Juan R.:

Bilge ha escrito:

Juan R.:

Bilge ha escrito:
Most people would disgree that a theory based on instantaneous
action at a distance is a better idea, since the instantaneous
part conflicts with observations.

Completely wrong!!!

Let me know when you plan to explain that statement.

To you? Newer. It is a waste of time. Just how attempt to explain to
Eric that

-c^2dt^2 = -d(ct)^2 and because x^0 = ct therefore, g_{00} = -1.
Independent of if c=1 or is not. Because if one does c=1

-1^2dt^2 = -d(1t)^2 and because x^0 = 1t therefore, g_{00} = -1 AGAIN.

If you're having a hard time understanding why no one takes you
seriously, the reason is that you reject science with kooky arguments.



As proven by many authors retarded LW potentials disagree with many
recent experimental data on Mercury forces on Hg, railguns, tokamaks
anomalies, etc. From a theoretical point of view, several authors have
proven in recent years that LW potentials are theoretically incorrect.
In the PRE article i cited above authors proved 1) LW potentials are
not complete solutions of Maxwell equations 2) The introductions of an
instantaneous action does the solutions complete.

My only criticism to that paper is the dualism concept which is solved
in my approach.

Is there some point to this superfluous digression or are you just
posting to read your own press?

The cite to paper was just for highligintg that your understanding of
relativity and instantaneous interactions is wrong

Colleague, have your hear about Wheeler/Feynmann theory of absortion?

Numbskull, have you heard about the nuclear optical model?
It has the same relevance to this thread as everything you've
posted - zero.


Nonsense!

Basically, both you and eugene suffer from the same misconceptions.
You believe the universe is _really_ galilean, deep down, and the
rest is a facade.

Another misconception! You have no idea guy!

Juan R.

Center for CANONICAL |SCIENCE)

Bilge ha escrito:

Juan R.:

Bilge ha escrito:

Juan R.:

Bilge ha escrito:
Most people would disgree that a theory based on instantaneous
action at a distance is a better idea, since the instantaneous
part conflicts with observations.

Completely wrong!!!

Let me know when you plan to explain that statement.

To you? Newer. It is a waste of time. Just how attempt to explain to
Eric that

-c^2dt^2 = -d(ct)^2 and because x^0 = ct therefore, g_{00} = -1.
Independent of if c=1 or is not. Because if one does c=1

-1^2dt^2 = -d(1t)^2 and because x^0 = 1t therefore, g_{00} = -1 AGAIN.

If you're having a hard time understanding why no one takes you
seriously, the reason is that you reject science with kooky arguments.

That simply is not true.



As proven by many authors retarded LW potentials disagree with many
recent experimental data on Mercury forces on Hg, railguns, tokamaks
anomalies, etc. From a theoretical point of view, several authors have
proven in recent years that LW potentials are theoretically incorrect.
In the PRE article i cited above authors proved 1) LW potentials are
not complete solutions of Maxwell equations 2) The introductions of an
instantaneous action does the solutions complete.

My only criticism to that paper is the dualism concept which is solved
in my approach.

Is there some point to this superfluous digression or are you just
posting to read your own press?

The cite to paper was just for highligintg that your understanding of
relativity and instantaneous interactions is wrong

Colleague, have your hear about Wheeler/Feynmann theory of absortion?

Numbskull, have you heard about the nuclear optical model?
It has the same relevance to this thread as everything you've
posted - zero.


Nonsense!

Basically, both you and eugene suffer from the same misconceptions.
You believe the universe is _really_ galilean, deep down, and the
rest is a facade.

Another misconception! You have no idea guy!

Juan R.

Center for CANONICAL |SCIENCE)

Bilge wrote:
Eugene Stefanovich:

Thank you for your kind offer.
There are basically 3 ways to do unitary transformation in quantum
mechanics.


