markwh04@yahoo.com
Aug15-07, 05:00 AM
On Aug 12, 6:48 pm, Eric Gisse <jowr...@gmail.com> wrote:
> Plus I'm having a hell of a time with interpreting the spin tensor.
Lately, I've approached the whole enterprise from a different
standpoint that might ultimately shed some light on matters here.
The simplest way to write down the kinematic and dynamic field laws is
to express the connection part of the field as a Lorentz gauge field,
as I mentioned before:
A = 1/2 Gamma^{ab} J_{ab}.
Then express the Lie bracket for the Lorentz gauge generators in
algebraic form as commutators:
[J_{ab}, J_{cd}] = J_{ab} J_{cd} - J_{cd} J_{ab}.
Combine this algebra with that for the differential forms
Gamma^{ab} = eta^{bc} Gamma^a_{dc} e^d.
So the product of A with itself will be
A^2 = 1/4 Gamma^{ab} J_{ab} Gamma^{cd} J_{cd}
= 1/4 Gamma^{ab} ^ Gamma^{cd} (J_{ab} J_{cd}).
Since the wedge product is anti-symmetric, then this may be
equivalently written as
A^2 = 1/8 (Gamma^{ab} ^ Gamma^{cd}) (J_{ab} J_{cd} - J_{cd} J_{ab})
= 1/8 (Gamma^{ab} ^ Gamma^{cd}) [J_{ab}, J_{cd}].
The Lie bracket for the Lorentz group is
[J_{ab}, J_{cd}] = eta_{bc} J_{ad} - eta_{bd} J_{ac} - eta_{ac}
J_{bd} + eta_{ad} J_{bc}.
So, as a result, this gives you the correct form for the quadratic
terms which, when combined with dA, gives you the curvature as the
corresponding field strength:
F = dA + A^2 = 1/2 R^{ab} J_{ab}.
The frame field and matter fields can be treated in the same way,
together. So, suppose you have a matter field expressed as a m-form
field,
q = (1/m!) q^{A}_{n1 ... nm} E_{A} e^{n1} ^ ... e^{nm}.
Here the matter field is assumed to have a field index A identifying
components with respect to some linear space whose basis is E_{A}.
The frame field is subsumed under this as a 1-form field
e = e^a p_a
where the "basis" p_a is sometimes likened to as the translation
generators of a Poincare' gauge field.
Assume the algebraic representation is extended so that the Lorentz
gauge generators act on the left. Thus, for the frame field, one has
J_{ab} (p_c) = eta_{bc} p_a - eta_{ac} p_b.
For a matter field, what you get will depend on what kind of index A
is. If it's a tensor index, then A is actually a composite consisting
of a number of space-time indices in the upper position and a number
in the lower position. For a scalar m-form valued field, A will reduce
to an empty slot and there will be no indices there.
With these items in mind, the torsion and covariant derivatives can be
written in algebraic form as
T = de + Ae, v = dq + Aq.
The covariant derivatives of "q" give you the gauge-covariant
"velocities" v associated with the field.
Thus, you have a set of equations relating field-strengths (and
"velocities") to gauge-potentials (and field components):
F = dA + A^2, T = de + Ae, v = dq + Aq. (1)
>From these follow the Bianchi identities, using the Leibnitz rule for
differential forms. Thus,
d(A^2) = (dA) A - A (dA), d(Ae) = (dA) e - A (de); d(Aq) = (dA)q -
A(dq);
while
d(dA) = 0, d(de) = 0, d(dq) = 0.
Thus, you find that
dF = (dA) A - A (dA) = (F - A^2) A - A (F - A^2) = FA - AF
dT = (dA) e - A (de) = (F - A^2) e - A (T - Ae) = Fe - AT
dv = (dA) q - A (dq) = (F - A^2) q - A (v - Aq) = Fq - Av.
