Integration by Parts of the Maxwell QED Action
Jay R. Yablon
I plan in the near future to pick up and simplify my work on Yang-Mills
theory where I left off in December 2008.
The ability to integrate the non-linear Yang-Mills action by parts, in
an exact fashion, is likely an indispensible step in being able to
ultimately do exact calculations in QCD without the compromises of
either perturbation theory or lattice gauge theory.
Linked below is a < three page review of how the pure, linear Maxwell
field is integrated by parts in QED:
http://jayryablon.files.wordpress.com/2009/09/integration-by-parts-of-the-maxwell-qed-action.pdf
This provides a useful "template" for the parallel, but much more
difficult, Yang-Mills integration by parts.
I would appreciate for now, if you can review this document, and advise
is anything is wrong or if there are points that should be made
differently or better clarified.
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Jay R. Yablon
posted a very good reply to my original post in this thread. If you are
at all interested, I urge you to look at his reply.
After studying Mark's reply, I posted a reply to sci.physics.research,
which ought to go up in the next day or two (you know how spr can
sometimes get a few days behind the posts), which is as follows:
Dear Mark,
Thank you for your reply of September 30. This was, clear, informative
and enjoyable to study. ;-)
You note at the outset that "(1) the exercise has nothing per se to do
with quantum theory. This is general and applies to field theory,
itself, irrespective of whether it's classical or quantum; (2) it can be
done without the need for any explicit expression for the Lagrangian."
I agree with you on (1), and I find (2) quite nifty.
In the filed linked below, below, I have retraced what you wrote, and at
the same time, added some additional information and raised several
questions to which I hope you and others will be able to reply.
http://jayryablon.files.wordpress.com/2009/10/reply-to-mhopkins.pdf
I have also in this link, updated the original post with which I started
the thread, to accommodate what I (hope I) learned from your reply.
Looking forward to further discussion.
Thanks,
Jay
Jay R. Yablon
Mark Hopkins over at sci.physics.research. Jay.
"Rock Brentwood" <mark...@yahoo.com> wrote in message
news:f7520c72-99fb-4856...@r31g2000vbi.googlegroups.com...
> On Oct 3, 3:06 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> I am studying your reply and hope to have more to say later, but can
>> you
>> please briefly comment on what, if anything, changes in your
>> development
>> for non-Abelian gauge theory.
>
> Actually, I worked out the exercise you're doing in general for gauge
> theory and I'll summarise the analysis and results below.
Hi Mark,
I have a few comments inline, and will have more to follow after I study
the details of your posting.
> The short answer is take the field components and add in an extra
> index for the Lie algebra. Then F_{mn} becomes F^a_{mn}. Note that
> this requires a bi-linear form k_{ab} to define quadratic combinations
> of the field components. This, in fact, generalizes the coefficient
> that appears in the Maxwell-Lorentz Lagrangian density:
> L = (epsilon c) root(-g) g^{mr} g^{ns} F_{mn} F_{rs}.
> That is, (epsilon c) generalizes to k_{ab} and the Maxwell-Lorentz
> Lagrangian generalizes directly to
> L = k_{ab} root(-g) g^{mr} g^{ns} F^a_{mn} F^b_{rs}.
>
> The field G^{mn} when defined as the derivative -dL/dF_{mn} for the
> Lagrangian above is
> G^{mn} = epsilon c root(-g) g^{mr} g^{ns} F_{rs} = epsilon c root(-
> g) F^{mn}
> which is a tensor density.
>
> Hence, one derives the *constitutive law* which relates F and G --
> this particular law being none other than the Lorentz relations:
> D = epsilon E; B = mu H.
>
> This generalizes to the gauge field G^{mn}_a. The components F_{mn}
> yield (E,B), while G^{mn} yield (D,H). For gauge theory extra indices
> appear as (E^a, B^a), (D_a, H_a). The Lorentz relations generalize to:
> "Lorentz-Yang-Mills Relations"
> D_a = k_{ab}/c E^b; B^a = k^{ab}/c H_b
> where k^{ab} is the inverse of the bi-linear form k_{ab} (so we have
> to assume there is an inverse).