1. You can apply the transformation to operators of observables
e.g., H' = U^{-1}HU and keep state vectors intact

2. You can apply transformation to state vectors |\Psi' = U |\Psi
and keep operators intact

3. You can apply transformation U to observables and inverse
transformation U^{-1} to state vectors.

In cases 1. and 2. the physics is changed, i.e., expectation values
are affected by the transformation. As you correctly pointed out in
your eq. (1),
cases 1. and 2. are equivalent. They are just two different ways
to say the same thing. In the case 3. there is no change in physics:
expectation values of observables do not change. This is a trivial
change of representation.

When I was talking about unitary dressing transformation, I was
talking about case 1. Your statement about the triviality of unitary
transformations refers to the case 3. We are talking about different
things.

I most certainly was _not_ talking about ``case 3.'' I clearly
operated with H and U^{-1}HU on the _same_ wavefunctions:


\Psi|H'|\Psi = \Psi|U^{-1}HU|\Psi (1)

Note the absence of primes on \Psi. Clearly, I can choose to apply
U to the state vectors to get \Psi' = U\Psi and write \Psi'|H|\Ps'.
Note there is no prime on that H. What you call ``case 3'' would
be something like, \Psi|U^{-1}U H U^{-1}U|\Psi = \Psi'|H'|\Psi'.

Agreed.
Then why did you say that unitary transformation of the Hamiltonian does
not change physics? If the state vector is kept the same then
expectation values of the original Hamiltonian

\Psi|H|\Psi

and the transformed Hamiltonian (your eq. (1)) are clearly different.
The physics has changed.

Eugene.

Bilge wrote:

Here's a really obvious problem with the action at a distance idea.
t
|
| .C A and B are simultaneous, so they occur
| at the ``same time,'' i.e., t = 0
+----+----+ x
A B
I now perform a lorentz transform,

t
| A and C occur at the ``same time,'' i.e., t = 0
|
|
+--------+-- x
A . C
B

If interactions are instantaneous, then obviously, at most only
one of the above can be correct, in which case, the system isn't
lorentz invariant, since the two choices for t = 0 differ only
by a lorentz transform. I guess that about sums up my opinion.


Your logic is fine except one important point. Lorentz (boost)
transformations
of space-time coordinates of events must depend on interaction
(that's the point missed in Einstein's special relativity and in your
diagrams). If they don't depend on interaction then, as can be easily
shown, you'll get into contradiction with the Poincare group
properties. Therefore, if events A and B are connected to each other
by interaction and by the cause-effect relationship, then they are
simultaneous in ALL reference frames.

This short argument looks like a handwaving, but it's not.
I wrote a book with detailed explanations of this point.
Hope you are having fun reading it.

Eugene.

In sci.physics Juan R. wrote:

[...]
The best attemp to reply my 'unortodox' view has been from specialist
Carlip. He has used a completely wrong metric with dimensions that
forces to us to change all of standard stuff -e.g. there is not EM four
currents in his nonstandard approach-. Finally he derives wrong
temporal dependence, wrong functional dependence of potentials,
incorrect equation of motion -moreover he just obtain the
nonrelativistic limit of the trajectory in a relativistic spacetime,
newer GR trajectory in a NONrelativistic spacetime-, he obtains zero
curvature -due to c^2 term into g_00 one has R = R_00 / g_00 -- 0- of
spacetime which reinforces my view that in the nonrelativistc regime
the causality structure of GR break -if gravity was spacetime curvature
then zero curvature would imply zero gravity which is wrong according
to Newtonian limit-.

Moreover, Carlip obtains all a couple of wrong results. for example he
obtains a nonzero 00-connection which implies that full physical
derivatives in the Newtonian limit are covariant ones WHICH is wrong.
In the Newtonian limit one, physical derivatives are partial and total
ones NEWER covariant ones.

According to Carlip derivatives as partial v / partial t that one find
in Newtonian textbooks are NON physical because he uses a non zero
00-connection.

Moreover, Carlip does not know what is the difference between a
potential and a field and he still unknow why Penrose (like other
specialists) has claimed that Ehlers boundary is unphysical. and
therefore Ehlers attempt to derive Newtonian limit of spacetime is
nonrigorous and experimentally unphysical, etc.

