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Integration by Parts of the Maxwell QED Action...
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Rock Brentwood...
Posted: Wed Sep 30, 2009 1:23 pm
Guest
On Sep 27, 2:52M-BM- pm, "Jay R. Yablon" <jyab... at (no spam) nycap.rr.com> wrote:
[quote:abb9e0b85c]Linked below is a < three page exercise showing how the pure, linear
Maxwell field is integrated by parts in QED:
[/quote:abb9e0b85c]
Well, several things: (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.

Let L be the Lagrangian density, so that the action is
S(U) = integral_U L d^4x
where d^4x = dx^0 ^ dx^1 ^ dx^2 ^ dx^3.

Let F_{mn} = d_m A_n - d_n A_m be the field tensor, where A_m is the
potential. Define the tensor densities G^{mn} = -dL/dF_{mn} and J^m =
dL/dA_m.

Then, the variational may be expressed as:
delta(L) = -1/2 delta(F_{mn}) G^{mn} + delta(A_m) J^m.

Anti-symmetry may be used to simplify the first term to
-1/2 delta(F_{mn}) G^{mn} = delta(d_n A_m) G^{mn}.

Integration by parts then yields for this term:
d_n delta(A_m) G^{mn}
= d_n (delta(A_m) G^{mn}) - delta(A_m) d_n G^{mn}.
Thus, the variational of the Lagrangian density becomes
delta(L)
= d_n(delta(A_m) G^{mn})
+ delta(A_m) (J^m - d_n G^{mn}).

This results in the total action variational
delta(S(U))
= integral_U d_n(delta(A_m) G^{mn}) d^4x
+ integral_U delta(A_m) (J^m - d_n G^{mn}) d^4x

The first integral is expressed in terms of the boundary dU of the
region U as
integral_{dU} (delta(A_m) G^{mn}) (d^3x)_n
where
(d^3x)_n = 1/6 epsilon_{nrst} dx^r ^ dx^s ^ dx^t.

So, the action principle states that if delta(A_m) is constrained to
be 0 on the boundary dU, then the variation delta(S(U)) will be 0.
This is what leads to the field equations:
d_n G^{mn} = J^m.

*More importantly* it also leads to the symplectic structure of the
field theory. That's what the boundary term is actually giving you.

So, the second application of the action integral invokes the
following principle: vary the action S(U) again, but this time
constrain the field to the solution space, while allowing the boundary
variational to run free. Then the total variational in S(U) becomes
delta S(U) = integral_{dU} (delta(A_m) G^{mn}) (d^3x)_n.

A Hamiltonian formulation is then specified by filling a given region
U by a time-like field d_t that has the property that the flow x |->
exp(t)x generated by d_t foliates the region U into a partition
U = union U_t (t in [t0,t1]).

Coordinates are chosen such that x^0 = t. Then, the variational for S
(U) becomes an integral
delta S(U) = W(t1) - W(t0)
where
W(t) = integral_{U_t} delta(A_m) G^{m0} dx^1 ^ dx^2 ^ dx^3.
gives you the "symplectic form" associated with the surface U_t and
"time" t.

This, in turn, determines the integrand G^{m0} as directly related to
the momentum conjugate to A_m.

Again, all this is in CLASSICAL field theory. All of it is BEFORE you
even begin to consider the quantization step.

In that light, it's also important to point out an all-too-common
subtlety that happens when people adopt the "throw out the boundary
terms".

And that subtlety is as follows:
Two Lagrangians may yield the same field equations but different
boundary terms. They produce the same field equations, but because the
boundary terms are different, their symplectic structures are
different. Consequently, their Hamiltonian formulations are NOT
equivalent and yield INEQUIVALENT quantum theories, when canonically
quantized.

So, the boundary terms matter and cannot be ignored.

The step you used, by the way, in dropping the fields off to 0 at
"infinity" is neither necessary nor warranted. The action principle
only requires that you restrict the variationals to 0 on the boundary
of the region of integration -- which is the process carried out
above.
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enders...
Posted: Fri Oct 02, 2009 5:38 am
Guest
Hello Rock,

Thank you for your elucidating comment! It strengthens my impression
that it is disadvantagous to consider the Lagrangian only. For the
absolute value of the Hamiltonian is fixed, if, in the stationary
case, it equals the total energy. Here, it is understood that the
latter is given by the maximum of energy which the system can deliver
to its environment. - What do you think?

Best wishes,
Peter
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Jay R. Yablon...
Posted: Fri Oct 02, 2009 10:06 pm
Guest
"Rock Brentwood" <markwh04 at (no spam) yahoo.com> wrote in message
news:961773b7-fef3-4f65-97bb-912937c5cea9 at (no spam) g31g2000yqc.googlegroups.com...
[quote:edd980f947]On Sep 27, 2:52M-BM- pm, "Jay R. Yablon" <jyab... at (no spam) nycap.rr.com> wrote:
Linked below is a < three page exercise showing how the pure, linear
Maxwell field is integrated by parts in QED:

Well, several things: (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.
[/quote:edd980f947]
Hi Mark,

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.

Thanks,

Jay

PS: Hi Peter -- next post down.

