Relation of g^uv = (1/2) {gamma^u,gamma^v} to gravitational fields

[Jay R. Yablon]

Dear friends,

The anomalous magnetic moments of the charged leptons are presently
understood in terms of perturbative corrections Lambda^u of the form
gamma^u --> Gamma ^u = gamma^u + Lambda^u introduced via the Dirac gamma
matrices gamma^u. (Gamma and Lambda being uppercase Greek symbols).

Because the metric tensor g^uv is related to these matrices by g^uv =
(1/2) {gamma^u,gamma^v}, so that a change in either of gamma^u or g^uv
is necessarily a change in the other, is there any reason why this
perturbative connection Lambda^u cannot be alternatively thought of
giving rise to a gravitational field?

Thank you.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm

[Igor Khavkine]

Jay R. Yablon wrote:
> Dear friends,
>
> The anomalous magnetic moments of the charged leptons are presently
> understood in terms of perturbative corrections Lambda^u of the form
> gamma^u --> Gamma ^u = gamma^u + Lambda^u introduced via the Dirac gamma
> matrices gamma^u. (Gamma and Lambda being uppercase Greek symbols).
>
> Because the metric tensor g^uv is related to these matrices by g^uv =
> (1/2) {gamma^u,gamma^v}, so that a change in either of gamma^u or g^uv
> is necessarily a change in the other, is there any reason why this
> perturbative connection Lambda^u cannot be alternatively thought of
> giving rise to a gravitational field?

Given a metric, we can construct the Clifford algebra generators (aka
Dirac gamma matrices). If you have a Clifford algebra whose generators
are labeled with space-time indices, then you can also reconstruct the
metric. That part you have correctly.

But why do you think that the anticommutators of the corrected
vertex factors Gamma^u will have anticommutators proportional to the
Clifford identity? That's a necessary condition for them to generate a
Clifford algebra.

Also, gravity is more than just a metric tensor. It is the metric
tensor coupled to matter in a specific way. For instance, in a
Lagrangian density with some scalar and and spinor fields, it enters as
follows (G - metric, @ - partial derivative, g - gamma matrix, A -
vector potential):

magnetic moment correction --+
v
L = G @phi @phi + psi g @psi + psi (g + lambda) A psi + ...
^ ^ ^
|__ gravity |__ gravity |__ gravity

When writing down the effective action (which is the only place where
the corrected vertex factor you refer to actually makes an appearance),
each coefficient shown above changes in a particular way compared to
its bare value. The G, g (in the psi-psi term), and g (in the
psi-A-psi) term will all change according to different rules. There is
no reason to expect them to conspire to change such that G stays equal
to the anticommutators of the g+lambda components.

Igor

[Jay R. Yablon]

> Given a metric, we can construct the Clifford algebra generators (aka
> Dirac gamma matrices). If you have a Clifford algebra whose generators
> are labeled with space-time indices, then you can also reconstruct the
> metric. That part you have correctly.
>
> But why do you think that the anticommutators of the corrected
> vertex factors Gamma^u will have anticommutators proportional to the
> Clifford identity? That's a necessary condition for them to generate a
> Clifford algebra.
>. . .
> Igor
>
Igor,

I already replied separately, and I don't know in what order my replies
will post. This post should be viewed second, after the prior reply.

Let me give a specific example of what I have in mind, which is in a
two-page pdf file linked below.

http://home.nycap.rr.com/jry/Papers/Schwarzschild-Clifford.pdf

In this file, I start with a metric, and to make it concrete, I start
with Schwarzschild. I construct the Clifford algebra generators. They
do anticummute proportional to the Clifford identity, and so meet the
"necessary condition." Then I use them in Dirac's eqaution and carry
all the way through to the gyromagnetic ratio, just to see what the end
result looks like.

Setting aside interpretation of the end result for the moment, I want to
know if there is anything wrong with carrying out this calculation in
the manner I have done.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web site: http://home.nycap.rr.com/jry/FermionMass.htm

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:1158091928.514623.68050@h48g2000cwc.googlegro ups.com...

