In article .com,
Juan R. wrote:

Gregory L. Hansen ha escrito:

Plane waves are solutions to the Dirac equation, and you think the Dirac
equation is not a wave equation?

Sorry, Gregory L. Hansen i cannot waste my time discussing
undergraduate level questions.

Uh, oh. You must be smarter than I am, and your work on Usenet must
be very important. I am intimidated.


The Dirac wave equation -which was *originally* proposed as a
relativistic generalization of Schrödinger wave equation- was
*abandoned* in R-QFT. The evolution equation in R-QFT is a
wave-funtional one, only defined for free fields -or particles-. This
is the reason that the only observable in R-QFT is the S-matrix and
observables derived from it. 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***wavefunction*** equation as is still claimed in many "outdated"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.

A wave equation that operates on an operator is still a wave equation.
--
"He who only sees business in business is a fool."

"Juan R." wrote in
oups.com:

To you? Newer.

I think you mean 'Never'.

My car is old, my wife's car is NEWER than mine.



--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

Eric Gisse:
Bilge wrote:

Actually, by insisting on covariance quite a number of results just fall
into your lap. Regardless of whether or not quantum field theory is the
last word (which is unlikely), the results you get as freebies from
covariance are remarkable.

What strikes me as interesting is how important covariance seems to be.

Precisely. But, what is more important is what it means physically,
because physics going all the way back to kepler and galileo exploits
the idea. The only real difference is the degree of naivete used to
implement the concept in a physical theory.

In SR, covariance is there by default otherwise the theory wouldn't
work.

Special relativity is probably the first theory to implement the
idea on purpose as the foundation of the theory rather than a theme
which runs through some ad hoc set of rules for lack of recognition.
Galileo and newton lacked the mathematical foundation and therefore
the physical intuition that accompanies familiarity to see that.
In fact, with nothing but variational calculus and galilean transforms,
you can derive all of newtonian mechnics plus obtain the conservation
laws which were regarded as ``true,'' but whose origins remained a
mystery until the early 20th century. Einstein's argument which yielded
E^2 = p^2 + m^2, was more of a shrewed guess based on the reduction
to the non-relativistic E = p^2/2m that a rigourous result. The rigor
only came from noether's work with hilbert on general relativity.

The basic physical concept of invariance and covariance is that
nature does not depend on our description of nature. If we choose
to use coordinates to describe something, there _must_ be a way
to translate the underlying physical content into a description
of the same physical phenomena any other observer might choose.
This is the fundamental difference between relativity and other so-called
``equivalent theories'' which rely on maxwell's equations or the
speed of light to define a velocity. (Einstein's reasons for including
the second postulate are historical and unfortunately still given
undue focus in introductory classes as an expedient.)

Einstein didn't know about weak and strong interactions, but suppose
he did. In that case, he wouldn't have benefitted from maxwell's
equations, since he would have one set of ``maxwell's'' equations
for each interaction. The equivalent maxwell equations for the strong
and weak force would be more complex, but one could imagine that each
of these would define a differentm unique ``constant'' velocity, under
which the equations might be invariant. That would natural in an ether
type theory. On the other hand, the important feature of relativity
is that it just replaces three euclidean dimesions with a 4-d spacetime
(hence the convention of setting c = 1 places time on equal footing
with space in which a line at 45 degrees to the x,y axes has a slope
of 1, rather than a special name). Now, it makes no difference what
theory you formulate. If you formulate a theory in spacetime, there
can be only one velocity which has a constant value, independent
of any observer. The only objects which propagate at that velocity
must be massless. Light propagates at that velocity because the photon
is massless, not because E&M is intrinsicaly special. The connection
between light and the other forces is the mass, which is an invariant.
The weak interaction appears weak because the W and Z are massive and
the force has a very short range, not because the interaction is
intrinsically weak. You can probably see that despite the superficial
resemblence of having the lorentz boosts in common, there is a definite
difference between relativity and ether-type theories in the way one
thinks about the underlying physics and how one would naturally try
to develop a theory to explain any newly discovered physics.

Quantities which are invariant are, by definition, telling you that
something is unobservable because it's the same quantity regardless
of which observer describes it. That makes invariants the natural
candidates for fundamental entities. Quantites which are covariant
are slightly different, in that they describe an invariant which
has an observer dependent _representation_. Since different rep-
resentations describe only the difference in the way different
observers describe the _same_ physics, the natural candidate for the
physics is in the symmetry group under which different descriptions
may be seen to be equivalent. (For the sake of completeness, the
technically correct jargon is torsor, but physicists, being less
pedantic than mathematicians, use the word group whether they mean
group or torsor.)