>
> When the Maxwell-Lorentz Lagrangian is generalized in this way, it
> becomes the Yang-Mills Lagrangian. Hence, there is the equation
> (Maxwell-Lorentz : Yang-Mills)
> = (Electromagnetism : Non-Abelian Gauge Field)
> = (Lorentz Relations : Lorentz-Yang-Mills Relations)
>
> The integration by parts is more involved since F^a_{mn} becomes a
> non-
> linear expression in the potentials A^a_m (which also have an extra
> Lie index a).
IMHO, you can bypass (or, more to the point, simplify) this added
complexity, if you employ a little "trick" that the Abelian field
strength tensor (&=partial derivative):
F_[uv] = &_[u A_v] = &_u A_v - &_v A_u (1)
is supplanted in non-Abelian theory with a field strength of exactly the
same form:
F_[uv] = D_[u G_v] = D_u G_v - D_v G_u (2)
where D_u = &_u - iG_u is the gauge-covariant derivative (scaled to
include the group charge strength g), and where the Yang-Mills gauge
field G_v is understood to be represented as an NxN matrix for SU(N),
whereby the extra index for the Lie algebra can be suppressed during
calculation. See also (1.1) of my reply just posted in this thread
earlier today, at:
http://jayryablon.files.wordpress.com/2009/10/reply-to-mhopkins.pdf
I have also previously worked out this exercise for gauge theory in
general, and while I will study your reply, I believe my approach is
much simpler because it maintains the clear analog to QED which is
represented by contrasting (1) and (2).
Specifically, the question that one might raise in comparing (1) and
(2), is whether this indicates a general Heuristic rule that you can
simply substitute:
&_u --> D_u = &_u - iG_u (3)
in passing over from Abelian to non-Abelian gauge theory (and moving the
four-vector fields A_u up to NxN matrices of four-vectors G_u). In
fact, this heuristic rule DOES survive the full, classical, Yang-Mills
integration by parts, as I have shown in sections 4, 5 and 6 of
http://jayryablon.files.wordpress.com/2008/12/yang-mills-paper-20.pdf,
culminating in equation (6.4).
Specifically, I find in (6.4), after integration by parts, that for
Yang-Mills theory: ($=integral):
-.5Tr(F^uvF_uv)= Tr $ G_u (g_uv D_s D_s - D^v D^u) G_v d^4x (4)
Aside from the trace and the factor of 2 that this introduces, this is
identical in form to the QED expression:
-.25(F^uvF_uv)= Tr $ A_u (g_uv &_s &_s - &^v &^u) A_v d^4x (5)
where one merely needs to substitute (3), heuristically, and introduce a
trace with a factor of 2.
I will also note that Hans deVries has not only validated the
correctness of my "trick" in (2) and (3) above, but has used this in
some of his own work, see, e.g., his equation (18) in
http://www.physics-quest.org/Non_Abelian_Lagrangian.pdf.
>
> The long answer is step back and assume nothing about the Lagrangian
> at the outset other than it conform to general symmetry principles --
> both in the case of electromagnetism and non-Abelian gauge field. The
> derivation of the field equations does not require specific knowledge
> of what the Lagrangian is. Only the constitutive laws require this
> knowledge -- and ultimately so does the Hamiltonian.
I do like your approach of working independently of a specific
expression for the Lagrangian, and so will study this closely.
> The analysis follows here:
>
> 1. The General Gauge Field Lagrangian.
> If you assume the Lagrangian density L is a gauge-invariant function
> of the field potentials and their derivatives then (by Utiyama's
> Theorem) it follows that it reduces to a function solely of the field
> strength. If there are extra fields involved then (by an extended
> version of Utiyama's Theorem) the dependence on the extra fields
> reduces to a dependence on the field components and their gauge-
> invariant derivatives. The latter is the only place an explicit
> dependence on the potential can appear in the Lagrangian.