> But why do you think that the anticommutators of the corrected
> vertex factors Gamma^u will have anticommutators proportional to the
> Clifford identity? That's a necessary condition for them to generate a
> Clifford algebra.

Is there any reason why they will *not* have anticommutators
proportional to the Clifford identity? Or, are you just saying that
there is no particular reason why they *should*, nor any reason why they
should not? I will ask this same question more precisely, below, in
response to the Lagrangian density you have laid out.

> Also, gravity is more than just a metric tensor. It is the metric
> tensor coupled to matter in a specific way. For instance, in a
> Lagrangian density with some scalar and and spinor fields, it enters
> as
> follows (G - metric, @ - partial derivative, g - gamma matrix, A -
> vector potential):
>
> magnetic moment correction --+
> v
> L = G @phi @phi + psi g @psi + psi (g + lambda) A psi + ...
> ^ ^ ^
> |__ gravity |__ gravity |__ gravity
>
> When writing down the effective action (which is the only place where
> the corrected vertex factor you refer to actually makes an
> appearance),
> each coefficient shown above changes in a particular way compared to
> its bare value. The G, g (in the psi-psi term), and g (in the
> psi-A-psi) term will all change according to different rules. There is
> no reason to expect them to conspire to change such that G stays equal
> to the anticommutators of the g+lambda components.
>
> Igor
>
OK, your Langrangian density helps to frame the question:

It drives me crazy to try to read equations with ^ _ @ -- etc. So, at
the link below is what I take to be your Lagrangian density written as
an easy-to-read equation:

http://home.nycap.rr.com/jry/Papers/Equation%201.jpg (1)

Now, my question is this: Start in flat spacetime with a Lagrangian
given by:

http://home.nycap.rr.com/jry/Papers/Equation%202.jpg (2)

Then introduce a perturbatively-corrected vertex factor Lambda^u so that
gamma^u --> gamma^u + Lambda^u. Is there any reason why the corrected
Lagrangian ought *not* look like the following?

http://home.nycap.rr.com/jry/Papers/Equation%203.jpg (3)

Or, are you just saying there is no reason why it should look like this
any more than it should not look like this, AND that the question of
whether the anticommutators of the corrected vertex factors Gamma^u will
have anticommutators proportional to the Clifford identity is what will
determine whether or not one ends up with a Lagrangian that looks like
(3)?

Thank you.

Jay.

[Igor Khavkine]

Jay R. Yablon wrote:
> "Igor Khavkine" <igor.kh@gmail.com> wrote in message
> news:1158091928.514623.68050@h48g2000cwc.googlegro ups.com...
>
> > But why do you think that the anticommutators of the corrected
> > vertex factors Gamma^u will have anticommutators proportional to the
> > Clifford identity? That's a necessary condition for them to generate a
> > Clifford algebra.
>
> Is there any reason why they will *not* have anticommutators
> proportional to the Clifford identity? Or, are you just saying that
> there is no particular reason why they *should*, nor any reason why they
> should not?

Is there any reason to believe that there *aren't* any positive
integers a,b,c that satisfy a^n + b^n = c^n, for n>2? If you are not
familiar with Fermat's Last Theorem and its history, you may think:
"Hey, this is a simple equation, there's got to be some solutions!"
But, as the celebrated proof of Fermat's theorem has shown, there exist
no such a,b,c. The fact is that the above equation is a restrictive
condition on triples of positive integers. Unless you have reason to
believe otherwise, you should never simply assume that any restrictive
condition always has solutions. So, instead of looking for reasons why
such a condition *shouldn't* hold, you should look for reasons why it
*should*.

Getting back on topic, if A and B are generic elements of a Clifford
algebra then their anticommutator {A,B} will be another generic element
of the algebra. And since the subspace of elements proportional to the
identity is only one-dimensional in a 2^n dimensional algebra (n -
space-time dimension), it is very unlikely that {A,B} will be contained
in it.