Briefly (since I don't have a lot of time to waste on cranks):

Juan R. has been shown mathematically rigorous derivations of the
Newtonian limit of general relativity. He doesn't like them because
they require a choice of coordinates to get the standard form of
Newton's equations. (General relativity is generally covariant,
while the form of Newton's equations that he likes isn't, but he
apparently believes that taking the limit c-infinity should
magically pick out a coordinate system.)

Juan R. uses a coordinate x^0=ct, where t is the Newtonian time,
and thinks this makes sense even in the limit c-infinity. Of
course, in this limit, t=x^0/c-0 for every finite value of x^0,
and all derivatives with respect to x^0 go to zero. He has
been told how to take the limit properly, even in his coordinates
(start with a large but finite value of c, multiply by an appropriate
power of c so that the c-infinity limit makes sense, and then take
the limit), and has been shown how this process gives the correct
Newtonian limit, but he doesn't like that, either.

Juan R. has apparently casually read a paragraph or two about
Cartan's formulation of Newtonian gravity as a spacetime theory
with a preferred time, and has misinterpreted what he read. In
particular, it is an easy calculation that in the Cartan formalism,
the spatial curvature at a fixed time is zero, but the spacetime
curvature is not; he has made the beginner's mistake of confusing
spatial and spacetime curvature (much as, in the post I'm replying
to here, he seems to confuse the scalar curvature with the curvature
tensor). See, for example, J. Christian, arxiv.org/abs/gr-qc/9701013.

Juan R. does not understand the role of boundary conditions. In
particular, he thinks that the need to impose boundary conditions
to obtain a Newtonian potential is somehow "unphysical" (basing this
largely, it seems, on an out-of-context quote of Christian). This
is again apparently related to his belief that the Newtonian limit
of a generally covariant theory should magically produce the right
coordinate system.

Juan R. does not think that the solution of the Poisson equation is
really the Newtonian potential. He also thinks that the Minkowski
metric should apply even to Newtonian gravity (!).

And, of course, Juan R. believes that his brilliant insights about
very elementary general relativity have somehow been missed by all
of the physicists who have worked on the subject for the last 90 years.

Steve Carlip

Juan R. wrote:

The Dirac equation is not longer a valid
***wavefunction*** equation as is still claimed in many "outdated"
'relativistic quantum mechanics' textbooks.

The Dirac equation of R-QFT (obtained from the QED Lagrangian) is not
the Dirac equation in the original sense of Dirac, because phi(x) there
is not a c-number wave function but a quantum operator.

Here I agree with Juan. Solutions of Dirac equation cannot be
interpreted as wave functions, because these solutions form a
non-unitary representation of the Poincare group. Therefore,
any probabilistic interpretation of these solutions would fail:
probabilities would not be preserved when the reference frame is
changed. Correct relativistic 1-particle wavefunctions and corresponding
equations are given by Wigner's construction of unitary irreducible
representations of the Poincare group.

Electron-positron quantum field does satisfy the Dirac equation,
however, there is no any physical meaning in quantum fields
(except as some formal linear combinations of particle creation
and annihilation operators), so there is no parallel whatsoever
between Dirac equation and (relativistic) Schroedinger equation.

Eugene.

Eugene Stefanovich:


Juan R. wrote:

The Dirac equation is not longer a valid
***wavefunction*** equation as is still claimed in many "outdated"
'relativistic quantum mechanics' textbooks.

The Dirac equation of R-QFT (obtained from the QED Lagrangian) is not
the Dirac equation in the original sense of Dirac, because phi(x) there
is not a c-number wave function but a quantum operator.

Here I agree with Juan. Solutions of Dirac equation cannot be
interpreted as wave functions, because these solutions form a

Sure they can and you're answering a different question that wasn't
asked, namely, is it possible to create a relativistically correct
version of non-relativistic quantum mechanics in terms of a relativistically
correct wavefunction. No. It's not. That's why field theory exists.
But, that wasn't the question asked.


non-unitary representation of the Poincare group. Therefore,
any probabilistic interpretation of these solutions would fail:
probabilities would not be preserved when the reference frame is
changed. Correct relativistic 1-particle wavefunctions and corresponding
equations are given by Wigner's construction of unitary irreducible
representations of the Poincare group.

