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Relativistic Quantum Mechanics?



 
 
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  #101  
Old November 10th 05 posted to sci.physics,sci.physics.relativity
Eric Gisse
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Default Relativistic Quantum Mechanics?


Juan R. wrote:

[snip]

New to me.

The linearized limit doesn't derive *that* definition of the potential
because it is assumed to be static. Read Wald again.

I don't know if GR can derive that limit or not, for I haven't tried
and it doesn't really interest me. But before criticising GR for not
deriving your preferred flavor of Newton, it would help if you realised
that what you use as the definition of a Newtonian potential is not the
standard definition.



ii)
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.


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?

[snip more of the same]


iii)
Between the couple of nonsense you wrote in last days i selected this

If all the connection coefficients are zero, which is what you desire,
then there is no acceleration. It is flat space.


Well, i proved just the contrary you claim. ALL connections are zero
and, however, acceleration is not zero, doing the standard curvature
interpretation of GR just wrong.


The hell you did. You just asserted that the coefficients were zero
even when they were manifestly not. Read Wald again.

Before saying stuff like "the standard interpretation of GR is wrong",
you might want to look into your own understanding.


It would be a waste of time atempt to explain to you why i followed
this way on my research; i.e. which were my motivations for ignoring
the geometric (curvature) interpretation. Curiously, people as Dirac,
Schrödinguer, Weinberg, or Feynmann between others, followed similar
thoughts.


*yawn*

Then why even mention it? I guess if you didn't, you wouldn't be able
to get in some name-dropping time.


I simply will say that S. Carlip just also proved here in
sci.physics.relativity on 7 Oct 2005 18:51:33 +0000 (UTC) that with ALL
connections zero, the aceleration is NOT zero. That is, Carlip (as i
did at beggining of 2005) proved just the CONTRARY to your recent
(wrong) claim.

The geodesic equation when (v/c) -- 0 is

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

Since x^0=ct, and g_{00} =1 - 2Gm/Rc^2

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


The c^2 term cancels out.

Incidently, you have one less thing to complain about. I thought the
c^2 term was still contained within the acceleration equation, but I
see that it in fact does cancel out after working your example out.
Nifty.

Keep going, you eventually arrive at d^2x^i/dt^2 = -
(partial/partial_x^i)*(GM/r), or - grad*phi.

Notice the lack of c^2.


acelleration is nonzero BUT for all index a, b, and c

\Gamma^a_{bc} -- 0 and spacetime looks flat...


WRONG! IT IS NOT ZERO! YOU JUST PROVED IT!

Do you not understand what you are doing when you manipulate the
geodesic equation?


As i already said in page 17 of

www.canonicalscience.com/stringcriticism.pdf

on April, the curvature interpretation of gravitation does not hold in
the nonrelativistic limit, doing the causal foundation of general
relativity misleading.


Except you just proved there are nonzero curvature components in this
particular case. Oops.


Precisely the belief on a 'curvature interpretation' is the basis of
the foundation -still unsolved- problems of GR: the (famous) problem of
energy, cosmological discrepancies, problem of reference frames, etc.


Energy is not a problem, nor are reference frames. Cosmological
discrepancies stem from GR not being used [simulations have been
Newtonian].

Why not complain about something that has merit?


The 'curvature interpretation' also has been one of barriers on a
proper quantization of GR and is the basis of the complete
***disconection*** between GR and the physics of the rest of
interactions.


The interpretation isn't the barrier.


Our approach unifies gravity with electromagnetism and can be unified
with other forces; it explains data that GR cannot explain, probably
solves the problem of energy, probably solves the problem of reference
frames, verifies the same solar systems test that GR, and obtains the
same experimental value for others tests, e.g. for binary pulsars BUT
without the use of real gravitational waves and EXPLAIN why.


Hahahahahaha.

GR and classical electromagnetism get along just fine even without your
assistance.

I see lots of conditional statements like "can be", "probably"...you
don't really know what your new theory does, assuming it isn't founded
upon several false premises nor contains severe logic gaps.


GR is a pure hidrodynamic theory. Our theory is more general and
permits us compute kinetics terms without appeal to Vlasov-Einstein
theory, etc. Moreover Vlasov-Einstein theory works just for conditional
averages. Our theory is more general...


Gibberish.


Moreover, the theory has been already quantized: there is not 'problem
of time' and explain why (is better than GD, HQG, Astherkar, and
related approaches), is not a perturbative approch (therefore is best
than string theory), there is not renormalization (therefore is best
that nonrenormalizable Feynman/Weinberg spin-two field theory), one can
obtain quantum g_{ab} (Wheeler geometrodynamics and similar approaches
cannot), a concept of spin naturally arises (remember that GR cannot
treat systems with spin and one needs the so-called Cartan extension),
at ultrashort distances dimension of space reduces to 2 (there some
link with most recent advances in triangulation thechniques), etc.