> > Also, gravity is more than just a metric tensor. It is the metric
> > tensor coupled to matter in a specific way. For instance, in a
> > Lagrangian density with some scalar and and spinor fields, it enters
> > as
> > follows (G - metric, @ - partial derivative, g - gamma matrix, A -
> > vector potential):
> >
> > magnetic moment correction --+
> > v
> > L = G @phi @phi + psi g @psi + psi (g + lambda) A psi + ...
> > ^ ^ ^
> > |__ gravity |__ gravity |__ gravity
> >
> > When writing down the effective action (which is the only place
> > where the corrected vertex factor you refer to actually makes an
> > appearance), each coefficient shown above changes in a particular
> > way compared to its bare value. The G, g (in the psi-psi term), and
> > g (in the psi-A-psi) term will all change according to different
> > rules. There is no reason to expect them to conspire to change such
> > that G stays equal to the anticommutators of the g+lambda
> > components.
> >
> > Igor
> >
> OK, your Langrangian density helps to frame the question:
>
> It drives me crazy to try to read equations with ^ _ @ -- etc.

That's unfortunate, because there is at least one benefit of the ASCII
only nature of Usenet. In case of mathematical notation, it forces you
to distill equations and formulas to their absolute essence to make
them readable. The latter should be the goal of any notation.

> [...]

The answer to the your new question is already present in my previous
reply. To help you decode it, I will place emphasis on the key words
that may have obscured its meaning.

> > ... [different terms of] the effective action ... will all change
> > according to different rules ... compared to [their] bare values.

So, what is an "effective action"? And how do the terms in the
effective action change compared to their bare value?

Finally, the actual answer to your question is negative. Again,
*generically* there is no reason for any of the coefficients of the
effective action to change in unison with the coefficient of
fermion-photon coupling. In fact, the same kind of coefficient will not
change in the same way between different fermions, as evidenced by
say different magnetic moments of different elementary particles.

Igor

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:1158423777.901910.251070@d34g2000cwd.googlegr oups.com...
.. . .
>> > Also, gravity is more than just a metric tensor. It is the metric
>> > tensor coupled to matter in a specific way. For instance, in a
>> > Lagrangian density with some scalar and and spinor fields, it
>> > enters
>> > as
>> > follows (G - metric, @ - partial derivative, g - gamma matrix, A -
>> > vector potential):
>> >
>> > magnetic moment correction --+
>> > v
>> > L = G @phi @phi + psi g @psi + psi (g + lambda) A psi + ...
>> > ^ ^ ^
>> > |__ gravity |__ gravity |__ gravity
>> >
>> > When writing down the effective action (which is the only place
>> > where the corrected vertex factor you refer to actually makes an
>> > appearance), each coefficient shown above changes in a particular
>> > way compared to its bare value. The G, g (in the psi-psi term), and
>> > g (in the psi-A-psi) term will all change according to different
>> > rules. There is no reason to expect them to conspire to change such
>> > that G stays equal to the anticommutators of the g+lambda
>> > components.
>> >
>> > Igor
>> >
.. . .
> The answer to the your new question is already present in my previous
> reply. To help you decode it, I will place emphasis on the key words
> that may have obscured its meaning.
>
>> > ... [different terms of] the effective action ... will all change
>> > according to different rules ... compared to [their] bare values.
>
> So, what is an "effective action"? And how do the terms in the
> effective action change compared to their bare value?
>
> Finally, the actual answer to your question is negative. Again,
> *generically* there is no reason for any of the coefficients of the
> effective action to change in unison with the coefficient of
> fermion-photon coupling. In fact, the same kind of coefficient will
> not
> change in the same way between different fermions, as evidenced by
> say different magnetic moments of different elementary particles.
>
> Igor
>

Thank you again for your reply Igor. I also want to take a moment to
thank you for what I have come to see as a real service to people who
are doing physics study and research in providing the feedback and
critiques that you do. For those of us who try as best as we can to
keep our ears open and understand your input, it is very helpful. And
since it is probably thankless for you most of the time, I do want to
again say thank you.

