|)> term and the |0>|1>|0>|1>
have different amplitudes.

Using my number uncertain state |U> = x|0> + y|1> instead of your truncated coherent state,
I calculate that these two outcomes have exactly the SAME AMPLITUDE.

This fact is the essence of the refutation..


Nick--

I was up all night calculating these terms
and I am pretty sure your scheme is refuted.

Using the Feynman rule the probabilities for these two distinguishable processes are indeed equal
and do not cancel but one process is linked to fringes in Alice's detectors
and the other process is linked to anti-fringes in Alice's detector.

An incoherent equally-weighted sum of fringes and anti-fringes = no interference.

Your error consists of dropping a term that seems to be harmlessly small.
When you restore this term, the scheme becomes an ordinary coincidence-triggered distance interference device.

Since you are more familiar with these sorts of calculations than I am,
I urge you to restore the missing term and recalculate.
I would be surprised if you do not agree
that KISS is refuted.

However your measurement scheme -- ambiguating the Fock states by mixing with states of uncertain photon number --
is very clever and may find some use in less-preposterous applications.

I really have enjoyed interacting with your and your KISS scheme.

Nick

On Feb 6, 2013, at 9:38 AM, Demetrios Kalamidas wrote:

Hi Nick,
 It is both a pleasure and an honor that you have analyzed my scheme to this extent and, thankfully, so far your hard analysis has not disproved it....and may have even generalized and strengthened the argument.
 If my idea is described in a mathematically valid way then, as you seem to point out, the experimental proposal is also a powerful test of the strength of "The Feynman Dictum", which, so far, has never failed.
Thanks Nick
Demetrios

On Wed, 6 Feb 2013 04:32:09 -0800
nick herbert <quanta@cruzio.com> wrote:
Demetrios--
I have been calculating my own version of your KISS proposal
using the state |U> = x|0> + y|1> instead of a coherent state with  alpha amplitude
as input to the beam splitter which you use. Using this state allows  me to avoid
approximations. But yours is a robust proposal and should be immune  to approximations.
Indeed I get the same result as you, making the approximation rx --->  0 to eliminate a small |0>|1> term
as do you. I calculate the amplitude of the quantum erasure term |1>| 0>|1>|0> to be -trxy.
Hence my result for the probability of the FTL effect is 1/2 |etrxy|^2
which is comparable to your 1/2|etralpha|^2.
So far so good. The KISS and KISS(U) calculations give compatible  results. FTL signaling seems secure.
------------
Next I decided to include the small term we both threw away. This  means calculating the amplitude for
the detector response |0>|1>|0>|1>.
Imagine my surprise when I discovered that this (also a quantum  erasure term by the way) amplitude is also trxy
with a plus sign!!!!!!!!!!!!
One might think that this term will exactly cancel your former  quantum erasure term and refute your KISS proposal.
But I do not think that's the way it works. According to the Feynman  rules you add amplitudes for indistinguishable paths,
and add probabilities for distinguishable paths. Since the |1>|0>|1>| 0> result is distinguishable from the |0>|1>|0>|1> result,
it seems that the proper thing to do here is add probabilities rather  than amplitudes. So not only do these two processes
not cancel but THEY DOUBLE THE SIZE OF YOUR FTL EFFECT.
At least that's the way I see it right now.
Seems like my attempt to refute your proposal is traveling in the  opposite direction.
Thanks for the fun.
Nick
Part 2
The more I think about this, the more I am convinced that this  calculation refutes KISS.
The amplitude for |1>|0>|1>|0> is "-trxy"and for |0>|1>|0>|1> is  "+trxy".
If you coincidence-trigger on the detector result |1>|0>|1>|0> you  get fringes.
If you coincidence-trigger on the detector result |0>|1>|0>|1> you  get anti-fringes.
If you do not coincidence-trigger you get an equal mixture of fringe  and anti-fringe.
QED: No FTL signaling.



I thought Nick said the two amplitudes were equal and opposite. If Demetrios is correct below I will be happy to retract my Eulogy for the demise of nonlocal entanglement signaling within ORTHODOX quantum theory as opposed to post-quantum extensions.

On Feb 6, 2013, at 10:07 AM, Demetrios Kalamidas <dakalamidas@sci.ccny.cuny.edu> wrote:

Hi to all,
I stated that in my previous message that "....thankfully, so far your hard analysis has not disproved it" but forgot to include the text of why I believe this:

The only way for "fringes" on the left wing of the experiment (caused by the |1>|0>|1>|0> term on the right) to be canceled by "anti-fringes" (caused by the |0>|1>|0>|1> term on the right) is if BOTH the |1>|0>|1>|0> term and the |0>|1>|0>|1> term had the SAME AMPLITUDE, and therefore the same probability of happening.

HOWEVER, in my scheme, the |1>|0>|1>|0> outcome is HEAVILY FAVORED when compared to the |0>|1>|0>|1> outcome because of the high asymmetry of the two beam splitters on the right.

In other words, even though the |0>|1>|0>|1> outcome may produce "anti-fringes", it has nowhere near the amplitude to cancel the "fringes" caused by the |1>|0>|1>|0> outcome....since the former outcome describes a right-going photon being reflected (extremely rare due to vanishing 'r') while the latter outcome describes a right-going photon being transmitted (very likely due to 't' approximately equal to 1).

Demetrios
Jack Sarfatti
KISS-OFF! ;-)
  • Jack Sarfatti Yes, Nick most likely the two terms cancel as you say at the end. The problem with all the attempts to derive entanglement signal nonlocality within orthodox quantum theory, is the neglect of relevant terms, which in the end as you show, cancel the result. I wrote at the beginning of this that such may happen here.

    Note, that this does not affect attempts as entanglement signal nonlocality using a more general nonlinear post-quantum theory as in Steven Weinberg's, Henry Stapp's and Antony Valentini's models.

    On Feb 6, 2013, at 4:32 AM, nick herbert <quanta@cruzio.com> wrote:

    Demetrios--

    I have been calculating my own version of your KISS proposal
    using the state |U> = x|0> + y|1> instead of a coherent state with alpha amplitude
    as input to the beam splitter which you use. Using this state allows me to avoid
    approximations. But yours is a robust proposal and should be immune to approximations.

    Indeed I get the same result as you, making the approximation rx ---> 0 to eliminate a small |0>|1> term
    as do you. I calculate the amplitude of the quantum erasure term |1>|0>|1>|0> to be -trxy.

    Hence my result for the probability of the FTL effect is 1/2 |etrxy|^2
    which is comparable to your 1/2|etralpha|^2.

    So far so good. The KISS and KISS(U) calculations give compatible results. FTL signaling seems secure.

    ------------

    Next I decided to include the small term we both threw away. This means calculating the amplitude for
    the detector response |0>|1>|0>|1>.

