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Jun 29, 2008, 10:08:35 PM6/29/08

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I have been looking over the following three links for Foldy-Wouthuysen

transformation from the Dirac-Pauli to the Newton-Wigner representation

of Dirac's equation:

transformation from the Dirac-Pauli to the Newton-Wigner representation

of Dirac's equation:

The first shows the calculation itself of this transformation:

I: http://www.physics.ucdavis.edu/~cheng/230A/RQM7.pdf

The second, an excellent and lucid exposition of the physics (why this

is of interest), is to be found at:

II:

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.27.3209&rep=rep1&type=pdf

The third, dealing with Zitterbewegung motion the the velcoty operator

in the Dirac-Pauli representation, is at:

III: http://en.wikipedia.org/wiki/Zitterbewegung.

What I would like to discuss, for the purpose of getting your reactions

is to whether I am on a sensible track, is the possibility that a

mechanism for generating fermion mass may be hidden in all of this.

I say this in particular because in the Dirac-Pauli representation, the

velocity operator is given by:

v^k = alpha^k (1)

where alpha^k=gamma^0gamma^k, see reference III. Further, the

eigenvalues of this velocity operator constrains the velocity of the

Fermion of be the speed of light, see reference II in the middle of page

3. This means that the fermion must be massless and luminous, in the

Dirac-Pauli representation. Why this is so, has long been a mystery,

and is thought not to make any sense, for obvious reasons.

Now, transform into the Newton-Wigner representation via

Foldy-Wouthuysen. The velocity operator in Newton-Wigner now takes the

classical form:

v = dx/dt (2)

where x is the position operator. But even more importantly,

Newton-Wigner permits a range of eigenvalues less than the speed of

light, and so, the fermions permitted by Newton-Wigner are massless and

sub-luminous.

Following this to its conclusion, this means that somewhere hidden in

the Foldy-Wouthuysen transformation, we have gone from a fermion which

is massless and luminous, to one which has a finite, non-zero rest mass

and travels at sub-luminous velocity. It seems, then, that it would be

important to specifically trace how the velocity operator of the

Dirac-Pauli representation with +/- c eigenvalues transforms into the

velocity operator (2) of Newton-Wigner which allows a continuous,

sub-luminous velocity spectrum, and at the same time, to trace through

how the rest mass goes from necessarily zero (with decoupled chiral

components), to non-zero with chiral couplings.

By doing so, perhaps one would find a mechanism for generating fermion

masses.

One contract to make here: think about how vector boson masses are

generated. One starts with a Lagrangian in which any mass term is

omitted. Then, via a well-knows technique, one breaks the symmetry and

reveals a boson mass. Perhaps the mystery of luminous velocity

eigenvalues in the Dirac-Pauli representation is telling us a similar

thing: Start out with a Dirac-Pauli Lagrangian in which the mass of the

fermion is zero, i.e., without a mass term. Then, the +/- c velocity

eigenvalues make sense. Transform that into the Newton-Wigner

representation. Somewhere along the line, a mass must appear, because a

subliminous velocity appears.

I will, of course, try to pinpoint how this all happens, if it does

indeed happen. But I would for now like some reactions as to the tree

up which I am barking.

Thanks,

Jay.

PS: Thanks to Neuropulp and Igor K. for turning me in this direction.

____________________________

Jay R. Yablon

Email: jya...@nycap.rr.com

co-moderator: sci.physics.foundations

Weblog: http://jayryablon.wordpress.com/

Web Site: http://home.nycap.rr.com/jry/FermionMass.htm

Jun 30, 2008, 4:21:58 AM6/30/08

to

Thus spake Jay R. Yablon <jya...@nycap.rr.com>

> But I would for now like some reactions as to the tree

>up which I am barking.

>

I'm afraid it is the wrong tree. Fermions are not massless. In fact mass

is a fundamental property as described in the Dirac equation. The

velocity operator in the Dirac-Pauli representation is not physically

meaningful. This is often thought to be because the precise measurement

of velocity would require a time trial between two exact position

measurements. An exact position measurement would require infinite

energy.

> But I would for now like some reactions as to the tree

>up which I am barking.

>

is a fundamental property as described in the Dirac equation. The

velocity operator in the Dirac-Pauli representation is not physically

meaningful. This is often thought to be because the precise measurement

of velocity would require a time trial between two exact position

measurements. An exact position measurement would require infinite

energy.

To get a more meaningful notion of velocity, we might look at an

electron travelling on a path (e.g. in a bubble chamber). But now we

have many, less exact, measurements of position. The formalisation of

quantum theory requires two measurements, initial and final. After

collapse of the wave function, the final measurement may then be taken

as the initial measurement for the next stage of the motion. Because of

the gauge freedom of qed, this is non-trivial. We have to remove phase

shifts in collapse. I show how to do this in

http://www.teleconnection.info/rqg/CEM

using what I have termed the field picture. Essentially this uses a

simplification of the Fouldy-Wouthuysen transformation which ignores

spin and antimatter, i.e. the Newton-Wigner representation is the exact

version of my simplified account. I think part of the value of my

account is that it casts light on why this particular representation

must be used to derive classical correspondences.

For more on the issue of the velocity operator, see Igor's and my

responses in the threads entitled "What is the velocity of a

relativistic electron?" on s.p.r. and on s.p.f. - on this occasion I

think Igor was more patient than I.

Regards

--

Charles Francis

moderator sci.physics.foundations.

charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and

braces)

Jun 30, 2008, 2:48:22 PM6/30/08

to

"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

news:DQmJ1uCR...@charlesfrancis.wanadoo.co.uk...

> Thus spake Jay R. Yablon <jya...@nycap.rr.com>

>> But I would for now like some reactions as to the tree

>>up which I am barking.

>>

> I'm afraid it is the wrong tree. Fermions are not massless. In fact

> mass

> is a fundamental property as described in the Dirac equation. The

> velocity operator in the Dirac-Pauli representation is not physically

> meaningful. This is often thought to be because the precise

> measurement

> of velocity would require a time trial between two exact position

> measurements. An exact position measurement would require infinite

> energy.

news:DQmJ1uCR...@charlesfrancis.wanadoo.co.uk...

> Thus spake Jay R. Yablon <jya...@nycap.rr.com>

>> But I would for now like some reactions as to the tree

>>up which I am barking.

>>

> I'm afraid it is the wrong tree. Fermions are not massless. In fact

> mass

> is a fundamental property as described in the Dirac equation. The

> velocity operator in the Dirac-Pauli representation is not physically

> meaningful. This is often thought to be because the precise

> measurement

> of velocity would require a time trial between two exact position

> measurements. An exact position measurement would require infinite

> energy.

Is there a reason why a velocity measurement would need to be so precise

as to cause infinite energy to be required? IMHO, this is not a good

reason to reject a velocity operator. Is there perhaps a different

reason? What am I missing here?

In the Standard Model, elementary fermions are massless without an

interaction with the Higgs field.

Best,

Fred Diether

Jun 30, 2008, 3:39:35 PM6/30/08

to

"FrediFizzx" <fredi...@hotmail.com> wrote in message

news:6csk1lF...@mid.individual.net...