>
> Considering only the case of a pure field Lagrangian, if you also
> assume it's Lorentz invariant then it reduces to a function of the
> Lorentz invariants of the field strength. In 4-D these invariants are:
> I^{ab} = -1/4 root(-g) g^{ms} g^{nr} F^a_{mn} F^b_{rs}
> J^{ab} = -1/8 epsilon^{mnrs} F^a_{mn} F^b_{rs}
> I^{abc} = 1/6 root(-g) g^{lm} g^{nr} g^{sk} F^a_{kl} F^b_{mn} F^c_
> {rs}
> I^{abc} = 1/12 epsilon^{lmnr} g^{sk} F^a_{kl} F^b_{mn} F^c_{rs}.
>
> Adopting the "Maxwell" nomenclature for the components
> phi = -A_0; A = (A_1, A_2, A_3)
> E = (F_{10}, F_{20}, F_{30}), B = (F_{23}, F_{31}, F_{12})
> these generalize to
> phi^a, A^a, E^a, B^a
> respectively. With the metric
> g = diag (c^2, -1, -1, -1)
> this yields
> I^{ab} = (1/2c) (E^a . E^b - c^2 B^a . B^b)
> J^{ab} = 1/2 (E^a . B^b + B^a . E^b)
> and more complex expressions for the cubic invariants, which you can
> work out.
>
> The conjugate fields are defined by
> D = (G^{01}, G^{02}, G^{03}), H = (G^{23}, G^{31}, G^{12})
> and source densities by
> rho = J^0, J = (J^1, J^2, J^3)
> and these generalize for non-Abelian fields to D_a, H_a, rho_a, J_a
> respectively; and are given by the general constitutive laws:
> D_a = dL/dE^a, H_a = -dL/dB^a
> rho_a = -dL/d(phi^a), J_a = dL/dA^a.
>
> The Lagrangian yields the following constitutive coefficients as its
> derivatives with respect to the Lorentz invariants:
> k_{ab} = dL/dI^{ab}
> theta_{ab} = dL/dJ^{ab}
> k_{abc} = dL/dI^{abc}
> theta_{abc} = dL/dJ^{abc}.
>
> This leads to the following relations:
> D_a = k_{ab}/c E^b + theta_{ab} B^b + (quadratic terms)
> H_a = c k_{ab} B^b - theta_{ab} E^b + (quadratic terms)
> The quadratic terms involve the coefficients k_{abc} and theta_{abc}.
> The coefficient theta_{ab} is parity non-symmetric and yield (up to
> proportion) the parameters of what is called the theta-vacuua.
>
> The variety of gauge fields most commonly studied (and frequently, but
> wrongly, taken to be synonymous with the term "gauge field") are the
> Yang-Mills fields. They can be defined by the following conditions:
>
> (0) The Lagrangian is analytic in the fields
> (1) The Lagrangian is homogeneous to the 2nd degree in the fields
> (i.e. no current associated with scale symmetry)
> (2) The coefficients k_{ab} define a positive-definite ADJOINT-
> INVARIANT metric for the gauge group.
>
> The requirements (0) and (1) get rid of the dependence on the cubic
> invariants and the coefficients k_{abc} and theta_{abc} drop out.
> Because of (0) the remaining coefficients k_{ab} and theta_{ab} become
> constants.
>
> The coefficients theta_{ab} can be eliminated then by redefining D and
> H by
> D_a <- D_a - theta_{ab} B^b
> H_a <- H_a + theta_{ab} E^b
> This does not affect the field equations...
>
> ... but it DOES affect the definition of the Hamiltonian! (and of the
> quantum theory that is derived canonically from it!!) ...
>
> For non-Abelian gauge fields, coupled to chiral fields, the different
> values of theta yield inequivalent state spaces and a set of vacuum
> states which possess the property that no two vacuum states have a
> coherent superposition with one another (and no two members of
> different state spaces can be coherently superposed with one another).
