I awoke with a start. So I want to record this before I forget the "dream."
OK consider Ibison's "point" approximation of the "Alcubierre" geodesic warp drive bubble history in an otherwise 1905 Special Relativity (SR) globally flat Minkowski space-time. In the usual way choose a Global Inertial Frame (GIF) allowing a spacelike foliation/slicing into 3D hypersurfaces. As I said the geodesic warp drive bubble "point" is a singularity relative to the SR theory in which RIGID T4 symmetry is broken replaced by ELASTIC T4(x) GCT symmetry. The SR observers outside the geodesic warp bubble are Ibison's AFO Minkowski observers who can be inertial geodesic or not -- doesn't matter.
But the geodesic warp drive history can be arbitrarily SPACELIKE TACHYONIC relative to the AOFs. That is ANYTHING GOES! Imagine a starting point at the t = 0 spacelike hypersurface at initial position P(0) where the tiny ship "electron" switches on its geodesic warp bubble. The ship arrives at the t hypersurface at some final position P'(t). Note that the proper time is the same for all possible warp bubble histories. Well this is precisely Feynman's quantum path integral picture of the propagator of a relativistic point particle that takes ALL POSSIBLE spacetime paths connecting P(0) to P'(t) including spacelike and BACKWARDS IN TIME (particle-antiparticle production and annihilation). In Bohm's picture this is the work of the Quantum Potential Q.
Therefore, Bohm's quantum potential Q for a "point particle" in otherwise gravity free globally flat Minkowski spacetime is equivalent to a microscopic geodesic warp drive bubble in which the Captain like in The Flying Dutchman takes all possible voyages from StarGate Portal P(0) to StarGate Portal P(t) where t can even be negative (backwards in time).
OK so Ibison and Davis are going to write a "paper" are they? Well it better not have this idea in it without mentioning that it's my idea given to me by ET perhaps? IN A WAKING DREAM. It's 5:47 AM PST Wed June 20, 2007 San Francisco.
On Jun 19, 2007, at 9:29 PM, Jack Sarfatti wrote:
2nd draft
What Michael Ibison and Eric Davis do not understand, do not grok about their claim is that like the spacetime singularity of a black hole in curved spacetime, their warp drive stringy world line is also a singularity of their global special relativity model. They do not understand how the group theory applies to their classical model. Each theory has a limited domain of validity specified by the symmetries of the theory. Those domains of validity are like circles of convergence of power series in complex function theory. The warp drive history is like a pole in the complex plane of w = w(z). The theory stops there and we need a larger theory. In a way this is Godel's incompleteness theorem of 1931 since any theory is a formal system. Progress in theoretical physics is like analytic continuation in the mathematics of complex functions of several complex variables.
On Jun 19, 2007, at 6:38 PM, Jack Sarfatti wrote:
What is this bait and switch? Ibison has simply parroted back my ideas as if he was saying that all the time? I will put my ideas he has fed back to me in red below.
RIGID SYMMETRY is GLOBAL SYMMETRY with a CONSERVED GLOBAL CHARGE.
RIGID SYMMETRY means there is Lie group G with n space-time independent parameters under which the global dynamical action of the system of the source fields is invariant.
In such a case, one can define local source current densities Ju that are locally conserved. This means that the globally flat Minkowksi spacetime SR 4-divergence of the currents vanish
J^au^,u = 0
u = 0,1,2,0 with 0 timelike inside the local light cone
a = 1 ... n
Given a spacelike slice of 4D spacetime the 3D integral of J0^a = Q^a the conserved charge i.e.
dQ/dx0 = 0 Emmy Noether's theorem of 1918 not known by Einstein really in 1916 in the most general form.
Now all this assumes that in addition to G that the dynamical action is invariant under RIGID T4 where a = 0,1,23 and the stress-energy current densities are now Tuv for a given set of classical fields
Total energy E = 3D space integral of T00
Pi = 3D space integral of T0i
i = 1,2,3
But you cannot do this in general in Einstein's general relativity because RIGID T4 is no longer a symmetry group of the dynamical action if real gravity fields are present, i.e. non-vanishing 4 rank GCT conformal curvature tensor fields in classical vacuum. No dark zero point Ricci source exotic vacuum energy - to keep it simple for now. This is basically the equivalence principle. The curved tetrad fields A^a are the compensating gauge connection fields when the 4 spacetime independent physical displacement parameters of T4 become arbitrary functions of the local coordinate chart. We now have a warped 4D "world crystal" where the parameters of the locally gauged NON-RIGID T4, i.e. T4(x) are the crystal distortions, and the gauge transformations are precisely Einstein's GCT's (General Coordinate Transformations).