Before proceeding further, I want to make certain I am absolutely clear
about what you are saying:

I understand you to say that just because the "effective action" based
on "psi (g + lambda) A psi" for the "fermion-photon coupling" is
perturbatively corrected from its "bare value" "psi g A psi," there is
no reason to expect that the "bare" term "psi g @psi" will similarly
become "corrected" to "psi (g + lambda) @psi," or that the "bare" metric
tensor G=.5{g,g} will become "corrected" to G=.5{g + lambda,g +
lambda}. Is this an accurate restatement?

I also understand you to say in your final sentence that the "different
magnetic moments of different elementary particles" are understood
*because* one does *not* change all of these coefficients in unison and
that were these to change in unison, we would not be able to explain why
different elementary particles have different magnetic moments. In sum,
you seem to me to say that the "different magnetic moments of different
elementary particles" are evidence that these coefficients do not /
cannot change in unison, because if they did change in unison, then
these fermions would all have the *same* magnetic moment. Is that
accurate? (I will note that in a separate post, I did ask about how the
differences among the electron, mu and tau lepton magnetic moments are
presently accounted for, but nobody has replied to that. I am still
interested in hearing about this, because my impression, studying Ryder,
is that this is a problem not fully-understood at present and I would
like to know what the "best" understanding is.)

Finally, when you refer to "different magnetic moments of different
elementary particles," I take it that you are thinking, in particular,
of the electron, mu and tau leptons, and the fifth-digit variations in
magnetic moments among these three particles. Is that also accurate?

Thanks,

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web site: http://home.nycap.rr.com/jry/FermionMass.htm

[Ken S. Tucker]

Igor Khavkine wrote:
...
> So, what is an "effective action"? And how do the terms in the
> effective action change compared to their bare value?
>
> Finally, the actual answer to your question is negative. Again,
> *generically* there is no reason for any of the coefficients of the
> effective action to change in unison with the coefficient of
> fermion-photon coupling. In fact, the same kind of coefficient will not
> change in the same way between different fermions, as evidenced by
> say different magnetic moments of different elementary particles.
> Igor

Evidentally Igor is forgetting the non-linearity of
the equations, specifically "positive feed-back",
and "negative feed-back", well known in basic
engineering.
To Igor's question, the measurement is constrained
by the harmonics, as defined 1st by de Broglie,
wherein all energy is a frequency.
Regards
Ken S. Tucker

[Igor Khavkine]

Jay R. Yablon wrote:

> Thank you again for your reply Igor. I also want to take a moment to
> thank you for what I have come to see as a real service to people who
> are doing physics study and research in providing the feedback and
> critiques that you do. For those of us who try as best as we can to
> keep our ears open and understand your input, it is very helpful. And
> since it is probably thankless for you most of the time, I do want to
> again say thank you.

Appreciation is always welcome. :-)

> Before proceeding further, I want to make certain I am absolutely clear
> about what you are saying:
>
> I understand you to say that just because the "effective action" based
> on "psi (g + lambda) A psi" for the "fermion-photon coupling" is
> perturbatively corrected from its "bare value" "psi g A psi," there is
> no reason to expect that the "bare" term "psi g @psi" will similarly
> become "corrected" to "psi (g + lambda) @psi," or that the "bare" metric
> tensor G=.5{g,g} will become "corrected" to G=.5{g + lambda,g +
> lambda}. Is this an accurate restatement?

Yes.

> I also understand you to say in your final sentence that the "different
> magnetic moments of different elementary particles" are understood
> *because* one does *not* change all of these coefficients in unison and
> that were these to change in unison, we would not be able to explain why
> different elementary particles have different magnetic moments. In sum,
> you seem to me to say that the "different magnetic moments of different
> elementary particles" are evidence that these coefficients do not /
> cannot change in unison, because if they did change in unison, then
> these fermions would all have the *same* magnetic moment. Is that
> accurate? (I will note that in a separate post, I did ask about how the
> differences among the electron, mu and tau lepton magnetic moments are
> presently accounted for, but nobody has replied to that. I am still
> interested in hearing about this, because my impression, studying Ryder,
> is that this is a problem not fully-understood at present and I would
> like to know what the "best" understanding is.)