    Imagine my surprise when I discovered that this (also a quantum erasure term by the way) amplitude is also trxy
    with a plus sign!!!!!!!!!!!!

    One might think that this term will exactly cancel your former quantum erasure term and refute your KISS proposal.

    But I do not think that's the way it works. According to the Feynman rules you add amplitudes for indistinguishable paths,
    and add probabilities for distinguishable paths. Since the |1>|0>|1>|0> result is distinguishable from the |0>|1>|0>|1> result,
    it seems that the proper thing to do here is add probabilities rather than amplitudes. So not only do these two processes
    not cancel but THEY DOUBLE THE SIZE OF YOUR FTL EFFECT.

    At least that's the way I see it right now.

    Seems like my attempt to refute your proposal is traveling in the opposite direction.

    Thanks for the fun.

    Nick

    Part 2
    The more I think about this, the more I am convinced that this calculation refutes KISS.

    The amplitude for |1>|0>|1>|0> is "-trxy"and for |0>|1>|0>|1> is "+trxy".

    If you coincidence-trigger on the detector result |1>|0>|1>|0> you get fringes.

    If you coincidence-trigger on the detector result |0>|1>|0>|1> you get anti-fringes.

    If you do not coincidence-trigger you get an equal mixture of fringe and anti-fringe.

    QED: No FTL signaling.


Jack Sarfatti This is hot. If the effect works it's the basis for a new Intel, Microsoft & Apple combined for those smart venture capitalists, physicists & engineers who get into it. This is as close as we have ever come since I started the ball rolling at Brandeis in 1960-61 & then in mid-70's see MIT Physics Professor David Kaiser's "How the Hippies Save Physics". I first saw this as a dim possibility in 1960 at Brandeis grad school and got into an intellectual fight about it with Sylvan Schweber and Stanley Deser. Then the flawed thought experiment published in the early editions of Gary Zukav's Dancing Wu Li Masters in 1979 - pictured in Hippies book tried to do what DK may now have actually done. That is, control the fringe visibility at one end of an entangled system from the other end without the need of a coincidence counter correlator after the fact. Of course, like Nick Herbert's FLASH at the same time late 70's, it was too naive to work and the nonlinear optics technology was not yet developed enough. We were far ahead of the curve as to the conceptual possibility of nonlocal retrocausal entanglement signaling starting 53 years ago at Brandeis when I was a National Defense Fellow Title IV graduate student.

Jack Sarfatti

about an hour ago near San Francisco
On Feb 5, 2013, at 12:28 PM, JACK SARFATTI <sarfatti@pacbell.net> wrote:

Thanks Nick. Keep up the good work. I hope to catch up with you on this soon. This may be a historic event of the first magnitude if the Fat Lady really sings this time and shatters the crystal goblet. On the Dark Side this may open Pandora's Box into a P.K. Dick Robert Anton Wilson reality with controllable delayed choice precognition technology. ;-)

On Feb 5, 2013, at 10:38 AM, nick herbert <quanta@cruzio.com> wrote:

Demetrios--

Looking over your wonderful paper I have detected one
inconsistency but it is not fatal to your argument.

On page 3 you drop two r terms because "alpha", the complex
amplitude of the coherent state can be arbitrarily large in
magnitude.

But on page 4 you reduce the magnitude of "alpha" so that
at most one photon is reflected. So now alpha cannot be
arbitrarily large in magnitude.

But this is just minor quibble in an otherwise superb argument.

This move does not affect your conclusion--which seems
to directly follow from application of the Feynman Rule: For distinguishable
outcomes, add probabilities; for indistinguishable outcomes, add amplitudes.

To help my own understanding of how your scheme works,
I have simplified your KISS proposal by replacing your coherent states with
the much simpler state |U> = x|0> + y|1>. I call this variation of your proposal KISS(U)

When this state |U> is mixed with the entangled states at the beamsplitters,
the same conclusion ensues: there are two |1>|1> results on Bob's side of the source
that cannot be distinguished -- and hence must be amplitude added.

The state |U> would be more difficult to prepare in the lab than a weak coherent state
but anything goes in a thought experiment. The main advantage of using state |U>
instead of coherent states is that the argument is simplified to its essence and needs
no approximations. Also the KISS(U) version shows that your argument is independent
of special properties possessed by coherent states such as overcompleteness and non-
orthogonality. The state |U> is both complete and orthogonal -- and works just as well
to prove your preposterous conclusion. --- that there is at least one way of making photon
measurements that violates the No-Signaling Theorem.

Thanks for injecting some fresh excitement into the FTL signaling conversation.

warm regards
Nick Herbert
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Jack Sarfatti On Feb 5, 2013, at 1:15 PM, Demetrios Kalamidas <dakalamidas@sci.ccny.cuny.edu> wrote:

Nope, no refutation I can think of so far....and I've tried hard.
Demetrios
...See More
33 minutes ago · Like

Joe Ganser Jack do you know a lot of people at CUNY? I take ph.d classes there.
26 minutes ago · Like

Joe Ganser I'm interested in who may do these sorts of topics in NYC
25 minutes ago · Like

Jack Sarfatti Daniel Greenberger!
9 minutes ago · Like · 1

a few seconds ago · Like

 

On Feb 5, 2013, at 1:15 PM, Demetrios Kalamidas <dakalamidas@sci.ccny.cuny.edu> wrote:

Nope, no refutation I can think of so far....and I've tried hard.
Demetrios

On Tue, 5 Feb 2013 13:09:28 -0800
nick herbert <quanta@cruzio.com> wrote:
Thanks, Demetrios. I understand now that alpha can be large
while alpha x r is made small. Also I notice that your FTL signaling scheme seems to work both ways. In your illustration the photons on the left side (Alice) are  combined at a 50/50 beam splitter so they cannot be used for which-way information. However if the 50/50 beamsplitter is removed, which-way info is present and the two versions of |1>|1> on the right-hand side (Bob) are now  distinguishable
and must be added incoherently, which presumably will give a  different answer and observably different behavior by Bob's  right-side detectors. So your scheme seems consistent -- FTL signals can be sent in either  direction.
This is looking pretty scary.
Do you happen to have a refutation up your sleeve
or are you just as baffled by this as the rest of us?
Nick

 

 

Therefore, Nick it is premature for you to claim that the full machinery of the Glauber coherent states, i.e. distinguishable over-complete non-orthogonality is not necessary for KISS to work. Let's not rush to judgement and proceed with caution. This technology, if it were to work is as momentous as the discovery of fire, the wheel, movable type, calculus, the steam engine, electricity, relativity, nuclear fission & fusion, Turing machine & Von Neumann's programmable computer concept, DNA, transistor, internet ...

