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% \filename{em2b.tex} \version{19-sep-2001<-20-nov-1997<-4-june-1997 <-24-feb-1995}
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\Newpage
\Typeout{Update with diagrams illustrating spatial curvature, synchronization
defect:}
\Section{Spatial curvature and torsion}
\Subsection{Definitions}
Having explored the spatial and spacetime connections, the next
step is to study the curvatures associated with both. The spatial versions
of the usual invariant formulas defining the torsion and Riemann curvature
tensors of a connection may be used to define the spatial torsion and
curvature tensors associated with the spatial part of the spatial connection.
Expressing them in terms of an observer-adapted frame
then gives the component definitions.
The spatial torsion definition gives no surprises, vanishing under the
assumption that the spacetime torsion is zero
\beq\eqalign{
& \del(u)\sub{X} Y- \del(u)\sub{Y} X -[X,Y](u) = \Tb(u)(X,Y) =0\ ,\cr
& \Tb(u)^a{}_{bc} = 2 \Gamma(u)^a{}_{[bc]} - C^a{}_{bc} =0\ ,\cr
}\eeq
where $X$ and $Y$ are spatial vector fields.
However, the rotation of the congruence leads to an additional temporal
derivative term in the spatial curvature formula which arises from the
temporal part of the commutator of two spatial vector fields.
Three different curvature tensors then result
from the choice of the Lie, Fermi-Walker or
co-rotating Fermi-Walker temporal derivative to
express the additional term
\beq\eqalign{
\big\{ [ \del(u)\sub{X}, \del(u)\sub{Y}] -\del(u)\sub{[X,Y]}\big\} Z
&= R\tem(u)(X,Y) Z
+ 2\omega(u)^\flat(X,Y) \del\tem(u) Z\ , \cr
& \qquad
{\scriptstyle {\rm tem} \,=\, {\rm lie, fw, cfw} } \ ,
}\eeq
where $X$, $Y$, and $Z$ are spatial vector fields. These three tensors,
the {\it Lie spatial curvature tensor} and the {\it Fermi-Walker spatial
curvature tensor} and the {\it co-rotating Fermi-Walker
spatial curvature tensor}
differ by the same kinematical terms as the temporal derivatives
themselves but reversed in sign
\beq\eqalign{
R\fw(u)(X,Y) Z &= R\lie(u)(X,Y) Z
+ 2\omega(u)^\flat(X,Y) k(u) \rightcontract Z\ ,\message{sign??}\cr
R\cfw(u)(X,Y) Z &= R\lie(u)(X,Y) Z
- 2\omega(u)^\flat(X,Y) \theta(u) \rightcontract Z\ ,\message{sign??}\cr
}\eeq
Expressing the defining identity
in an observer-adapted frame leads to the formula
\beq\eqalign{
R\tem(u)^a{}_{bcd} &= \Gamma(u)^a{}_{db,c} - \Gamma(u)^a{}_{cb,d}
-\Gamma(u)^a{}_{fb} C^f{}_{cd} \cr
&\qquad +\Gamma(u)^a{}_{cf} \Gamma(u)^f{}_{db}
-\Gamma(u)^a{}_{df} \Gamma(u)^f{}_{cb}
- 2 C\tem(u)^a{}_b \omega(u)_{cd} \ .\cr
}\eeq
The three curvature tensors are related to each other by
\beq\eqalign{
R\fw(u)^a{}_{bcd} &= R\lie(u)^a{}_{bcd} + 2 k(u)^a{}_b \omega(u)_{cd}\ ,\cr
R\cfw(u)^a{}_{bcd} &= R\lie(u)^a{}_{bcd}
- 2 \theta(u)^a{}_b \omega(u)_{cd}\ .\cr
}\eeq
These are clearly antisymmetric in their final pair of indices, but the
remaining symmetries characterizing
an ordinary curvature tensor must be checked.
The formula for the components of these spatial curvature tensors
in terms of the spatial connection components
and the spatial structure functions differs from the usual one only in the
last term. By choosing an observer-adapted frame whose spatial frame
undergoes the corresponding transport, this term is set to zero.
For the Lie spatial curvature tensor
this term is absent in a spatially-comoving spatial frame,
and hence in the discussions of Zel'manov and Ferrarese who
rely on such frames.
%%%%%%%%%
\Subsection{Algebraic symmetries}
An alternative way to obtain these definitions which reveals the symmetry
properties of the first pair of indices relies on the spatial
Ricci identity for a spatial vector field or 1-form,
which for zero torsion have the form
\beq\eqalign{
[\del(u)_\gamma, \del(u)_\delta] Z^\alpha &= 2Z^\alpha{}_{||[\delta\gamma]}\cr
&=R\tem(u)^\alpha{}_{\beta\gamma\delta} Z^\beta
+ 2\omega(u)_{\gamma\delta} \del\tem(u) Z^\alpha\ ,\cr
[\del(u)_\gamma, \del(u)_\delta] Z_\alpha &= 2 Z_{\alpha||[\delta\gamma]}\cr
&= - Z_\beta R\tem(u)^\beta{}_{\alpha\gamma\delta}
+ 2\omega(u)_{\gamma\delta} \del\tem(u) Z_\alpha\ ,\cr
&\qquad {\scriptstyle {\rm tem} \,=\, {\rm lie, fw, cfw} } \ .
}\eeq
Since the spatial metric has zero ordinary or co-rotating Fermi-Walker
temporal
derivative, the two Ricci identities for a vector field and a 1-form
are clearly related in a consistent way by index shifting
as long as the corresponding
curvature tensor has the usual antisymmetry property in
the first pair of indices
\beq
R\fw(u)^{(\alpha\beta)}{}_{\gamma\delta}
= R\cfw(u)^{(\alpha\beta)}{}_{\gamma\delta}
= 0 \ .
\eeq
This implies that the symmetric part of the Lie
spatial curvature tensor is not zero
\beq
R\lie(u)^{(\alpha\beta)}{}_{\gamma\delta} =
2 \theta(u)^{\alpha\beta} \omega(u)_{\gamma\delta}\ ,
\eeq
which is exactly the condition that shifting indices in
these Ricci identities be consistent when expressed in terms of that
curvature and the Lie derivative term.