[ Mod. note: 86 extra quoted lines snipped. -ik ]
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Jay R. Yablon...
Posted: Sun Oct 04, 2009 11:41 pm
Guest
"Rock Brentwood" <markwh04 at (no spam) yahoo.com> wrote in message
news:961773b7-fef3-4f65-97bb-912937c5cea9 at (no spam) g31g2000yqc.googlegroups.com...
[quote:2ee642af60]On Sep 27, 2:52M-BM- pm, "Jay R. Yablon" <jyab... at (no spam) nycap.rr.com> wrote:
Linked below is a < three page exercise showing how the pure, linear
Maxwell field is integrated by parts in QED:

Well, several things: (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.
.. . .[/quote:2ee642af60]

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
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Rock Brentwood...
Posted: Mon Oct 05, 2009 8:10 pm
Guest
Jay R. Yablon wrote:
[quote:9acf861df3]Dear Mark,

Thank you for your reply of September 30. This was, clear, informative
and enjoyable to study. Wink
[/quote:9acf861df3]
The "short answer" part of my second reply was to directly follow up
on your question. The "long answer" part was something I was going to
add in, independently, because you got me interested in this again.

These are some additional notes that fill in some of the loose ends;
including 1: the matrix form of the dual basis, 2: the actual
appearance of the integration by parts in component form. In addition,
there is a comment on Maxwell's 19th century precursor to GSW
unification -- which was one of the first places an (Abelian) gauge
theory arose.

This is arranged in 6 somewhat long footnotes.

If you can incorporate this kind of material and, additionally, fill
in the details on the theorems mentioned (Utiyama's Theorem, the
extended version of Utiyama's Theorem for extra fields coupled to
gauge fields, the generic form for adjoint-invaraint metrics) it might
even serve as the groundwork for an interesting monograph --
especially so, if this is used as a launching point to do a more
careful and systematic study of the prospective quantizations of the
classical field theories derived.

Most people who do quantization start too far downstream,
automatically assuming the homogeneous quadratic Lagrangian; while we
already know in quantum field theory that scale invariant (which is
equivalent to 2nd order homogeneity in the Lagrangian) is broken --
that's what renormalization really pertains to. This is pulled back
classically into a non-trivial Lagrangian; one that is more general
than the naive quadratic Lagrangian.

So, it's interesting to see what all those extra "constitutive
coefficients" (besides theta) would actually leads to; and what kind
of Physics the quantized theories contain.

Note 1: Errata
First, correcting a few errors:
[quote:9acf861df3]the Maxwell-Lorentz Lagrangian density:
L = (epsilon c) root(-g) g^{mr} g^{ns} F_{mn} F_{rs}.
... generalizes directly to
L = k_{ab} root(-g) g^{mr} g^{ns} F^a_{mn} F^b_{rs}.
[/quote:9acf861df3]
add factors of -1/4 to each.

[quote:9acf861df3]In 4-D these invariants are:
I^{ab} = -1/4 root(-g) g^{ms} g^{nr} F^a_{mn} F^b_{rs}
[/quote:9acf861df3]
g^{mr} g^{ns} not g^{ms} g^{nr}

[quote:9acf861df3]For simple gauge groups...
k_{ab} = delta_{ab}/g^2...
Then the action integral becomes:
S_U = integral_U (1/g^2) (-k_{ab} F^a ^ *F^b).
[/quote:9acf861df3]
That should be
integral_U (-1/2 k_{ab} F^a ^ *F^b).
which works out to
integral_U (-1/(2g^2)) delta_{ab} F^a ^ *F^b

[quote:9acf861df3]G = D_a.dS - H_a.dr ^ dt
J = rho_a dV - J_a . dS ^ dt
should read[/quote:9acf861df3]
G_a = ..., J_a = ...

[quote:9acf861df3]G = 1/2 G^{mn}_a e_{mn}
J = J^m e_m
should read[/quote:9acf861df3]
G_a = 1/2 G^{mn}_a e_{mn}, J_a = J^m_a e_m

Note 2: Adjoint-Invariant Metrics & Coupling Coefficients
For a simple gauge group, as was mentioned, the general form for an
adjoint-invariant positive definite metric k_{ab} is
k_{ab} = kappa_{ab}/g^2
where kappa is the Killing metric.

For a semi-simple gauge group G = G1 x G2 x ... x Gn, the most general
form for an adjoint-invariant positive-definite bilinear form is:
k_{ab} = kappa^1_{ab}/g1^2 + kappa^2_{ab}/g2^2 + ... + kappa^n_{ab}/
gn^2
where kappa^1, kappa^2, ..., kappa^n are the Killing metrics the
simple groups G1, G2, ..., Gn and the coefficients g1, g2, ..., gn are
the coupling coefficients associated with the respective groups.

If the gauge group is combined with an Abelian gauge group (i.e., U(1)
^n for some power n) the most general metric has non-zero cross-
components both between the U(1)'s and with the U(1)'s and the groups
G1, G2, ..., Gn.

This can be reduced to a normal form by replacing the U(1) basis by a
suitable mixture of the U(1)'s combined with the basis elements of G1,
G2, ..., Gn -- the process is a variation of the Gram-Schmidt
orthogonalization procedure.