In GR, covariance is there because it reduces to SR at "local"
distances, and it is there globally due to the covariant derivative and
various other tensorial objects.

Here's a more accurate way to describe the relationship between
general relativity and special relativity. In special relativity,
its possible to define a globally inertial frame, since the existence
of such a frame is assumed implicitly in the derivation. Since all
inertial frames are related by a poincare transform, observers are
are only free to define coordinates for which different choices of
inertial frames are related by a lorentz transform, globally.
Inertial frames are therefore special in special relativity. General
relativity elminates that restriction, since one can show no such
global frame exists. That allows different observers
to define inertial frames locally such that a lorentz transform
connects different choices for inertial frames _at_ _a_ _point_.
Whether or not you can use special relativity depends upon how
well you can approximate the region you want to describe by
extending the inertial frame at a point to an inertial frame
over some region about a point.

An analogy would be the case where two observers begin traveling
from the north pole to the south pole along different meridians
at a constant velocity. Each observer will define a locally inertial
frame, but the only two points at which they each agree on their
relative velocity as being a constant, will be at the north and
south poles. Both descriptions are correct, yet each will view
the other as accelerating at various times.

It sounds like, on a simple level, if you demand covariance and by
extention special relativity and work within those bounds what you get
is QFT.

Pretty much. By assuming relativity, you are going to get _a_ field
theory of some sort. You cannot formulate a relativistic theory in
terms of potential energy functions, ala newtonian mechanics, and
vice versa. If you add quantum mechanics to the picture, then your
field theory is a quantum mechanical field theory. It's also useful
to note that the probabilistic aspect of quantum theory is the only
way to describe an invariant vacuum. Random fluctuations cannot be
made less random by any mathematical algorithm, so the one feature
all observers agree upon is that no physics can depend upon some
hypothetical interaction of anything with the vacuum.


What covariance buys you is a restriction on the possible physics
which can be realized in an observer independent theory. The really
important contribution special relativity made to physics was that
it clearly pointed out that physics should not just be observer
independent, but that all of the physics is in uncovering and elimimating
the observer dependent part from the part that looks identical to
every observer.

Now I really want that Dover QFT book.

There are two really good introductions that come to mind. If you
understand the dirac equation, then ``Gauge Theories of the Strong,
Weak and Electromagnetic Interactions,'' by chris quigg is excellent.
He has a knack for clearly connecting the mathematics to the physical
consequences. The second, is, ``Particles, Fields and Geometry,''
by bjoern felsager. It presumes less background and is written at
the level of an advanced undergradute/beginning graduate student.
It covers a great deal more using more modern notation and terminology.
It covers exterior algebra and differential forms as well as some
topics like instantons, solitons and path integrals, which arent
usually included in a such a general textbook. If you have never
encountered relativistic quantum mechanics, this would be my choice
for a text. Most field theory books are rather terse and preseume a
lot of bacground, which can be frustrating if you're trying to understand
the point but find the level of math assumed, gets in the way.


(1) Obtain the lorentz or galilean transforms. Those are the possible
candidates for a description of spacetime which aren't obviously
wrong.

(2) rule out galilean transforms from experimental observation
(for example, galilean transforms preserve mass locally)

Is this the same process that yields 4 types of transforms that occurs
when you assume the principle of relativity but don't explicitly assume
the existance of c?

The only assumptions involved are that of a 4-d spacetime manifold
(in both cases). The galilean group is simple a different group of
transformations that represents the c-\infty limit of the lorentz
transforms If c s finite, then c = 1.

[...]
So, from the brief sketch above you can see that covariance is
central to field theory (even though the above was pretty terse
and won't substitute for a textbook). In particular, you should
note that we didn't need to say anything at all about what \Psi
is. Every result followed from a requirement on how it must transform
such that everything remains covariant.

Terse doesn't even scratch the surface but it is the best introduction
I have seen. I suppose thats why I need a book :O)

Unfortunately, it represents the consolidation of several textbooks
and courses worth of material into the best physical summary I could
come up with, without reducing it to the level of an article from
``New Scientist.'' However, having to grind through the details is a
lot more interesting if you can see the point in advance.

http://store.yahoo.com/doverpublicat...486442284.html

Do you have any opinions on this particular book? I have had good
experiences with various other Dover books in the past so I don't think
I will be steered wrong by grabbing a Dover on a lark.