So let's look at this simply. Suppose the AFO (Asymptotically Flat Observer) outside the warp bubble in its FUTURE LIGHT CONE receives long waves only that are much bigger than the warp bubble with the ship inside it. The bubble is a "point" to such probes. The history of the warp bubble is a SINGULAR STRING WORLD LINE like a cut for a complex function w = f(z) in the complex z-plane. That is 1905 SR with RIGID T4 symmetry breaks down on the string replaced by NON-RIGID ELASTIC T4(x) that is a larger symmetry group. The A^a warped tetrad fields are zero outside the singular warp string. They are not zero in the string. If you think of vortices A^a are like the vorticity of the vortex core.
OK let 3G(ti) & 3G(tf) be the initial and final 3 Geometries (hypersurfaces) connected by the warp bubble string. The ship leaves at ti and arrives at tf in some localized foliation. Now ANYTHING GOES, there is no reason why the Pu of the matter fields of the ship and its occupants need to be conserved, i.e. need to be the same on both initial and final spacelike hypersurfaces. They CAN BE, they NEED NOT BE! It's completely arbitrary subject to the whim of the captain of the ship because the warp history does not obey RIGID T4 NO PROPELLANT NEED APPLY!
NO SYMMETRY --> NO CONSERVATION!
LOCAL CONSERVATION of T4 CURRENTS does not imply GLOBAL CONSERVATION of T4 CHARGES on 3D spacelike hypesurfaces when T4 is violated inside the 4D spacetime region where the WARP BUBBLE is.
NO PROPELLANT NEED APPLY!
Note that for internal symmetries and RIGID T4 the above does not apply because there is no 4D equivalence principle for internal symmetries and their conservation laws.
Then we have spontaneous symmetry breakdown - another story.
On Jun 19, 2007, at 4:52 PM, Dr. Eric Davis wrote:
What Michael relates below is completely correct. It couldn't be better articulated.
Eric
snip----------------------------------
Bullshit. I have said no such thing. Are you being intentionally obstreperous?
Let me repeat what you are saying back to you so you cannot keep misrepresenting
me:
Zero 4D covariant 4-acceleration means on-board g-force.
NO Mike you got it wrong. That's NOT what I wrote. I will give you benefit of the doubt that it's a typo.
What I wrote a JILLION TIMES seemingly to no avail was JUST THE OPPOSITE
Zero 4D covariant 4-acceleration means ZERO on-board g-force. NO g-force.
The Einstein GR geodesic equation for this is
DP^u/ds = 0
Assuming dm/ds = 0 i.e. no actual ejection of rest mass - and this include a jet of HFGW (High Frequency Gravity Waves), or photons, or atoms, or anything, then
DP^u/ds = mcdV^u/ds - mc(Connection)^uvwV^vV^w = On-board g-force = 0
I mean here 4-force obviously.
P^u = mcV^u
m = rest mass of COM of ship
V^u = dx^u/ds
ds = dt/(gamma)
d/ds = (gamma)d/dt
gamma = (1 - (v/c)^2)^-1/2
Now
mcdV^u/ds is the kinematical "Euclidean" APPARENT "Newtonian" g-force (generalized to 1905 SR) computed from radar bounces et-al
mc(Connection)^uvwV^vV^w is, if you like sci-fi, the curved (possibly torsioned) spacetime "inertial compensation" term. That is the term we must control with dark zero point energy to metric engineer warp and wormhole for controlled geodesic dogfights without on-board g-force and NO TIME DILATION!
Such 'motion' does not require propellant. Indeed, there must be no propellant
in order for such motion to take place.
Parrots back what I have repeatedly written in the public record.
I will assume that you are with me so far, and move on:
Obviously Mike you and Eric must be under the control of an Evil Alien or a Malevolent Fairy or some self-professed Skeptic perhaps?
Amazing Randi
This is your bait and switch where you take my words and throw it back at me as if you are "educating" me! You could sell used cars in Tijuana with that shtick. ;-)
The above statement are assumed valid for the craft and it's passengers. They
are valid because of the assumed metric bubble enveloping the craft.
So you parrot back what I have written.
At some point the metric bubble flattens out to flat space (I am working in
coordinates such that g_00=1, g_0i=0.)
Put a box around that metric-bubble-plus-craft. Call that box a 'particle'.
IF that particle has mass, that is: if the ADM adjusted mass of the ship is
non-zero, then that particle must conserve momentum IN FLAT SPACE;
it cannot accelerate as measured against that flat space without propellant -
again as defined by the flat space observer.