Once this reduction has been made, the same form above for the metric
can be extended to include the U(1) factors.

When applied to the electroweak field, what this does is reduce the
Maxwell U(1) to a combination of this and the isospin SU(2) part of
the field, producing the hypercharge U(1).

Thus, the metric associated with the electroweak gauge field is
k_{ab} = kappa^1_{ab}/g1^2 + kappa^2_{ab}/g2^2
where kappa^1, g1 are associated NOT with the Maxwell U(1), but the
hypercharge U(1); while kappa^2 and g2 are associated with the isospin
SU(2).

The mixing coefficient is directly related to the electroweak mixing
angle.

This idea -- along with the idea of Abelian gauge fields -- goes back
to (and before) Maxwell's time. I'll say more on that below in Note 6.

Note 3: Integration by Parts, Component-Wise
The explicit integration by parts involves the following:
delta(L) = -1/2 delta(F^c_{mn}) G_c^{mn} + delta(A^a_m) J_a^m
with
F^c_{mn} = d_m A^c_n - d_n A^c_m + f^c_{ab} A^a_m A^b_n.

The non-linear part of this gives you
-1/2 f^c_{ab} delta(A^a_m A^b_n) G_c^{mn}.
and reduces to
-1/2 f^c_{ab} (delta(A^a_m) A^b_n G_c^{mn} + A^a_m delta(A^b_n) G_c^
{mn})
By anti-symmetry of G and f, reduces to
-f^c_{ab} delta(A^a_m) A^b_n G_c^{mn}.
(This is the step that runs parallel to the point in the other
derivation where the Jacobi identity was used).

The linear part gives you
-1/2 delta(d_m A^c_n - d_n A^c_m) G_c^{mn}.
By anti-symmetry of G, this reduces to
delta(d_n A^c_m) G_c^{mn} = d_n(delta A^c_m) G_c^{mn}.

This is where the integration by parts applies. You end up getting
(with a change in index):
d_n(delta(A^a_m) G_a^{mn}) - delta(A^a_m) d_n(G_a^{mn})

Thus, the total variational is:
delta(L) = d_n(delta(A^a_m) G_a^{mn})
+ delta(A^a_m) (J_a^m - d_n(G_a^{mn}) - f^c_{ab} A^b_n G_c^{mn}).

The field law is thus
d_n(G_a^{mn}) + f^c_{ab} A^b_n G_c^{mn} = J_a^m
while the symplectic structure is derived from the boundary
variational:
delta(S_U) = integral_{dU} (delta(A^a_m) G_a^{mn} (d^3x)_n)
or for a 3+1 decomposition of U, as described before,
delta(S_U) = K(t1) - K(t0)
where
K(t) = integral_{dU_t} (delta(A^a_m) G_a^{m0} dx ^ dy ^ dz).
This makes the field -D_a the conjugate of the 3-vector potential A^a,
while the potential phi^a has a conjugate of 0 (i.e. the "Gaussian
constraint" comes to play here).

Note 4: Differential Forms as the "Original" Language of Maxwell
Equations
It bears pointing out that this is not a "repackaging" or
"modernization" of Maxwell's theory or equations. In fact, the exact
opposite is true: it's the form that Maxwell used from the very
beginning (in his 1861 and 1864 papers)!

Apart from the novelty of combining the space and time parts, with the
addition of the ^dt factor to E.dr, H.dr and J.dS, this is how Maxwell
originally expressed these quantities before adopting the vector
notation (in its 19th century guise of quaternions) in his treatise.
Even in the treatise, he placed emphasis on the expression of field
quantities in terms of differential forms -- including using the anti-
commutativity property of the Grassmann product.

The extra ^dt factor, though historically connected with the
appearance Minkowski geometry actually has nothing whatsoever to do
Minkowski space, Special Relativity or anything of the like. The
natural appearance of 4-dimensionality in the expressions for the
field quantities and equations has nothing to do with spacetime
signature nor with the distinction between relativity and non-
relativistic theory -- it is just as imminent in the Galilean form of
the field equations.

That's because both the Galilean form and Lorentzian form of the
equations are identical: dA = F, dG = J. In fact, these equations are
entirely independent of causal structure or spacetime signature
because they are invariant under a group that contains the
diffeomorphism group as a proper subgroup.

The reason for the natural appearance of the 4-dimensionality is
precisely because of the diffeomorphism invariance of the equations.

The only place the Galilean and Lorentzian versions of the theory
differ is in the constitutive law. There is no constitutive law for a
Galilean vacuum (by "vacuum" meaning a medium that is both isotropic
and boost-invariant). Rather, the only constitutive law is for the
general isotropic medium. This exists in both Galilean and Lorentzian
form as:
D + alpha G x H = epsilon(E + G x B) + theta(B - alpha G x E)
H - G x D = (1/mu) (B - alpha G x E) - theta(E + G x B).