On Feb 5, 2013, at 12:18 PM, Demetrios Kalamidas <dakalamidas@sci.ccny.cuny.edu> wrote:

Hi Nick,

 And thanks much for your careful examination of my scheme....however, there appears to be a misunderstanding.
 Let me explain:

"On page 3 you drop two r terms because "alpha", the complex amplitude of the coherent state can be arbitrarily large in magnitude."

I drop the two terms in eq.5b because they are proportional to 'r'....and 'r' approaches zero. However, the INITIAL INPUT amplitude, 'alpha', of each coherent state can be as large as we desire in order to get whatever SMALL BUT NONVANISHING AND SIGNIFICANT product 'r*alpha', which is related to the terms I retain.

In other words, for whatever 'r*alpha' we want, lets say 'r*alpha'=0.2, 'r' can be as close to zero as we want since we can always input a coherent state with large enough initial 'alpha' to give us the 0.2 amplitude that we want.

So, terms proportional to 'r' are vanishing, while terms proportional to 'r*alpha' are small but significant and observable.
You state:

"But on page 4 you reduce the magnitude of "alpha" so that at most one photon is reflected. So now alpha cannot be arbitrarily large in magnitude."

The magnitude of 'alpha' is for the INITIAL coherent states coming from a3 and b3, BEFORE they are split at BSa and BSb. It is this 'alpha' that is pre-adjusted, according to how small 'r' is, to give us an appropriately small reflected magnitude, i.e. 'r*alpha'=0.2, so that the "....weak coherent state containing at most one photon...." condition is reasonably valid.

Demetrios


On Feb 5, 2013, at 12:28 PM, JACK SARFATTI <sarfatti@pacbell.net> wrote:

Thanks Nick. Keep up the good work. I hope to catch up with you on this soon. This may be a historic event of the first magnitude if the Fat Lady really sings this time and shatters the crystal goblet. On the Dark Side this may open Pandora's Box into a P.K. Dick Robert Anton Wilson reality with controllable delayed choice precognition technology. ;-)

On Feb 5, 2013, at 10:38 AM, nick herbert <quanta@cruzio.com> wrote:

Demetrios--

Looking over your wonderful paper I have detected one
inconsistency but it is not fatal to your argument.

On page 3 you drop two r terms because "alpha", the complex
amplitude of the coherent state can be arbitrarily large in
magnitude.

But on page 4 you reduce the magnitude of "alpha" so that
at most one photon is reflected. So now alpha cannot be
arbitrarily large in magnitude.

But this is just minor quibble in an otherwise superb argument.

This move does not affect your conclusion--which seems
to directly follow from application of the Feynman Rule: For distinguishable
outcomes, add probabilities; for indistinguishable outcomes, add amplitudes.

To help my own understanding of how your scheme works,
I have simplified your KISS proposal by replacing your coherent states with
the much simpler state |U> = x|0> + y|1>. I call this variation of your proposal KISS(U)

When this state |U> is mixed with the entangled states at the beamsplitters,
the same conclusion ensues: there are two |1>|1> results on Bob's side of the source
that cannot be distinguished -- and hence must be amplitude added.

The state |U> would be more difficult to prepare in the lab than a weak coherent state
but anything goes in a thought experiment. The main advantage of using state |U>
instead of coherent states is that the argument is simplified to its essence and needs
no approximations. Also the KISS(U) version shows that your argument is independent
of special properties possessed by coherent states such as overcompleteness and non-
orthogonality. The state |U> is both complete and orthogonal -- and works just as well
to prove your preposterous conclusion. --- that there is at least one way of making photon
measurements that violates the No-Signaling Theorem.

Thanks for injecting some fresh excitement into the FTL signaling conversation.

warm regards
Nick Herbert



  • On Feb 3, 2013, at 12:42 PM, JACK SARFATTI <sarfatti@pacbell.net> wrote:

    Fred, I think you are making an error here. The vacuum |0> is as good a state as |1> in Fock space for a given mode-radiation oscillator. DK's eq. 1 is a FOUR PHOTON state - two REAL PHOTONS & TWO VIRTUAL PHOTONS

    Note also that Glauber coherent states use |0> in an fundamental way.

    quantum optics interferometer experiments use the |0> states e.g. papers by Carlton Caves

    http://info.phys.unm.edu/~caves/

    http://info.phys.unm.edu/~caves/research.html

    http://info.phys.unm.edu/~caves/talks/talks.html


    Search Results
    [PDF] Quantum-limited measurements: One physicist's crooked path from ...
    www.phys.virginia.edu/Announcements/Seminars/.../S1466.pd...
    File Format: PDF/Adobe Acrobat - Quick View
    physicist's crooked path from quantum optics to quantum information. I. Introduction. II. Squeezed states and optical interferometry. III. ... Carlton M. Caves ...
    [PDF] Quantum metrology - University of New Mexico
    info.phys.unm.edu/~caves/talks/qmetrologylectures.pdf
    File Format: PDF/Adobe Acrobat - Quick View
    Carlton M. Caves. Center for Quantum ... Ramsey interferometry, cat states, and spin squeezing. Carlton M. ... Weinstein, and N. Mavalvala, Nature Physics 4, ...



    On Feb 3, 2013, at 12:26 PM, fred alan wolf <fawolf@ix.netcom.com> wrote:

        Nick and Demetrios, basic quantum physics tells me that eq. 1 of
    KISS is a 4-photon state. That is my point. Let the Hamiltonian go. Ergo, to
    claim it as 2-photon state cannot be correct. Eq. 1 says something about
    phases as well.  If I write a quantum wave function as a sum over i of
    |ai>|bi>|ci>|di> then there must be 4 objects, not two, regardless of how
    large is i.  Even if |ai> is a sum of possibilities such as (|A1>+|A2>) and
    similarly for the bi, ci and di states, I still can't get this to reduce to
    a sum over two particle states.  Nicht wahr?     So I am confused how you both seem to see this as OK as far as
    quantum physics is concerned.

        Jack, do you or do you not see my point?   
    Best Wishes,

    Fred Alan Wolf Ph.D.  aka Dr. Quantum ®
     
    Jack Sarfatti Hi all,

    I'll quickly respond to Fred's question. The state in eq.1 is perfectly legitimate and has been experimentally realized already.
    In this scheme it is tacitly assumed that the source S is a down-conversion source, since this is by far the main way in which entangled photon pairs are created. These sources need a pump to stimulate the nonlinear medium (i.e. down-conversion crystal).
    Usually about one in every million pump photons are split into an entangled pair, each photon of which comes out at a specific angle and energy. The way to create two photons in modes a1a2 is to have the pump come from the bottom and pass upward; the way to create two photons in modes b1b2 is the BACK-REFLECT the same pump downward through the crystal again.
    So,each run of the experiment is ONE DOUBLE-PASS of the pump through the crystal....most of the times you get nothing and, to very good approximation, the rest of the time you get one pair created (either in a1a2 or b1b2)....Of course there is also the far smaller amplitude of creating two pairs (one in a1a2 and one in b1b2, or two in a1a2, or two in b1b2)....according to the expansion of the Hamiltonian....but these are negligible terms and do not affect the outcomes in all these entanglement experiments.
    Demetrios
  • Jack Sarfatti On Feb 3, 2013, at 11:48 AM, JACK SARFATTI <sarfatti@pacbell.net> wrote:

    I agree with Nick.