Let us know when you derive something already known, like lensing. Or
something unknown but testable.


Now we are working in topological properties of spacetime at short
distances. Interestingly, into one topological regime one finds stable
configurations of 2 and 3 'pre-particles', this is some kind of QCD?
What is the link with Finkelstein pre-spinors and Coxeter groups?


**** if I know, I don't know anything about QCD. Why not ask someone
who knows?


Canonical science approach to gravitation is, of course, not a finished
theory but it is a very interesting approach.


Not if the way you think is any indication.



Juan R.

Center for CANONICAL SCIENCE)


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  #102  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Bilge
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Default Relativistic Quantum Mechanics?

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.

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?

[*lots of unsupported assertions snipped*]


One can formulate quantum field theory, e.g., QED, without using fields
as basic concepts. This sounds rather controversial, but it can be done.


Sure. You can rename anything and then claim you've eliminated the
concept. That doesn't make it so. It does, however, make it easier
to make the mistake of treating what is supposed to be unobservable
in field theory as observable.


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.

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.

  #103  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Bilge
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Default Relativistic Quantum Mechanics?

Eugene Stefanovich:
Bilge wrote:
Eugene Stefanovich:


Bilge wrote:

Eugene starts with the same theory. All he does is
perform a unitary transform. Changing representations doesn't
change any physics.

Your statements are formally correct. However they misrepresent the
idea of the dressing transformation. "Changing representation"
implies that you transform unitarily both operators of
observables and state vectors. Then, of course, physics is not
changed.

In the "dressing transformation" only the Hamiltonian is transformed.
State vectors are not touched. Then physics IS different.


No, it isn't. However, you have provided yet another illustration of
of where you attribute physics to unobservable phases:

Consider two Hamiltonians H and H' that are related by a unitary
transformation (U does not commute with H)
H' = U H U^{-1}



Which means that if H and H' describe the same system, then at time
t = 0,

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

= \Phi|H|\Phi \Phi = U\Psi

Obviously, H' acting on \Psi is the same as H acting on \Psi' and
since [H,U] != 0, by hypothesis, there is no \Psi which are simultaneous
eigenstates of both H and H'. Or equivalently,

H' = U^{-1}HU = H + U^{-1} [H,U]

(H' - H) = U^{-1} [H,U]

Take an arbitrary state vector |Psi at time t=0 and consider its
time evolution described by the two Hamiltonians


|Psi(t) = exp(iHt) |Psi
|Psi'(t) = exp(iH't) |Psi


So what? Is that your so-called proof? \Psi can't be an eigenvector of
both H and H'.

so unitary transformation of the Hamiltonian DOES change physics.


No, it changes your description of the physics. You just failed to
include the transformation between H and H'.

That's exactly what is done in the dressing transformation approach:
A unitary transformation U is found that transforms the Hamiltonian
of QED H (with infinite counterterms) to a finite well-defined
Hamiltonian H' which can be used for time evolution calculations.


You keep restating the obvious and avoiding the point.


I am sorry, it seems that we have very different ideas about quantum
mechanics. I am not going to buy your interpretation,


What interpretation? What I wote above regarding the unitary
transforms you gave is mathematical fact. It's the definition
of a unitary transform. You do realize that,

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

are just different ways to say the same thing, righ? If not, I
refer you to any quantum mechanics textbook, and if you need a
page number, I'll find one in schiff for you.

and I am sure
you will not accept my arguments either. I suggest we stop this
fruitless discussion and switch to something more productive.

Eugene.



  #104  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Bilge
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Default Relativistic Quantum Mechanics?

Juan R., jargon thrower:

Bilge, it is easy, you simply are arrogantly attacking people without
subtanting your discourse.


If you have something to say, post it. I already told you I'm not
going to play 20 questions or try to read your mind. If you want
to go on endless digressions, find someone else to jump through
your hoops.

In fact you are still not replied to


Case in point - If you have something to say, just get to your
point.

With is the relativistic interaction between two electrons at second
order in c and e?


I also ask tou you, what is the relativisitic quantum equation for one
electron?


You have just cited a page, please to wrote down the equation or just
write its name. (Dirac?)



Are you
claiming that QED is a relativisitc quantum MECHANICS?


Please write also the relativistic uncertainty relations.

Please prove also why (2) light which propagates at `c', and (3)
electromagnetic interactions which propagate instantaneously are
incompatible.
Or at least proves that PRE 1996 53(5) 5373-5381 is wrong.

Write also the diference between a field and a potential?

Juan R.