Electron-positron quantum field does satisfy the Dirac equation,
however, there is no any physical meaning in quantum fields
(except as some formal linear combinations of particle creation
and annihilation operators), so there is no parallel whatsoever
between Dirac equation and (relativistic) Schroedinger equation.

Eugene.

Eugene Stefanovich:
Bilge wrote:

Note the absence of primes on \Psi. Clearly, I can choose to apply
U to the state vectors to get \Psi' = U\Psi and write \Psi'|H|\Ps'.
Note there is no prime on that H. What you call ``case 3'' would
be something like, \Psi|U^{-1}U H U^{-1}U|\Psi = \Psi'|H'|\Psi'.

Agreed.
Then why did you say that unitary transformation of the Hamiltonian does
not change physics?

Because it hasn't changed. Obviously, H' isn't diagonaliazed by \Psi,
since you imposed the constraint that H and U don't commute. Since,

[H,U] = HU - UH,

U^{-1}HU = U^{-1} (UH + [H,U]) = H + U^{-1}[H,U]

= H + U^{-1} dU/dt

which implies that U^{-1}dU/dt = 0, if H and H' are equivalent. Since
\Psi describes both positive and negative energy states, I think you'll
find that by separating that expression into even and odd terms via,


U^{-1}dU/dt = \exp(iS)(d/dt)\exp(-iS)

= \exp(-iS) { (d/dt)[1 - iS - S^2/2! + iS^3/3! + ... ] }

= \exp(-iS) { -( idS/dt + (1/2!) (SdS/dt + (dS/dt)S) + ...)}

and iterating, you'll get a foldy-wouthuysen transform.

If the state vector is kept the same then expectation values of the
original Hamiltonian

\Psi|H|\Psi

and the transformed Hamiltonian (your eq. (1)) are clearly different.
The physics has changed.

They aren't different. You can easily see that what you've written
is equivalent to,

\Psi|H|\Psi = \Psi'|H|\Psi'

So, all you are doing is constructing a different set of states, since,

\Psi' = U\Psi = (1 - iS - S^2/2! +...)\Psi'

and, so

\Psi|U^{-1}H|\Psi = \Psi|(1 + iS +...) H (1 - iS + ...)|\Psi

= \Psi|H|\Psi + ...

So what?

Eugene Stefanovich:


Bilge wrote:

Here's a really obvious problem with the action at a distance idea.
t
|
| .C A and B are simultaneous, so they occur
| at the ``same time,'' i.e., t = 0
+----+----+ x
A B
I now perform a lorentz transform,

t
| A and C occur at the ``same time,'' i.e., t = 0
|
|
+--------+-- x
A . C
B

If interactions are instantaneous, then obviously, at most only
one of the above can be correct, in which case, the system isn't
lorentz invariant, since the two choices for t = 0 differ only
by a lorentz transform. I guess that about sums up my opinion.


Your logic is fine except one important point. Lorentz (boost)
transformations
of space-time coordinates of events must depend on interaction
(that's the point missed in Einstein's special relativity and in your
diagrams). If they don't depend on interaction then, as can be easily
shown, you'll get into contradiction with the Poincare group

I've explained why you're wrong and so have a number of others.
The reason you beieve that is because you've ignored the gauge
invariance and turned two unphysical degrees of freedom into
two real ones. A coordinate transformation cannot increase the
number of real degrees of freedom.

properties. Therefore, if events A and B are connected to each other
by interaction and by the cause-effect relationship, then they are
simultaneous in ALL reference frames.

This short argument looks like a handwaving, but it's not.
I wrote a book with detailed explanations of this point.
Hope you are having fun reading it.

Eugene.

ha escrito:
Briefly (since I don't have a lot of time to waste on cranks):

Carlip, if you have some time, could you explain here to "physicists"
why

there is not c^2 extra term in Caroll online notes?