Putting these together, you get the Bianchi identities and the
"homogeneous" field equations for the field q:
dF + AF - FA = 0, dT + AT = Fe, dv + Av = Fq. (2)
The Bianchi identity (dT + AT = Fe) is the one that gives you R_{abcd}
+ R_{acdb} + R_{adbc} = 0 when the torsion is absent.
To determine the dynamics, assume the Lagrangian is a function of only
the potentials and covariant derivatives:
L = L(A, e, q, F, T, v).
This here is the Lagrangian 4-form. So, written in terms of the
Lagrangian density *L, it would be:
L = *L e^1 ^ e^2 ^ e^3 ^ e^4.
But it's best to work with the 4-form, instead of with the density.
As per the previous discussion, there will be no explicit dependence
on A, but it'll be kept in for now. The total variational in the
Lagrangian (using D as ASCII for "delta") will then factor out to
something of the form:
DL = DA . S + De . P + Dq . f - DF . U - DT . X + (-1)^m Dv . p.
Here, the indices attached to e and q (e^a, q^A) will be complemented
by indices attached to P and f (P_a, f_A). Similarly for the other
terms. Therefore, it is assumed that the expression above involves a
contraction on these indices. Thus, for instance,
De . P = De^a ^ P_a, Dq . f = Dq^A ^ f_A.
This introduces the dynamic fields S, P, f, U, X and p. The fields S,
P are 3-form currents (for spin and momentum). The field f is (4-m)
form valued. The fields U, X are 2-forms and function as
"superpotentials" for S and P. The field X will ultimately be involved
in giving you the extra "orbital" contribution to the spin current 3-
form J. The field p is a (3-m) form.
To define covariant derivatives for these fields, the action of the
Lorentz group has to be defined for the corresponding bases. Thus, the
fields written out fully are:
S = 1/2 S_{ab} j^{ab}, U = 1/2 U_{ab} j^{ab}
P = P_a p^a, X = X_a p^a
f = f_A E^A, p = p_A E^A,
where j^{ab} is the dual to the Lie basis J_{ab}, p^a and E^A are
duals to p_a and E_A.
It is assumed that the algebraic representation is defined in such a
way that the contractions satisfy the same cyclicity is satisfied that
would be satisfied for the trace. Thus, for instance,
p_c . p^d J_{ab}
= J_{ab} p_c . p^d
= (eta_{bc} p_a - eta_{ac} p_b) . p^d
= eta_{bc} delta^d_a - eta_{ac} delta^d_b,
so that
p^d J_{ab} = (eta_{bc} p^c) delta^d_a - (eta_{ac} p^c) delta^d_b.
When differential forms are involved, the cyclicity condition has to
account for transposing the terms involved in the various wedge
products. Thus, one has
A q . p = - q . p A,
since the 1-form A passes over the m-form q and (3-m) form p, for a
total of 3 transpositions.
For pure gravity with any of the Lagrangian terms I previously
described, L = L(e, F), so the only non-zero terms are P and U. The 3-
form P is the Einstein tensor(!), up to sign. The Euler-Lagrange
equation for the 2-form U involve the spin tensor, which will be
derived from U. It will lead to 0 for the torsion.
With the Lagrangians I previously described, the components U_{ab} of
the 2-form U = 1/2 U_{ab} J^{ab} will more or less just be the 2-forms
constructed from the frame, itself. So, the plane elements play the
role as the superpotential U.
The variation satisfies the ordinary Leibnitz rule, so for instance
D(A^2) = DA A + A DA.
The variationals can thus be worked out explicitly and partially
integrated:
DF . U = D(dA + A^2) . U = d(DA).U + (DA A.U + A DA.U).
Noting that
d(DA.U) = d(DA).U - DA.dU,
partial integration yields
D(dA).U = d(DA).U = d(DA.U) + DA.dU.
Thus,
DF.U = d(DA.U) + DA.dU + DA A.U + A DA.U
= d(DA.U) + DA.(dU + AU - UA).