The commutator of the Lie derivative with index shifting generates
the expansion tensor term
\beq
\Lie(u)\sub{u} [ P(u)_{\alpha\beta} Z^\beta ] -
P(u)_{\alpha\beta} \Lie(u)\sub{u} Z^\beta
= [\Lie(u)\sub{u} P(u)_{\alpha\beta}] Z^\beta
= 2\theta(u)_{\alpha\beta} Z^\beta \ .
\eeq
These equivalent conditions on the symmetric part of the
curvature tensors are in turn equivalent to the
spatial Ricci identity for the spatial metric itself,
modified by the rotation term from the noncommutivity of the
spatial derivatives
\beq\eqalign{
0= [\del(u)_\gamma,\del(u)_\delta] P(u)_{\alpha\beta}
&= -2 R\lie(u)_{(\alpha\beta)\gamma\delta}
+ 2\omega(u)_{\gamma\delta} \Lie(u)\sub{u} P(u)_{\alpha\beta} \cr
&= -2 R\fw(u)_{(\alpha\beta)\gamma\delta}
= -2 R\cfw(u)_{(\alpha\beta)\gamma\delta} \ .\cr
}\eeq
The relation between the curvature tensors is therefore equivalent to
\beq\eqalign{
R\fw(u)_{\alpha\beta\gamma\delta} &=
R\lie(u)_{[\alpha\beta]\gamma\delta}
+ 2\omega(u)_{\alpha\beta} \omega(u)_{\gamma\delta} \ ,\cr
R\cfw(u)_{\alpha\beta\gamma\delta} &=
R\lie(u)_{[\alpha\beta]\gamma\delta} \ .\cr
}\eeq
\message{check?? finish here}
The component formula for the Lie spatial curvature tensor is identical
with the usual frame component formula for an actual Riemannian 3-manifold,
at least in a spatially-comoving frame.
Thus any algebraic identity which follows directly from the latter formula
with no index shifting (which can lead to additional expansion tensor terms
when derivatives are involved)
continues to hold for the Lie spatial curvature
tensor. The usual cyclic
symmetry of the torsionfree Bianchi
identity of the first kind therefore holds
\beq
3 R\lie(u)^\alpha{}_{[\beta\gamma\delta]} =
R\lie(u)^\alpha{}_{\beta\gamma\delta}
+ R\lie(u)^\alpha{}_{\gamma\delta\beta}
+ R\lie(u)^\alpha{}_{\delta\beta\gamma} = 0 \ ,
\eeq
as one may directly verify by expressing it in terms of the component
formula and noting that all terms cancel in pairs. This implies
that the other spatial curvature tensors do not
satisfy the usual form of the first Bianchi identity
\beq\eqalign{
R\fw(u)^\alpha{}_{[\beta\gamma\delta]} &=
\fraction13 [ R\fw(u)^\alpha{}_{\beta\gamma\delta}
+ R\fw(u)^\alpha{}_{\gamma\delta\beta}
+ R\fw(u)^\alpha{}_{\delta\beta\gamma}] \cr
&= 2 k(u)^\alpha{}_{[\beta} \omega(u)_{\gamma\delta]} \ ,\cr
R\cfw(u)^\alpha{}_{[\beta\gamma\delta]}
&= - 2 \theta(u)^\alpha{}_{[\beta} \omega(u)_{\gamma\delta]} \ .\cr
}\eeq
%%%%%%%%%%%%%%%
\Subsection{Symmetry-obeying spatial curvature}
One can isolate the kinematical terms in the spatial curvature which
arise from the commutator of spatial derivatives by making the second
derivative terms in the curvature formula explicit. In this
observer-adapted frame component derivation one needs the
following relations
\beq\eqalign{
0 &= \del(u)_c h_{ab} = \partial_c h_{ab}
- \Gamma(u)_{acb} - \Gamma(u)_{bca}\ ,\cr
0 &= \del(u)_c h^{ab} = \partial_c h^{ab}
+ \Gamma(u)^a{}_c{}^b + \Gamma(u)^b{}_c{}^a\ , \cr
}\eeq
reflecting the covariant constancy of the spatial metric and its inverse.
Then one has
\beq\eqalign{
R\lie(u)_{abcd}
&= h_{ae}[ \partial_c \Gamma(u)^e{}_{db} - \partial_d \Gamma(u)^e{}_{cb}]\cr
&\qquad + \Gamma(u)_{ace} \Gamma(u)^e{}_{db}
- \Gamma(u)_{ade} \Gamma(u)^e{}_{cb} -C^e{}_{cd} \Gamma(u)_{aeb}
-2C_{a\top b} \omega(u)_{cd}\cr
&= [ \partial_c \Gamma(u)_{adb} - \partial_d \Gamma(u)_{acb}]
+ h_{ae}[ \partial_c h^{ef} \Gamma(u)_{fdb}
- \partial_d h^{ef} \Gamma(u)^e{}_{cb}] + \ldots \cr
&= [ \partial_c \Gamma(u)_{adb} - \partial_d \Gamma(u)_{acb}] \cr
&\qquad - [\Gamma(u)_{ace} + \Gamma(u)_{eca}]\Gamma(u)^e{}_{db}
+ [\Gamma(u)_{ade} + \Gamma(u)_{eda}]\Gamma(u)^e{}_{cb} + \cdots \cr
&= [ \partial_c \Gamma(u)_{adb} - \partial_d \Gamma(u)_{acb}] \cr
&\qquad - \Gamma(u)_{eca} \Gamma(u)^e{}_{db}
+ \Gamma(u)_{eda}\Gamma(u)^e{}_{cb}
-C^e{}_{cd} \Gamma(u)_{aeb}
-2C_{a\top b} \omega(u)_{cd} \ .\cr
}\eeq
Now one can substitute for the covariant gammas using the derivative
formulas, but the calculation is much easier in a spatially-holonomic
($C^a{}_{bc}=0$) spatially-comoving ($C^a{}_{\top b} =0$) spatial frame
\beq\eqalign{
R\lie(u)_{abcd} &= \half [ h_{ab,dc} - h_{bd,ac} + h_{da,bc}
-h_{ab,cd} + h_{bc,ad} - h_{ca,bd} ] \cr
&\quad -\Gamma(u)^e{}_{ca} \Gamma(u)_{edb}
+\Gamma(u)^e{}_{da} \Gamma(u)_{ecb} \cr
&= \half [\partial_c,\partial_d] h_{ab}
+ \half [ h_{ad,bc} - h_{ac,bd} + h_{bc,ad} - h_{bd,ac} ]
+ \cdots \cr
&= 2\theta(u)_{ab} \omega(u)_{cd}
+ \half [ h_{ad,bc} - h_{ac,bd} + h_{bc,ad} - h_{bd,ac} ]
+ \cdots \ .\cr
}\eeq
Now one sees explicitly the expansion rotation factor responsible for
the symmetric part of the curvature,
but the remainder of the expression
(chosen by Cattaneo-Gasperini as the spatial curvature tensor \Cite{1961})
still has hidden kinematical terms in the antisymmetric part of the
second derivative terms.