I have detected this error in several physicists and students of
physics, and i think that would be a good thing if you explain this to
them. For instance, my personal explanation to Eric Gisse was as follow

*************************
The element of line in the flat spacetime of SR (using both trace +2
and summation conventions) is

ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 = g_{ab} dx^a dx^b

As clearly explained in the page 2 of Sean Carroll basic online manual
(pancake.uchicago.edu/~carroll/notes/grtinypdf.pdf), the spacetime
dimensions are

x^0 = ct (this is EXACTLY equation 2 of Carroll manual)

x^1 = x

x^2 = y

x^3 = z

and the SR metric is g_{ab} = (-1 +1 +1 +1) which is, EXACTLY, Carroll
equation 3.

If one takes a system of units with c=1

ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 = g_{ab} dx^a dx^b

x^0 = t (this is EXACTLY equation 2 of Carroll manual with the
asumption c=1)

x^1 = x

x^2 = y

x^3 = z

and the SR metric continues being the SAME g_{ab} = diag(-1 +1 +1 +1)

The metric does NOT contain any c^2 term as you incorrectly write and
this is independent of choice of system of units, with either c the
unity or NOT.

*************************

Several physicists and students continue thinking that "metric is
wrong" and that correct metric is (-c^2 +1 +1 +1) when one reintroduces
"c". They argue that Carroll is using a system of units with c=1 and
therefore the metric (Carroll equation 3) is (-1 +1 +1 +1) ONLY in that
case.

I did an experiment with people with no idea of physics. I wrote the
whole ds and after g_{ab} (dx^a) (dx^b) doing

x^0 = ct (this is EXACTLY equation 2 of Carroll manual)

x^1 = x

x^2 = y

x^3 = z

and being g_{11} = g_{22} = g_{33} = 1, I asked what is the value of
g_{00}? All young guys, old womans (without superior studies), etc.
said "-1" NOBODY said "c^2". How is possible that several "physicists"
and students of physics cannot obtain the correct reply and people from
street can? Are those the physicists of future?

I also explained to Eric Gisse what would be his error

*************************

Your (undergraduate-level) MISTAKE may be that you see the element of
line of SR

ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 = g_{ab} dx^a dx^b

and you believe that the "-c^2" term is the g_{00} component of the
metric and that g_{00} may be -1 only when c=1. But that is your
personal error!

Since -c^2dt^2 = -d(ct)^2, the element of line really is

ds^2 = -d(ct)^2 + dx^2 + dy^2 + dz^2

ds^2 = g_{00} (dx^0)^2 + g_{11} (dx^1)^2 + g_{22} (dx^2)^2 + g_{33}
(dx^3)^2

because x^0 = ct (equation 2 of Carroll) and, therefore, g_{00} = -1

*************************

But he continues saying that lack a c^2 term... and adds that "-1" is
only valid when c=1

That is because the metric is using units where c =1. It is explained
on the page you cite. You are familiar with units, right?

Since this appears to be a common error (i already count 6 guys here in
sci.physics.relativity, in a few days) perhaps you would wrote some
words about it.


Juan R. has been shown mathematically rigorous derivations of the
Newtonian limit of general relativity. He doesn't like them because
they require a choice of coordinates to get the standard form of
Newton's equations. (General relativity is generally covariant,
while the form of Newton's equations that he likes isn't, but he
apparently believes that taking the limit c-infinity should
magically pick out a coordinate system.)

This is not true. The basic linear textbooks derivation is wrong. This
is explained in Wald that in the linear regime a=0 and, thus, others
textbooks are wrong. Then Wald argues that in the NON-linear regime one
obtains Newton equation.

But it is not as i already explained. Moreover, Wald is doing the slow
motion, weak field limit, which is NOT the Newtonian limit. Moreover
all in Wald 'derivation' is /ad hoc/ mathematically non-rigorous, and
based in several aditional premises.

This is the reason that rigorous research does not follow textbook
derivation. "There is at least two reasons why the simple textbook
extractions of the Newtonian limit are not rigorous" (Schutz. The
Newtonian limit).

At least!

Then people choose a diferent way via NC theory or similar. Some
authors as Ehlërs have done some advanced work on the topic. Did they
obtain a derivation of Newtonian limit? Of course NOT.

i) Imagine that Ehlërs boundary is correct. Even in that artificial
case, Perry and Bohun write

"We illustrate how the Ehlers' formal mathematical deffinition of the
Newtonian limit requires additional information to successfully
determine the Newtonian limit."