In Lorentzian form, alpha = 1/c^2 and this is the Einstein-Laub-
Minkowski form of the constitutive relations for an isotropic medium.
The velocity G determines the velocity of the observer relative to the
(generally unique) frame in which isotropy holds. In Galilean form,
alpha = 0, and this reduces to the Maxwell-Thomseon form of the
constututive laws:
D = epsilon(E + G x B), B = mu(H - G x D)
apart from the theta term. The extra G x B is present in Maxwell -- he
added it as part of the definition of E, itself. The term -G x D was
noted by Thomson (and tacitly by Maxwell in the 1860's, during the
period before he fully distinguished B from H).

The constitutive law is boost-invariant (i.e. G drops out) only if
where alpha = mu epsilon (i.e. where the produce mu epsilon reduces to
1/c^2). This is a necessary, BUT NOT SUFFICIENT condition.

That is, only the Lorentzian version has a "vacuum" (i.e. isotropic,
boost-invariant) form for the constitutive laws; but the "not
sufficient" part means that EVEN IN Lorentzian spacetime, there can
STILL be a dependence on G even for a vacuum. That occurs precisely
when mu epsilon G^2 = 1.

Note 5:
When the field law rendered in matrix form with differential forms,
this should yield the result originally stated:
dG + A^G - G^A = J.

In fact, the matrix form can be recovered by matching this with the
field equation. First, the field equations can be rendered using
differential forms, by defining
G_a = 1/2 G_a^{mn} e_{mn}
e_{mn} = 1/2 root(-g) epsilon_{mnrs} dx^r ^ dx^s
J_a = J_a^m e_m
e_m = 1/6 root(-g) epsilon_{mnrs} dx^n ^ dx^r ^ dx^s

Algebraically, writing
e = 1/24 root(-g) dx^m ^ dx^n ^ dx^r ^ dx^s = root(-g) dt ^ dx ^ dy ^
dz
we have
dx^m ^ e_n = delta^m_n e
dx^m ^ e_{nr} = delta^m_r e_n - delta^m_n e_r
dx^m ^ dx^n ^ e_{rs} = 1/2 (delta^m_r delta^n_s - delta^m_s
delta^n_r) e

So, we can write (using the anti-symmetry of G):
dG_a = 1/2 d_r G_a^{mn} dx^r ^ e_{mn}
= 1/2 d_r G_a^{mn} (delta^r_n e_m - delta^r_m e_n)
= d_n(G_a^{mn}) e_m.
and
A^b ^ G_c = 1/2 A^b_r G_c^{mn} dx^r ^ e_{mn}
= 1/2 A^b_r G_c^{mn} (delta^r_n e_m - delta^r_m e_n)
= A^b_n G_c^{mn} e_m,
again using the anti-symmetry of G.

Thus, the field law is written with differential forms as:
dG_a + f^c_{ab} A^b ^ G_c = J_a.

This is to be matched against the matrix form
dG + A ^ G - G ^ A = J,
which requires the components for the matrix form of G and (in turn)
the matrix form for the dual basis {Y^a}.

The trace condition
Tr(Y^a Y_b) = delta^a_b
implies that the matrix components of the dual basis should satisfy
the condition:
delta^d_a = f^c_{ab} (Y^d)^b_c.
There is an explicit solution in terms of the Killing metric (and its
dual) provided the Killing metric has an inverse. This occurs
precisely when the gauge group is simple or semi-simple.

The matrix forms of G and J are thus
G^b_c = G_d (Y^d)^b_c, J^b_c = J_d (Y^d)^b_c
and satisfy the identities
f^c_{ab} G^b_c = G_a, f^c_{ab} J^b_c = J_a.
Substituting into the field law, this yields
f^c_{ab} dG^b_c + f^e_{cd} f^c_{ab} A^b ^ G^d_e = f^c_{ab} J^b_c.
The Jacobi identity applied to the non-linear term on the left yields
f^c_{bd} f^e_{ac} A^b ^ G^d_e + f^c_{da} f^e_{bc} A^b ^ G^d_e
or
f^e_{ac} A^c_d ^ G^d_e + f^c_{da} A^e_c ^ G^d_e
= f^e_{ac} A^c_d ^ G^d_e - f^c_{ad} G^d_e ^ A^e_c
or with relabelling of indices
f^c_{ab} A^b_d ^ G^d_c - f^c_{ab} G^b_d ^ A^d_c
Thus we have
f^c_{ab} (dG + A^G - G^A)^b_c = f^c_{ab} J^b_c
and after cancelling out the factor f^c_{ab} (by multiplying by (Y^a)
^b_c and summing), we get
dG + A^G - G^A = J.

Note 6: Maxwell's "GSW" Electrogravitational Unification Model
The mixing angle + Maxwell issue raised in Note 2:

What Maxwell did in his treatise is relay one of the numerous
approaches to modelling positive and negative charges around at the
time. It also seemed to entail electrogravitational unification via an
electrogravitational mixing angle. The model is entirely analogous to
the GSW unification of electromagnetism and the weak force.

The metric enters directly into the force law for two charges. For
electromagnetism the force law reads:
F = ee'/(4 pi epsilon r^2) = (mu/(4pi)) ee'/r^2.
For a gauge field, the charge e generalizes to e = e_a Y^a and the
force law (making use of the correspondence mu c <-> k^{ab})
generalizes to
k^{ab}/(4 pi c) e_a e'_b/r^2.