I'm not sure why you get the impression from Ryder that magnetic
moments are not well understood. Do you have a reference from the text
that you can cite?

In my understanding, there are two things that have to be understood in
order to theoretically calculate the magnetic moment of any particle,
one is kinematical and the other dynamical. First, there is the
definition of the magnetic moment in terms of the fermion-photon
coupling verted. I believe it is defined as one of the terms in a
vectorial decomposition of the vertex in the limit of small incoming
momenta. I don't remember the details off hand, so I can't reproduce
them here. But this definition is standard and well agreed upon.
Second, there is the issue of calculating the actual fermion-photon
vertex factor. This is done by evaluating and summing all Feynman
graphs with one external photon line and two external fermion lines.
This is also well understood, and gives well defined finite answers at
each order in perturbation theory after renormalization. The main
difficulty is in the large number of graphs that need to be taken into
account once we go to higher perturbative orders. I'm not aware of any
other difficulties in this calculation.

> Finally, when you refer to "different magnetic moments of different
> elementary particles," I take it that you are thinking, in particular,
> of the electron, mu and tau leptons, and the fifth-digit variations in
> magnetic moments among these three particles. Is that also accurate?

Yes, QFT corrections come into the eps term in the magnetic moment mu =
q/2m (1 + eps), where q is the particle charge and m is the particle
mass. Here neglecting the differences between q and m for different
particles, the eps term will be different for each particle since it
will correspond to a summation of different Feynman graphs in each
case.

Hope this helps.

Igor

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:1158091928.514623.68050@h48g2000cwc.googlegro ups.com...
> Jay R. Yablon wrote:
>> Dear friends,
>>
>> The anomalous magnetic moments of the charged leptons are presently
>> understood in terms of perturbative corrections Lambda^u of the form
>> gamma^u --> Gamma ^u = gamma^u + Lambda^u introduced via the Dirac
>> gamma
>> matrices gamma^u. (Gamma and Lambda being uppercase Greek symbols).
>>
>> Because the metric tensor g^uv is related to these matrices by g^uv =
>> (1/2) {gamma^u,gamma^v}, so that a change in either of gamma^u or
>> g^uv
>> is necessarily a change in the other, is there any reason why this
>> perturbative connection Lambda^u cannot be alternatively thought of
>> giving rise to a gravitational field?
>
> Given a metric, we can construct the Clifford algebra generators (aka
> Dirac gamma matrices). If you have a Clifford algebra whose generators
> are labeled with space-time indices, then you can also reconstruct the
> metric. That part you have correctly.
>
> But why do you think that the anticommutators of the corrected
> vertex factors Gamma^u will have anticommutators proportional to the
> Clifford identity? That's a necessary condition for them to generate a
> Clifford algebra.
>
> Also, gravity is more than just a metric tensor. . . .

Igor,

I have focused on the latter part of this discussion in some replies,
but I want to make sure I also understand exactly what you are saying
when you say "why do you think that the anticommutators of the corrected
vertex factors Gamma^u will have anticommutators proportional to the
Clifford identity?"

I have been studying the Schwinger calculation carefully and closely.
If sigma_uv denotes the object constructed from antisymmetric [gamma,
gamma] combinations, and p, p' designate the ingoing and outgoing
electron four-momenta, and alpha is the EM coupling, then it looks to me
like the form of the Schwinger "correction" Lambda is:

Lambda = (alpha/2pi)(i sigma (p'-p)/2m) (1)

or, alternatively, decomposed:

Lambda = (alpha/2pi)gamma - (alpha/2pi)((p'-p)/m) (2)

In both, the full vertex factor is:

Gamma ^u = gamma^u + Lambda^u (3)

Looking specifically at (2), when you say "anticommutators proportional
to the Clifford identity?" do you mean that although the term
(alpha/2pi)gamma in (2) anticommutes proportionally to the Dirac
Clifford algebra generators gamma, the second term (alpha/2pi)((p'-p)/m)
does NOT? And, is this the source of the issue you have identified
here?