From Quevedo equation they write

"One /cannot/ obtain this information from the /deffinition/ of Ehlers.
It must be added through a physical argument."

Emphasis in the original.

ii) But as it is well-known Ehlërs boundary is INcorrect. Christian
remarks

"However, physical evidence clearly suggests that we are not living in
an 'island universe' (cf. Penrose 1996, 593-594) - i.e., universe
is not 'an island of matter surrounded by emptiness' (Misner et al.
1973, 295). Therefore, a better procedure of recovering the Newtonian
theory from Einstein's theory is not to impose such a strong and
unphysical global boundary condition, but, instead, to require that
only the weaker condition on the curvature tensor, (5.16), is
satisfied."

BUT curiously equation (5.16) CANNOT be derived from GR alone, and one
may introduce it /ad hoc/. If one does not introduce that /ad hoc/
equation -which cannot be *derived* from GR- then one is unable to
obtain a Newtonian-like limit.

*Derivation* means that all is contained in the original equations. If
one needs know first NG for next obtaining a restriction to unphysical
limits of GR, then that is not a derivation. It is an /ad hoc/ fitting
of GR to NG using NG for obtaining the fit. This, of course, implies
that GR alone is NOT sufficient, since from pure GR equations one can
obtain both the *correct* and the *incorrect* answer in the NC
approach.

Moreover, as proven in a number of very rigorous papers the weak NC
approach is not correct. In particular, Earman (2002) observes that NC
cannot be derived from an action principle, hence it cannot be
formulated as a constrained Hamiltonian system.

Etc.

TODAY there is NOT still derivation of NG from GR (even if one uses
aditional asumptions or equations). Derivation means "/derivation/".
Moreover, "/derivation of NG/" does NOT mean "obtaining of Newton
second law for one-body". All of Newton physics is not contained in the
second law.

Juan R. uses a coordinate x^0=ct, where t is the Newtonian time,

This is NOT true. In 'my' approach t is not Newtonian time. I already
explained this to you.

and thinks this makes sense even in the limit c-infinity.

I already explain this point, you just misunderstand it. I already said
that x^0 ***coordinate*** of spacetime collapse, which is correct
because t in Newtonian physics is NOT a dimension. This is one of your
errors via chossing a (t, x) spacetime in the Newtonian limit. I
recover a 3-dimensional space in the limit, you obtain a 4-dimensional
spacetime. Your approach is not correct.

course, in this limit, t=x^0/c-0 for every finite value of x^0,
and all derivatives with respect to x^0 go to zero. He has
been told how to take the limit properly, even in his coordinates
(start with a large but finite value of c, multiply by an appropriate
power of c so that the c-infinity limit makes sense, and then take
the limit), and has been shown how this process gives the correct
Newtonian limit, but he doesn't like that, either.

That is not true. I said time ago that one obtains Newtonian-like
equation with zero curvature, doing the curvature interpretation of
gravity incorrect. In fact, i remember to you verifing my point a =
-grad(phi) with a ZERO connection (for x^0 = ct choosing).

Juan R. has apparently casually read a paragraph or two about
Cartan's formulation of Newtonian gravity as a spacetime theory
with a preferred time, and has misinterpreted what he read. In
particular, it is an easy calculation that in the Cartan formalism,
the spatial curvature at a fixed time is zero, but the spacetime
curvature is not; he has made the beginner's mistake of confusing
spatial and spacetime curvature (much as, in the post I'm replying
to here, he seems to confuse the scalar curvature with the curvature
tensor). See, for example, J. Christian, arxiv.org/abs/gr-qc/9701013.

Nothing of that is true :-)

1) I know what is spacetime and space curvature. In fact, i said to you
that due that covariant derivative is the physical one in the spliting
into "flat space" more "potential" the 'gauge' is not fixed and one
needs either boundary conditions or adittional equation for deriving
the Newtonian limit, that is for doing phi(NC) = phy(NG).