    On Feb 3, 2013, at 11:25 AM, nick herbert <quanta@cruzio.com> wrote:

    No need for Hamiltonians, Fred.
    The KISS proposal is as simple as LEGOs.
    Every part of it is something
    THAT HAS ALREADY BEEN DEMONSTRATED IN A LAB.

    Kalamidas has put these existing Legos together
    in an imaginative way that seems to permit
    superluminal signaling.

    But probably does not.

    If you, Fred, are waiting for a Hamiltonian formulation
    of this experiment you will be waiting for a long time
    and will have essentially disconnected yourself
    from the KISS adventure.

    Nick Herbert
    KISS = Kalamidas's Instant Signaling Scheme
    ---- end of Nick's message above, I wrote:
    OK there are two separate issues here.

    Question 1: Fred if DK's wave function

    Could be made, then do you agree with DK's logic for the rest of the paper.

    I think the above wave function is perfectly legitimate in principle although whether one can make it in the lab is another question.

    (1) is perfectly sensible in quantum field theory in Fock space.

    There are four radiation oscillators with two real photons and two zero point photons distributed among them. The vacuum states |0> are legitimate states.

    Question 2. Accepting (1) is DK's logic etc. correct? I think Nick Herbert is working on that question.

    I personally am still thinking about the whole thing looking at Mandel as well and trying to understand the whole thing better.

    My previous work on the Glauber state distinguishable non-orthogonality loop hole in the no-signaling belief is generally compatible with the spirit of what DK is proposing. I mean

    On Feb 3, 2013, at 9:53 AM, fred alan wolf wrote:

    Guys and girls,

    I don't believe this will work simply because to my knowledge there is no foundation based on quantum physics which supports this initial supposedly 2-particle quantum wave function. What Hamiltonian does it solve? You can always invent quantum wave functions (which are not connected to reality) but to claim this one (which apparently uses 4 photons not 2) has solved the ftl problem is simply bad physics as I see it. If I am wrong here, will somebody explain how this quantum wave function is a two body quantum wave function? Can you show me the Hamiltonian it is solution for?

    Best Wishes,

    Fred Alan Wolf Ph.D. aka Dr. Quantum
  1. Thanks Nick. What would Santa do without you in his workshop? ;-)
    Looks good. Remember I have been stressing the relevance of Glauber coherent states.
    They are obviously distinguishably non-orthogonal & over-complete.


    On Feb 2, 2013, at 1:48 PM, nick herbert <quanta@cruzio.com> wrote:

    Demetrios--

    Congratulations again on your clever FTL-signaling scheme.

    I am busy constructing (on my white board) your thought experiment
    using my own notation.

    First: I hope you do not mind the acronym I have chosen for this project = KISS

    KISS = Kalamidas's Instant Signaling Scheme.

    Second: It has become conventional to imagine these signals sent between Alice and Bob.
    So everything on left side should be labeled "A" and on the right side "B".

    Since A and B photons are delivered into two (entangled) modes, I have chosen to label these modes U and D (for Up and Down). In this labeling convention the basic entangled state vector |ES> becomes

    |ES> = |1>(AU)|0>(AD)|1>(BU) |0>(BD)  + |0>(AU)|1>(AD)|0>(BU)|1>(BD)

    or (dropping the subscripts)

    |ES> = |1>|0>|1>|0> + |0>|1>|0>|1>

    which is essentially your (unnormalized) EQ 1.

    Also it is conventional for beam-splitter modes to be labeled 1, 2, 3, 4
    where 1 and 2 are inputs and 3 and 4 are outputs.

    So for my thought experiment I will label the 4 modes of Bob's two beam splitters U and D
    as |U1>, |U2>, |U3>, |U4> and |D1>, |D2>, |D3> and |D4> with a similar convention for the 50/50 beamsplitter encountered by Alice's photons.

    I like your clever use of coherent states to muddle the which-way question. But instead of inputting coherent states at  Bob's beamsplitters U and D, I will be inputting the coherent XYZ states |BU> and |BD>

    where |BU> = x|0> + y|1> + z|2>

    and |BD> has a similar definition.

    These are truncated coherent states sufficient to produce the ambiguities you claim will lead to coincidence-less, Bob-controllable interference in Alice's 50/50 beamsplitter and are easier to calculate than the infinite sums of real coherent states.

    Thanks for the opportunity to return to the algebra of few photons on an asymmetric beam splitter. And for the chance to reformulate your clever KISS experiment in terms that make sense to me.

    I am always looking for (high quality) work to do.

    And your KISS proposal is both of high quality and within my modest abilities for calculating quantum outcomes.

    warm regards
    Nick Herbert
    http://quantumtantra.blogspot.com
     
    If this paper proves correct in the lab, it vindicates my struggle since 1960 or so that MIT Physics Professor David Kaiser has recorded for history in his book "How the Hippies Saved Physics." This will be a science-technology revolution worth billions if not trillions of dollars for visionary venture capitalists.
    "Proposal for a feasible quantum-optical experiment to test the validity of the no-signaling theorem
    Demetrios A. Kalamidas
    4 Raith USA, 2805 Veterans Memorial Hwy, Ronkonkoma, New York 11779, USA (dakalamidas@sci.ccny.cuny.edu)
    Received November 29, 2012; accepted January 17, 2013;
    posted January 24, 2013 (Doc. ID 180742)
    Motivated by a proposal from [Phys. Scr. T76, 57 (1998)] for superluminal signaling and inspired by an experiment
    from [Phys. Rev. Lett. 67, 318 (1991)] showing interference effects within multiparticle entanglement without
    coincidence detection, we propose a feasible quantum-optical experiment that purports to manifest the capacity
    for superluminal transfer of information between distant parties." © 2013 Optical Society of America
    OCIS codes: 270.4180, 270.5290, 270.5565, 270.5585.
     