Thanks,

Jay.

[mihai cartoaje]

Jay R. Yablon wrote:
> Dear friends,
>
> The anomalous magnetic moments of the charged leptons are presently
> understood in terms of perturbative corrections Lambda^u of the form
> gamma^u --> Gamma ^u = gamma^u + Lambda^u introduced via the Dirac gamma
> matrices gamma^u. (Gamma and Lambda being uppercase Greek symbols).
>
> Because the metric tensor g^uv is related to these matrices by g^uv =
> (1/2) {gamma^u,gamma^v}, so that a change in either of gamma^u or g^uv
> is necessarily a change in the other, is there any reason why this
> perturbative connection Lambda^u cannot be alternatively thought of
> giving rise to a gravitational field?

If we suppose that gamma'^j are linear combinations of Dirac gamma
matrices, then the equation,

{gamma'^j, gamma'^k} = 2 g^jk

has 10 equations for 16 unknowns. If you are to use the gamma' in a
theory, you have to verify that the physics are independent of an
arbitrary choice of the 6 remaining unknowns.

If you search this group, you shall find a candidate generalisation of
the Dirac equation. I ignore if everything works well with it.

[Igor Khavkine]

Jay R. Yablon wrote:

> I have focused on the latter part of this discussion in some replies,
> but I want to make sure I also understand exactly what you are saying
> when you say "why do you think that the anticommutators of the corrected
> vertex factors Gamma^u will have anticommutators proportional to the
> Clifford identity?"
>
> I have been studying the Schwinger calculation carefully and closely.
> If sigma_uv denotes the object constructed from antisymmetric [gamma,
> gamma] combinations, and p, p' designate the ingoing and outgoing
> electron four-momenta, and alpha is the EM coupling, then it looks to me
> like the form of the Schwinger "correction" Lambda is:
>
> Lambda = (alpha/2pi)(i sigma (p'-p)/2m) (1)
>
> or, alternatively, decomposed:
>
> Lambda = (alpha/2pi)gamma - (alpha/2pi)((p'-p)/m) (2)
>
> In both, the full vertex factor is:
>
> Gamma ^u = gamma^u + Lambda^u (3)
>
> Looking specifically at (2), when you say "anticommutators proportional
> to the Clifford identity?" do you mean that although the term
> (alpha/2pi)gamma in (2) anticommutes proportionally to the Dirac
> Clifford algebra generators gamma, the second term (alpha/2pi)((p'-p)/m)
> does NOT? And, is this the source of the issue you have identified
> here?

First, I don't understand how you got your equations (1) and (2). Using
Weinberg's QFT v.1 book as a reference, I'm looking at equation
(11.3.29):

Gamma^u = gamma^u F_1 + i/2 sigma^uv (p'-p)_v F_2,

where F_1 and F_2 are scalar structure factors that depend only on the
square of the momentum transfer. Also F_1 ~ 1 + O(alpha) and F_2 ~
O(alpha). Some combination of these O(alpha) terms gives the correction
to the magnetic moment. However, this equation looks neither like your
(1) nor (2).

And instead of wondering whether the anticommutators of Gamma^u are
proportional to the identity or not, why don't you just check. Have you
tried to compute the anticommutators? The Clifford algebra is
16-dimensional, with a basis given by the identity, gamma^u, products
of two, three, and four gamma matrices. So, every algebra element has a
canonical form when expressed in this basis. Calculate the
anticommutator and express it in this basis, then you'll know whether
it's proportional to the identity or not.

Igor

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:1158508200.365538.52290@k70g2000cwa.googlegro ups.com...
.. . .
>> (I will note that in a separate post, I did ask about how the
>> differences among the electron, mu and tau lepton magnetic moments
>> are
>> presently accounted for, but nobody has replied to that. I am still
>> interested in hearing about this, because my impression, studying
>> Ryder,
>> is that this is a problem not fully-understood at present and I would
>> like to know what the "best" understanding is.)
>
> I'm not sure why you get the impression from Ryder that magnetic
> moments are not well understood. Do you have a reference from the text
> that you can cite?
>
Page 345: "for the muon, agreement between experiment and the purely
electromagentic calculation is not so good, but beither is it expected
to be, since there are contricutions from hadrons, W and X bosons, and
the Higgs boson."