Making these explicit leads to the expression
\beq\eqalign{
R\lie(u)_{abcd} &= 2 \theta(u)_{ab} \omega(u)_{cd} \cr
&\quad +[ -\theta(u)_{ad} \omega(u)_{bc}
+\theta(u)_{ac} \omega(u)_{bd}
-\theta(u)_{bc} \omega(u)_{ad}
+\theta(u)_{bd} \omega(u)_{ac} ]\cr
&\quad + \half [ h_{ad,(bc)} - h_{ac,(bd)} + h_{bc,(ad)} - h_{bd,(ac)} ]
+ \cdots \ .\cr
}\eeq
As noted by Ferrarese \Cite{1965},
the final part of the formula following the explicit kinematical terms
has all the algebraic symmetries of the corresponding identical coordinate
formula on a Riemannian manifold and therefore has all of the algebraic
symmetries of the usual covariant Riemann tensor.
This {\it symmetry-obeying spatial curvature tensor} has the expression
\beq\eqalign{
R\sym (u)_{abcd} &=
\half [ h_{ad,(bc)} - h_{ac,(bd)} + h_{bc,(ad)} - h_{bd,(ac)} ] \cr
&\quad -\Gamma(u)^e{}_{ca} \Gamma(u)_{edb}
+\Gamma(u)^e{}_{da} \Gamma(u)_{ecb} \cr
}\eeq
in a spatially-holonomic spatially-comoving frame, and is related to the Lie
spatial curvature tensor and other curvature tensors (independent of the frame) by
\beq\eqalign{
R\sym (u)^{ab}{}_{cd} &= R\lie(u)^{ab}{}_{cd}
- 2 \theta(u)^{ab} \omega(u)_{cd}
-4 \theta(u)^{[a}{}_{[c} \omega(u)^{b]}{}_{d]} \cr
&= R\lie(u)^{[ab]}{}_{cd}
-4 \theta(u)^{[a}{}_{[c} \omega(u)^{b]}{}_{d]} \cr
&= R\cfw(u)^{ab}{}_{cd}
-4 \theta(u)^{[a}{}_{[c} \omega(u)^{b]}{}_{d]} \cr
&= R\fw(u)^{ab}{}_{cd}
-4 \theta(u)^{[a}{}_{[c} \omega(u)^{b]}{}_{d]}
- 2\omega(u)^{ab}\omega(u)_{cd}\ . \cr
}\eeq
Since both $R\lie(u)_{abcd}$ and $R\sym (u)_{abcd}$ satisfy the usual
cyclic symmetry of the torsionfree Bianchi
identity of the first kind,
the two kinematical difference terms together must also, as
can be easily verified. Absorbing the first of these into the Lie spatial
curvature to yield its antisymmetric part (in the first pair of indices)
then explains why that antisymmetric part alone
does not satisfy this cyclic symmetry.
%%%%%%%%%%
\Subsection{Spatial Ricci tensors and scalar curvatures}
Neither one of the two kinematical difference terms between the
Lie and symmetry-obeying spatial curvature tensors
contribute to the curvature scalar
\beq
R\sym (u)^{ab}{}_{ab} = R\lie(u)^{ab}{}_{ab}=
R\cfw(u)^{ab}{}_{ab}
= R\fw(u)^{ab}{}_{ab} - 2 \omega(u)^{ab} \omega(u)_{ab}
\eeq
but they do make a difference in the spatial Ricci and Einstein tensors.
It does not make much sense to define the Lie spatial Ricci tensor by taking
the trace of $R\lie(u)^{ab}{}_{cd}$
by contracting either one or the other of the
indices of the first index pair against one from the second antisymmetric
pair since two different tensors result which differ by a kinematical term
associated with the symmetric part of the first index pair.
The trace using instead the antisymmetric part of the first index pair
does result in a unique Lie spatial Ricci tensor,
as does the trace of the Fermi-Walker spatial curvature as well as
the trace of
the ``symmetry-obeying" spatial curvature tensor, but only the last Ricci
tensor is symmetric
\beq\eqalign{
R\lie(u)^a{}_b &= R\lie(u)^{[ca]}{}_{cb} \ ,\cr
R\fw(u)^a{}_b &= R\lie(u)^a{}_b + 2 \omega(u)^{ca} \omega(u)_{cb} \ ,\cr
R\sym (u)^a{}_b &= R\sym (u)^{ca}{}_{cb} \ . \cr
}\eeq
The relation between the Lie and the symmetry-obeying
Ricci tensors is
\beq\eqalign{
R\sym (u)^a{}_b
&= R\lie(u)^a{}_b -4 \theta(u)^{[c}{}_{[c} \omega(u)^{a]}{}_{b]}\cr
&= R\lie(u)^a{}_b - \theta(u)^c{}_c \omega(u)^a{}_b
+ \theta(u)^c{}_b \omega(u)^a{}_c
- \theta(u)^{ca} \omega(u)_{bc}\ ,\cr
}\eeq
so
\beq
0= R\sym (u)_{[ab]} = R\lie(u)_{[ab]}
-\theta(u)^c{}_c \omega(u)_{ab}
- 2\theta(u)^c{}_{[a} \omega(u)_{b]c} \ ,
\eeq
hence
\beq\eqalign{
R\tem (u)_{[ab]} &=
\theta(u)^c{}_c \omega(u)_{ab}
+ 2\theta(u)^c{}_{[a} \omega(u)_{b]c} \ ,\cr
&\qquad {\scriptstyle {\rm tem} \,=\, {\rm lie, fw, cfw} } \ .\cr
}\eeq
This antisymmetric part of the spatial Ricci tensor comes from the
lack of pair interchange symmetry in the corresponding
spatial curvature tensor.