I actually scoped out the calculation last night, watching what happens

to the rest mass when applying a Foldy-Wouthuysen Transformation. It

does indeed appear possible to take what is a zero-mass fermion in the

Dirac-Pauli representation, and turn it into a massive fermion in the

Newton-Wigner representation. I am going away for vacation shortly, but

if I have the time, I will try to post the calculationn for comment

before I leave.

Jay

Jun 30, 2008, 8:58:19 PM6/30/08

to

. . .

>

> Is there a reason why a velocity measurement would need to be so

> precise as to cause infinite energy to be required? IMHO, this is not

> a good reason to reject a velocity operator. Is there perhaps a

> different reason? What am I missing here?

>

> In the Standard Model, elementary fermions are massless without an

> interaction with the Higgs field.

>

> Best,

>

> Fred Diether

Fred,

>

> Is there a reason why a velocity measurement would need to be so

> precise as to cause infinite energy to be required? IMHO, this is not

> a good reason to reject a velocity operator. Is there perhaps a

> different reason? What am I missing here?

>

> In the Standard Model, elementary fermions are massless without an

> interaction with the Higgs field.

>

> Best,

>

> Fred Diether

I did a calculation of what happens to the mass matrix during the

transformation from the Dirac-Pauli representation to the Newton-Wigner

representation via Foldy-Wouthuysen. This is shown in:

http://jayryablon.wordpress.com/files/2008/06/foldy-wouthuysen.pdf

Not sure where to go from there, but I'll be away the rest of the week

on vacation, so I'll take another look when I return.

Best,

Jay.

Jul 1, 2008, 6:01:38 AM7/1/08

to

Thus spake FrediFizzx <fredi...@hotmail.com>>uCRkI...@charlesfrancis.wanadoo.co.uk...

>> Thus spake Jay R. Yablon <jya...@nycap.rr.com>

>>> But I would for now like some reactions as to the tree

>>>up which I am barking.

>>>

>> I'm afraid it is the wrong tree. Fermions are not massless. In fact

>>mass

>> is a fundamental property as described in the Dirac equation. The

>> velocity operator in the Dirac-Pauli representation is not physically

>> meaningful. This is often thought to be because the precise

>>measurement

>> of velocity would require a time trial between two exact position

>> measurements. An exact position measurement would require infinite

>> energy.

>

>Is there a reason why a velocity measurement would need to be so

>precise as to cause infinite energy to be required?

>> Thus spake Jay R. Yablon <jya...@nycap.rr.com>

>>> But I would for now like some reactions as to the tree

>>>up which I am barking.

>>>

>> I'm afraid it is the wrong tree. Fermions are not massless. In fact

>>mass

>> is a fundamental property as described in the Dirac equation. The

>> velocity operator in the Dirac-Pauli representation is not physically

>> meaningful. This is often thought to be because the precise

>>measurement

>> of velocity would require a time trial between two exact position

>> measurements. An exact position measurement would require infinite

>> energy.

>

>Is there a reason why a velocity measurement would need to be so

>precise as to cause infinite energy to be required?

Yes. We are talking here of instantaneous velocity. Physically this

would mean taking the limit of two measurements separated by an

infinitesimal time interval - as a result, perfect accuracy would be

required.

>In the Standard Model, elementary fermions are massless without an

>interaction with the Higgs field.

I think you mean Bosons, but in any case I have never seen any reason to

think the Higgs mechanism is more than unholy metaphysics. To me, it is

based on a notion of the vacuum which is a contradiction in terms, and

which introduces an extraordinary complexity in nature at the point of

fundamental physics where I would expect the fundamental building blocks

of matter to be simple. Moreover, the reasons given that such a

mechanism might be necessary for the interpretation of fundamental law

completely overlook the study of foundations, and imv they are false.

Jul 1, 2008, 6:15:13 AM7/1/08

to

Thus spake FrediFizzx <fredi...@hotmail.com>

>Is there a reason why a velocity measurement would need to be so

>precise as to cause infinite energy to be required? IMHO, this is not

>a good reason to reject a velocity operator. Is there perhaps a

>different reason?

>Is there a reason why a velocity measurement would need to be so

>precise as to cause infinite energy to be required? IMHO, this is not

>a good reason to reject a velocity operator. Is there perhaps a

>different reason?

I should have added to my last post, yes there is a different, and much

more subtle, reason. I touch upon it in my reply to Jay, but fully

understanding it requires a very comprehensive understanding of the

relation between quantum mechanics and quantum field theory. It is more

than I can give justice to in a post. Ultimately both reasons boil down

to the same basic fact, that alpha = dx/dt, cannot be taken to be a

meaningful velocity operator in the Dirac-Pauli representation.

Jul 1, 2008, 6:15:09 AM7/1/08

to

The fermion is massive its mass is m.

> Why this is so, has long been a mystery, and is thought not to make any

> sense, for obvious reasons.

Is not longer a mystery. This is studied in my paper "Chubykalo and

Smirnov-Rueda dualism: Foundation and generalizations" which is now under

revision.

The reason which the velocity in Dirac equation is c is explained by the

fact that equation was constrained to be *linear* in momentum. This is

also the basis for the known difficulties with Dirac *wave* equation.

Alternative relativistic quantum mechanics correct Dirac troubles by using

a non-linear propagator for fermions was *quadratic* in momenta.

> Now, transform into the Newton-Wigner representation via

> Foldy-Wouthuysen. The velocity operator in Newton-Wigner now takes the

> classical form:

>

> v = dx/dt (2)

>

> where x is the position operator.

This *is* the general definition. For Dirac quantum wave theory gives

v = dx/dt = [H,x] = +- c

> But even more importantly,

> Newton-Wigner permits a range of eigenvalues less than the speed of

> light, and so, the fermions permitted by Newton-Wigner are massless and

> sub-luminous.

No, because the Newton-Wigner is *not* the actual velocity of the fermion

but only the velocity of an *average* fermion [1].

(\blockquote [emphasis mine]

We have seen that the *standard* velocity operator gives the electron

speed as equal to the velocity of the light.

)

Next he discusses a *different* velocity operator (his eq. 7.41) and adds

(\blockquote [emphasis mine]

These two can be *reconciled* if we identify (7.41) with the *average*

velocity of the particle.

)

Then the *full* motion of the electron can be divided into two parts

(\blockquote [emphasis mine]

Firstly there is the *average* velocity (7.41) and secondly there is very

rapid oscillatory motion which ensures that if an instantaneous

measurement of the *velocity* of the electron could be done it would give

c.

)

The author of [1] splits the "velocity of the electron" into two parts,

say

v = <v> + Dv

he identifies <v>, the *average* velocity, with

(\blockquote

the average motion of the particle, i.e. that given by classical

relativistic formulae

)

Thus <v> does not describe the full velocity of the *quantum* particle.

And the author gives a value c for actual velocity. A simple

transformation cannot change that fact.

In my paper I also analyze the question of the mass of an *free* electron

traveling at c. In Dirac equation the *free* electron is massive and so

far as I know this does not violate *any* known experiment or theory.