Center for CANONICAL |SCIENCE)

  #105  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Bilge
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Posts: 13,439
Default Relativistic Quantum Mechanics?

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?

In the

e + e = 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 = |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?

You and eugene share the same misconception. You want to define
particles on constant slices of time, but you expect to obtain solutions
inconsistent with that choice. Obviously, I can choose all sorts of ways
to define t = 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. 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.

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?

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.

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.


  #106  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Eric Gisse
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Posts: 17,703
Default Relativistic Quantum Mechanics?


Juan R. wrote:

[snip]


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).


Why do insist on 'nice' solutions to multibody problems?

You are making the same argument as before.



Juan R.

Center for CANONICAL |SCIENCE)


  #107  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Eric Gisse
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Posts: 17,703
Default Relativistic Quantum Mechanics?


Bilge wrote:

[snip]


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?


Is the notion of covariance fundamental to QFT also?

If so, it is something I will keep in mind.


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.


Nor are there exact solutions in general relativity, to my knowledge,
that handle multiple particles.

  #108  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Bilge
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Posts: 13,439
Default Relativistic Quantum Mechanics?

Gregory L. Hansen:

What's-his-name, author of a Dover book on classical field theory written
with the purpose of being extensible to quantum field theories, wrote
about a delayed action at a distance theory. I had trouble understanding
how a theory could be both action at a distance and delayed. Just
wondering if you know anything about that and have comments on it.


Without knowing for sure what the theory is, the opinion based on
my impression of what you described woud be ``misguided'' or ``thumb-
twiddling.'' One could ask the same question about the spatial axes,
i.e., how come everything that happens at a different time doesn't
happen in the same place?

I would think the simplest solution is to just take relativity at
face value. I mean, what's wrong with simply choosing the most obvious
explanation:

Events A and B are connected by a null ray. What is their separation
is spacetime? Well, ds^2 = 0 and d\tau^2 = 0, which makes it rather
evident that the proper distances and times separating the events are
zero. I don't see that interactions which take place over an interval
of zero to be action at a distance. Even better - all observers will
agree that the events are separated by zero proper time and proper
distance. Evidently, the description of a particle (apart from its
quantum numbers) depends the coordinates we choose to describe it.
Quantum numbers like mass, spin, charge, etc. have no coordinate
dependence, so trying to describe them as classical ``things''
doesn't work too well.


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.



  #109  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Bilge
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Posts: 13,439
Default Relativistic Quantum Mechanics?

Eric Gisse:

Bilge wrote:

[snip]


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?


Is the notion of covariance fundamental to QFT also?


Yes, but...

Let's distinguish between the formalism and the practical application.
In the formal sense quantum field theory takes the covariance to the extreme.
Field theory applies noether's theorem to everything with a vengence.
Noether's theorem says that every continuous symmetry of the lagrangian
corresponds to a conserved current, which ends up being where gauge bosons
originate.

If so, it is something I will keep in mind.


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.

It turns out that if you start with the assumptions of (1) special relativity,
(2) the quantum mechanical prescription for replacing classical variables
with quantum operators, (e.g. E - i\hbar d/dt, p - -i\hbar\grad), the
following sequence of derivations follows from insisting on invariance under
spacetime displacements and little else.

(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)

(3) Insist that the lagrangian be invariant under spacetime displacements.
(It's sufficiant to just assume L satisfies the euler-lagrange eqns
and that it depends on some field and the first derivatives of the
fields. This is the equivalent of using generalized coordinates
and momenta as done in classical physics carried over to special
relativity). That gives a conserved four-momentum via noether's
theorem.

(4) Since the conserved four-momentum is E^2 - p^2 = m^2, try obtaining
a linear verison which is covariant, i.e., E = a.p + bm by solving
for a and b. Doing that gives a set of 4 - 4x4 matrices, a_i, b.
Inserting i\hbar d/dt for E and -i\hbar\grad for p. We now have
a candidate for a relativistic wave equation (well, a second
candidate, the first being the second order equation from which
this one was derived.) The wave equation just obtained is the
dirac equation. (Really, it takes no more than a couple of pages
to derive everything so far).

(5) The equation derived is now, i(d/dt)\Psi = (-i a.\grad + bm/hbar)\Psi
which after multiplying through by b gives the dirac equation in
in standard form. Since b^2 = 1 (actually, the 4x4 identity), we
define the dirac matrices \gamma^0 = b, \gamma^i = ba^i. Then,
writing p_u rather than (E,p), we get (\gamma^u p_u - m)\Psi = 0.

And finally by defining the slash notation to mean a scalar
product of a four vector and dirac matrix, p/ == \gamma^u p_u,
The dirac equation is compactly written, (p/ - m)\Psi = 0.