Similarly, we obtain
DT.X = D(de + Ae).X
= d(De.X) + De.dX + DA.eX - De.X A
= d(De.X) + De.(dX - XA) + DA.(eX).
and
Dv.p = D(dq + Aq).p
= d(Dq.p) - (-1)^m Dq.dp + DA.q p - Dq. p A
= d(Dq.p) + DA.(qp) + Dq.(-pA - (-1)^m dp).
Combining these results, we get:
DL = d(-DA.U - De.X + (-1)^m Dq.p)
+ DA.(S - dU - AU + UA - eX - (-1)^m qp)
+ De.(p - dX + XA)
+ Dq.(f - dp - (-1)^m pA).
The boundary term gives you the canonical 1-form for the field theory
theta = -DA.U - De.X + (-1)^m Dq.p
out of which all the conservation laws (and ultimately, even the
Poisson-brackets and quantization) would proceed.
The remaining terms give you, as Euler-Lagrange equations, the dynamic
laws:
dU + AU - UA = J, dX - XA = P, dp + (-1)^m pA = f. (3)
where
J = S + eX + (-1)^m qp.
The decomposition
J -> S + eX
excluding the matter field, is the spin-orbit decomposition.
Since the Lagrangian (as per the previous discussion) has no explicit
dependence on A, S would be 0. The only contribution to the intrinsic
spin, therefore, would be from the matter field,
S* = (-1)^m qp.
Thus, the spin-orbit decomposition would be:
J = S* + eX.
Differentiating these equations leads to the conservation laws,
dJ + AJ - JA = FU - UF, dP + PA = -XF, df - (-1)^m fA = -pF. (4)
The first equation involving S would give you something more complex
which ultimately yields the equation for the anti-symmetric part of
the matter stress tensor.
I haven't quite yet linked this to the spin tensor, but the spin
tensor should be involved somewhere when you write out the first
equation replacing the angular momentum 3-form J by the spin 3-form
S*.
For pure gravity, equations (3) and (4) reduce to:
dU + AU - UA = 0, P = 0
and
FU - UF = 0,
with J = S = 0 and X = 0.
> Plus I'm having a hell of a time with interpreting the spin tensor.
Lately, I've approached the whole enterprise from a different
standpoint that might ultimately shed some light on matters here.
The simplest way to write down the kinematic and dynamic field laws is
to express the connection part of the field as a Lorentz gauge field,
as I mentioned before:
A = 1/2 Gamma^{ab} J_{ab}.
Then express the Lie bracket for the Lorentz gauge generators in
algebraic form as commutators:
[J_{ab}, J_{cd}] = J_{ab} J_{cd} - J_{cd} J_{ab}.
Combine this algebra with that for the differential forms
Gamma^{ab} = eta^{bc} Gamma^a_{dc} e^d.
So the product of A with itself will be
A^2 = 1/4 Gamma^{ab} J_{ab} Gamma^{cd} J_{cd}
= 1/4 Gamma^{ab} ^ Gamma^{cd} (J_{ab} J_{cd}).
Since the wedge product is anti-symmetric, then this may be
equivalently written as
A^2 = 1/8 (Gamma^{ab} ^ Gamma^{cd}) (J_{ab} J_{cd} - J_{cd} J_{ab})
= 1/8 (Gamma^{ab} ^ Gamma^{cd}) [J_{ab}, J_{cd}].
The Lie bracket for the Lorentz group is
[J_{ab}, J_{cd}] = eta_{bc} J_{ad} - eta_{bd} J_{ac} - eta_{ac}
J_{bd} + eta_{ad} J_{bc}.
So, as a result, this gives you the correct form for the quadratic
terms which, when combined with dA, gives you the curvature as the
corresponding field strength:
F = dA + A^2 = 1/2 R^{ab} J_{ab}.
The frame field and matter fields can be treated in the same way,
together. So, suppose you have a matter field expressed as a m-form
field,
q = (1/m!) q^{A}_{n1 ... nm} E_{A} e^{n1} ^ ... e^{nm}.