The usual experiments for v<c use bound electrons, and SR theory arguments

*start* from a kinetic term of kind

H = (\sqrt m^2 + p^2)

However, I agree that is the more speculative part of my work and warn

readers about that.

[1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).

Chapter 7.

--

Center for CANONICAL |SCIENCE)

http://canonicalscience.org

Jul 1, 2008, 6:22:17 AM7/1/08

to

>I actually scoped out the calculation last night, watching what happens

>to the rest mass when applying a Foldy-Wouthuysen Transformation. It

>does indeed appear possible to take what is a zero-mass fermion in the

>Dirac-Pauli representation, and turn it into a massive fermion in the

>Newton-Wigner representation.

>to the rest mass when applying a Foldy-Wouthuysen Transformation. It

>does indeed appear possible to take what is a zero-mass fermion in the

>Dirac-Pauli representation, and turn it into a massive fermion in the

>Newton-Wigner representation.

The Foldy-Wouthuysen transformation is a unitary transformation, and

cannot change the value of observable quantities.

As for the apparent change in the velocity operator, please study

carefully the page to which I referred you

http://www.teleconnection.info/rqg/CEM

The resolution is to consider the definability of instantaneous velocity

as an observable, and on the subtleties introduced into quantum field

theory because of local gauge symmetry. These matters really are quite

extraordinarily subtle. It is not possible to apply a naive approach

based simply on formula bashing. We can only use quantum field operators

to construct observable operators in a particular picture (the field

picture, or, more strictly, the Newton-Wigner representation), because

otherwise field operators and wave functions do not remain in phase. We

cannot simply write down a Hermitian operator in the Dirac-Pauli

representation (in this case alpha) and call it an observable.

Observable operators must be calculated from Hermitian operators

appearing in the interaction Hamiltonian. These involve the fields.

Ignoring this leads to errors in the application of Ehrenfest's theorem,

false correspondences between operators and observables, and to a raft

of false results.

Jul 1, 2008, 6:42:59 AM7/1/08

to

Thus spake Juan R. González-Álvarez <juanR...@canonicalscience.com>

>The fermion is massive its mass is m.

>The fermion is massive its mass is m.

We agree on that much :-)

>

>The reason which the velocity in Dirac equation is c is explained by the

>fact that equation was constrained to be *linear* in momentum. This is

>also the basis for the known difficulties with Dirac *wave* equation.

There are no difficulties with the Dirac equation. Any problems in qed

are much deeper and more subtle, and have to do with interactions.

>

>Alternative relativistic quantum mechanics correct Dirac troubles by using

>a non-linear propagator for fermions was *quadratic* in momenta.

What theories? Are you talking of formulations of qft based on the

Klein-Gordon equation, which are known both to contain much more serious

foundational problems?

>

>> Now, transform into the Newton-Wigner representation via

>> Foldy-Wouthuysen. The velocity operator in Newton-Wigner now takes the

>> classical form:

>>

>> v = dx/dt (2)

>>

>> where x is the position operator.

>

>This *is* the general definition. For Dirac quantum wave theory gives

>

> v = dx/dt = [H,x] = +- c

As has been thoroughly explained, both here and on s.p.r, this merely

shows that the definition v=dx/dt is not satisfactory.

>

>> But even more importantly,

>> Newton-Wigner permits a range of eigenvalues less than the speed of

>> light, and so, the fermions permitted by Newton-Wigner are massless and

>> sub-luminous.

>

>No, because the Newton-Wigner is *not* the actual velocity of the fermion

>but only the velocity of an *average* fermion [1].

being the only meaningful form of velocity.

>

>(\blockquote [emphasis mine]

> We have seen that the *standard* velocity operator gives the electron

> speed as equal to the velocity of the light.

>)

>

>Next he discusses a *different* velocity operator (his eq. 7.41) and adds

>

>(\blockquote [emphasis mine]

> These two can be *reconciled* if we identify (7.41) with the *average*

> velocity of the particle.

>)

>

>Then the *full* motion of the electron can be divided into two parts

>

>(\blockquote [emphasis mine]

> Firstly there is the *average* velocity (7.41) and secondly there is very

> rapid oscillatory motion which ensures that if an instantaneous

> measurement of the *velocity* of the electron could be done it would give

> c.

>)

>

The hypothesis of oscillatory motion is not right. Such measurements are

not possible in practice, and it ignores the real issue which has to do

with the relationship between qft and rqm.

Jul 1, 2008, 8:43:47 PM7/1/08

to

news:IJFhYVGr...@charlesfrancis.wanadoo.co.uk...

> Thus spake FrediFizzx <fredi...@hotmail.com>

>>Is there a reason why a velocity measurement would need to be so

>>precise as to cause infinite energy to be required? IMHO, this is not

>>a good reason to reject a velocity operator. Is there perhaps a

>>different reason?

>

> I should have added to my last post, yes there is a different, and

> much

> more subtle, reason. I touch upon it in my reply to Jay, but fully

> understanding it requires a very comprehensive understanding of the

> relation between quantum mechanics and quantum field theory. It is

> more

> than I can give justice to in a post. Ultimately both reasons boil

> down

> to the same basic fact, that alpha = dx/dt, cannot be taken to be a

> meaningful velocity operator in the Dirac-Pauli representation.

> Thus spake FrediFizzx <fredi...@hotmail.com>

>>Is there a reason why a velocity measurement would need to be so

>>precise as to cause infinite energy to be required? IMHO, this is not

>>a good reason to reject a velocity operator. Is there perhaps a

>>different reason?

>

> I should have added to my last post, yes there is a different, and

> much

> more subtle, reason. I touch upon it in my reply to Jay, but fully

> understanding it requires a very comprehensive understanding of the

> relation between quantum mechanics and quantum field theory. It is

> more

> than I can give justice to in a post. Ultimately both reasons boil

> down

> to the same basic fact, that alpha = dx/dt, cannot be taken to be a

> meaningful velocity operator in the Dirac-Pauli representation.

Charles,

You refer to "meaningful velocity operator in the Dirac-Pauli

representation."

This is the crux of my concern. "Meaningful" is an imprecise word. I

suspect what you mean is that it is not an "observable operator." Yet,

when transformed into the Newton-Wigner representation via

Foldy-Wouthuysen, it seems that it does become an "observable operator,"

or at least "meaningful" in some sense that I'd like to see better

defined.

Perhaps this is a good context to pose the questions:

1) What, precisely, makes alpha = dx/dt "meaningless" (or not an

observable operator) in one representation, yet allows to to be

transformed into a "meaningful" ("observable operator"?) in another

representation?

2) What is your definition of "meaningful"?

3) More generally, what makes an operator in general, into an

"observable" operator? How does that apply here to the velocity

operator in the two representations under consideration?

4) You have made reference in this thread, twice, to

http://www.teleconnection.info/rqg/CEM, in reference to "the apparent

change in the velocity operator." But that discussion is presented in

the context of electromagnetic gauge theory, which includes the vector

gauge potential A^u. I have here been considering Dirac's equation for

a *free particle*, without an external EM field, so it is not clear how

that discussion applies here. It is precisely for an EM field, that the

Newton-Wigner representation is much less advantageous that Dirac-Pauli.

