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# Might Foldy-Wouthuysen Transformations contain a Hidden Fermion Mass Generation Mechanism?

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### Jay R. Yablon

Jun 29, 2008, 10:08:35 PM6/29/08
to
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:

The second, an excellent and lucid exposition of the physics (why this
is of interest), is to be found at:

The third, dealing with Zitterbewegung motion the the velcoty operator
in the Dirac-Pauli representation, is at:

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

### Oh No

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.

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)

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

### FrediFizzx

Jun 30, 2008, 2:48:22 PM6/30/08
to
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

> 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

### Jay R. Yablon

Jun 30, 2008, 3:39:35 PM6/30/08
to

"FrediFizzx" <fredi...@hotmail.com> wrote in message
news:6csk1lF...@mid.individual.net...
Just so, Fred.

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

### Jay R. Yablon

Jun 30, 2008, 8:58:19 PM6/30/08
to

"FrediFizzx" <fredi...@hotmail.com> wrote in message
news:6csk1lF...@mid.individual.net...
. . .

>
> 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,

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.

### Oh No

Jul 1, 2008, 6:01:38 AM7/1/08
to
Thus spake FrediFizzx <fredi...@hotmail.com>

>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message news:DQmJ1

>> 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.

### Oh No

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?

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.

### Juan R.

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

[1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
Chapter 7.

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

### Oh No

Jul 1, 2008, 6:22:17 AM7/1/08
to
Thus spake Jay R. Yablon <jya...@nycap.rr.com>
>
>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.

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.

### Oh No

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.

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.

### Jay R. Yablon

Jul 1, 2008, 8:43:47 PM7/1/08
to

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

> 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?

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.

### Oh No

Jul 2, 2008, 3:59:51 AM7/2/08
to
Thus spake Jay R. Yablon <jya...@nycap.rr.com>
>
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message news:IJFhY

>> 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).

>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.

### Cl.Massé

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

> I think you mean Bosons,

Also fermions.

--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.

### Cl.Massé

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.)

### FrediFizzx

Jul 2, 2008, 2:16:40 PM7/2/08
to
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

> Thus spake FrediFizzx <fredi...@hotmail.com>
>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
>>news:DQmJ1
>>> 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.

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

Best,

Fred Diether

### Oh No

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
>> Thus spake FrediFizzx <fredi...@hotmail.com>
>>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
>>>news:DQmJ1
>>>> 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

>
>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.
>
>
>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
>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.

### Hans de Vries

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:
>
> Email: jyab...@nycap.rr.com

> 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:

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

### Juan R.

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].

### Cl.Massé

Jul 3, 2008, 12:45:24 PM7/3/08
to
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de

> 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.

### Juan R.

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.

### Hans de Vries

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.
>

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:

### Juan R.

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

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.

Thanks by those very informative links!

### Hans de Vries

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

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

### Mirror Space

Jul 7, 2008, 2:58:24 AM7/7/08
to
> > >http://www.physics-quest.org/onthe Dirac equation:
> Regards, Hans- Hide quoted text -
>
> - 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

### Cl.Mass�

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...

> 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.

### Jay R. Yablon

Jul 8, 2008, 12:45:52 AM7/8/08
to

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

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

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

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.

### FrediFizzx

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...

> 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

### Juan R.

Jul 8, 2008, 8:30:54 AM7/8/08
to

Hi Hans,

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

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.

[1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
Chapter 7.

--
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

### Juan R.

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]
> 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.

### David Rutherford

Jul 8, 2008, 7:55:51 PM7/8/08
to

Jay R. Yablon wrote:
>
> While nobody today knows the specific mechanism by which fermion mass is
> generated

I can account for the mass of the electron. It's _entirely_
electromagnetic in origin. Please see "4/3 Problem Resolution"
under "Applications" at

http://www.softcom.net/users/der555

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"

### Hans de Vries

Jul 9, 2008, 4:40:20 PM7/9/08
to
On Jul 8, 2:30 pm, "Juan R." González-Álvarez

<juanREM...@canonicalscience.com> wrote:
>
> Hi Hans,
>
> 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.
>

Hi, Juan

The velocity components *are* independent in the relativistic theory

As I said, Paul makes a bit of a blooper here.

First: The square of an operator doesn't give you the
square of the observable. Like you have in classical
physics:

(dx/dt)^2 = v^2, but (d/dt)^2 x = acceleration

It is also very dangerous to say that if the velocity
component operators don't commute that the velocity
components them self "don't commute". It's wrong in
this case.