In the case of a nonrotating observer congruence, all four
spatial curvature tensors coincide and are equivalent to the Riemann curvature
tensor of the Riemannian 3-manifolds which result from the induced metric
on the hypersurfaces orthogonal to the congruence.
In the general rotating case, one may use any one of the four distinct
tensors, although one may be better suited than the others in a particular
context in terms of absorbing extra kinematical terms.
In the decomposition of the Einstein equations, it is clearly preferable
to have a symmetric spatial Ricci or Einstein tensor, so the symmetry-obeying
choice of spatial curvature tensor is perhaps the cleanest separation of
spatial curvature from kinematical terms which occur there.
In the special case of an observer congruence with vanishing expansion tensor,
as occurs in stationary spacetimes if the congruence is taken along the
Killing trajectories, the Lie, co-rotating Fermi-Walker
and symmetry-obeying choices agree
but differ from the Fermi-Walker choice
by a term quadratic in the rotation.
%%%%%%%%%%%
\Subsection{Pair interchange symmetry}
The pair interchange symmetry of an ordinary curvature tensor follows
from the antisymmetry of the first index pair together with the
cyclic symmetry of the Bianchi identity of the first kind.
The symmetry-obeying spatial curvature tensor has all of these
symmetries
\beq
R\sym(u)_{\alpha\beta\gamma\delta}
= R\sym(u)_{\gamma\delta\alpha\beta}\ .
\eeq
The relationship between the symmetry-obeying and Lie spatial curvatures
then implies the following identity for the latter curvature
\beq\eqalign{
R\lie(u)^{[\alpha\beta]}{}_{\gamma\delta} -
R\lie(u)_{[\gamma\delta]}{}^{\alpha\beta}
&= R\sym(u)^{\alpha\beta}{}_{\gamma\delta} -
R\sym(u)_{\gamma\delta}{}^{\alpha\beta} \cr
&\qquad + 4 \theta(u)^{[\alpha}{}_{[\gamma} \omega(u)^{\beta]}{}_{\delta]}
- 4 \theta(u)_{[\gamma}{}^{[\alpha} \omega(u)_{\delta]}{}^{\beta]} \cr
&= 8 \theta(u)^{[\alpha}{}_{[\gamma} \omega(u)^{\beta]}{}_{\delta]} \cr
&= R\fw(u)^{\alpha\beta}{}_{\gamma\delta} -
R\fw(u)_{\gamma\delta}{}^{\alpha\beta} \cr
&= R\cfw(u)^{\alpha\beta}{}_{\gamma\delta} -
R\cfw(u)_{\gamma\delta}{}^{\alpha\beta} \cr
}\eeq\message{check??}%
or equivalently
\beq\eqalign{
R\lie(u)&^{\alpha\beta}{}_{\gamma\delta} -
R\lie(u)_{\gamma\delta}{}^{\alpha\beta} \cr
&= 8 \theta(u)^{[\alpha}{}_{[\gamma} \omega(u)^{\beta]}{}_{\delta]}
+ 2 \theta(u)^{\alpha\beta} \omega(u)_{\gamma\delta}
- 2 \theta(u)_{\gamma\delta} \omega(u)^{\alpha\beta} \ .\cr
}\eeq
The trace of the first of these relations
provides another way to obtain the antisymmetric
part of the Lie spatial Ricci tensor.
%%%%%%%%%%%%
\Subsection{Spatial covariant exterior derivative}\label{sec:sc}%
The differential form approach to curvature may also be developed in the
spatial context. Just as in the spacetime context, the
{\it {spatial covariant exterior derivative}} defined
for {\it spatial tensor-valued
differential forms} is very useful in concisely presenting many identities.
This operator interpolates between the spatial exterior derivative
for scalar-valued (i.e., ordinary) differential forms and the spatial covariant
derivative for tensor-valued zero-forms, and is entirely analogous
to the corresponding covariant exterior derivative on spacetime, as described
for a general manifold in appendix A.
Suppose $S^{\alpha\ldots}_{\ \ \beta\ldots\gamma_1\ldots\gamma_p}$
is a spatial tensor which is antisymmetric in a group of $p$ covariant indices.
One may think of it as a tensor-valued $p$-form, i.e., when its $p$ covariant
arguments are evaluated, one is left with a tensor with the remaining
index structure. One has in invariant notation
\beq\eqalign{
S^{\alpha\ldots}_{\ \ \beta\ldots} &=
{1\over p!} S^{\alpha\ldots}_{\ \ \beta\ldots\gamma_1\ldots\gamma_p}
\omega^{\gamma_1}\wedge \cdots \wedge \omega^{\gamma_p}\ ,\cr
S &= e_\alpha \otimes \cdots \otimes \omega^\beta \otimes \cdots \otimes
S^{\alpha\ldots}_{\ \ \beta\ldots} \ ,\cr
}\eeq
and the spatial covariant exterior derivative is then defined by
\beq\eqalign{
D(u) S &= e_\alpha \otimes \cdots \otimes \omega^\beta \otimes \cdots \otimes
D(u) S^{\alpha\ldots}_{\ \ \beta\ldots} \ ,\cr
&=
e_\alpha \otimes \cdots \otimes \omega^\beta \otimes \cdots \otimes
d(u) S^{\alpha\ldots}_{\ \ \beta\ldots} \cr
&\qquad
+ \del(u) \wedge
[ e_\alpha \otimes \cdots \otimes \omega^\beta \otimes \cdots ]\otimes
S^{\alpha\ldots}_{\ \ \beta\ldots} \ ,\cr
}\eeq
where the notation $\del \wedge$ is meant to
indicate the wedge product of the additional covariant derivative index
with the $p$-form.
The first line of the above definition emphasizes the sloppy classical
notation of not distinguishing the components of the covariant derivative
from the derivative of the components. The former interpretation is
assumed to hold for the occurrence in this formula.