It would be beneficial, for a free fermion, to see a trace of precisely

how that velocity and position operators go from meaningless in one

representation, to meaningful in another.

5) You state separately in this thread, that "The Foldy-Wouthuysen

transformation is a unitary transformation, and cannot change the value

of observable quantities." So, I again reiterate the above questions:

how does Foldy-Wouthuysen manage to -- if you will -- "rehabilitate" the

velocity and position operators, but not do anything of consequence to

the mass or other "observable" quantities.

6) If velocity cannot be defined as an observable absent infinite

energy to take instantaneous measurements, then what makes it meaningful

in one representation and meaningless in the other?

7) I still am not comfortable with the treatment of time as a parameter

that in a way that does not Lorentz transform in a transparent manner

with the space operators which are of an altogether different character.

Someone can probably give us an extended treatise on why this is so --

but I then have to observe that physics formulated in this fashion is

far from any ideal of being "simple."

Jay.

Jul 2, 2008, 3:59:51 AM7/2/08

to

>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message news:IJFhY

>VGrJf...@charlesfrancis.wanadoo.co.uk...>> Thus spake FrediFizzx <fredi...@hotmail.com>

>>>Is there a reason why a velocity measurement would need to be so

>>>precise as to cause infinite energy to be required? IMHO, this is not

>>>a good reason to reject a velocity operator. Is there perhaps a

>>>different reason?

>>

>> I should have added to my last post, yes there is a different, and

>>much

>> more subtle, reason. I touch upon it in my reply to Jay, but fully

>> understanding it requires a very comprehensive understanding of the

>> relation between quantum mechanics and quantum field theory. It is

>>more

>> than I can give justice to in a post. Ultimately both reasons boil

>>down

>> to the same basic fact, that alpha = dx/dt, cannot be taken to be a

>> meaningful velocity operator in the Dirac-Pauli representation.

>

>Charles,

>

>You refer to "meaningful velocity operator in the Dirac-Pauli

>representation."

>

>This is the crux of my concern. "Meaningful" is an imprecise word. I

>suspect what you mean is that it is not an "observable operator."

Yes. It is not possible to construct an experiment to determine

eigenvalues of this operator.

>Yet, when transformed into the Newton-Wigner representation via Foldy-

>Wouthuysen, it seems that it does become an "observable operator," or

>at least "meaningful" in some sense that I'd like to see better

>defined.

>

>Perhaps this is a good context to pose the questions:

>

>1) What, precisely, makes alpha = dx/dt "meaningless" (or not an

>observable operator) in one representation, yet allows to to be

>transformed into a "meaningful" ("observable operator"?) in another

>representation?

Precisely, it is local gauge freedom in qed, whereas in qm we only have

global gauge freedom (phase invariance)

>

>2) What is your definition of "meaningful"?

as above

>

>3) More generally, what makes an operator in general, into an

>"observable" operator? How does that apply here to the velocity

>operator in the two representations under consideration?

We must be able to actually do a physical measurement to determine the

value of the observable. This means we cannot consider a non-interacting

theory, because we must make the particle interact with the apparatus.

We must then express the observable in terms of things we find in the

interaction, viz, the field operators (or their creation & annihilation

parts).

>4) You have made reference in this thread, twice, to http://www.teleco

>nnection.info/rqg/CEM, in reference to "the apparent change in the

>velocity operator." But that discussion is presented in the context of

>electromagnetic gauge theory, which includes the vector gauge potential

>A^u. I have here been considering Dirac's equation for a *free

>particle*, without an external EM field, so it is not clear how that

>discussion applies here. It is precisely for an EM field, that the

>Newton-Wigner representation is much less advantageous that Dirac-

>Pauli. It would be beneficial, for a free fermion, to see a trace of

>precisely how that velocity and position operators go from meaningless

>in one representation, to meaningful in another.

We can ignore the external potential, and just consider the momentum

operator P, which should commute with the velocity operator. I tried

applying Ehrenfest's theorem to calculate the acceleration, and did not

get the Lorentz force. I traced the problem to phase shifts between the

field operators defined on the non-interacting space, and the wave

function defined in the interacting space. By applying the field

picture, which is effectively the Newton-Wigner representation, I

eliminated these phase shifts. I applied Ehrenfest's theorem again, and

the Lorentz force dropped out quite simply. I am pretty sure we have the

same problem here. (more after your next q).

>5) You state separately in this thread, that "The Foldy-Wouthuysen

>transformation is a unitary transformation, and cannot change the value

>of observable quantities." So, I again reiterate the above questions:

>how does Foldy-Wouthuysen manage to -- if you will -- "rehabilitate"

>the velocity and position operators, but not do anything of consequence

>to the mass or other "observable" quantities.

Consider the forms I give for the momentum operator, just below the

heading Momentum in the Interacting Theory

The usual abbreviated form is simply id^a, but I also give the more

long-winded form

i Integral_dx |x> d^a <x|

I express all observables in this form in

http://www.teleconnection.info/rqg/Observables

This was done originally simply for mathematical rigour - I had no idea

at the time of the implications for observables in qed. The common

abbreviated form is literally just that, an abbreviation for an operator

whose full expression is the long form (the abbreviation is an operator

on a particular representation using wave functions. It is not an

operator on ket space). The requirement that observables derive from

interactions means that the terms |x> and <x| must be the creation and

annihilation operators which appear in the interaction, not just the

basis states.

Now the problem is that the creation and annihilation operators act on

the non-interacting space, whereas the wave function evolves in the

interacting space. There is a phase shift between the two. Normally the

phase shift is invisible, but when we act with a differential operator

it becomes visible. This is not allowable, so it means in practice that

the abbreviated form of an observable breaks down. We need to keep sight

of the behaviour of the creation and annihilation operators, but more

significantly we need to fix the theory so that the phases of the

creation and annihilation operators (for which there is local gauge

symmetry) remains the same as the phases of the wave function (for which

there is only global symmetry).

The Foldy-Wouthuysen transformation does exactly what is required in

fixing the phase shifts. As a result, the abbreviated forms of the

operators which we commonly use are strictly only correct in the Newton-

Wigner representation, or in my simplified treatment, the field picture.

>6) If velocity cannot be defined as an observable absent infinite

>energy to take instantaneous measurements, then what makes it

>meaningful in one representation and meaningless in the other?

All right, I'll put my hand up. What I actually said was "This is *often

thought to be* because the precise measurement of velocity would require

a time trial between two exact position measurements. An exact position

measurement would require infinite energy." I gave the usual glib

explanation. In my heart of hearts I don't think this is actually the

correct explanation. I have found the issues surrounding local gauge

freedom to be just about the most subtle aspect of qed, and in

particular in determining the relationship between relativistic quantum

mechanics and qft. In my defence, I would point out that most treatments

give up at this point, and simply replace rqm with qft. I have never

been happy with that.

The real issue is that Ehrenfest's theorem applies to quantum mechanics.