>
> The PERMEABILITY epsilon_{ab} and PERMITTIVITY mu^{ab} can be defined
> by
> epsilon_{ab} = k_{ab}/c
> mu^{ab} = k^{ab}/c
> where k^{ab} is the inverse of k_{ab}.
>
> For simple gauge groups, it is a general result that condition (2)
> implies the coefficients can be reduced to a multiple of the Killing
> metric. Adopting a suitable basis for the gauge group's Lie algebra,
> this leads to a metric of the form
> k_{ab} = delta_{ab}/g^2.
> The coefficient g is then deemed the "coupling coefficient" of the
> gauge field.
>
> Thus, for Maxwell', g^2 corresponds to mu c and 1/g^2 to epsilon c.
>
> Thus, the general form derived for the Lagrangian density is
> L = delta_{ab}/g^2 I^{ab}
> = -1/4 delta_{ab}/g^2 root(-g) g^{mr} g^{ns} F^a_{mn} F^b_{rs}.
>
> 2. Derivation of the Yang-Mills Lagrangian
> Integration by parts is best NOT done component-wise, but rather in
> the language of differential forms. So, the following will set this
> problem up.
I also prefer wherever possible to use differential forms. They are
especially transparent, when the time comes to integrate over boundaries
and apply Stokes' / Gauss' theorem
> For what's to follow, I'll denote the Lagrangian density by #L and use
> L to denote, instead, the Lagrangian 4-form:
> L = #L dt ^ dx ^ dy ^ dz.
>
> The field strengths can be captured by the 2-form:
> F^a = 1/2 F^a_{mn} dx^m ^ dx^n.
> In 3+1 form this reduces to
> F^a = B^a.dS + E^a.dr ^ dt
> where
> dr = (dx, dy, dz)
> dS = (dy ^ dz, dz ^ dx, dx ^ dy).
>
> The Hodge dual can be written as
> *F^a = 1/2 F^a_{mn} e^{mn}
> where
> e^{mn} = 1/2 root(-g) epsilon^{mn}_{rs} dx^r ^ dx^s
> with epsilon the Levi-Civita tensor normalized as
> epsilon_{0123} = 1
> and
> epsilon^{mn}_{rs} = g^{mt} g^{nu} epsilon_{turs}.
>
> Then the action integral becomes:
> S_U = integral_U (1/g^2) (-k_{ab} F^a ^ *F^b).
>
> This is usually cast in terms of a "trace" operator. This arises as
> follows.
>
> The general expression for the Killing metric is
> kappa_{ab} = f^d_{ac} f^c_{db}
> where the f's are the structure coefficients. For a Lie algebra basis
> {Y_a} they are given by
> [Y_a, Y_b] = f^c_{ab} Y_c.
>
> So, the expression for the coeffcients k_{ab} for a simple gauge group
> actually has the following general form
> k_{ab} = (1/g^2) f^d_{ac} f^c_{db}
> rather than delta_{ab}/g^2.
>
> In the "adjoint representation", the basis itself can be represented
> as a matrix with the components
> (Y_a)^c_b = f^c_{ab}.
> Then, the field adopts the following matrix form
> F^c_b = f^c_{ab} F^a
> and the Lagrangian reduces to
> S_U = integral_U (1/2g^2) F^d_c ^ *F^c_d
> or, since g is a constant here,
> S_U = 1/(2g^2) integral_U Trace(F ^ *F).
>
> 3. Integration by Parts via Differential Forms
> With the Lagrangian density #L replaced by the Lagrangian 4-form L,
> the expression for the variational of #L
>
> delta(#L) = -1/2 G^{mn} delta(F_{mn}) + J^m delta(A_m)
>
> can be rewritten as
>
> delta(L) = -delta(F) ^ G + delta(A) ^ J.
>
> This requires expressing G and J as differential forms. In 3-vector
> notation this would be given by:
> G = D_a.dS - H_a.dr ^ dt
> J = rho_a dV - J_a . dS ^ dt
> where dV = dx ^ dy ^ dz.