Hence, suppose we model positive and negative charges as two KINDS of
charges (p and n, both positive) with electric charge e being given up
to proportion by p - n.

The corresponding gauge field is a U(1) x U(1) Abelian field. The
force at unit distance for unit positive charges p, p' and unit
negative charges n, n' would have the forms
p-p' => F1
n-n' => F2
p-n => -F3
where F1 ~ F2 ~ F3 > 0; where up to proportion, the gauge group metric
would be the matrix given by the inverse of the following matrix
k^{pp} = F1, k^{pn} = -F3 = k^{np}, k^{nn} = F2.

This can be normalized to orthogonal form by the incorporation of a
suitable mixing angle, defining the dual basis by
Y^e = Y^p cos T + Y^n sin T
Y^g = Y^n cos T - Y^p sin T

The corresponding components of the charge vector are then
e_e = p cos T - n sin T ~ electric charge
e_g = n cos T + p sin T ~ particle number

The resulting metric is then
k^{ee} = k(Y^e, Y^e)
= k(Y^p cos T + Y^n sin T, Y^p cos T + Y^n sin T)
= F1 cos^2 T - F3 sin 2T + F2 sin^2 T;
and similarly,
k^{eg} = k^{ge} = (F2 - F1) sin 2T/2 - F3 cos 2T
k^{gg} = F1 sin^2 T + F3 sin 2T + F2 cos^2 T.

To eliminate the cross-component and get k^{eg} = 0 = k^{ge} requires
that
F1 - F2 = 2Z cos 2T, F3 = -Z sin 2T,
Z > 0 with Z^2 = ((F2 - F1)/2)^2 + F3^2
so that T will be close to 45 degrees.

Upon substitution, this yields for the remaining components
k^{ee} = (F1 + F2)/2 + Z
k^{gg} = (F1 + F2)/2 - Z
This yields a coupling k^{ee} ~ 2Z and k^{gg} ~ 0.

If F1 F2 < F3^2, then k^{gg} < 0 and the g-component of the force will
be attractive and coupled to particle number, while the e-component of
the force will satisfy the conditions usually ascribed to the
electromagnetic force.

The result, as per Maxwell, was an early account of an
electrogravitational unification involving a mixing angle T whose non-
trivial feature would have been that it deviated from 45 degrees.

In modern guise, this model does NOT represent gravity. The "particle
number" degree of freedom is essentially baryon number, while
"electric charge" would now be hypercharge. So, what you're actually
talking about here is a U(1) x U(1) unification that breaks down into
baryon number and electric charge.

So, instead of representing gravity, it actually represents the "5th
force" that's coupled to the difference of baryon and lepton number.
This is the only force that can be consistently added to the U(1) x SU
(2) x SU(3) spectrum.
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Jay R. Yablon...
Posted: Mon Oct 05, 2009 8:46 pm
Guest
"Rock Brentwood" <markwh04 at (no spam) yahoo.com> wrote in message
news:f7520c72-99fb-4856-bf4a-130676014d49 at (no spam) r31g2000vbi.googlegroups.com...
[quote:803b4b8015]On Oct 3, 3:06 am, "Jay R. Yablon" <jyab... at (no spam) 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.
[/quote:803b4b8015]
Hi Mark,

I have a few comments inline, and will have more to follow after I study
the details of your posting.

[quote:803b4b8015]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).
[/quote:803b4b8015]
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 (1Cool in
http://www.physics-quest.org/Non_Abelian_Lagrangian.pdf.

[quote:803b4b8015]
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.
[/quote:803b4b8015]
I do like your approach of working independently of a specific
expression for the Lagrangian, and so will study this closely.

[quote:803b4b8015]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.
[/quote:803b4b8015]
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

[quote:803b4b8015]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.
[/quote:803b4b8015]
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

[quote:803b4b8015]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.
[/quote:803b4b8015]
Back to top  
Rock Brentwood...
Posted: Tue Oct 06, 2009 12:07 pm
Guest
Some comments on your article and reply:

(0) Even with keeping the Lagrangian general, apart from symmetry
principles and general assumptions (like that it be analytic), the
relations between the fields F_{mn} and their conjugates G^{mn} assume
the same general form -- with all the specifics locked up into the
constitutive coefficients.

This shoudl be recognized for what it is: the classical version of
renormalization. The constitutive coefficients are the analogue of the
renormalization coefficients. The Taylor series expansions of the
coefficients (which follows, if the Lagrangian is analytic) in terms
of the Lorentz invariants yields coefficients that play the analogue
of the beta coefficients of renormalization group theory.

What the exact correspondence between the two approaches is, and
between their respective coefficients, I do not yet know.

On your PDF reply:
(1) There is no sign reversal or sign error in (1.5). In fact, your
"correction" introduced the sign errors in (1.6) and afterwards. Your
"Euler-Lagrange" equation has the wrong sign; and the expressions you
rewrote G^{mn} as now have their signs backwards, which means the
boundary term now has the wrong sign.

The Euler-Lagrange equations are
dL/dA_m = d_n(dL/d(d_nA_m))
When the functional dependence of the Lagrangian L = L(A_m, d_mA_n)
reduces to a dependence of the form L = L(A_m, F_{mn}) then the second
term becomes d_n(dL/d(F_{nm})) - which is backwards from what you
wrote.