BUT and this is point you missed, the boundaries are ***unphysical***
(see Christian) and the aditional constrain equations are NOT derived
from GR and need to be invoked as a new ***postulate****.

2) Even doing all of that, one is not deriving correct Newtonian
equation. This is reason even today Christian has not obtained the
correct Newtonian limit. So recent as 2001, Christian has presented a
NEW version of NCG formulated explicitly as a constrained Hamiltonian
system. Until now all approaches that you claim "reduce to NG" were
incorrect because they lack the correct Hamiltonian formulation for
several bodies. BUT still his derivation -being an advance- is not
complete...

3) About curvature tensor and curvature scalar i think that i know what
both are and this is the reason that i explicitely said "R" or "scalar
curvature" on my writtings :-)

Do you find some post mine where is said that R = 0 implies tensor
curvature zero? did i say that you obtained zero curvature tensor with
your x^0 = t choosing? :-)

Juan R. does not understand the role of boundary conditions. In
particular, he thinks that the need to impose boundary conditions
to obtain a Newtonian potential is somehow "unphysical" (basing this
largely, it seems, on an out-of-context quote of Christian). This
is again apparently related to his belief that the Newtonian limit
of a generally covariant theory should magically produce the right
coordinate system.

I NEWER said that the need to impose boundary conditions was somehow
"unphysical" I said that boundary you are claiming is UNPHYSICAL,
which is a different thing :-)

Out of context quote? I will offer to you the whole quote (from page 25
of work i already cited several times):

"However, physical evidence clearly suggests that we are not living in
an 'island universe' (cf. Penrose 1996, 593-594) - i.e., universe
is not 'an island of matter surrounded by emptiness' (Misner et al.
1973, 295). Therefore, a better procedure of recovering the Newtonian
theory from Einstein's theory is not to impose such a strong and
unphysical global boundary condition, but, instead, to require that
only the weaker condition on the curvature tensor, (5.16), is
satisfied."

Could you explain us here that you understand by "Therefore, a better
procedure of recovering the Newtonian theory from Einstein's theory
is NOT to impose such a STRONG and UNPHYSICAL global boundary
condition"?

I already explained to you that astronomers verified that 'island
asumption' is experimentally false since does not correspond to OUR
universe. I already said to you that ***smart*** people like Penrose or
***real*** specialists (sorry to say this but you are not one of
recognized specialists on Newtonian limits) on Newtonian limit as
Christian clearly state that boundary IS UNPHYSICAL.

Your error (which is not Penrose or Christian one) is that you do not
know what is the diference between an Newtonian 'potential' and an LW
'field'.

The limit R-- infinite on Newtonian potential IS verified by
experiments, the limit R-- infinite on a LW field is NOT verified by
experiments. Therein Christian EXPLICIT rejection of Ehlërs UNPHYSICAL
boundary.

Juan R. does not think that the solution of the Poisson equation is
really the Newtonian potential.

That is not true. In fact i remember to me citing in the past the
CORRECT solution of Poisson equations for you improve your knowledge. I
remember citing equation (25) of the 2nd volume of the Handbook of
molecular physics and quantum chemistry; John Wiley & Sons Ltd: West
Sussex, 2003; Chapter 21, equation 25

Grad^2 U = 4 pi delta(x-y)

solving it the non-relativistic potential is U = U(R) NOT your wrong U
= U(x
t) which is also the criticism of authors of PRE 1996 53(5), 5373.

I remember citing several articles (for example (PRA 2002 65, 0341041)
where appears that Newtonian potential is phy = phy(R), NOT your wrong
phy(x, t).

In above very rigorous article in Phys Rev E one can see the correct
Poisson equation and how its **correct** solution, phy = phy(R), is
explicitely studied, NOT your wrong phy(x, t). Moreover authors prove
that your wrong choosing phy(x, t) violates causality and has some
other trouble doing it incorrect.

That you wrong phy(x, t) has experimental and thereotical problems is
studied in a number of others published articles. i have no time for
citing all of them, the list is large...