>
> Component-wise, this is:
> G = 1/2 G^{mn}_a e_{mn}
> J = J^m e_m
> where
> e_{mn} = 1/2 root(-g) epsilon_{mnrs} dx^r ^ dx^s
> e_m = 1/6 root(-g) epsilon_{mnrs} }dx^n ^ dx^r ^ dx^s.
>
> This generalizes to non-Abelian gauge fields to
> delta(L) = -delta(F_a) ^ G^a + delta(A_a) ^ J^a.
>
> To integrate by parts, you first need the expression for the field
> strength in terms of the potentials. The potentials may be written as
> the 1-forn which, in 3-vector form would be:
> A^a = A^a_i dx^i - phi^a dt.
>
> The field strength in terms of the potentials is
> F^a_{mn} = d_m A^a_n - d_n A^a_m + f^a_{bc} A^b_m A^c_n.
> When written as differential forms, this becomes
> F^a = 1/2 F^a_{mn} dx^m ^ dx^n
> = dA^a + 1/2 f^a_{bc} A^b ^ A^c
>
> where the matrix representation is used:
> A^a_c = f^a_{bc} A^b.
>
> Converting this fully to the matrix representation, using also the
> matrix form for the potentials
> A^c_b = f^c_{ab} A^a
> you get
> F^c_b = f^c_{ab} F^a = dA^c_b + 1/2 f^c_{ab} f^a_{de} A^d ^ A^e.
>
> The last term can be converted using the Jacobi identity
> f^c_{ab} f^a_{de} = -f^c_{ad} f^a_{eb} - f^c_{ae} f^a_{bd}
> to get
> F^c_b = dA^c_b + 1/2 A^c_a ^ A^a_b - 1/2 A^a_b ^ A^c_a.
>
> Supressing the matrix indices, this becomes
> F = dA + A ^ A.
>
> Thus, for the variational of the Lagrangian 4-form L, you get:
> delta(L) = -Tr(delta(dA + A^A) ^ G) + Tr(delta(A) ^ J).
>
> The integration by parts is on delta(dA) = d(delta A) and results in
> (all inside the trace operator Tr()):
> -d (delta A) ^ G = d(-delta A ^ G) - delta A ^ dG,
> (noting that delta A is a 1-form so an extra - is acquired jumping
> over it).
>
> For the non-linear term, you have
> -delta(A ^ A) ^ G = -delta(A) ^ A ^ G - A ^ delta(A) ^ G.
> Since the trace operator is cyclic, the last term can be written as
> Tr(-A ^ delta(A) ^ G) = Tr(delta(A) ^ G ^ A)
> (again, noting that delta(A) is a 1-form and is jumping over the 3-
> form G ^ A).
>
> The result is the variational:
> delta(L) = -d (Tr(delta A ^ G))
> + Tr(delta(A) ^ (J - dG - A ^ G + G ^ A)).
>
> Thus, the field law is in matrix form:
> dG + A ^ G - G ^ A = J.
I will study this too. I am VERY in interested in the field equation
for non-Abelian gauge theory, because of the issue of quark confinement.
Based on the MIT bag model, we may regard the "boundary" within which
quarks are confined as one in which the four components of the current
J^u=0. And, in non-Abelian theory, the differential equation:
dG + A ^ G - G ^ A = J = 0!!!
based on your derivation above, yields non-trivial solutions for the
gauge field G at which the current components all vanish. If one then
expresses G(x) as a function of spacetime coordinates, one can obtain
expressions for the geometric surfaces of "no charge and no current
flux," i.e., confinement. As you pointed out last time, however, this
is all CLASSICAL derivation. The question I have been struggling with,
is what happens to all of this once we move over to quantum theory?
And, whether one can make the transition to quantum simply by using
expectation values, or whether something more complicated is arises.
I will be back with more as I review your details
Thanks again,
Jay
> To restore this to component form requires the matrix form for the
> dual basis Y^a. Asusming
> Tr(Y^a Y_b) = delta^a_b
> you can find what the components of Y^a ought to be. I won't do that
> here.
>