So, defining J^m as dL/dA_m and G^{mn} as -dL/dF_{mn} yields the field
equations d_nG^{mn} = J^m.

(2) In reply to: "the second application of the action integral
invokes the following principle: vary the action S(U) again, but this
time constrain the field to the solution space, while allowing the
variational to run free…" you said "as a general rule, we must have
delta S(U) = delta integral_U L d^4 x = 0 [for the action
principle]".

This comment isn't talking about the action principle, but the
application of the action integral to a second principle. That second
principle is where you get the symplectic structure for the field
theory from.

(3) The boundary dU of a space-time region U can be entirely
spacelike. In that case the integral
integral_{dU} (... (d^3x)_m)
is of the form
integral_{dU} (... dx ^ dy ^ dz)
with respect to a suitable coordinatization; involving *no other*
volume 3-forms (d^3x)_m than (d^3x)_0 = dx ^ dy ^ dz.

In particular, a time-like field spanning U which produces a foliation
U_t never includes "side boundaries" if the timelike field has U as a
compact support (and is differentiable). In the discussion of the
reduction to a 3+1 foliation of U, I forgot to mention that U should
be restricted to compact regions and the timelike fields to compact
support on U.

What this means is that the foliation will always take the form in
which each U_t shares the same boundary H (the "horizon") and the
boundary of U reduces to the 3-chain U_{t1} - U_{t0}, if [t0, t1] is
the interval that the foliation spans.

This applies independent of whether the underlying spacetime is flat
or curved, and whether there even is an asymptotic infinity to push
anything off to or not.

It is also independent of whether the underlying spacetime is globally
hyperbolic or not - provided the region U is globally hyperbolic as a
submanifold (i.e. that it have a time-like foliation).

However, there are issues raised if the embedding of the region U in
the space-time is not causally convex. A spacelike region (even one
that's globally hyperbolic) can be embedded in a spacetime in such a
way that one part of it lies in the future light cone from another
part - even if the underlying spacetime, itself, is globally
hyperbolic (even if it's flat Minkowski space!).

Finally, even when you have an asymptotic infinity; if you push
everything "off to infinity" without paying proper attention to H
during the limiting process, you will lose the information about the
tacit dependence of everything on H -- and (along with it) a lot of
relevant physics. This includes all issues, subtleties, anomalies,
etc. related to global topology.

(4) On the inequivalence issue:
There should be no confusion about the comment "different Lagrangians,
even when they produce the same field equations, yield different
Hamiltonians and therefore, upon canonical quantization inequivalent
quantizations." It means just that.

The question of whether the canonical quantization associated with one
Lagrangian agrees with that associated with another Lagrangian is
entirely independent of whether the Lagrangians produce the same Euler-
Lagrange equations or not (classically OR quantum-theoretically).

Therefore, discussion of the Euler-Lagrange equations has no bearing
on the issue.

A canonical quantization is done with respect to a Hamiltonian. A
Hamiltonian is derived from the Lagrangian by a Legendre
transformation. The Legendre transformation makes explicit reference
to the components of the canonical momenta (i.e. to the symplectic
structure of the underlying dynamics).

Two Lagrangians with the same field equations generally differ in
their symplectic structure, yielding different canonical momenta, and
thus yielding different Hamiltonians. Different Hamiltonians yield
different, and generally inequivalent, canonical quantizations.

Moreover, the inequivalence is actually of the classical symplectic
structures generated by the two Lagrangians, not just of the
quantizations. So, it's not a quantum issue. It's a classical issue.

You don't see the inequivalence classically as such, because the
inequivalence shows up as the mutual incoherence of the state spaces
associated with the respective dynamics (and it shows up as different
commutators). Classically this entails nothing new, since every two
states already reside in separate coherent subspaces and the
commutators are all already 0 and so are likewise unaffected.

In a quantized theory, on the other hand, the inequivalence means that
whatever parameters are used to index the different coherent subspaces
are essentially classical and *unquantizable* variables. The most
notable example of this is the theta parameter in a chiral gauge
theory.

The symplectic structures associated two Lagrangians need not even
have the same constraints. So, even the *method* by which the
Hamiltonian is canonically quantized can be affected by the choice of
a Lagrangian; since the structure of the constraints determines how to
quantize the Hamiltonian dynamics.

In the Hamiltonian dynamics of gauge theory (both classical and
quantized), the selection of an appropriate Lagrangian is not only a
major issue; it's THE major issue.

(5) Finally, the answer to how far you can get to not know the
explicit form of the Lagrangian for the path integral method is -- not
very far. In fact, the assumption of the specific form of the
Lagrangian is built right into the method. The moment you employ the
gaussian trick, you're already fixing the Lagrangian to be quadratic
homogeneous in the field strengths (i.e. the Maxwell-Lorentz
Lagrangian or its non-Abelian generalization, the Yang-Mills
Lagrangian). At the very least you're doing something analogous to a
"mean field" expansion about a quadratic homogeneous Lagrangian.