The definition of the dx/dt as a velocity operator, and the calculation

of its eigenvalues, use Ehrenfest's theorem. But to do observations we

have to interact with the particle. Interactions require field

operators, and observables must be formulated using creation and

annihilation operators. The treatment using just quantum mechanics

suppresses phase, and as a result, Ehrenfest's theorem breaks down for

observables described by differential operators. The correct definition

of a velocity operator requires that we also take into account phase

shifts. In the Newton-Wigner representation the phase shift between

wave function and field operator is eliminated, so everything works

fine, but in other representations the velocity operator has a different

form.

>7) I still am not comfortable with the treatment of time as a

>parameter that in a way that does not Lorentz transform in a

>transparent manner with the space operators which are of an altogether

>different character. Someone can probably give us an extended treatise

>on why this is so -- but I then have to observe that physics

>formulated in this fashion is far from any ideal of being "simple."

I think the treatment I have given of relativistic quantum mechanics,

starting from measurement a a particular time, is just such a treatise.

I would have given this answer more quickly had I not been trying to

answer within the context of usual treatments, because think I make more

of the fact that measurement is measurement at given time than is

normal, and, as I say, usually treatments of qft are divorced from

foundations of quantum mechanics. I must admit, I felt a bit dumb when

Igor pointed out that this asymmetry is always present, because actually

I knew that too.

Mathematically qed is not simple. It is stacked with subtleties, such as

we have been discussing. Nonetheless it seems to me that what it

describes physically is simple, i.e. a system containing two types of

particle, electrons and photons, each individually obeying simple

equations, together with a simple interaction between those particles.

The complexities arise because we can have any number of each of these

particles, and in almost any configuration, and because of the non-

physical nature of phase.

Jul 2, 2008, 12:51:57 PM7/2/08

to

Thus spake FrediFizzx <fredi...@hotmail.com>

>>In the Standard Model, elementary fermions are massless without an

>>interaction with the Higgs field.

"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de

news:Q7LH0mEQ...@charlesfrancis.wanadoo.co.uk...

> I think you mean Bosons,

Also fermions.