Let me modify what I said before: Why is it that we would have this
problem with the muon but not the electron? The electron also interacts
with "hadrons, W and X bosons, and the Higgs boson." No differently
than the electron. Why would we be able to obtain good agreement for
the electron but not the muon? What is considered to make the muon
different from the electron (or those in turn from the tauon) in terms
of which Feynman graphs are summed?
.. . .
>> Finally, when you refer to "different magnetic moments of different
>> elementary particles," I take it that you are thinking, in
>> particular,
>> of the electron, mu and tau leptons, and the fifth-digit variations
>> in
>> magnetic moments among these three particles. Is that also accurate?
>
> Yes, QFT corrections come into the eps term in the magnetic moment mu
> =
> q/2m (1 + eps), where q is the particle charge and m is the particle
> mass. Here neglecting the differences between q and m for different
> particles, the eps term will be different for each particle since it
> will correspond to a summation of different Feynman graphs in each
> case.

Again, my question: What is considered to make the muon different from
the electron (or those in turn from the tauon) in terms of which Feynman
graphs are summed? Don't they all have identical interactions with at
least the hadrons and W and X bosons (if not the Higgs boson where the
coupling is proportional to mass). The only difference in these
particles as far as we know is their mass. Is the mass thought to be
what is responsible for there being different graphs so that they all
have slightly different magnetic moments?

Jay

[sr]

Jay Yablon asked:

> I did ask about how the differences among the
> electron, mu and tau lepton magnetic moments are
> presently accounted for, but nobody has replied
> to that.

If you look on the Particle Data Group website
(pdg.lbl.gov) there's a page of "reviews, tables and
plots", and one of the more recent PDF documents
available there reviews the status of the muon's
anomalous magnetic moment. [That website is a
good source whenever one is wondering about
the "current best understanding" of various
topics in particle physics.]

> I am still interested in hearing about
> this, because my impression, studying Ryder,
> is that this is a problem not fully-understood
> at present and I would like to know what the
> "best" understanding is.

The muon's anomalous magnetic moment has been
calculated very accurately (to 4-loops, plus an estimate
for 5-loops). The calculation involves not just photon
contributions, but also electroweak and hadronic
diagrams. I.e: one must use the full Standard Model,
not merely QED. The higher order terms depend on the
muon mass, hence we get a different answer compared to
the electron.

The calculated value agrees extremely well with experiment.
So there's simply no need to bring gravitation into it.
The Standard Model does a very good job of predicting
the observed value.

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:1158728994.691348.273450@m73g2000cwd.googlegr oups.com...

>> I have been studying the Schwinger calculation carefully and closely.
>> If sigma_uv denotes the object constructed from antisymmetric
>> [gamma,
>> gamma] combinations, and p, p' designate the ingoing and outgoing
>> electron four-momenta, and alpha is the EM coupling, then it looks to
>> me
>> like the form of the Schwinger "correction" Lambda is:
>>
>> Lambda = (alpha/2pi)(i sigma (p'-p)/2m) (1)
>>
>> or, alternatively, decomposed:
>>
>> Lambda = (alpha/2pi)gamma - (alpha/2pi)((p'-p)/2m) (2)
>>
>> In both, the full vertex factor is:
>>
>> Gamma ^u = gamma^u + Lambda^u (3)
>>
>> Looking specifically at (2), when you say "anticommutators
>> proportional
>> to the Clifford identity?" do you mean that although the term
>> (alpha/2pi)gamma in (2) anticommutes proportionally to the Dirac
>> Clifford algebra generators gamma, the second term
>> (alpha/2pi)((p'-p)/2m)
>> does NOT? And, is this the source of the issue you have identified
>> here?
>
> First, I don't understand how you got your equations (1) and (2).
> Using
> Weinberg's QFT v.1 book as a reference, I'm looking at equation
> (11.3.29):
>
> Gamma^u = gamma^u F_1 + i/2 sigma^uv (p'-p)_v F_2,
>
> where F_1 and F_2 are scalar structure factors that depend only on the
> square of the momentum transfer. Also F_1 ~ 1 + O(alpha) and F_2 ~
> O(alpha). Some combination of these O(alpha) terms gives the
> correction
> to the magnetic moment. However, this equation looks neither like your
> (1) nor (2).