In an observer-adapted frame, the component formula is directly analogous
to the general one. If $S$ is a spatial tensor-valued differential form
of type $\sigma$ (describing the tensor-valued index structure) then
its spatial covariant exterior derivative is
\beq\eqalign{
S &= e_a \otimes \cdots \otimes \omega^b \otimes \cdots \otimes
S^{a\ldots}_{\ \ b\ldots} \ ,\cr
D(u) S^{a\ldots}_{\ \ b\ldots} &=
d(u) S^{a\ldots}_{\ \ b\ldots}
+ [\sigma(\bfomega(u))\wedge S ]^{a\ldots}_{\ \ b\ldots}\ , \cr
D(u) S &= e_a \otimes \cdots \otimes \omega^b \otimes \cdots \otimes
D(u) S^{a\ldots}_{\ \ b\ldots} \ ,\cr
}\eeq
where
\beq
\bfomega(u)= e_a \otimes \bfomega(u)^a{}_b \otimes \omega^b \ ,\qquad
\bfomega(u)^a{}_b = \Gamma(u)^a{}_{cb}\omega^c
\eeq
is the matrix of spatial connection 1-forms thought of as a
$1\choose1$-``tensor"-valued 1-form and the wedge product is meant
between the connection 1-form and the $p$-form indices.
By introducing the spatial identity
tensor as a vector-valued 1-form
\beq
\bfvartheta(u) = P(u) = e_a \otimes \omega^a \ ,\qquad
\bfvartheta(u)^a = \omega^a\ ,
\eeq
one obtains the vector-valued
spatial torsion 2-form as its spatial covariant exterior
derivative
\beq
\bfTheta(u) = D(u) \bfvartheta(u) =
d(u) \bfvartheta(u) + \bfomega(u) \rightcontract \wedge \bfvartheta(u)
\eeq
where the unusual notation $\rightcontract\wedge$
necessary for the index-free notation
indicates a right contraction between the adjacent tensor-valued indices
and an exterior product between the differential form indices.
In the index-notation one has explicitly
\beq\eqalign{
\bfTheta(u)^a &= D(u) \bfvartheta(u)^a =
d(u) \bfvartheta(u)^a + \bfomega(u)^a{}_b \wedge \bfvartheta(u)^b \cr
&= \half \Tb(u)^a{}_{bc} \omega^b\wedge \omega^c =0\ .\cr
}\eeq
Similarly the Lie spatial curvature tensor
thought of as a $1\choose1$-tensor-valued 2-form has the definition
\beq\eqalign{
\bfOmega\lie(u) &= d(u) \bfomega(u)
+ \bfomega(u) \rightcontract \wedge \bfomega(u)
- 2\bfC\lie(u) \omega(u)^\flat \cr
\bfOmega\lie(u)^a{}_b &= d(u) \bfomega(u)^a{}_b
+ \bfomega(u)^a{}_c \wedge \bfomega(u)^c{}_b
- 2 C\lie(u)^a{}_b \omega(u)_{cd} \cr
&= \half R\lie(u)^a{}_{bcd} \, \omega^c\wedge \omega^d\ ,\cr
}\eeq
where the unfamiliar extra terms are necessary for the spatial Lie
derivatives which enter into this geometry in the case of nonzero rotation.
The spatial Ricci identities for spatial tensors may then be written in the
language of tensor-valued differential forms as
\beq
D(u)^2 S = \sigma(\bfOmega\lie(u)) \wedge S
+ 2 \omega(u)^\flat \wedge \Lie(u)\sub{u} S \ .
\eeq
The spatial Bianchi identity of the first kind is easily
derived in this notation
\beq\eqalign{
0 &= D(u) \bfTheta(u) = D(u)^2 \bfvartheta(u) \cr
&= \bfOmega\lie(u)\rightcontract \wedge \bfvartheta(u)
+ 2 \omega(u)^\flat \wedge \Lie(u)\sub{u} \bfvartheta(u)\ ,\cr
}\eeq
but since
\beq
\Lie(u)\sub{u} \bfvartheta(u) =
\Lie(u)\sub{u} P(u) =0\ ,
\eeq
one has
\beq
\bfOmega\lie(u)^a{}_b \wedge \omega^b
= \half R\lie(u)^a{}_{[bcd]} \omega^b \wedge \omega^c \wedge \omega^c
=0\ ,
\eeq
which when expressed in components gives the usual form of the identity.
Similarly the symmetric part of the spatial curvature comes from the
Ricci identity for the spatial metric viewed as a
``tensor-valued 0-form"
\beq\eqalign{
0&=D(u)^2 P(u)_{\alpha\beta} \cr
&= -2\bfOmega\lie(u)_{(\alpha\beta)} +
2\omega(u)^\flat \wedge \Lie(u)\sub{u} P(u)_{\alpha\beta}
\ , \cr
}\eeq
which leads to the previous result.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\Newpage
\Section{The symmetrized curl operator for symmetric spatial 2-tensors}
\label{sec:Scurl}
For a spatial vector-valued 1-form, the spatial dual of the spatial covariant
exterior derivative leads to a new vector valued 1-form which might naturally
be called the spatial curl of the tensor.
\beq
[\curl(u) T]^\alpha{}_\beta = [ \dualp{u} D(u) T^\alpha ]_\beta
= \eta_\beta^{\gamma\delta} \del(u)_\gamma T^\alpha{}_{\delta} \ ,
\eeq
If the spatial tensor $X^\alpha{}_\beta$ is antisymmetric, with
spatial dual $[\dualp{u}X]^\alpha$,
then this curl may be re-expressed as
\beq
[\curl(u) X]^{\alpha}{}_{\beta}
= \del(u)_\beta [\dualp{u}X]^\alpha
- \delta^\alpha{}_\beta \del(u)_\gamma [\dualp{u}X]^\gamma
\eeq
with trace $-2 \div \dualp{u} X$.
If one begins
with a symmetric such tensor, the result is automatically
is tracefree as a tensor
\beq\eqalign{
[\curl(u) S]^\alpha{}_\beta = [ \dualp{u} D(u) S^\alpha ]_\beta
&= \eta_\beta^{\gamma\delta} \del(u)_\gamma S^\alpha{}_{\delta} \ ,\cr
[ \dualp{u} D(u) S^\alpha ]_\alpha
&= \eta^{\alpha\gamma\delta} \del(u)_\gamma S_{\alpha\delta} = 0\ ,\cr
}\eeq
and for a pure trace such tensor $T^\alpha{}_\beta = f\delta^\alpha{}_\beta$,
the result is the spatial dual of the exterior derivative of the scalar
coefficient (dual of the vector ``$\curl\grad$"),
modulo sign and index shifting
\beq
[\curl T]^{\alpha\beta}
= - \eta(u)^{\alpha\beta\gamma} \del(u)_\gamma f \ .
\eeq
This operator can also be symmetrized, leading to
an operator will be called the $\Scurl$ operator.