From reading the url and looking at the url with the table of contents,
it looks like a reasonable book and covers a lot the main ideas. Without
seeing the book, I can't say anything about how well the author succeeds.
But, even in the case where the execution was terrible, I've never found
any textbook to a complete waste (other than cohen-tannoudji's two-volume
set on quantum mechanics). So, I'd go ahead and get it. Even if you find
it less than what you expeted now, you might find it much better with some
additional background.

[...]
the invariance holds: 4 momenta, (i.e., E and p_i), 3 spatial rotations
and 3 boosts. You cannot choose all 10 to be simultaneous observables
in _any_ representation. (you can choose at most 7 and only then if
you work in light-cone coordinates. Otherwise, you can choose at most,
6).

I was going to ask 'why', but I figure I would be better served by
asking how does one know how many transformations are allowable or
observables that are observable at one time?

That's not the right way to think of observables. The term
``observable'' ostensibly refers to a quantity you can actually
measure in an experiment, which in quantum mechanics, corresponds
to the eigenvalues of hermitian operators. So, given a number of
observables, A,B,C,.., what you want to know is which of those
are compatible with each other in the same measurement. Mathenatically,
it means finding subsets of those operators which all have common
eigenvectors, i.e., are simultaneously diagnolizable in some
basis. It's straight forward to determine. If \Psi is an eigenvector
of A, then A\Psi = a\Psi, where a is an eigenvalue. Since \Psi is
unchanged, then BA\Psi = Ba\Psi = ba\Psi, where b is an eigenvalue
of B. Similarly, if you perform the operation in reverse, you should get,
AB\Psi = ab\Psi, so that if (AB - BA)\Psi = 0, both A and B are
simultaneously observable, i.e., A and B commute: [A,B] = 0. For
p and x, p = -i\hbar\grad, so

[p,x] = (px - xp)\Psi = -i\hbar\grad.(x\Psi) - (-i\hbar x.\grad\Psi))

= -i\hbar (\grad.x) - x.\grad\Psi + x.\grad\Psi

= -i\hbar

So, p and x are simultaneously observable. If you have a group, then
the various elements of the group are related by the structure constants
of the group. Angular momenta have the relation, [L_x, L_y] = i\hbar L_z,
(and cyclic permutations of x,y, z). So, only one of L_x, L_y, L_z
can be measured in any measurement.

In my GR text [It should be ok, it was in the context of SR], the
notion of isometries was detailed. While the finite number of
invariants and transformations was mentioned, I have never seen them
constrained before.

Then again, Carroll probably skipped it. But I don't see it in either
Wald or MTW.

It's more likely to be found in a quantum field theory text, since
the poincare group is fundamental to quantum field theory.

Or...is the only reason this is occuring is because of the way the
theory is quantified?

Assuming the theory is correct, then what the theory is telling you
is that there is only so much information there to measure, so you
have to decide between incompatible measurements. The way one ``quantifies''
the theory, so to speak, is to choose a set of quantities that tells
you everything there is to know about the measurements.

[...]
There is a crank for every level of physics. Every little bit I learn
shines the light on a new person in here, or perhaps the same one in a
different way, who obviously got "stuck" somewhere along the line.

At the bottom, we have Don1 and crew who can't grasp units.

Then we have the people who can't grasp abstract things like algebra.
Like Jim Greenfield.

Next, the people who just can't let go of Newton. Like Henri Wilson.

Henriii doesn't even understand newton. He once argued that energy
wasn't conserved in newton mechanics because the energies of two
moving blocks were different depending on the reference frame of
the observer.

Now, Juan R, who among other things, does not understand what he is
doing when he is manipulating the geodesic equation.

I'm not sure what his deal is. He seems to make a big deal out of
things which are irrelevant for the sake of disagreeing and appearing
rigorous. An analogy would be someone who argues against a theory
of superconductors because if fails to describe superconductors at
infinite temperature for some hypothetical gedanken material. That
is sort of backwards. Since there is no such material, one has to
consider the failure to hold in such a limiting case to be a feature.
If E&M and the weak interaction become a single electroweak interaction
at high energy, then one would have to wonder how qed could be correct
if it didn't become unphysical in some limit as a standalone theory.

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

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'.

Juan R.:

Bilge ha escrito:

Right. Juan is deliberately misconstruing the technical difficulties
to be major theoretical problems and being hypocritical about it, too.
The same technical difficulties exist in classical physics. In addition,
classical physics has severe theoretical difficulties if one tries to
create a theory of the electron, for example.