In a recent Solvay conference, i find exactly the same potential for
quantum mechanics. (2.1) of Adv Chem Phys 1997, 99, 1. The potential
has dependence V(|q_i -q_j|) that is EXACTLY i wrote U(R) because R
= |q_i -q_j|. NOT your wrong phy(x, t). authors use that potential for
both EM and gravitational studies.

In Phys Lett A 1988 128(3,4) 123, authors (one of them one of most
respected particle physicists) study a relativistic generalization of
Newtonian-like potentials. Note the explicit dependence V = V(rho) in
equation (3). doing c-- infinite (they explicitely do in other
published paper) one EXACTLY obtains V = V(R), with 'MY' implicit time
dependence R = R(t). They do not obtain your WRONG (x, t) dependence.
They also worked in gravity.

In Prigogine, I. Non-equilibrium statistical mechanics 1962, John Wiley
and sons, we can see that for classical systems the dependence is again
V(R(t)) equation 1 of chapter 4. Remember that Prigogine was one of
leaders in the field of statistical mechanics and Nobel Prize 1977.

I also remember i cited Goldstein manual on mechanics and his discusion
of the two-body problem. In the chapter 3 of Spanish version, Goldstein
discusses the Lagrangian and explicitely does U (energy) function of R,
dR/dt, and higgher order derivatives of **** R **** /in the general
case/. After he works the Newtonian prescription where, of course, only
enters R. The Newtonian potential for GRAVITY is just V = V(R) NOT your
wrong V(x, t).

Etc.

You continue thinking that a Newtonian potential is phy = phy(x, t)
which is not only wrong as proven in many manuals, handbooks, and
published articles it is (sorry by words) an authentic absurdity.

He also thinks that the Minkowski
metric should apply even to Newtonian gravity (!).

I remember that YOU proved here in spr that setting x^0=ct, then
g_{00}=1 - 2Gm/rc^2, and the geodesic equation looks like

d^2x^i/dt^2 = -\partial_i U

Now explain to me what is the c-- infinite limit of

g_{00} = 1 - 2Gm/rc^2,

and

g_{00} = [-1 + 2Gm/rc^2]^(-1)

YES (you are correct!), it is

g_{00} --1

and

g_{rr} -- -1

which is ***flat*** Minkoski spacetime (1 -1 -1 -1).

Let me note that your derivation contained a small (but conceptually
important) imprecision. Seting x^0 = ct one cannot (must not) write
geodesic equation with respect to "t". One may do that respect to tau,
because in Newtonian physics time is not a dimension is an evolution
parameter. Therefore, your derivation would be more correct if you
repeat it again taking now tau as the correct concept of time in
Newtonian limit.

That is doing

d^2x^i/d(tau)^2 = -\Gamma^i_{00}(dx^0/d(tau))^2

and NOT

d^2x^i/(dx^0)^2 = -(1/2)\partial_i g_{00}

as you did!

That was the reason that your obtained

(1/c^2)d^2x^i/dt^2 = -(1/c^2)\partial_i U

and said

"It's true that both sides go to zero as c-infinity, but for a trivial
reason -- they have a common factor of 1/c^2. If you do the sensible
thing and multiply both sides by c^2 before taking the limit, you get
the usual Newtonian limit."

But there is not reason for writting the equation with respect to x^0
(remeber that dimension collapse) one may write with respect to
universal lambda parameter of the trajectory (for example tau). In my
approach, i do not need multiply by c...

Remember I obtain a 3-space parametrized by a evolution universal
parameter tau (in the c-- infinite limit it concides with
Galilean/Newton concept of time). You obtain a 4-spacetime where time
is a coordinate which is not Newton prescription. In some sense you are
working a 'post-Minkoskian' approach i am working in the
'post-Newtonian' one inspirated by NC thoughts.

And remember that i am obtaining a=/=0 with a pure flat spacetime
R_{ab} = 0 R = 0, \Gamma^i_{ab} = 0, etc.

And, of course, Juan R. believes that his brilliant insights about
very elementary general relativity have somehow been missed by all
of the physicists who have worked on the subject for the last 90 years.

Steve Carlip

Ah this may be the problem, not the physics or the math, just the 90
years.


Juan R.

Center for CANONICAL |SCIENCE)