You can do things like this if the input and output of
an operator are of the same type as is the case with
rotation operators and boost operators. The product
of two rotation operators is a new rotation operator.
no problem. In other cases on has to be very careful.

> 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.
>

Feynman only uses alpha^2=1 as a quick trick to obtain
the magnitude of the eigen values of alpha (=+/- 1) which is
not quite the same (link 2, page 47)

He then concludes that the eigenvelocities are +/- c without
actually applying the operator. Note that -c and +c operate
on the left and right chiral components respectively which
actually *do* propagate at the speed of light but "average"
out to the actual speed v

(see my previous post for links to the general proof of the
light like behaviour)

> 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.
>

Many of the problems associated with the Dirac theory
are just caused by doing the wrong math in my opinion.
Such as is the case with the velocity operator....

For the same reason I do think that the FW-transform is a
mess. ElectroWeak unification has shown that the chiral
representation leads to an interpretation which is much
closer to physics as the large/small (particle/anti-particle)
component interpretation of the old days.

Regards, Hans
http://www.physics-quest.org/

### Hans de Vries

Jul 9, 2008, 4:50:33 PM7/9/08
to
On Jul 7, 8:58 am, Mirror Space <Ino...@yahoo.com> wrote:
>
>   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

Hi, andy.

I'm not in that stage yet... But anyway. if you're not one of
the really "big-shots" then you need some public exposure first.

Regards, Hans

### pellis

Jul 21, 2008, 10:37:25 AM7/21/08
to
On Jul 2, 9:17 pm, Hans de Vries <Hans.de.Vries...@gmail.com> wrote:
> On Jun 30, 4:08 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:

With all the recent activity on Foldy-Wouthuysen in this group at
present, could someone (Hans?) satisfy the Wikipedia request for an
article on this topic:

http://en.wikipedia.org/wiki/Wikipedia:Requested_articles/Natural_sciences#Physics

(see section Physics/QM)

and

http://en.wikipedia.org/wiki/Foldy-Wouthuysen_transformation

I'm sure this would help some of us who might like to be better able

Thanks - P

### Jay R. Yablon

Jul 21, 2008, 1:34:06 PM7/21/08
to
Per below:

I started a Wiki article for this, which I will try to beef up over the
next few days. I think it would be a good exercise for me.

Jay.

"pellis" <pel...@london.edu> wrote in message

### shal...@prizatech.com

Aug 11, 2014, 2:40:03 AM8/11/14
to
Hi,

University of Sydney has published the solution of Klein Gordon equation http://www.physics.usyd.edu.au/ugrad/hons/hons_courses/RQM7.pdf for the hydrogen atom has a fundamental mistake in equations and the solution. The derivation of equation 7.8 from 7.7 is wrong and sign of right hand side of the equation is completely reversed. The results agree with the result of Dirac equation but if the mathematics is wrong at such a fundamental level, there is something big wrong in the way University of Sydney is solving it.

On Sunday, June 29, 2008 7:08:35 PM UTC-7, Jay R. Yablon 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:
> Email: jya...@nycap.rr.com

### FrediFizzx

Aug 11, 2014, 2:50:03 AM8/11/14
to
<shal...@prizatech.com> wrote in message
> Hi,
>
> University of Sydney has published the solution of Klein Gordon equation
> hydrogen atom has a fundamental mistake in equations and the solution. The
> derivation of equation 7.8 from 7.7 is wrong and sign of right hand side
> of the equation is completely reversed. The results agree with the result
> of Dirac equation but if the mathematics is wrong at such a fundamental
> level, there is something big wrong in the way University of Sydney is
> solving it.
===================

### John Heath

Aug 22, 2015, 1:00:05 PM8/22/15
to
On Monday, June 30, 2008 at 8:58:19 PM UTC-4, Jay R. Yablon wrote:
> "FrediFizzx" <fredi...@hotmail.com> wrote in message
> news:6csk1lF...@mid.individual.net...
> . . .
> >
> > 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,
>
> 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.

Hi Jay

>From :

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

You said:

In the third line we 1 2 b = , and also b = - b which flips the sign in the b term. In the final
line, we use 1 2 = and p p = ( + + ) + + = p 2 2 2 2 2 2 2 / / x y

This did not copy and paste properly so I will translate

we use a^2=1 and p^2/p = (px^2 + py^2 + pz^2)/(px^2 + py^2 + pz^2)^.5 = p

Allow me to sprinkle a little Koide magic on this.

(px + py + pz )/( px^2 + py^2 + pz^2 )^.5 = p

We know ahead of time if p = 2/3 then x y z will render the lepton line.

Is there a way to hammer a round peg into a square hole to use the Koide magic?

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