Juan R. as many other people is able to distingish technical
difficulties from fundamental flaws on some of theories currently
accepted. Some people cannot not the differences but is not my problem.

It may be really diffiicult explain the difference between "technical
difficulties" and "wrong axioms" to anyone that still believe that the
Drac equation is a wave equation.

For you, it will be impossible, since your compreshension of physics
is limited to playing word games with jargon.

Juan R.:

Bilge ha escrito:

Juan R.:

Where is there a complete bound-state theory in Weinberg manual for
example?

R-QFT clearly states that only possible observables are those derived
from S-matrix, which is only valid for independent particles (remember
the cluster decomposition principle). In rigor R-QFT only deal with
free fields.

QED is only part of a quantum field theory called the standard
model. The stardard model is believed to be asymptotically free.
What makes you think that _any_ theory of E&M which is completely
independent of the weak and strong interaction at any energy
could be consistent with the unification of all three forces?
If you believe qed should require the inclusion of qcd at high
enough energies, why would you expect qed to be valid in that
regime without including qcd? If you don't think qcd is required
at any energy, how do you explain the fact that the muon magnetic
moment is correctly predicted by including qcd corrections?

Perhaps you forget i said R-QFT. I was not particularizing for QED :-).

Perhaps I'm not interested in your stupid word games and don't
really care.


In the

e + e =3D 2 photon

scattering. R-QFT only can study the wavefunction of the electrons or
the photons when are not interacting. That is when the wavefunction
factorizes |12 =3D |1|2.

Well, gee... Did you ever stop to consider the possibilty that
if you _define_ electrons, photons, etc., as free particles,
then the interacting particles just might not be separable into
free particle wavefunctions plus an interaction?

The only physical states in R-QFT are free fields or if you prefer free
particles. There is not bound relativistic quantum states on R-QFT.
There is not definition of interacting particles in R-QFT.

You and eugene share the same misconception. You want to define
particles on constant slices of time,

False

but you expect to obtain solutions
inconsistent with that choice. Obviously, I can choose all sorts of ways
to define t =3D 0. The choice is not unique. You appear to have some
underlying belief in a galilean universe. If you want to solve the
equations in a form which is manifestly covariant, use point form to write
the hamiltonian.

WRONG

Then you'll have an eigentime for the hamiltonian. Eugene
is essentially working in instant form (where all of the particles are
quantized on a single time slice) then interpreting his results as if they
had been obtained in point form, in which the causal relationship is
explicit. Unfortunately, the point form is a lot harder to work with,
which is why there's not been much done in that area compared with
instant form.

In an atom or molecule you can claim that electrons are infinitely
separated and |12 is NOT |1|2. All test of QED are for nonboundend
states for example scattering two two electrons in acellerator physics
(which is an ONE-body problem), hidrogen atoms or hidrogenic ions
He^(++). In fact, recent test of two electrons in bound states has been
a failure.

Irrelevant babbling. Apparently, you think a rebuttal consists of
inserting a bunch of irrelevant nonsense.

Curious reply. Precisely people working in bound states has proven as
one may generalize R-QFT if one want study NON-free particles. One of
generalizations is the abandonment of the Hilbert-fock space!

Yes one can obtain a Dirac equation for a single particle. What is the
corresponding equation for two particles. It cannot be derived from
field theory and equations proposed in literature /ad hoc/, for example
two-body covariant are not rigorous and not complete.

Naturally. You insist on a formalism that cannot be covariant under
what you insist upon, so what do you expect?

????

Bethe-Salpeter and others are incorrect. At one hand, one claims that
two body state is a 16 component wavefunction. At the other hand in the
interaction regime one uses propagators derived from formals series of
QED which clearly state that there is not two body wavefunction for the
two electrons.

Why you think that R-QFT states that only scattering states are
observables?

Oh. Gee... I guess you've never been on the inside of a physics
laboratory, but let's see if we can shed some light on this conundrum.
Solid state Si detectors, for example, come in a range of sizes, but just
to get a 15-order of magnitude approximation, lets say that it's circular
with a planar area of about 500 mm^2 x 1000 microns thick. Let's say we
are interested in the details of the interaction at the order of a compton
wavelength. For an electron, that's about 386 fm.

Wow! There is no way to fit my detectors aound a couple of electrons
at that distance, not to mention alignment difficulties or the kind of
naievete required to actually postulate such a scenario to motivate
even a gedanken experiment). So, first we note the detectors are a long
way from all of the excitement. Also, the particles involved in the
interaction aren't initially very close either. We might even be inclined
to treat the particles as not interacting initially or at the time they
strike the detector. Gee, what to do?