The path integral method, itself, is not well-founded: the Osterwalder-
Schrader Theorem (sp?) - which is the basis of the path integral
formalism - only applies to flat-spacetimes. There is no known
formulation of path integrals to curved spacetimes. More general, more
robust formalisms should be used that don't suffer this defect.

You may endeavor to limit your exercise to flat spacetime (and further
limit it to only those foliations of flat spacetime that have flat
foliation layers), but then the question is: "To what end, other than
self-edification? In particular, to what end that is not already
superseded by more robust approaches?" There are already methods that
generalize both perturbation theory and renormalization theory (and,
thus, also effective field theory) to curved spacetimes.
Back to top  
Rock Brentwood...
Posted: Wed Oct 07, 2009 6:07 am
Guest
[[Mod. note -- A strict interpretation of our newsgroup charter would
probably forbid discussions (such as this posting) which aren't about
physics per se, but rather about data formats & software for
writing/publishing physics (& other scientific) papers. However, I
think this topic is very much a part of the scientific life of anyone
involved in scientific research, and (judging from the many past
discussions on this topic), it's of interest to many s.p.r readers.
-- jt]]

Jay R. Yablon wrote:
[quote:d3d2651bac]http://jayryablon.files.wordpress.com/2009/10/reply-to-mhopkins.pdf
[/quote:d3d2651bac]
I've written this up as a PDF as well. Drop me a note and I'll send a
copy; you're free to archive it.

One general note for you and other readers of this group. I use
Equation Edit which comes with the Microsoft Office and prefer it to
other math packages because it renders much more cleanly.

The best way to go with publishing, however, is NEITHER this, nor the
(defunct and stillborn) MathML, nor necessarily TeX. There is in more
recent times an open international standard under the ISO (which was
also subject to open public review) called "Open Document". The focal
point of this effort was Sun, who has put Open Office on the net as
freely downloadable software. Others, including a large and growing
number of governments (including NATO, itself) have made Open Document
their standard, and some have issued a committment to use this as a
vehicle for getting people into the digital world.

Most notable in this list is the state of Massachusetts, which kinda
puts the heavy involvement of MIT in the superseded MathML effort in
proper perspective.

Open Office's math language is directly human-readable and (unlike
MathML) useable without the need for intervention by specialized code-
generating software. The Word clone ("Writer") and Equation Edit
analogue ("Math") are simultaneously WYSIWYG and text-based.

Microsoft only lately relented and buckled under the pressure and put
a text-based language in their Office 2007 and provided facilities to
retrofit it in earlier Office. Both Equation Edit and Open Office's
Math (as well as GUI-based TeX editors and, to a lesser degree,
MathML) tend to use the same set of widgets and templates; so it's all
converging onto a common dialect.

If you want to improve your PDF's, you should download Open Office and
use that. The Word clone (Writer) exports PDF, though it does not
import it.

Equation Edit from Microsoft works with Writer, by the way, and Math
works inside of Microsoft Office; so it's all interoperable. But
Nature and Science have dropped direct support for Equation Edit,
because the facility can sometimes render straight into graphics (the
DOCX format), and apparently nobody there has the ability to reverse-
translate the graphics to a linear text-based language (the exercise
is not that difficult! It's much less than full-fledged OCR, since the
input for the graphic is limited to the set of facilities provided in
the Equation Edit menu and are easy to parse out of the graphics --
somebody should set up a reverse-convertor).

Open Office also reads MathML (as does the freely downloadable Web
browser Opera).
Back to top  
Jay R. Yablon...
Posted: Mon Oct 12, 2009 12:22 am
Guest
"Rock Brentwood" <markwh04 at (no spam) yahoo.com> wrote in message
news:94d92ab5-a4c1-4451-aade-2c224be438bc at (no spam) g6g2000vbr.googlegroups.com...

Hi Mark,

I have some preliminary questions regarding the end of your point (4)
and point (5), as excerpted below:

[quote:643bacd44d]Some comments on your article and reply:
.. . .
(4) On the inequivalence issue:
.. . .

In the Hamiltonian dynamics of gauge theory (both classical and
quantized), the selection of an appropriate Lagrangian is not only a
major issue; it's THE major issue.

(5) Finally, the answer to how far you can get to not know the
explicit form of the Lagrangian for the path integral method is -- not
very far. In fact, the assumption of the specific form of the
Lagrangian is built right into the method. The moment you employ the
gaussian trick, you're already fixing the Lagrangian to be quadratic
homogeneous in the field strengths (i.e. the Maxwell-Lorentz
Lagrangian or its non-Abelian generalization, the Yang-Mills
Lagrangian). At the very least you're doing something analogous to a
"mean field" expansion about a quadratic homogeneous Lagrangian.

The path integral method, itself, is not well-founded: the
Osterwalder-
Schrader Theorem (sp?) - which is the basis of the path integral
formalism - only applies to flat-spacetimes. There is no known
formulation of path integrals to curved spacetimes. More general, more
robust formalisms should be used that don't suffer this defect.

You may endeavor to limit your exercise to flat spacetime (and further
limit it to only those foliations of flat spacetime that have flat
foliation layers), but then the question is: "To what end, other than
self-edification? In particular, to what end that is not already
superseded by more robust approaches?" There are already methods that
generalize both perturbation theory and renormalization theory (and,
thus, also effective field theory) to curved spacetimes.