Conserve momentum where-when? This is your error! Of course when the metric bubble vanishes it will need propellant in the usual Newtonian and SR way. SO WHAT?
Look when the warp bubble flattens out as you put it, then of course obviously there are global inertial frames GIFs where
(Connection Field) = 0 at all points in the flat region
and Tuv^,v(Ship) = 0
etc. so momentum of the COM of ship is LOCALLY CONSERVED in usual SR -> Galilean fashion when there is no anomalous warp bubble. SO WHAT ELSE IS NEW?
Please think about this before replying. It is entirely possible that the motion
of the craft is geodesic and there is no propellant as seen from on board the
craft. Whilst, AT THE SAME TIME, the flat space observer for whom
craft-plus-metric-bubble = particle, there must be propellant if that box has
ordinary acceleration.
But the box does not have ordinary acceleration inside the warp bubble - that's the point!
It would be nice if you defined your terms.
I mean dV^u/ds as "ordinary acceleration". Indeed that would be what radar sees if the bubble is small compared to wavelength of signal, but that's not what we have in general. It's the other way around so we need the Terrell type calculations. Follow the null geodesics from different points on the ship to the AFO's. Not a trivial simulation even for Alcubierre's toy model I bet. Why don't you try it?
Yes: ORDINARY ACCELERATION! Why? Because they can only
see flat space coordinates plus a 'particle', the interior details of which do
not concern them.
See above.
They are only interested in the fact that the 'particle' (as a
unit) has a non-zero mass as referred (unambiguously) to their coordinate
system. To their level of acuity, Covariant Acceleration of 'particle' =
Ordinary Acceleration of 'particle'.
That is only true if (CONNECTION) = 0 at every future light cone origin along the ship's world line where the ship's future light cones intercept AFO's detectors. That is only when warp bubble is not there! Draw a spacetime diagram.
Please observe the fact that I am referring to the craft-plus-metric-bubble as a
unit = 'particle'. That unit requires propellant to accelerate. You, by
contrast, keep referring to the craft. The craft DOES NOT NEED PROPELLANT to
accelerate (if, as assumed, it moves on a geodesic). Please take time to
understand this distinction before replying!
Einstein said, make it simple, but not simpler than is possible. You are making it more complicated than is needed. So where is this phantom meta-propellant of yours to come from? It's not required for Einstein's GR and beyond to Einstein-Cartan with torsion. You got turtles on turtles. "It's turtles all the way down." Hawking in "A Brief History of Time" intro.
Basically you have obliterated the difference between
Tuv^,v(ship) = 0 NO WARP BUBBLE
Tuv,^;v(ship) = 0 WARP BUBBLE
Where Tuv is a finite world tube of support for the material fields of the ship inside the warp bubble.
There is no AOF global P^u conservation before warp and after warp needed. Strict Pu global conservation is only true for RIGID T4 symmetry which is VIOLATED inside the warp bubble! It's NOT LIKE S-Matrix theory. Your analogical argument is not good. This is one reason why "quantum gravity" is elusive.
This is pretty trivial stuff Jack. You seem to be so beguiled by metric bubbles
and warp drives that you cannot hear what I am saying.
So far you have not said anything real. You have invoked some ghost propellant not required by GR or its extensions.
Once you see what I am
saying rather than what you THINK I am saying, no doubt you will say that it is
obvious and not worthy of debate. It is obvious! It isn't worthy of debate! But
since you are having difficulty, Eric and & have decided to write it up as a
short paper - just in case there are others out there who cannot see the wood
for the trees. We'll see you in court!
By all means write a paper, but don't leave my name out of it as if I don't exist, and let me check that you cite me correctly I mean that you don't misrepresent my ideas.
To give you the benefit of the doubt Michael & Eric, your verbal arguments are too vague. You have not shown enough math to pin down your meaning. Hence I say "not even wrong" - Pauli
In a nut shell - the "point" warp bubble is a kind of SINGULARITY for the AFOs in which RIGID T4 symmetry is VIOLATED. Hence, by Noether's theorem you cannot conclude that GLOBAL Pu for the ship is CONSERVED in time on spacelike slices.
You do not get the RIGID GLOBAL SYMMETRY implies and is implied by GLOBAL CONSERVATION.
The symmetry group here is T4.
The warp bubble VIOLATES T4 invariance of the warp bubble-ship field action. If you think of warp bubble + ship inside it as a POINT, then what you have is a SINGULARITY - a world tube singularity in which T4 is violated.
NO SYMMETRY --> NO CONSERVATION!
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