    "Numerous experiments to date, mainly in the quantum-optical domain, seem to strongly support the notion of an inherent nonlocality pertaining to certain multiparticle quantum mechanical processes. However, with apparently equal support, this time from a theoretical perspective, it is held that these nonlocal “influences” cannot be exploited to produce superluminal transfer of information between distant parties. The theoretical objection to superluminal communication, via quantum mechanical multiparticle entanglement, is essentially encapsulated by the “no-signaling theorem” [1]. So, it is within this context that we present a scheme whose mathematical description leads to a result that directly contradicts the no-signaling theorem and manifests, using only the standard quantum mechanical formalism, the capacity for superluminal transmission of information."
  1. If this paper proves correct in the lab, it vindicates my struggle since 1960 or so that MIT Physics Professor David Kaiser has recorded for history in his book "How the Hippies Saved Physics." This will be a science-technology revolution worth billions if not trillions of dollars for visionary venture capitalists.
    "Proposal for a feasible quantum-optical experiment to test the validity of the no-signaling theorem
    Demetrios A. Kalamidas
    4 Raith USA, 2805 Veterans Memorial Hwy, Ronkonkoma, New York 11779, USA (dakalamidas@sci.ccny.cuny.edu)
    Received November 29, 2012; accepted January 17, 2013;
    posted January 24, 2013 (Doc. ID 180742)
    Motivated by a proposal from [Phys. Scr. T76, 57 (1998)] for superluminal signaling and inspired by an experiment
    from [Phys. Rev. Lett. 67, 318 (1991)] showing interference effects within multiparticle entanglement without
    coincidence detection, we propose a feasible quantum-optical experiment that purports to manifest the capacity
    for superluminal transfer of information between distant parties." © 2013 Optical Society of America
    OCIS codes: 270.4180, 270.5290, 270.5565, 270.5585.
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    • Jack Sarfatti On 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 Sarfatti PS
      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

On Feb 1, 2013, at 8:10 PM, nick herbert <quanta@cruzio.com> wrote:

Demetrios--

I'm already assembling a thought experiment in my head.
The nicest thing about thought experiments is that
all the sources and detectors are ideal
and work perfectly every time.

If we can't find a flaw using thought experiments
then physicists in every optics lab on Earth
will stampede
to be the first to observe
the Kalamidas Effect.

And reap the rewards.

Nick


On Feb 1, 2013, at 7:30 PM, Demetrios Kalamidas wrote:

Nick, you state:
"Although all of the parts of this experiment are
possible the whole experiment itself would be quite difficult."
  It would indeed be a technically challenging experiment, on the order of complexity of Zeilinger's recent Canary Islands teleportation stuff, IF the required distance to achieve the superluminality condition is sought for....
HOWEVER, if this bizarre effect is observed in just a table-top version, on the order of one meter, it will be extremely strong evidence that the same effect will be seen even if we stretch out the left and right wings to 10s or 100s of miles....there is no change in the nature of the set-up by doing this.
   The superluminality condition in my set-up is achieved when the distance is large enough for an observer, on the left, to statistically distinguish a "1" from a "0" before a classical and luminal signal gets there (and that is just a function of the efficiency of my scheme and technology).
Demetrios



On Fri, 1 Feb 2013 19:02:04 -0800
nick herbert <quanta@cruzio.com> wrote:
Well it isn't going to work.
But we may learn something
by seeing where it fails.
Although all of the parts of this experiment are possible
the whole experiment itself would be quite difficult.
Thought experiments are easier and cheaper
and don't need any hardware
except the human mind
and some paper and pencils.
So real experiment is premature.
On Feb 1, 2013, at 6:55 PM, JACK SARFATTI wrote:
Nose to the grindstone Nick!
I await your penetrating analysis.
If this worked it would be a Brave New World. ;-)
I have been preoccupied with Jim Woodward's Star Ship book and have  only been giving this partial attention.

On Feb 1, 2013, at 6:44 PM, nick herbert <quanta@cruzio.com> wrote:

Demetrios--

Clever. And of course--if it works -- there exist an optimum  product alpha x r that maximizes the Kalamidas Effect.
I can't offhand refute it but now that I understand what you're doing
I will certainly try.

Thanks, Jack, for sending me this clever FTL scheme.

Nick



On Feb 1, 2013, at 6:00 PM, Demetrios Kalamidas wrote:

Hi Nick,

  Yes....you got the main point of what I'm trying to do.
In Mandel's experiment, the "two halves" in which an idler photon  can exist are collapsed into a single path such that the origin  of the idler is "in principle" impossible to determine....we  don't even need any detectors in that idler path to destructively  register a photon.
  I am doing an analogous action by "blurring" each of the two  halves (modes a2 and b2),in which a right-going photon can exist,  with an indefinite photon number so that again, albeit in a less  efficient and more noisy way, we cannot SOMETIMES tell, even in  principle, if that right-going photon existed in mode a2 or in  mode b2.
  The "sometimes" part is, namely, the outcome |1>a2'|1>b2'  since it could be that: the photon in a2' came from the entangled  pair while the photon in b2' came from a weak coherent state !OR!  the photon in a2' came from a weak coherent state while the  photon in b2' came from the entangled pair.
  We DO NOT NEED ANY DETECTORS on the right wing of the experiment, as in Mandel's set-up. In my scenario, the possible  outcomes in modes a2' and b2' (in terms of photon number) are:  01,10,11,02,20,12,21 which are all "in principle"  distinguishable, with the only caveat being that the outcome "11"  has the special effect of erasing the path information of a left- going photon (which in turn leads to a small amount of  interference on the left).
Demetrios



On Fri, 1 Feb 2013 15:56:00 -0800
nick herbert <quanta@cruzio.com> wrote:
Demetrios--
I am trying to understand your device.
You seem to be trying to erase the "which path" info
without combining the two possible paths.
How are you doing this?
For clarity I assume your detectors are perfect
and measure the Number of Photons in
each two-photon entangled event.
In any ordinary experiment that number (or either side a or b)  must  be One.
And where that One Photon ends up can indicate
Which Path or Which Interference Pattern depending on design.
Both of these designs involve coincidence detection.
If I understand your proposal
you attempt to erase the which-path info
by adding (via a biased beam splitter)
a coherent state to each possibility channel.
Since coherent states possess an indefinite photon number
the number of photons that appear at the detectors is also  indefinite
and the observer cannot decide
which path the photon took
no matter what the reading of the detectors.
Is this how your device works?
Nick
On Jan 30, 2013, at 4:51 PM, nick herbert wrote:
Each single photon of the pair is produced in a SUPERPOSITION
of a and b directions. Observation of "which path" can collapse  the
superposition into either a or b but (in conventional experiments)
these collapses (in the absence of coincidence signals) appear
to occur at random.

Destroying the path information by conventional means
(say, combining a and b in a beam splitter) does not
produce interference by itself but can do so if coincidence
signals are introduced.

DAK claims that by adding coherent states to the separated
halves of the superposition, that he can destroy "which path"
information in a manner that produces "weak interference"
without resorting to coincidence signals.