\beq
[\Scurl(u) T]^{\alpha\beta}
= \eta^{\gamma\delta(\alpha} \del(u)_\gamma T^{\alpha)}{}_{\delta} \ ,
\eeq
It maps
the symmetric spatial vector-valued 1-forms
to tracefree symmetric such forms and annihilates their pure trace part
\beq\eqalign{
[\Scurl(u) S]^{\alpha\beta}
&= \eta^{\gamma\delta(\alpha} \del(u)_\gamma S^{\beta)}{}_\delta \cr
&= [\Scurl(u) S\TF]^{\alpha\beta} \ .\cr
}\eeq
The generalized
Cotton-York tensor \Cite{Cotton 18??, Eisenhart 19??, York 1971}
is obtained by applying this operator to the
spatial Ricci or Einstein tensors or their tracefree parts
\Typeout{symmetric parts??}
\beq
C\tem(u)^{\alpha\beta} = -2[\Scurl R\tem(u)]^{\alpha\beta}
= -2 \eta(u)^{\gamma\delta (\alpha}
\del(u)_\gamma R\tem(u)^{\beta}){}_\delta \ .
\eeq
\Typeout{(to finish)}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\Newpage
\Section{Splitting spacetime curvature}
The spacetime curvature tensor
may be split into its representative spatial fields directly or
one may first split it algebraically into its irreducible parts: the Weyl
curvature tensor, the tracefree Ricci tensor and the scalar curvature,
and then split into their representative spatial fields.
Both forms of the splitting are convenient for certain applications.
These splittings are most easily represented in terms of components with
respect to an observer-adapted frame.
%%%%%%%%%%%
\Subsection{Splitting definitions}
\Typeout{Discuss sign Conventions for electric and magnetic parts of WEYL!}
Because of the symmetries of the Riemann tensor, its splitting leads
to three independent spatial fields which are more efficiently
packaged by taking the spatial dual on the remaining pairs of antisymmetric
indices, as discussed in exercises 14.14 and 14.5 of Misner, Thorne and
Wheeler \Cite{1973}.
Since $\four R_{\alpha\beta\gamma\delta}$ is symmetric in its two pairs
of antisymmetric indices, one can split each pair into its electric and
magnetic parts leading to either both electric or both magnetic
or one of each for a total of three independent spatial fields.
The magnetic spatial index pairs may then be converted into single
spatial indices by the spatial dual, resulting in three second rank
spatial tensors. The last such field (mixed electric and magnetic parts)
transforms as an oriented tensor under a change of spatial basis
(due to the single spatial dual) and so defines the magnetic part of
the Riemann curvature tensor, while the remaining two transform as
ordinary tensors under such transformations (due to an even number of
spatial duals) and so define the electric part, in analogy with the
electric and magnetic vector fields defined from the electromagnetic 2-form.
The two fields which represent the electric part will be called
the electric-electric electric part and the magnetic-magnetic electric part
in an obvious though somewhat lengthy terminology.
These three fields are
\beq\eqalign{
\Escr(u)^a{}_b &= \four R^a{}_{\top b \top}\ ,\cr
\Fscr(u)^a{}_b &= \fraction14 \eta(u)^{acd}\eta(u)_{bfg} \four R^{fg}{}_{cd}
=[\dual\,\four R\dual\,]_{b \top}{}^a{}_\top
=[\dual\,\four R\dual\,]^a{}_{\top b \top} \ ,\cr
\Hscr(u)^a{}_b &= \half \four R^a{}_{\top cd} \eta(u)^{cd}{}_b
= - [\four R\dual\,]^a{}_{\top b \top} \ ,\cr
& \Escr(u)_{[ab]} =0= \Fscr(u)_{[ab]}\ , \quad \Hscr(u)^a{}_a =0\ .\cr
}\eeq
Here the sign of the electric-electric part has been changed to conform
with the convention established by
Ehlers \Cite{1962, 1993} and Ellis \Cite{1971}.
The spatial fields which represent the Ricci tensor, its tracefree part,
and the Einstein tensor are related to the electric and magnetic parts
of the Riemann curvature tensor in the following way
\begincomment%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\beq\eqalign{
\four R^a{}_b &= -\Escr(u)^a{}_b - \Fscr(u)^a{}_b
+ \delta^a{}_b \Fscr(u)^c{}_c \ ,\quad &
\four Q^a{}_b &= -\Escr(u)^a{}_b - \Fscr(u)^a{}_b
+ \half \delta^a{}_b (\Escr(u)^c{}_c + \Fscr(u)^c{}_c) \ ,\quad &
\four G^a{}_b &= -\Escr(u)^a{}_b - \Fscr(u)^a{}_b
+ \delta^a{}_b \Escr(u)^c{}_c
\ ,\cr
\four R^\top{}_\top &= \Escr(u)^c{}_c\ , &
\four Q^\top{}_\top &=\half( \Escr(u)^c{}_c -\Fscr(u)^c{}_c) \ , &
\four G^\top{}_\top &= -\Fscr(u)^c{}_c\ , \cr
\four R &= 2(\Escr(u)^c{}_c +\Fscr(u)^c{}_c) \ , &
\four Q &=0 \ , &
\four G &= -(\Escr(u)^c{}_c + \Fscr(u)^c{}_c)\ , \cr
\four R^\top{}_a &= &
\four Q^\top{}_a &= &
\four G^\top{}_a &= \half \eta_{abc}\Hscr(u)^{bc}\ . \cr
}\eeq
\endcomment%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\beq\eqalign{
\four R^a{}_b &= -\Escr(u)^a{}_b - \Fscr(u)^a{}_b
+ \delta^a{}_b \Fscr(u)^c{}_c \ ,\cr
\four Q^a{}_b &= -\Escr(u)^a{}_b - \Fscr(u)^a{}_b
+ \half \delta^a{}_b (\Escr(u)^c{}_c + \Fscr(u)^c{}_c) \ ,\cr
\four G^a{}_b &= -\Escr(u)^a{}_b - \Fscr(u)^a{}_b
+ \delta^a{}_b \Escr(u)^c{}_c \ ,\cr
}\eeq
\beq\meqalign{
\four R^\top{}_\top &= -\Escr(u)^c{}_c\ , \quad &
\four R &= 2(-\Escr(u)^c{}_c +\Fscr(u)^c{}_c) \ , \cr
\four Q^\top{}_\top &= -\half( \Escr(u)^c{}_c + \Fscr(u)^c{}_c) \ , \quad &
\four Q &=0 \ , \cr
\four G^\top{}_\top &= -\Fscr(u)^c{}_c\ , \quad &
\four G &= 2 (\Escr(u)^c{}_c - \Fscr(u)^c{}_c)\ , \cr
}\eeq
and
\beq
\four R^\top{}_a =
\four Q^\top{}_a =
\four G^\top{}_a = \eta_{abc}\Hscr(u)^{bc}\ .