--

~~~~ clmasse on free F-country

Liberty, Equality, Profitability.

Jul 2, 2008, 12:53:06 PM7/2/08

to

Jay R. Yablon wrote on Sun, 29 Jun 2008 20:08:35 -0600:

>> I say this in particular because in the Dirac-Pauli representation, the

>> velocity operator is given by:

>>

>> v^k = alpha^k (1)

>>

>> where alpha^k=gamma^0gamma^k, see reference III. Further, the

>> eigenvalues of this velocity operator constrains the velocity of the

>> Fermion of be the speed of light, see reference II in the middle of page

>> 3. This means that the fermion must be massless and luminous, in the

>> Dirac-Pauli representation.

"Juan R. González-Álvarez" <juanR...@canonicalscience.com> a écrit dans le

message de news:pan.2008.07...@canonicalscience.com...

> The fermion is massive its mass is m.

Yes, it is the instantaneous velocity, which changes all the time

(Zitterbewegung = shaking motion) so that the smoothed velocity is lower

than c. And what makes the velocity change is precisely the mass term.

>> Why this is so, has long been a mystery, and is thought not to make any

>> sense, for obvious reasons.

The only covariant velocity distribution is in one dimension and with only

two possible (i.e. with non zero probability) values +c and -c. The

explanation is that the Dirac equation is covariant, and consequently the

velocity operators in different directions don't commute.

> Is not longer a mystery. This is studied in my paper "Chubykalo and

> Smirnov-Rueda dualism: Foundation and generalizations" which is now under

> revision.

>

> The reason which the velocity in Dirac equation is c is explained by the

> fact that equation was constrained to be *linear* in momentum. This is

> also the basis for the known difficulties with Dirac *wave* equation.

>

> Alternative relativistic quantum mechanics correct Dirac troubles by using

> a non-linear propagator for fermions was *quadratic* in momenta.

Not sure.

>> Now, transform into the Newton-Wigner representation via

>> Foldy-Wouthuysen. The velocity operator in Newton-Wigner now takes the

>> classical form:

>>

>> v = dx/dt (2)

>>

>> where x is the position operator.

This definition entails two (infinitely close) times, so that it is no

longer

instantaneous velocity, it is the smoothed velocity. (I think that term is

more appropriate than average, which corresponds to a finite time interval.)

Jul 2, 2008, 2:16:40 PM7/2/08

to

"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

news:Q7LH0mEQ...@charlesfrancis.wanadoo.co.uk...> Thus spake FrediFizzx <fredi...@hotmail.com>

>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

>>news:DQmJ1

>>uCRkI...@charlesfrancis.wanadoo.co.uk...

>>> Thus spake Jay R. Yablon <jya...@nycap.rr.com>

>>>> But I would for now like some reactions as to the tree

>>>>up which I am barking.

>>>>

>>> I'm afraid it is the wrong tree. Fermions are not massless. In fact

>>>mass

>>> is a fundamental property as described in the Dirac equation. The

>>> velocity operator in the Dirac-Pauli representation is not

>>> physically

>>> meaningful. This is often thought to be because the precise

>>>measurement

>>> of velocity would require a time trial between two exact position

>>> measurements. An exact position measurement would require infinite

>>> energy.

>>

>>Is there a reason why a velocity measurement would need to be so

>>precise as to cause infinite energy to be required?

>

> Yes. We are talking here of instantaneous velocity. Physically this

> would mean taking the limit of two measurements separated by an

> infinitesimal time interval - as a result, perfect accuracy would be

> required.

Well, that is true in the classical case also. We get infinite stuff

all the time from infinitesimals. Why can't we do an average velocity?

http://modelingnts.la.asu.edu/pdf/ZBW_I_QM.pdf

See the above starting around eq. 67.

>>In the Standard Model, elementary fermions are massless without an

>>interaction with the Higgs field.

>

>

> I think you mean Bosons, but in any case I have never seen any reason

> to

> think the Higgs mechanism is more than unholy metaphysics. To me, it

> is

> based on a notion of the vacuum which is a contradiction in terms, and

> which introduces an extraordinary complexity in nature at the point of

> fundamental physics where I would expect the fundamental building

> blocks

> of matter to be simple. Moreover, the reasons given that such a

> mechanism might be necessary for the interpretation of fundamental law

> completely overlook the study of foundations, and imv they are false.

I mean elementary fermions also. Please see the following link,

http://en.wikipedia.org/wiki/Standard_model_%28mathematical_formulation%29#The_Higgs_field

It doesn't really have to have anything to do with vacuum notions.

Rather that the vacuum is filled with something we don't quite

understand fully yet. Hopefully the LHC with give us much more clues

about this. I fully expect that they will eventually find evidence of a

Higgs-like field of some kind. Though, I expect the Higgs bosons to be

composites.

Best,

Fred Diether

Jul 2, 2008, 4:00:30 PM7/2/08

to

Thus spake FrediFizzx <fredi...@hotmail.com>

>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

>news:Q7LH0mEQ...@charlesfrancis.wanadoo.co.uk...

>> Thus spake FrediFizzx <fredi...@hotmail.com>

>>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

>>>news:DQmJ1

>>>uCRkI...@charlesfrancis.wanadoo.co.uk...

>>>> Thus spake Jay R. Yablon <jya...@nycap.rr.com>

>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

>news:Q7LH0mEQ...@charlesfrancis.wanadoo.co.uk...

>> Thus spake FrediFizzx <fredi...@hotmail.com>

>>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

>>>news:DQmJ1

>>>uCRkI...@charlesfrancis.wanadoo.co.uk...

>>>> Thus spake Jay R. Yablon <jya...@nycap.rr.com>

>>>Is there a reason why a velocity measurement would need to be so

>>>precise as to cause infinite energy to be required?

>>

>> Yes. We are talking here of instantaneous velocity. Physically this

>> would mean taking the limit of two measurements separated by an

>> infinitesimal time interval - as a result, perfect accuracy would be

>> required.

>

>Well, that is true in the classical case also. We get infinite stuff

>all the time from infinitesimals. Why can't we do an average velocity?

It's not really the way qm is formulated. qm gives us a probability

amplitude from an initial state to a final state. When we make the final

state the initial state for the next part of the motion, we have to

reformulate the theory. That is where the phase shift comes in. However,

as I said, this is the usual answer, and is really only intended to say

that it does not matter that v is not a good velocity operator in Dirac

theory. I don't really accept it myself. See my post to Jay on what I

think the real answer is.

>

>http://modelingnts.la.asu.edu/pdf/ZBW_I_QM.pdf

>

>See the above starting around eq. 67.

It looks to me that one has to make too many auxiliary assumptions.

>>>In the Standard Model, elementary fermions are massless without an

>>>interaction with the Higgs field.

>>

>>

>> I think you mean Bosons, but in any case I have never seen any reason

>> to

>> think the Higgs mechanism is more than unholy metaphysics. To me, it

>> is

>> based on a notion of the vacuum which is a contradiction in terms, and

>> which introduces an extraordinary complexity in nature at the point of

>> fundamental physics where I would expect the fundamental building

>> blocks

>> of matter to be simple. Moreover, the reasons given that such a

>> mechanism might be necessary for the interpretation of fundamental law

>> completely overlook the study of foundations, and imv they are false.

>

>I mean elementary fermions also. Please see the following link,

>

>http://en.wikipedia.org/wiki/Standard_model_%28mathematical_formulation

>%29#The_Higgs_field

>It doesn't really have to have anything to do with vacuum notions.

>Rather that the vacuum is filled with something we don't quite

>understand fully yet. Hopefully the LHC with give us much more clues

>about this. I fully expect that they will eventually find evidence of a

>Higgs-like field of some kind. Though, I expect the Higgs bosons to be

>composites.

>

I'm afraid I still don't think this whole approach makes sense. To me a

vacuum is the absence of anything, and that is how I formulate theory. I

see no reason for particles to acquire mass. That is a fundamental

property, imv. I don't think the LHC will come up with anything on

Higgs, but we will wait and see.

Jul 2, 2008, 4:17:59 PM7/2/08

to

On Jun 30, 4:08 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:

> I have been looking over the following three links for Foldy-Wouthuysen

> transformation from the Dirac-Pauli to the Newton-Wigner representation

> of Dirac's equation:

>

> The first shows the calculation itself of this transformation:

>

> I:http://www.physics.ucdavis.edu/~cheng/230A/RQM7.pdf

>

> The second, an excellent and lucid exposition of the physics (why this

> is of interest), is to be found at:

>

> II:http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.27.3209&rep=...> I have been looking over the following three links for Foldy-Wouthuysen