Here the matter field is assumed to have a field index A identifying
components with respect to some linear space whose basis is E_{A}.
The frame field is subsumed under this as a 1-form field
e = e^a p_a
where the "basis" p_a is sometimes likened to as the translation
generators of a Poincare' gauge field.
Assume the algebraic representation is extended so that the Lorentz
gauge generators act on the left. Thus, for the frame field, one has
J_{ab} (p_c) = eta_{bc} p_a - eta_{ac} p_b.
For a matter field, what you get will depend on what kind of index A
is. If it's a tensor index, then A is actually a composite consisting
of a number of space-time indices in the upper position and a number
in the lower position. For a scalar m-form valued field, A will reduce
to an empty slot and there will be no indices there.
With these items in mind, the torsion and covariant derivatives can be
written in algebraic form as
T = de + Ae, v = dq + Aq.
The covariant derivatives of "q" give you the gauge-covariant
"velocities" v associated with the field.
Thus, you have a set of equations relating field-strengths (and
"velocities") to gauge-potentials (and field components):
F = dA + A^2, T = de + Ae, v = dq + Aq. (1)
>From these follow the Bianchi identities, using the Leibnitz rule for
differential forms. Thus,
d(A^2) = (dA) A - A (dA), d(Ae) = (dA) e - A (de); d(Aq) = (dA)q -
A(dq);
while
d(dA) = 0, d(de) = 0, d(dq) = 0.
Thus, you find that
dF = (dA) A - A (dA) = (F - A^2) A - A (F - A^2) = FA - AF
dT = (dA) e - A (de) = (F - A^2) e - A (T - Ae) = Fe - AT
dv = (dA) q - A (dq) = (F - A^2) q - A (v - Aq) = Fq - Av.
Putting these together, you get the Bianchi identities and the
"homogeneous" field equations for the field q:
dF + AF - FA = 0, dT + AT = Fe, dv + Av = Fq. (2)
The Bianchi identity (dT + AT = Fe) is the one that gives you R_{abcd}
+ R_{acdb} + R_{adbc} = 0 when the torsion is absent.
To determine the dynamics, assume the Lagrangian is a function of only
the potentials and covariant derivatives:
L = L(A, e, q, F, T, v).
This here is the Lagrangian 4-form. So, written in terms of the
Lagrangian density *L, it would be:
L = *L e^1 ^ e^2 ^ e^3 ^ e^4.
But it's best to work with the 4-form, instead of with the density.
As per the previous discussion, there will be no explicit dependence
on A, but it'll be kept in for now. The total variational in the
Lagrangian (using D as ASCII for "delta") will then factor out to
something of the form:
DL = DA . S + De . P + Dq . f - DF . U - DT . X + (-1)^m Dv . p.
Here, the indices attached to e and q (e^a, q^A) will be complemented
by indices attached to P and f (P_a, f_A). Similarly for the other
terms. Therefore, it is assumed that the expression above involves a
contraction on these indices. Thus, for instance,
De . P = De^a ^ P_a, Dq . f = Dq^A ^ f_A.
This introduces the dynamic fields S, P, f, U, X and p. The fields S,
P are 3-form currents (for spin and momentum). The field f is (4-m)
form valued. The fields U, X are 2-forms and function as
"superpotentials" for S and P. The field X will ultimately be involved
in giving you the extra "orbital" contribution to the spin current 3-
form J. The field p is a (3-m) form.
To define covariant derivatives for these fields, the action of the
Lorentz group has to be defined for the corresponding bases. Thus, the
fields written out fully are:
S = 1/2 S_{ab} j^{ab}, U = 1/2 U_{ab} j^{ab}
P = P_a p^a, X = X_a p^a
f = f_A E^A, p = p_A E^A,
where j^{ab} is the dual to the Lie basis J_{ab}, p^a and E^A are
duals to p_a and E_A.