\eeq
It is also very useful to split the Riemann curvature tensor into
its irreducible parts under a change of frame
\beq\eqalign{
\four R^{\alpha\beta}{}_{\gamma\delta} &=
\four C^{\alpha\beta}{}_{\gamma\delta}
+2 \four Q^{[\alpha}_{\ \ [\gamma} \delta^{\beta]}_{\ \ \delta]}
+\fraction1{12} \four R \delta^{\alpha\beta}_{\ \ \gamma\delta}\ ,\cr
\four Q^\alpha{}_\beta &=
\four R^\alpha{}_\beta -\fraction14 \delta^\alpha{}_\beta \four R\ .\cr
}\eeq
This formula may be expressed in terms of the Ricci tensor $\four
R^\alpha{}_\beta $ or the Einstein tensor $\four G^\alpha{}_\beta $ in place of
the tracefree part of the Ricci tensor $\four Q^\alpha{}_\beta $, leading to
different coefficients for the scalar curvature term.
The Weyl tensor $\four C_{\alpha\beta\gamma\delta}$ can itself be
split into its family of spatial fields in the same fashion as the Riemann
tensor, but since it is self-dual $-\dual \four C \dual = \four C$,
the two electric parts of the Riemann tensor collapse to a single electric
part of the Weyl tensor
%
\beq\eqalign{
E(u)^a{}_b &= \four C^a{}_{\top b \top}
= \half [-\Escr(u) + \Fscr(u)]\TF {}^a{}_b \cr
&= \half [-\Escr(u)^a{}_b + \Fscr(u)^a{}_b
+\fraction13 \delta^a{}_b (\Escr(u)^c{}_c -\Fscr(u)^c{}_c )] \ ,\cr
H(u)^{ab} &= \half \four C^{\top a}{}_{cd} \eta(u)^{bcd}
= [\dual{}\four C]^{\top a}{}_{\top}{}^b
= [\four C\dual\,]^{\top a}{}_{\top}{}^b \cr
&=
% \Hscr(u)^{ab} - \half\eta^{abc} \four G^\top{}_c
\Hscr(u)^{(ab)}\ ,\cr
}\eeq
where $\TF$ indicates the tracefree part.
These are both symmetric and tracefree.
This decomposition is apparently due to Trumper \Cite{1961} but
the sign convention is different.
\Typeout{check this sign problem: weyl electric/magnetic parts.}
%%%%%%%%%%%%%%%%%%
\Subsection{Spacetime duality and curvature}
The origin of the decomposition of the Riemann tensor into Weyl, tracefree
Ricci and the curvature scalar lies in the behavior of the Riemann tensor
under the independent duality transformations one may perform on its two
pairs of antisymmetric indices. First note that the dual of the dual of a
2-form yields the sign-reversal of the 2-form due to the Lorentz signature
of the spacetime metric
\beq\eqalign{
\rightdual F_{\alpha\beta} &= \half \four \eta_{\alpha\beta}{}^{\gamma\delta}
F_{\gamma\delta} \ ,\cr
\dual{} \rightdual F_{\alpha\beta} &= \fraction14 \eta_{\alpha\beta}{}^{\mu\nu}
\eta_{\mu\nu}{}^{\gamma\delta} F_{\gamma\delta}
= -\fraction12 \delta_{\ \ \alpha\beta}^{\gamma\delta} F_{\gamma\delta}
= -F_{\alpha\beta}\ .\cr
}\eeq
For this reason it is convenient to introduce a new duality operation
``$\,\hook\,$" called ``hook" \Cite{Taub 19??}
whose repeated action is the identity operation
\beq\eqalign{
\hook F_{\alpha\beta} &= i \, \rightdual F_{\alpha\beta}\ ,\cr
\hook \hook F_{\alpha\beta} &= F_{\alpha\beta}\ .\cr
}\eeq
One can then look for eigenvectors of this new
duality operation with eigenvalues 1 and $-1$ (rather than $i$ and $-i$ for the
original duality operation).
A 2-form $F$ is then said to be self-dual if $\hook F= F$ and anti-self-dual
if $\hook F= - F$, either of
which can be true only for a complex-valued 2-form. For example,
if $F$ is a real 2-form, then $Z=F + \hook F$ is self-dual while its
complex conjugate $\bar Z$ is anti-self-dual, and $F=\half (Z + \bar Z)$
is the decomposition of the real 2-form into its complex self-dual and
anti-self-dual parts.
One can extend the hook operation to rank 4 tensors
$B_{\alpha\beta\gamma\delta}$ like
the curvature tensor which are symmetric in
two pairs of antisymmetric indices. Such tensors may be interpreted as symmetric
bilinear functions on 2-forms (bi-vectors). Any such tensor can be decomposed
into its real self-dual and anti-self-dual parts under the two-sided hook
operation (a hook on the left pair of indices and another on the right pair)
\beq\eqalign{
B= S + A\ , \qquad & S = \half (B + \hook B \righthook)\ , \quad
\hook S \righthook = S\ ,\cr
& A = \half (B - \hook B \righthook)\ , \quad
\hook A \righthook = -A\ ,\cr
}\eeq
For the Riemann curvature tensor this yields
\beq\eqalign{
\four R^{\alpha\beta}{}_{\gamma\delta} &=
S^{\alpha\beta}{}_{\gamma\delta} + A^{\alpha\beta}{}_{\gamma\delta}\ ,\cr
&\ S^{\alpha\beta}{}_{\gamma\delta} = \four C^{\alpha\beta}{}_{\gamma\delta}
+ \fraction1{12} R \delta^{\alpha\beta}_{\ \gamma\delta} \ ,\cr
&\ A^{\alpha\beta}{}_{\gamma\delta} =
2 \four Q^{[\alpha}_{\ \ [\gamma} \delta^{\beta]}_{\ \ \delta]}
\ ,\cr
}\eeq
defining the Weyl tensor as the tracefree part of the self-dual part of
the Riemann tensor and the curvature scalar as
the trace (apart from a constant),
while the tracefree Ricci tensor determines the anti-self-dual part.