> transformation from the Dirac-Pauli to the Newton-Wigner representation

> of Dirac's equation:

>

> The first shows the calculation itself of this transformation:

>

> I:http://www.physics.ucdavis.edu/~cheng/230A/RQM7.pdf

>

> The second, an excellent and lucid exposition of the physics (why this

> is of interest), is to be found at:

>

> co-moderator: sci.physics.foundations

> Weblog:http://jayryablon.wordpress.com/

> Web Site:http://home.nycap.rr.com/jry/FermionMass.htm

Jay,

If you apply the velocity operator to the Dirac field as I did here:

http://www.physicsforums.com/showthread.php?t=219205

Then you simply get its speed v: the correct answer.

If you apply another operator in the same way, for instance, the

energy or momentum operators then you'll get the right answers

as well.

Regards, Hans

Jul 3, 2008, 10:23:25 AM7/3/08

to

Cl\.Massé wrote on Wed, 02 Jul 2008 10:53:06 -0600:

> The only covariant velocity distribution is in one dimension and with

> only two possible (i.e. with non zero probability) values +c and -c. The

> explanation is that the Dirac equation is covariant, and consequently

> the velocity operators in different directions don't commute.

Stuckelberg theory is also covariant and the velocity predicted is not c.

The reason which Dirac gives c may be traced to its Hamiltonian, being

*linear* in p.

Differentiating something linear on p cannot yield a function of p,

it may yield a constant. This may be obvious. By relativistic constraints

that constant is c for Dirac Hamiltonian.

>>> v = dx/dt (2)

>>>

>>> where x is the position operator.

>

> This definition entails two (infinitely close) times, so that it is no

> longer

> instantaneous velocity, it is the smoothed velocity. (I think that term

> is more appropriate than average, which corresponds to a finite time

> interval.)

As explained in [1] the equation (2) gives the instantaneous velocity

of the electron and would not be confused with the 'average' or 'smoother'

velocity given after applying the FW.

This latter velocity only represents electron motion in some average or

smooth sense over scales of space and time greater than certain

characteristic quantity, of order 10^(-13) seconds [1].

Jul 3, 2008, 12:45:24 PM7/3/08

to

"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de

news:x1ZvX6Aa...@charlesfrancis.wanadoo.co.uk...

news:x1ZvX6Aa...@charlesfrancis.wanadoo.co.uk...

> Precisely, it is local gauge freedom in qed, whereas in qm we only have

> global gauge freedom (phase invariance)

There is also a local gauge freedom in QM, it is linked to the freedom of

the potential. The momentum operator contains the potential so that its

eigenvalues don't depend on the gauge.

Jul 3, 2008, 5:19:11 PM7/3/08

to

Your 2d Dirac equation method corresponds to computing one of

*components* of the velocity v. E.g. in [1] it is also computed the

component v_x and showed it is less than c (of order |v|) because each

wave function component gives a v/2c contribution. But as explained in

[1] this does not imply that total (\vect v) was of magnitude < c.

One may take into account certain commutativity oddities associated to

H_dirac when computing the total (\vect v) from each component v_x, v_y,

v_z.

The most easy way to compute the magnitude is from <v^2>. If you do that

you will find c^2. From here follows the well-known result that |v| = c.

See [1] for details and further discussion.

Jul 5, 2008, 1:10:56 PM7/5/08

to

On Jul 3, 11:19 pm, "Juan R." González-Álvarez

<juanREM...@canonicalscience.com> wrote:

>

> Your 2d Dirac equation method corresponds to computing one of

> *components* of the velocity v. E.g. in [1] it is also computed the

> component v_x and showed it is less than c (of order |v|) because each

> wave function component gives a v/2c contribution. But as explained in

> [1] this does not imply that total (\vect v) was of magnitude < c.

>

> One may take into account certain commutativity oddities associated to

> H_dirac when computing the total (\vect v) from each component v_x, v_y,

> v_z.

>

> The most easy way to compute the magnitude is from <v^2>. If you do that

> you will find c^2. From here follows the well-known result that |v| = c.

> See [1] for details and further discussion.

>

> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).

> Chapter 7.

>

<juanREM...@canonicalscience.com> wrote:

>

> Your 2d Dirac equation method corresponds to computing one of

> *components* of the velocity v. E.g. in [1] it is also computed the

> component v_x and showed it is less than c (of order |v|) because each

> wave function component gives a v/2c contribution. But as explained in

> [1] this does not imply that total (\vect v) was of magnitude < c.

>

> One may take into account certain commutativity oddities associated to

> H_dirac when computing the total (\vect v) from each component v_x, v_y,

> v_z.

>

> The most easy way to compute the magnitude is from <v^2>. If you do that

> you will find c^2. From here follows the well-known result that |v| = c.

> See [1] for details and further discussion.

>

> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).

> Chapter 7.

>

Hi, Juan,

I happen to know the passage of Paul's book you are quoting here. I

really

like his book but here he makes a bit of a blooper.. The point is that

<v^2>

is not the same as <v>^2. In fact he shows so in the text himself.

In the 2d dirac equation v has only 1 component but if you do the same

for

the 4d version than one get's the three components v_x,v_y and v_z if

you

apply the sigma_x, _y and _z matrices. This is of course just the

calculation of the vector-current divided by the normalization.

Note that there are three velocity operators:

v_x = [H,x], v_y = [H,y], v_z = [H,z]

Which use the sigma_x, sigma_y and sigma_z matrices respectively,

just as required to get the right results.

Regards, Hans.

http://www.physics-quest.org/ on the Dirac equation:

http://physics-quest.org/Book_Chapter_Dirac.pdf

http://physics-quest.org/Book_Chapter_Dirac_Fields.pdf

http://physics-quest.org/Book_Chapter_Gordon_Decomposition.pdf

Jul 6, 2008, 7:05:26 AM7/6/08

to

Hans de Vries wrote on Sat, 05 Jul 2008 11:10:56 -0600:

> On Jul 3, 11:19 pm, "Juan R." GonzÃ¡lez-Ã lvarez

Hi Hans,

Yes, as pointed above, you can compute components of v. Now the problem is

when you take the 1D result (2D Dirac) and assume that you can generalize

it to 3D dimensions (4D Dirac) giving a total velocity

v = (v_x + v_y + v_z) < c

but, as remarked in my previous post, you cannot do that.

This is also explained in [1]. See the page 207, where Paul Strange

notices that the components of the velocity v_x, v_y, and v_z *don't

commute*.

However, you can still compute the magnitude of the vector. <v^2> gives

c^2, you may check that. From here follows the well-known result that

|v| = c

for Dirac Hamiltonians.

This result is also obtained and explained by Feynman in his famous

monograph on QED

http://www.amazon.com/Quantum-Electrodynamics-Advanced-Book-Classics/

dp/0201360756

Instead computing v^2, Feynman directly obtains the eigenvalues of \alpha

and thus the eigenvalues of v are +- c from the operator. He also offers

arguments on why Dirac electrons may move at c.

My argument is that particles described by Hamiltonians linear on momentum

p travel to c. This is the case for photons (H = pc) and Dirac electrons

(H = \alpha pc).

> http://www.physics-quest.org/ on the Dirac equation:

Revise the links in your site. E.g. in "Chapter 6: The Chern-Simons EM

spin and axial current density", "Chapter" links to one pdf and "6" links

to other different. Also in "Chapter 13" link.

Some users may be clicking in the wrong place and obtaining a different

pdf they would.

> http://physics-quest.org/Book_Chapter_Dirac.pdf

> http://physics-quest.org/Book_Chapter_Dirac_Fields.pdf

> http://physics-quest.org/Book_Chapter_Gordon_Decomposition.pdf

Thanks by those very informative links!

Jul 6, 2008, 6:04:41 PM7/6/08

to

On Jul 6, 1:05 pm, "Juan R." González-Álvarez> >http://www.physics-quest.org/on the Dirac equation:

>

> Revise the links in your site. E.g. in "Chapter 6: The Chern-Simons EM

> spin and axial current density", "Chapter" links to one pdf and "6" links

> to other different. Also in "Chapter 13" link.

>

> Some users may be clicking in the wrong place and obtaining a different

> pdf they would.

>

> >http://physics-quest.org/Book_Chapter_Dirac.pdf

> >http://physics-quest.org/Book_Chapter_Dirac_Fields.pdf

> >http://physics-quest.org/Book_Chapter_Gordon_Decomposition.pdf

>

> Thanks by those very informative links!

>

> --

> Center for CANONICAL |SCIENCE) http://canonicalscience.org

>

> Revise the links in your site. E.g. in "Chapter 6: The Chern-Simons EM

> spin and axial current density", "Chapter" links to one pdf and "6" links

> to other different. Also in "Chapter 13" link.

>

> Some users may be clicking in the wrong place and obtaining a different

> pdf they would.

>

> >http://physics-quest.org/Book_Chapter_Dirac.pdf

> >http://physics-quest.org/Book_Chapter_Dirac_Fields.pdf

> >http://physics-quest.org/Book_Chapter_Gordon_Decomposition.pdf