On Jan 30, 2013, at 2:30 PM, Demetrios Kalamidas wrote:

Hi guys,
....and thanks for the interest in my idea....and SORRY! Fred  for  not getting back to you, I've been traveling all last  week and  this week for my job....I'm responding from an MIT  computer right  now (as I'm working).

Let me try to quickly clarify some points:
  The source S produces only SINGLE PAIRS of photons, with a   photon pair created in modes a1a2 !OR! b1b2.
   In Mandel's experiment, it is the overlap of the two idler  modes causes erasure of the 'which-way' info for a  signal photon.  I wanted to find an 'unfolded' version of this  concept so that  space-like separation could be achieved.
  The method that, I purport, does the job of erasing the  'which- way' info for a left-going photon (that could be in  EITHER mode a1  OR in mode b1) is that the corresponding modes, a2 and b2, are  'mixed' with weak coherent states (each  having at most one photon)  such that, sometimes, we'll get  one photon in each of the two  output modes, a2' and b2', and  this makes it impossible to tell  where each of these two  photons came from.  If the math is valid,  this procedure  leads to a small amount of 'pure state' on the left wing of  the experiment....as opposed to the completely mixed state   that would arise if the coherent states were absent and only  the  two-photon state from S was present.
  I'll try to keep up with any further comments, questions,  and  discussions.
Demetrios



On Wed, 30 Jan 2013 13:03:37 -0800
JACK SARFATTI <adastra1@me.com> wrote:
PS
OK the two coherent state inputs replace Mandel's idler photons.  So when you include a3 & b3 with the original pair  from S you  have 4-photon states in the Hilbert space two of  them are Glauber  states and the original pair are Fock states.
Begin forwarded message:
On Jan 30, 2013, at 12:56 PM, JACK SARFATTI   <sarfatti@pacbell.net> wrote:
Wait a second, he has 4 photons s1, i1, s2, i2 - at least in  the  Mandel experiment
However, you & Fred are right, Kalamidas's picture is  confusing  it seems to show only two photons, but he cites  Mandel, so does  he actually have 4 photons - two signal &  two idler like Mandel?  On Jan 30, 2013, at 12:41 PM, nick  herbert <quanta@cruzio.com>  wrote:
Fred Wolf is right. Like the original EPR this is a TWO- PARTICLE experiment -- one particle going to the left and  one  particle going to the right in each elemental  emission. If  DAK's argument depends on seeing this as a 4- particle  experiment, then DAK is certainly WRONG.
Nick Herbert
On Jan 29, 2013, at 10:22 AM, JACK SARFATTI wrote:
Thanks Fred.
I hadn't thought to check out his starting point Eq. 1 I only  looked at Eq. 6. These experiments are tricky. I  have not yet  understood the details. Hopefully Nick &  others will chime in.  Begin forwarded message:
From: "fred alan wolf" <fawolf@ix.netcom.com>
Subject: RE: PPS Demetrios A. Kalamidas's new claim for superluminal entanglement communication looks obvious at second sight
Date: January 28, 2013 11:11:31 PM PST
To: "'JACK SARFATTI'" <sarfatti@pacbell.net>
            Of course it is wrong for some serious and perhaps not so obvious reason.  He has confused a four  photon  state with an entanglement of two entangled (two)  particle  states. He approached me and I explained why it  was wrong.  Here is my explanation sent to him to which he has not  responded:
“Thanks for the paper.  Following Zeilinger’s paper   (attached) I am having some trouble understanding your  eq. 1.  If I understand it correctly you are using a  path  entanglement scheme similar to the one illustrated in  Zeilinger’s attached paper (p S290).  Therefore I  think you  should have  a1 entangled with b2 and a2  entangled with b1.  We would get e.g., (|a1>|b2>+ |b1>| a2>)/Ö2. Given that |a1> =  (|0>+exp(iphi)|1>)/Ö2, and  similarly for a2, b1, and b2, I  fail to see how you get your eq. 1, which seems to be some  kind of mixed four  photon state.”     Best Wishes,
Fred Alan Wolf Ph.D.  aka Dr. Quantum ®

So Jim means frame invariant rest mass m by "inertia" in the quote below. He does not mean the global geodesic pattern solution of Einstein's GR field equations in matter
Guv + (8piG(index of refraction)^4/c^4)Tuv = 0.
Now Jim claims a bare mass cosmological screening.
m is from Higgs field, + Quantum Chromodynamica + Low Energy Nuclear + Atomic + Molecular + Solid State Physics.
Jim claims
m* = m(Cosmological Screening Factor) = mC
C =/= c = speed of light in vacuum.
and his Mach Effect Thruster claims to control C.
Personally I think, MY BIAS, this is astrology and that his effect in the lab is simply bad measurement. However, I may be wrong, and since I am writing my own book and my review of his book will be the in it - I continue.
We now know that Einstein was naive about the Origin of Inertia, Mach even more so. We have come a long way since those pre-quantum theory days.
The Origin of Inertia is
1) Higgs vacuum coherent Glauber state field post-inflation
2) SU(3) strong force Quantum Chromodynamics & it's low energy tail - Nuclear Physics
3) Atomic, Molecular, Solid State Many Body Problem - models of binding energies on different scales.
Inertia m is a FREE PARAMETER in Einstein's GR which is a local gauge theory of the T4 group.
The origin of inertia needs the U1, SU2 & SU3 internal groups + Higgs-Goldstone-Anderson "More is different" emergence of order in the spontaneous breakdown of symmetries in the ground states of complex many particle systems both real and virtual.
to be continued
Like · · Share
Jack Sarfatti There is no way to change the internal energies by a large fraction without destroying the material!

I fundamentally disagree with Jim Woodward .The inertia of the starship has nothing to do with real warp drive or the construction of a benign wormhole star gate time travel machine.