Hi Mark,[/quote:643bacd44d]

Let's write the path integral in the general form ($=integral, == means
"defined as"):

Z = $D psi exp i$[d^4x L] == C exp i[W] (1)

Here, psi is the field, L is the Lagrangian density, and W is an
"amplitude expression." (Perhaps there is a more precise name that W is
given.) S=$[d^4x L] is the action.

This, in effect, is a *definition* of W. If one has some Lagrangian
density L (recognizing your statement that "the selection of an
appropriate Lagrangian is not only a major issue; it's THE major
issue"), then the only question left is to do the mathematics (which can
be formidable if not daunting) to obtain an exact expression for W as
defined in (1).

For example, suppose with used as our Lagrangian density the Einstein
Hilbert action:

S = $ [(.5/kappa)R + L_m] sqrt(-g) d^4x (2)

where L_m == L_matter, see
http://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action. This
corresponds wholly to the Einstein equation, it is non-linear, it takes
account of gravitational fields and accounts for curved spacetime, and I
assume you would regard this as an "appropriate Lagrangian" and if not
would explain why not.

IF one was able to then use (2) in all of the gory non-linear detail
that is hidden in (2), and be able to do the mathematics so as to
ascertain an EXACT expression for W in (1), then W would be the exact
amplitude expression for gravitational integrations and would inherently
impose quantum constraints upon gravitational theory which would enable
us to understand the manner in which gravitation is quantized. Or, at
least this seems to me to be so, in principle.

Now, let's look at the "Gaussian trick," which uses the mathematical
Gaussian identity:

$ d psi exp[-V(psi) - .5 K psi^2 + J psi]
= C exp[-V(delta/delta J)] exp[.5 K^-1 J^2] (3)

Of course, this contains quadratic terms homogeneous in the field
strengths and so may be akin to the drunk looking for the quarter under
the lamp post because "that's where the light is." But the use of the
terms V(psi) and V(delta/delta J) do provide the mathematical means to
solve the path integral (1) for W *exactly*, even if we cannot put the
solution into closed forms and have to deal with a double infinite
series from which we extract our Greens functions and take our Feynman
diagrams. Thus, these terms do not restrict us to any particular
Lagrangian form *other than* that we have the Lagrangian density terms
..5 K psi^2 + J psi off of which to do an expansion. Once we have .5 K
psi^2 + J psi in our Lagrangian density, we can solve the path integral
exactly, in principle, no matter what other terms may be in there.

In gravitational theory:

T_uv = -2 (delta L_m / delta g^uv) + g_uv L_m (4)

so that multiplying through by g^uv:

4 L_m = g^uv T_uv + 2 g^uv (delta L_m / delta g^uv) (5)

and the g^uv T_uv gives us the J psi term needed for the Gaussian trick.
Certainly, there are terms in L with psi^2 ~ g_uv ^2.

Thus, unless you can argue that having at least the terms .5 K psi^2 + J
psi in a Lagrangian density, plus any and all other terms of any order
in the fields, is so badly limiting as to exclude real Lagrangian
densities of real physical interest which do not have the terms .5 K
psi^2 + J psi, is seems to me that (1)-(3) above do provide a useful
vehicle -- in principle -- to perform a path integral quantization of
all known theories of physical interest, including gravitational theory
in curved spacetime, even if *calculating* the precise mathematical
solution or consolidating the solution into a closed form is still a
daunting challenge.

I'd appreciate your thoughts on these reflections.

Thanks,

Jay

PS: You wrote elsewhere in the thread "I've written this up as a PDF as
well. Drop me a note and I'll send a copy; you're free to archive it."
Did you get my email?
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Jay R. Yablon...
Posted: Tue Oct 13, 2009 8:48 am
Guest
"Rock Brentwood" <markwh04 at (no spam) yahoo.com> wrote in message
news:94d92ab5-a4c1-4451-aade-2c224be438bc at (no spam) g6g2000vbr.googlegroups.com...
[quote:073c5fffe8]Some comments on your article and reply:
.. . .[/quote:073c5fffe8]

[quote:073c5fffe8]On your PDF reply:
(1) There is no sign reversal or sign error in (1.5). In fact, your
"correction" introduced the sign errors in (1.6) and afterwards. Your
"Euler-Lagrange" equation has the wrong sign; and the expressions you
rewrote G^{mn} as now have their signs backwards, which means the
boundary term now has the wrong sign.

The Euler-Lagrange equations are
dL/dA_m = d_n(dL/d(d_nA_m))
When the functional dependence of the Lagrangian L = L(A_m, d_mA_n)
reduces to a dependence of the form L = L(A_m, F_{mn}) then the second
term becomes d_n(dL/d(F_{nm})) - which is backwards from what you
wrote.

So, defining J^m as dL/dA_m and G^{mn} as -dL/dF_{mn} yields the field
equations d_nG^{mn} = J^m.
[/quote:073c5fffe8]
Yes, I now see that. I too quickly took d_n(dL/d(F_{nm})) to be the
term d_n(dL/d(d_nA_m)) in the Euler Lagrange equation.

I will update the reply to correct that oversight.

Jay
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