Irrelevant. This is the reason which serious people is working in a
bound state generalization of R-QFT, especially in relativistic quantum
chemistry and molecular physics, where particles are not infinitely
separated.

But wait -- By using quantum theory with a liberal helping of
common sense, we have a solution. Since we are free to choose a
representation for our states any way we choose, subject to unitarity
constraints, viola, we can break the evolution of the interaction into
pieces:

lim \exp(iH_0t')\exp(-iH(t'-t)\exp(-iH_0t)

By taking the limits for t',t to +/- \infty. repectively, we have
defined the S-matrix. Obviously, in the limit, the interaction H,
specifies all of the S-matrix elements.

The standard definition of the S-matrix is non-rigorous. It can be also
proven that the full evolution is not well-defined unless interaction H
was exactly zero. Therein the well-known claim that only free fields
are *well-defined* on R-QFT.

Still today, nobody has found the correct, consistent, and complete
relativistic equation for N-bodies (perhaps our center has already done
but are cheking details).

Nobody has found the nonrelativistic equations for N bodies either.
If you think so, write down the exact equations of motion for three
bodies interacting under a 1/r (newtonian) potential.

This triviality already suggest your weak understanding of relativistic
physics. The exact equations of motion in non-relativisitc mechanics
are very well known. There is some diffuclties on solving them due to
singular points in phase space, but modern powerfull thecniques (some
inspired in chaos theory) solve them. Divergences in the three body
problem are solved via theory of Poincar=E9 resonances using certain
functional generalizations developed in last 50 years in the topic.

Juan R.

Center for CANONICAL |SCIENCE)

Juan R.:
Bilge:
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.


Sorry to be so hard but the "obvious problem with the action at a
distance idea" is only in your head.

Obviously, that can't be true. Since my ``obvious problem with
action at a distance'' was posted right above your reply, it's
at least one place other than my head. In fact, it's in a place
where you could have posted a valid objection rather than blatantly
erroneous comment you did post.




Juan R.

Center for CANONICAL |SCIENCE)

Gregory L. Hansen ha escrito:

In article .com,
Juan R. wrote:

Gregory L. Hansen ha escrito:

Plane waves are solutions to the Dirac equation, and you think the Dirac
equation is not a wave equation?

Sorry, Gregory L. Hansen i cannot waste my time discussing
undergraduate level questions.

Uh, oh. You must be smarter than I am, and your work on Usenet must
be very important. I am intimidated.

It is not that; simply i have no time. Read basic literature.


The Dirac wave equation -which was *originally* proposed as a
relativistic generalization of Schrödinger wave equation- was
*abandoned* in R-QFT. The evolution equation in R-QFT is a
wave-funtional one, only defined for free fields -or particles-. This
is the reason that the only observable in R-QFT is the S-matrix and
observables derived from it. 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.

A wave equation that operates on an operator is still a wave equation.

The Dirac equation -in RQFT- is an identity for operator phi(x). The
equation of evolution is Schrödinger-like one for wave*functionals* of
field configurations on spacetime.

Juan R.

Center for CANONICAL |SCIENCE)

bz ha escrito:

"Juan R." wrote in
oups.com:

To you? Newer.

I think you mean 'Never'.

My car is old, my wife's car is NEWER than mine.

Sorry was a typo


Juan R.

Center for CANONICAL |SCIENCE)




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

Bilge ha escrito:

Juan R.:
Bilge:
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.


Sorry to be so hard but the "obvious problem with the action at a
distance idea" is only in your head.

Obviously, that can't be true. Since my ``obvious problem with
action at a distance'' was posted right above your reply, it's
at least one place other than my head. In fact, it's in a place
where you could have posted a valid objection rather than blatantly
erroneous comment you did post.

:-) Nice reply, but has you perhaps detected that the important part on
above reply was on "obvious problem" not in the description?

Yes, i could post a vlaid objection but for what?

If you are unable to understand some of most basic stuff that is
already available in literature, it is may be difficult explain to you
some advanced questions. Basically because you lack the basic
understanding of several ideas that are being currently debated in
specialized literature.

Clear! after you claim that i am using 'jargon'.

Simply i want to state that your criticism on 'action at a distance
theory' in this thread is both completely wrong and outdated. Proof is
available on literature including several high-level journals on math
and physics.

Read literature on the topic (NOT basic textbooks!) and when ready then
try again in this (or other thread).


Juan R.

Center for CANONICAL |SCIENCE)