(6) From that you can gets lots of things. For example, the dirac
matrices anticommute to give you the metric tensor and commute
to give the spin tensor. Every potential which is possible in
the dirac theory can be written as a combination of dirac matrices.
There are only five. You also get anti-particles and particles
with spin-1/2.

(7) The most compelling reason for taking field theory seriously, is
that at this point, one can derive E&M from scratch, without
assuming much of anything else. In lagrangian form, the dirac
equation is L = (\Psi+)(p/ - m)\Psi, where \Psi+ means hermitian
conjugate of \Psi. Since phases cannot be observables in quantum
theory, the dirac lagrangian must be invariant under a change of
phase, \Psi - \exp(-iS)\Psi. Furthermore, we want to require that
different observers be allowed to choose their phase conventions
locally, so that the phase S is a function of x^u, S(x^u). This
means that observers don't need to check with the rest of the
universe to decide if they've measured an electron.

(8) The result is that the dirac lagrangian as written will
not be invariant under a change of phase (i.e., a gauge
transformation). The reason is that p_u take the derivative
of \exp(-iS)\Psi to give p_u\Psi - i\Psi \d_u S, so we have
an extra term. That extra term is the electromagnetic field.
In order that the new lagrangian be invariant, the field
must transform covariantly, which happens to be precisely
the transformation properties we need for identifiying the
term with the A_u in maxwell's equations and with the photon.
The tranformation above is a U(1) tranformation.

(9) If we require covariance under a more complex phase transformation,
where the S is now a matrix instead of a function, we get more
particles. For SU(2) we get the weak interaction and for SU(3),
we get qcd.

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.

The other aspect of field theory is formally quantizing the stuff
above, which so far has been manifestly covariant and required knowing
nothing at all about the fields. In order to quantize the fields, you
change over to the hamiltonian formalism. Unfortunately, if you choose
the usual cartesian coordinates, the manifest covariance is broken,
since you are choosing some frame in which to quantize the theory.
Special relativity is poincare invariant, so there are 10 conserved
quantities as a result of the 10 possible transformations under which
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). That's ok, so long as you don't disregard the fact that the choice
rules out using some of those 10 as quantum observables. (This is not
unlike having to choose between working in momentum space or configuration
space. You can choose either, but your choice can't let you measure
both p and x). That is the origin of eugene's misconceptions. What he
calls a physical effect is an artifact of choosing to quantize the
qed lagrangian in a particular way and then promptly disregardig the
limitations of that choice on his onservables.

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.


Nor are there exact solutions in general relativity, to my knowledge,
that handle multiple particles.


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.

  #110  
Old November 10th 05 posted to sci.physics,sci.physics.relativity,rec.org.mensa
Eric Gisse
external usenet poster
 
Posts: 17,703
Default Relativistic Quantum Mechanics?


Bilge wrote:
Eric Gisse:

Bilge wrote:

[snip]


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?


Is the notion of covariance fundamental to QFT also?


Yes, but...

Let's distinguish between the formalism and the practical application.
In the formal sense quantum field theory takes the covariance to the extreme.
Field theory applies noether's theorem to everything with a vengence.
Noether's theorem says that every continuous symmetry of the lagrangian
corresponds to a conserved current, which ends up being where gauge bosons
originate.

If so, it is something I will keep in mind.


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.

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

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.

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.

Now I really want that Dover QFT book.


It turns out that if you start with the assumptions of (1) special relativity,
(2) the quantum mechanical prescription for replacing classical variables
with quantum operators, (e.g. E - i\hbar d/dt, p - -i\hbar\grad), the
following sequence of derivations follows from insisting on invariance under
spacetime displacements and little else.

(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?

[snip]


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)

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.


The other aspect of field theory is formally quantizing the stuff
above, which so far has been manifestly covariant and required knowing
nothing at all about the fields. In order to quantize the fields, you
change over to the hamiltonian formalism. Unfortunately, if you choose
the usual cartesian coordinates, the manifest covariance is broken,
since you are choosing some frame in which to quantize the theory.
Special relativity is poincare invariant, so there are 10 conserved
quantities as a result of the 10 possible transformations under which
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?

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.

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


That's ok, so long as you don't disregard the fact that the choice
rules out using some of those 10 as quantum observables. (This is not
unlike having to choose between working in momentum space or configuration
space. You can choose either, but your choice can't let you measure
both p and x). That is the origin of eugene's misconceptions. What he
calls a physical effect is an artifact of choosing to quantize the
qed lagrangian in a particular way and then promptly disregardig the
limitations of that choice on his onservables.

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.


Nor are there exact solutions in general relativity, to my knowledge,
that handle multiple particles.


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.


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.

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

 




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