to be continued

"in a manner that produces "weak interference" without resorting to coincidence signals."
Yes Nick, but is it true? - is the 64 trillion dollar question. ;-)
On Jan 30, 2013, at 4:51 PM, nick herbert <quanta@cruzio.com> wrote:
Each single photon of the pair is produced in a SUPERPOSITION
of a and b directions. Observation of "which path" can collapse the
superposition into either a or b but (in conventional experiments)
these collapses (in the absence of coincidence signals) appear
to occur at random.
Destroying the path information by conventional means
(say, combining a and b in a beam splitter) does not
produce interference by itself but can do so if coincidence
signals are introduced.
DAK claims that by adding coherent states to the separated
halves of the superposition, that he can destroy "which path"
information in a manner that produces "weak interference"
without resorting to coincidence signals.
On Jan 30, 2013, at 2:30 PM, $ wrote:
Hi guys,
....and thanks for the interest in my idea....and SORRY! Fred for not getting back to you, I've been traveling all last week and this week for my job....I'm responding from an MIT computer right now (as I'm working).
Let me try to quickly clarify some points:
The source S produces only SINGLE PAIRS of photons, with a photon pair created in modes a1a2 !OR! b1b2.
In Mandel's experiment, it is the overlap of the two idler modes causes erasure of the 'which-way' info for a signal photon. I wanted to find an 'unfolded' version of this concept so that space-like separation could be achieved.
The method that, I purport, does the job of erasing the 'which-way' info for a left-going photon (that could be in EITHER mode a1 OR in mode b1) is that the corresponding modes, a2 and b2, are 'mixed' with weak coherent states (each having at most one photon) such that, sometimes, we'll get one photon in each of the two output modes, a2' and b2', and this makes it impossible to tell where each of these two photons came from. If the math is valid, this procedure leads to a small amount of 'pure state' on the left wing of the experiment....as opposed to the completely mixed state that would arise if the coherent states were absent and only the two-photon state from S was present.
I'll try to keep up with any further comments, questions, and discussions.
Demetrios
On Wed, 30 Jan 2013 13:03:37 -0800
JACK SARFATTI <adastra1@me.com> wrote:
PS
OK the two coherent state inputs replace Mandel's idler photons. So when you include a3 & b3 with the original pair from S you have 4-photon states in the Hilbert space two of them are Glauber states and the original pair are Fock states.
Begin forwarded message:
On Jan 30, 2013, at 12:56 PM, JACK SARFATTI <sarfatti@pacbell.net> wrote:
Wait a second, he has 4 photons s1, i1, s2, i2 - at least in the Mandel experiment
However, you & Fred are right, Kalamidas's picture is confusing it seems to show only two photons, but he cites Mandel, so does he actually have 4 photons - two signal & two idler like Mandel?

On Jan 30, 2013, at 12:41 PM, nick herbert <quanta@cruzio.com> wrote:
Fred Wolf is right. Like the original EPR this is a TWO-PARTICLE experiment -- one particle going to the left and one particle going to the right in each elemental emission. If DAK's argument depends on seeing this as a 4-particle experiment, then DAK is certainly WRONG.
Nick Herbert


On Jan 29, 2013, at 10:22 AM, JACK SARFATTI wrote:
Thanks Fred.
I hadn't thought to check out his starting point Eq. 1 I only looked at Eq. 6. These experiments are tricky. I have not yet understood the details. Hopefully Nick & others will chime in. Begin forwarded message:


From: "fred alan wolf" <fawolf@ix.netcom.com>
Subject: RE: PPS Demetrios A. Kalamidas's new claim for superluminal entanglement communication looks obvious at second sight
Date: January 28, 2013 11:11:31 PM PST
To: "'JACK SARFATTI'" <sarfatti@pacbell.net>
Of course it is wrong for some serious and perhaps not so obvious reason. He has confused a four photon state with an entanglement of two entangled (two) particle states. He approached me and I explained why it was wrong. Here is my explanation sent to him to which he has not responded:
“Thanks for the paper. Following Zeilinger’s paper (attached) I am having some trouble understanding your eq. 1. If I understand it correctly you are using a path entanglement scheme similar to the one illustrated in Zeilinger’s attached paper (p S290). Therefore I think you should have a1 entangled with b2 and a2 entangled with b1. We would get e.g., (|a1>|b2>+ |b1>|a2>)/Ö2. Given that |a1> = (|0>+exp(iphi)|1>)/Ö2, and similarly for a2, b1, and b2, I fail to see how you get your eq. 1, which seems to be some kind of mixed four photon state.” Best Wishes,
Fred Alan Wolf Ph.D. aka Dr. Quantum


On Jan 29, 2013, at 12:51 AM, jfwoodward@juno.com wrote:

Yes Paul, it is possible to treat inertia, and inertial forces in particular, in GRT just as one does in Newtonian mechanics -- that is, inertia (and its measure, mass-energy) is a primary quality of matter not requiring further explanation. 
Jack: "Inertia", "Inertial" is used in self-contradictory ways, often by the same author in the same paper or text book.

Wheeler & Co seem to use it consistently. In the book "Gravity & Inertia" by the "origin of inertia" they do not mean the rest mass "m" of the observed test particle in Newton's second law F = ma in inertial frames. They mean the pattern of force free geodesics. Note, that in non-inertial frames

F +   Ffictitious = ma'
where  F is the external non-gravity force spacelike 3-vector, a is the proper acceleration 3-vector in the inertial frame. In a rotating non-inertial frame for example in the v/c << 1 limit with rotation 3D pseudo-vector w with apparent velocity v' and radial vector r' in the non-inertial frame.

 Ffictitious ~ -mw x w x r' - 2m w x v' - GMmr'/r^3 - mdw/dt x r'

= 0 in an inertial frame


The apparent acceleration (aka "kinematical") of the test particle in the non-inertial frame is a'

In the formalism of Einstein,  Ffictitious is contained in the Levi-Civita connection term of the covariant derivative.

Case 1 test particle on a geodesic

F = 0, a = 0

Here the fictitious force is really fictitious as far as the test particle m is concerned, though it may be caused be real forces on the detector/observer.

Case 2 test particle is off-geodesic - case of uniform circular motion, with M = 0

Specializing to the now non-inertial co-rotating rest frame of the test particle

F =/= 0, a =/= 0,  dw/dt = 0, a' = 0, v' = 0

All we have left is

F = -mw x w x r'

Where F is the real radially inward non-gravity centripetal force, with an equal and opposite centrifugal force on the source of F.

Case 3  gravitational geodesic orbit - no real force on the test particle.

 -mw x w x r' - 2m w x v' - GMmr'/r^3 - mdw/dt x r' = ma'

m cancels out of the equation.

w and v' are the rotation pseudo-vector of the source mass M and relative velocity of the detector with the test particle of mass m.

The proper acceleration of the test particle a = 0.


Jim: That is why I included Abraham Pais's quote of Einstein on Mach's principle that is the header for Chapter 2 in the book.  But I think you will agree that the discussion of the past year and a half about inertia has been about substantially more than just semantic disagreements.  Unless there is some backsliding, we now seem all to be talking about the same thing, though there is still no agreement on the CAUSE of inertia and inertial "effects" [that is, inertial reaction (third law) forces].