>

> Thanks by those very informative links!

>

> --

> Center for CANONICAL |SCIENCE) http://canonicalscience.org

Hi, Juan.

The role of c is much easier to see in the chiral

representation. Both the left and right chiral currents

propagate at c while the electron as a whole

has a velocity v anywhere between +c and -c.

The two chiral components are helicity eigen-

states because the continue to propagate at c

regardless of the overall speed v.

If v=0 then the two chiral currents are equal but

opposite and their mutual coupling keeps the

particle as a whole localized and at rest.

The general proof that J_L and J_R transform

light-like is given in section 13.4 of:

http://physics-quest.org/Book_Chapter_Dirac_Fields.pdf

This (being complicated by the spinor algebra) in

4d is just an extension of the 2d dirac equation

which is much simpler.

This is in fact how I work towards the Dirac equation:

http://physics-quest.org/Book_Chapter_Dirac.pdf

-------- ( Step 1 ) ----------

The classical 2nd order wave equation for a mass-

less particle is linearized into a 100% equivalent

2 component equation with a left moving and a right

moving component. (see section 11.1, we restrict

the wave function to 2d here)

-------- ( Step 2 ) ----------

The left and right moving components which move

at c (zero mass) are now coupled together by the

mass m. The two components still travel at c

but the particle as a whole moves with v.

Figure 11.4 shows a computer simulation of

the above. What we have obtained here is just

the 2d Dirac equation.

-------- ( Step 3 ) ----------

Replacing the two components by two spinors,

as in equation (11.21) and (11.22), then leads

to the full 4d Dirac equation in the Chiral

representation.

The two chiral components still propagate at c

in case of the 4d Dirac equation regardless

of the speed v of the electron as a whole.

Regards, Hans

Jul 7, 2008, 2:58:24 AM7/7/08

to

>

> - Show quoted text -

Dear Hans! It is interesting where are you going to publish your

book ?

I'm quite puzzled, if big names like "Wiley and Sons" would allow to

expose

to public partial book's content ?

Best, andy

Jul 7, 2008, 3:35:07 PM7/7/08

to

"Juan R. Gonz�lez-�lvarez" <juanR...@canonicalscience.com> a �crit dans le

message de news:pan.2008.07...@canonicalscience.com...

message de news:pan.2008.07...@canonicalscience.com...

> Stuckelberg theory is also covariant and the velocity predicted is not c.

That's not the same thing. I spoke about a statistical distribution of

instantaneous velocity, not a state velocity.

Jul 8, 2008, 12:45:52 AM7/8/08

to

"Hans de Vries" <Hans.de....@gmail.com> wrote in message

news:67f78ea8-8dbf-48e6...@b1g2000hsg.googlegroups.com... . . .

>

> Jay,

>

> If you apply the velocity operator to the Dirac field as I did here:

>

> http://www.physicsforums.com/showthread.php?t=219205

>

> Then you simply get its speed v: the correct answer.

>

> If you apply another operator in the same way, for instance, the

> energy or momentum operators then you'll get the right answers

> as well.

>

> Regards, Hans

>

Good to hear from you! I see that while I was headed to cape May NJ for

a few days at the shore, you started a whole new branch to the thread.

I will study the chapter from your book to be found at

http://chip-architect.com/physics/Book_Chapter_Dirac.pdf, in some depth.

I will also pass along some ideas which I developed at the ocean last,

which is always a good place for regaining perspective.

I very much like your approach based on separate left-right chiral

components linked by mass.

While nobody today knows the specific mechanism by which fermion mass is

generated (and by which one can come to thereby understand the values of

the fermion masses which challenges all serious physicists by

perpetually taunting us with hard experimental data which nobody has yet

explained), I think it should be clear that whatever that mechanism

turns out to be, one will need to start off with a Lagrangian in which a

fermion mass term is omitted entirely (as is done with vector boson

masses), which means that m=0 in the underlying Dirac Lagrangian. Then,

one will need to find some mechanism by which a fermion mass term

suddenly becomes "revealed" (also as is done with vector bosons masses).

The approach you take starts in exactly the right place, because it

considers chiral projections which, when decoupled from one another,

necessarily have zero mass and travel at the speed of light consistent

with the Dirac velocity "alpha" operator. How the coupling of

left-to-right then arises, is at the centre of the fermion mass

question.

I know you spend a lot of time playing with masses from a

"numerological" approach (and I do not mean this pejoratively unless one

thinks that Planck misbehaved when he did "whatever it takes" to explain

blackbody spectrum). But, have you, in the course of your assembling

the subject in your chapter on Dirac's equation, especially given your

chiral starting point, ever come upon a hint as to a possible mechanism

for generating fermion mass without numerological considerations? If

so, what have you been able to glean along what I consider to be your

very correct path?

Thanks,

Jay.

Jul 8, 2008, 4:28:42 AM7/8/08

to

"Jay R. Yablon" <jya...@nycap.rr.com> wrote in message

news:6dg66tF...@mid.individual.net...

news:6dg66tF...@mid.individual.net...

> I know you spend a lot of time playing with masses from a

> "numerological" approach (and I do not mean this pejoratively unless

> one thinks that Planck misbehaved when he did "whatever it takes" to

> explain blackbody spectrum). But, have you, in the course of your

> assembling the subject in your chapter on Dirac's equation,

> especially given your chiral starting point, ever come upon a hint

> as to a possible mechanism for generating fermion mass without

> numerological considerations? If so, what have you been able to

> glean along what I consider to be your very correct path?

Jay, what don't you like about the Higgs or a Higgs-like mechanism for

generating masses? I thought you were on to something long time ago

relating the masses to vev. It is just a matter of finding the right

geometry of interactions with a Higgs-like field for me.

Best,

Fred Diether

Jul 8, 2008, 8:30:54 AM7/8/08

to

Hi Hans,

I read your reply and the pdfs but you did not address the complaints

about the commutativity of the operators, neither you computed <v^2>.

Your "chiral representation" does not change the properties of \alpha,

thus the commutative properties for v_x, v_y, and v_z are the same ones

that are noticed in page 207 of [1].

As explained in [1] you cannot simply add each component to built the

(\vect v). Or, said in another way,

(\vect v) = (v_x + v_y + v_z)

is not an observable in Dirac theory.

You seem to ignore these points when generalizing the 2D chiral (1

velocity component) to the 4D chiral (3 components).

How non-commutative operators v_x, v_y, v_z give a commutative observable

(\vect v)? You don't explain this in your book.

Also, as computed in Feynman [2] and in Strange [1], <v^2> gives c^2,

which implies (|v| = c). This is also the case for your (11.20) on

http://physics-quest.org/Book_Chapter_Dirac.pdf

Introducing the v^2 operator into your (11.20) gives c^2. This result

implies (|v| = c), as the own Dirac recognized, or see also [1,2]. How do

you interpret that result as saying |v| is not c?

In [1,2] it is explained that correct results are

(|v| = c) and (|v|^2 = c^2)

Also you did not commented in the well-known point that Hamiltonians

linear in p may give particles traveling at c. The photon Hamiltonian is

the more popular case.

All solutions/corrections to Dirac Hamiltonian I know introduce non-

linear corrections on p: Stuckelberg, F-W transformation, RQD...

Those non-linear Hamiltonians yield |v| < c.

It seems you also ignore all other difficulties traditionally associated

to Dirac theory,

http://en.wikipedia.org/wiki/Relativistic_dynamics

or at least I don't see discussion in the chapter of your book you cited

above.

What are your comments on all of this?

[1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).

Chapter 7.

[2] http://www.amazon.com/Quantum-Electrodynamics-Advanced-Book-Classics/

dp/0201360756

--

Center for CANONICAL |SCIENCE)

http://canonicalscience.org

======================================= MODERATOR'S COMMENT:

NB: Fock ('Principles of Qquantum Theory') considered |v| = c not to be the last word

Jul 8, 2008, 11:45:07 AM7/8/08

to

"Juan R." González-Álvarez wrote on Tue, 08 Jul 2008 06:30:54 -0600:

> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).

> Chapter 7.

>

> [2]

> http://www.amazon.com/Quantum-Electrodynamics-Advanced-Book-Classics/

> dp/0201360756

Some moderator wrote

> ======================================= MODERATOR'S COMMENT:

> NB: Fock ('Principles of Qquantum Theory') considered |v| = c not to be

> the last word

What were Fock arguments? Did he proved that |v| < c for Dirac theory.

All literature I know agrees that |v| = c for an electron described by

Dirac Hamiltonian. And Dirac agreed on this also.

Jul 8, 2008, 7:55:51 PM7/8/08