It is assumed that the algebraic representation is defined in such a
way that the contractions satisfy the same cyclicity is satisfied that
would be satisfied for the trace. Thus, for instance,
p_c . p^d J_{ab}
= J_{ab} p_c . p^d
= (eta_{bc} p_a - eta_{ac} p_b) . p^d
= eta_{bc} delta^d_a - eta_{ac} delta^d_b,
so that
p^d J_{ab} = (eta_{bc} p^c) delta^d_a - (eta_{ac} p^c) delta^d_b.
When differential forms are involved, the cyclicity condition has to
account for transposing the terms involved in the various wedge
products. Thus, one has
A q . p = - q . p A,
since the 1-form A passes over the m-form q and (3-m) form p, for a
total of 3 transpositions.
For pure gravity with any of the Lagrangian terms I previously
described, L = L(e, F), so the only non-zero terms are P and U. The 3-
form P is the Einstein tensor(!), up to sign. The Euler-Lagrange
equation for the 2-form U involve the spin tensor, which will be
derived from U. It will lead to 0 for the torsion.
With the Lagrangians I previously described, the components U_{ab} of
the 2-form U = 1/2 U_{ab} J^{ab} will more or less just be the 2-forms
constructed from the frame, itself. So, the plane elements play the
role as the superpotential U.
The variation satisfies the ordinary Leibnitz rule, so for instance
D(A^2) = DA A + A DA.
The variationals can thus be worked out explicitly and partially
integrated:
DF . U = D(dA + A^2) . U = d(DA).U + (DA A.U + A DA.U).
Noting that
d(DA.U) = d(DA).U - DA.dU,
partial integration yields
D(dA).U = d(DA).U = d(DA.U) + DA.dU.
Thus,
DF.U = d(DA.U) + DA.dU + DA A.U + A DA.U
= d(DA.U) + DA.(dU + AU - UA).
Similarly, we obtain
DT.X = D(de + Ae).X
= d(De.X) + De.dX + DA.eX - De.X A
= d(De.X) + De.(dX - XA) + DA.(eX).
and
Dv.p = D(dq + Aq).p
= d(Dq.p) - (-1)^m Dq.dp + DA.q p - Dq. p A
= d(Dq.p) + DA.(qp) + Dq.(-pA - (-1)^m dp).
Combining these results, we get:
DL = d(-DA.U - De.X + (-1)^m Dq.p)
+ DA.(S - dU - AU + UA - eX - (-1)^m qp)
+ De.(p - dX + XA)
+ Dq.(f - dp - (-1)^m pA).
The boundary term gives you the canonical 1-form for the field theory
theta = -DA.U - De.X + (-1)^m Dq.p
out of which all the conservation laws (and ultimately, even the
Poisson-brackets and quantization) would proceed.
The remaining terms give you, as Euler-Lagrange equations, the dynamic
laws:
dU + AU - UA = J, dX - XA = P, dp + (-1)^m pA = f. (3)
where
J = S + eX + (-1)^m qp.
The decomposition
J -> S + eX
excluding the matter field, is the spin-orbit decomposition.
Since the Lagrangian (as per the previous discussion) has no explicit
dependence on A, S would be 0. The only contribution to the intrinsic
spin, therefore, would be from the matter field,
S* = (-1)^m qp.
Thus, the spin-orbit decomposition would be:
J = S* + eX.
Differentiating these equations leads to the conservation laws,
dJ + AJ - JA = FU - UF, dP + PA = -XF, df - (-1)^m fA = -pF. (4)
The first equation involving S would give you something more complex
which ultimately yields the equation for the anti-symmetric part of
the matter stress tensor.
I haven't quite yet linked this to the spin tensor, but the spin
tensor should be involved somewhere when you write out the first
equation replacing the angular momentum 3-form J by the spin 3-form
S*.
For pure gravity, equations (3) and (4) reduce to:
dU + AU - UA = 0, P = 0
and
FU - UF = 0,
with J = S = 0 and X = 0.