Jack: Jim your above remark is mostly metaphysics not physics. Your remark on the "CAUSE of inertia" is not Popper falsifiable. Again, do you mean "m" or "geodesic pattern"? Probably the former. The cause of Newton's third law is not a mystery. It's mainstream physics Noether's "first" theorem applied to the translation group T3 in an isolated non-dissipative system.
Noether's (first) theorem states that any differentiable symmetry of the action of a physical system has a corresponding conservation law. The theorem was proved by German mathematician Emmy Noether in 1915 and published in 1918.[1] The action of a physical system is the integral over time of a Lagrangian function (which may or may not be an integral over space of a Lagrangian density function), from which the system's behavior can be determined by the principle of least action.
Noether's theorem has become a fundamental tool of modern theoretical physics and the calculus of variations. A generalization of the seminal formulations on constants of motion inLagrangian and Hamiltonian mechanics (developed in 1788 and 1833, respectively), it does not apply to systems that cannot be modeled with a Lagrangian alone (e.g. systems with a Rayleigh dissipation function). In particular, dissipative systems with continuous symmetries need not have a corresponding conservation law. ...
Examples
Time invariance
For illustration, consider a Lagrangian that does not depend on time, i.e., that is invariant (symmetric) under changes t → t + δt, without any change in the coordinates q. In this case,N = 1, T = 1 and Q = 0; the corresponding conserved quantity is the total energy H[5]

Translational invariance
Consider a Lagrangian which does not depend on an ("ignorable", as above) coordinate qk; so it is invariant (symmetric) under changes qk → qk + δqk. In that case, N = 1, T = 0, andQk = 1; the conserved quantity is the corresponding momentum pk[6]

In special and general relativity, these apparently separate conservation laws are aspects of a single conservation law, that of the stress–energy tensor,[7] that is derived in the next section.
Rotational invariance
The conservation of the angular momentum L = r × p is analogous to its linear momentum counterpart.[8] It is assumed that the symmetry of the Lagrangian is rotational, i.e., that the Lagrangian does not depend on the absolute orientation of the physical system in space. For concreteness, assume that the Lagrangian does not change under small rotations of an angle δθ about an axis n; such a rotation transforms the Cartesian coordinates by the equation

Since time is not being transformed, T=0. Taking δθ as the ε parameter and the Cartesian coordinates r as the generalized coordinates q, the corresponding Q variables are given by

Then Noether's theorem states that the following quantity is conserved,

In other words, the component of the angular momentum L along the n axis is conserved.
If n is arbitrary, i.e., if the system is insensitive to any rotation, then every component of L is conserved; in short, angular momentum is conserved.

http://en.wikipedia.org/wiki/Noether's_theorem

Jack: NO NEED FOR ASTROLOGY!

Jim: As far as I am concerned, this is real progress.  I gave up hope of even getting this far a year ago.  That is why my participation in this discussion over the past few months has been minimal. The other part of the issue of inertia, on which I still hold out no hope of getting understanding and agreement, is that we now -- 90 years after Einstein said what was reported by Pais in the mentioned quote -- know a lot more about cosmology.  Indeed, we know that critical cosmic matter density obtains, and accordingly that space is flat at cosmic scale AS MATTERS OF FACT. 
Jack: This is irrelevant it seems to me.

Jim: That means that, as Jack is now calling it, Sciama's "screening factor" -- the coefficient of time derivative of the gravitomagnetic vector potential in the gravielectric field equation (in the approximation where only the g_oo and g_oi potentials need be considered) is one.  And that means that inertial "effects" are accounted for as the gravitational action of chiefly cosmic matter (where "matter" is everything that gravitates).

Jack: Jim, I think your sentences here are very ambiguous.  First of all g00 and g0i are local observer frame-dependent. They are not invariant geometric objects. Also the universe on the large scale is not rotating so g0i = 0 in the usual representations. For example, in the usual comoving observer representation g00 = 1 and g0i = 0, so what happened to your theory here?

General metric for LIF geodesic co-moving observers

The FLRW metric starts with the assumption of homogeneity and isotropy of space. It also assumes that the spatial component of the metric can be time-dependent. The generic metric which meets these conditions is

where  ranges over a 3-dimensional space of uniform curvature, that is, elliptical space, Euclidean space, or hyperbolic space. It is normally written as a function of three spatial coordinates, but there are several conventions for doing so, detailed below.  does not depend on t — all of the time dependence is in the function a(t), known as the "scale factor".
[edit]Reduced-circumference polar coordinates
In reduced-circumference polar coordinates the spatial metric has the form

k is a constant representing the curvature of the space. There are two common unit conventions:
k may be taken to have units of length−2, in which case r has units of length and a(t) is unitless. k is then the Gaussian curvature of the space at the time when a(t) = 1. r is sometimes called the reduced circumference because it is equal to the measured circumference of a circle (at that value of r), centered at the origin, divided by 2π (like the r of Schwarzschild coordinates). Where appropriate, a(t) is often chosen to equal 1 in the present cosmological era, so that  measures comoving distance.
Alternatively, k may be taken to belong to the set {−1,0,+1} (for negative, zero, and positive curvature respectively). Then r is unitless and a(t) has units of length. When k = ±1, a(t) is the radius of curvature of the space, and may also be written R(t).
A disadvantage of reduced circumference coordinates is that they cover only half of the 3-sphere in the case of positive curvature—circumferences beyond that point begin to decrease, leading to degeneracy. (This is not a problem if space is elliptical, i.e. a 3-sphere with opposite points identified.)
http://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

Jack: Not only that, but our future universe is asympotically de Sitter which in the static LNIF representation is

Static coordinates

We can introduce static coordinates  for de Sitter as follows:



where  gives the standard embedding the (n−2)-sphere in Rn−1. In these coordinates the de Sitter metric takes the form:

Note that there is a cosmological horizon at .
http://en.wikipedia.org/wiki/De_Sitter_space


Jack to Jim: Indeed you did not seem to know that g00 = 0 is a condition for a horizon. Note that here g0i = 0.


Jack: Machian Screening Factor?

Analogy to Debye Screening

Debye length
From Wikipedia, the free encyclopedia
In plasmas and electrolytes the Debye length (also called Debye radius), named after the Dutch physicist and physical chemist Peter Debye, is the scale over which mobile charge carriers (e.g. electrons) screen out electric fields. In other words, the Debye length is the distance over which significant charge separation can occur. A Debye sphere is a volume whose radius is the Debye length, in which there is a sphere of influence, and outside of which charges are screened. The notion of Debye length plays an important role in plasma physics, electrolytes and colloids (DLVO theory).
http://en.wikipedia.org/wiki/Debye_length

For off-geodesic motion F = ma Newton's 2nd law

Jim based on an obscure model of Sciama & Berry? (who denies it in email to me) says

F = ma should be replaced by

F = (cosmological screening factor) ma
Exactly what it is and how to compute it using real Einstein GR as opposed to a completely unjustified EM analogy is not clear to me. I hope it is clear in Jim's book?

Suppose it is really there? Even so, it says nothing about m itself. You can think of m as the cosmologically bare mass of the test particle. m is still determined locally by Higgs field, + QCD + standard nuclear, atomic, molecular, solid state physics of binding energies.

Next step is to look at the Hoyle-Narlikar Wheeler-Feynman theory of gravity.