I checked Weinberg (11.3.29). My (1) and (2) above are special cases of
(11.3.29), where F_1 and F_2 are based on the EM Schwinger correction,
only. (See and contrast Ryder, eqs. (9.136) and (9.138).) I like
Weinberg (11.3.29) because it is perfectly general.

By the way, your term "i/2 sigma^uv (p'-p)_v" has an sic: it is either
"i/2 [gamma^u,gamma^v] (p'-p)_v" or "sigma^uv (p'-p)_v" but not "i/2
sigma^uv (p'-p)_v."

My main point is that Gamma^u = gamma^u F_1 + i/2 [gamma^u,gamma^v]
(p'-p)_v F_2, and that the anticommutation properties are determined by
gamma^u (which will commute with the identity) and sigma^uv (p'-p)_v =
i/2 [gamma^u,gamma^v] (p'-p)_v (which will not).

> And instead of wondering whether the anticommutators of Gamma^u are
> proportional to the identity or not, why don't you just check. Have
> you
> tried to compute the anticommutators?

Yes, I have, done a preliminary calculation. Using only the Schwinger
correction, the anticommutators {Gamma,Gamma}, to first order in alpha,
turn out to be equal to the Minkowski metric n_uv = {gamma,gamma}. The
terms containing sigma^uv (p'-p)_v which are not proportional to gamma^u
cancel one another precisely, again, to first order in alpha. I am
wondering what the meaning of this might be. Second order in alpha
*does* leave some residual terms not proportional to the gamma^u. Just
thinking out loud, I am wondering whether, perhaps, one might want to
impose a *condition* that these cancellations occur at all orders in
alpha as they do in first order using the Schwinger correction, which
might then drive what the terms look like at each order, and give us a
canonical way of generating higher order terms which will reproduce what
we get from calculating loop diagrams.

Jay.

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:1158728994.691348.273450@m73g2000cwd.googlegr oups.com...
> And instead of wondering whether the anticommutators of Gamma^u are
> proportional to the identity or not, why don't you just check. Have
> you
> tried to compute the anticommutators? The Clifford algebra is
> 16-dimensional, with a basis given by the identity, gamma^u, products
> of two, three, and four gamma matrices. So, every algebra element has
> a
> canonical form when expressed in this basis. Calculate the
> anticommutator and express it in this basis, then you'll know whether
> it's proportional to the identity or not.
>
> Igor
>

Dear Igor,

I just did the full calculation you suggested, using (11.3.29) from
Weinberg. It is linked in the following three-page pdf file.

http://home.nycap.rr.com/jry/Papers/Anticommutator%20Calculation.pdf

I welcome your comments (or those of anyone else) on this calculation.

Very truly yours,

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm

[Jay R. Yablon]

>
> By the way, your term "i/2 sigma^uv (p'-p)_v" has an sic: it is either
> "i/2 [gamma^u,gamma^v] (p'-p)_v" or "sigma^uv (p'-p)_v" but not "i/2
> sigma^uv (p'-p)_v."

Actually, Igor, I need to correct the above; your original post was are
correct; I misread the meaning of Weinberg's notation for the sigma^uv.
The calculation I posted at
http://home.nycap.rr.com/jry/Papers/Anticommutator%20Calculation.pdf
has a factor that is off by -1/8, that is, all the F_2^2 terms need to
be multiplied by -1/8 to be correct. Corrected repost will follow in
due course.

Jay.