The two-sided (ordinary) dual of the Riemann tensor is useful enough to
give its own name following the notation of Misner, Thorne and Wheeler
\Cite{1973}
\beq\eqalign{
\four\Gb^{\alpha\beta}{}_{\gamma\delta} &=
[\dual \four R \dual\,]^{\alpha\beta}{}_{\gamma\delta} = -
[\hook \four R \righthook]^{\alpha\beta}{}_{\gamma\delta} =
-\four S^{\alpha\beta}{}_{\gamma\delta}
+\four A^{\alpha\beta}{}_{\gamma\delta} \cr
&= -\four C^{\alpha\beta}{}_{\gamma\delta} + 2 \four Q^{[\alpha}{}_{[\gamma}
\delta^{\beta]}{}_{\delta]}
- \fraction1{12} \four R\delta^{\alpha\beta}{}_{\gamma\delta} \ ,\cr
}\eeq
or
\beq\eqalign{
\four C^{\alpha\beta}{}_{\gamma\delta} &=
-\four \Gb^{\alpha\beta}{}_{\gamma\delta} + 2 \four Q^{[\alpha}{}_{[\gamma}
\delta^{\beta]}{}_{\delta]}
- \fraction1{12} \four R\delta^{\alpha\beta}{}_{\gamma\delta} \ ,\cr
&= -\four \Gb^{\alpha\beta}{}_{\gamma\delta} + 2 \four G^{[\alpha}{}_{[\gamma}
\delta^{\beta]}{}_{\delta]}
+ \fraction16 \four R\delta^{\alpha\beta}{}_{\gamma\delta} \ .\cr
}\eeq
The Einstein tensor is the trace of the two-sided dual of the Riemann tensor
in the same way that the Ricci tensor is the trace of the Riemann tensor
itself
\beq
\four \Gb^{\alpha\beta}{}_{\alpha\delta} = \four G^\beta{}_\delta\ ,\qquad
\four\Gb^{\alpha\beta}{}_{\alpha\beta} = - \four R\ .
\eeq
%%%%%%%%%%%%%%%
\Subsection{Evaluation of splitting fields}
The actual splitting of the Riemann tensor for the congruence point of view
is obtained by simply evaluating the standard component formula for
an observer-adapted frame and
has been given originally by Cattaneo-Gasperini \Cite{1961}
and later by Ferrarese \Cite{1963, 1965, 1987, 1988}.
Using the convenient the abbreviation
$k(u)\rightcontract k(u) = k(u)^2$ one has
\beq\eqalign{
\Escr(u)^a{}_b &= \four R^a{}_{\top b \top}
= \Lie(u)\sub{u} k(u)^a{}_b - [k(u)^2]^a{}_b
+ [ \del(u)_b + a(u)_b] a(u)^a \ ,\cr
& \phantom{ = \four R^a{}_{\top b \top}}
= \del\fw(u)\sub{u} k(u)^a{}_b - [k(u)^2]^a{}_b
+ [ \del(u)_b + a(u)_b] a(u)^a \ ,\cr
\Hscr(u)^{ab} &= - \del(u)_{[c} k(u)^a{}_{d]} \eta^{bcd}
- 2a(u)^a \omega(u)^b\ ,\cr \message{check}
}\eeq
(taking note of (\ref{eq:LieFWK})) and
\beq\eqalign{
\four R^{ab}{}_{cd} &= 2 k(u)^a{}_{[c} k(u)^b{}_{d]}
+2 k(u)^{ab} \omega(u)_{cd} + R\lie(u)^{ab}{}_{cd}\ ,\cr
&= 2 k(u)^a{}_{[c} k(u)^b{}_{d]} + R\fw(u)^{ab}{}_{cd}\ ,\cr
&= \{ 2 \theta(u)^a{}_{[c} \theta(u)^b{}_{d]} \}
+ \{ 2 \omega(u)^a{}_{[c} \omega(u)^b{}_{d]}
+ 2 \omega(u)^{ab} \omega(u)_{cd} \}
+ \{ R\sym(u)^{ab}{}_{cd} \}\ .\cr
}\eeq
Each of the terms delimited by curly brackets in this last expression
individually has all of the usual symmetries of a curvature tensor.
Taking the two-sided spatial dual of this expression yields
the manifestly symmetric
magnetic-magnetic electric part of the Riemann tensor
\beq\eqalign{
\Fscr(u)^a{}_b &= [ \theta(u)^2 - \Theta(u) \theta(u)]^a{}_b
-\half \delta^a{}_b [\Tr \theta(u)^2 - \Theta(u)^2 ]
+3\omega(u)^a \omega(u)_b - G\sym(u)^a{}_b\ ,\cr
\Fscr(u)^c{}_c &= -\half [\Tr \theta(u)^2 - \Theta(u)^2]
+ 3 \omega(u)^c \omega(u)_c +\half R\sym(u)^c{}_c\ ,\cr
}\eeq
where the two-sided
spatial dual of the spatial curvature yields the corresponding
sign-reversed spatial Einstein tensor.
The above expression for the electric-electric electric part of the Riemann
tensor is not manifestly symmetric
\beq\eqalign{
\Escr(u)_{ab} &=
\Lie(u)\sub{u} k(u)_{ab} + [\theta(u)^2 - \omega(u)^2]_{ab} \cr
& \qquad + 2 \theta(u)_{c(a} \omega(u)^c{}_{b)}
+[ \del(u)_b + a(u)_b] a(u)_a \ ,\cr
\Escr(u)_{(ab)} &=
- \Lie(u)\sub{u} \theta(u)_{ab} + [\theta(u)^2 - \omega(u)^2]_{ab} \cr
& \qquad - 2 \theta(u)_{c(a} \omega(u)^c{}_{b)}
+ [ \del(u)_{(a} + a(u)_{(a}] a(u)_{b)} \ ,\cr
\Escr(u)_{[ab]} &=
\Lie(u)\sub{u} \omega(u)_{ab}
+ \del(u)_{[b} a(u)_{a]} = 0 \ ,\cr
\Escr(u)^c{}_c &=
-[\Lie(u)\sub{u} +\Theta(u)] \Theta(u) - \Tr \theta(u)^2 + \Theta(u)^2
+ 2 \omega(u)^c \omega(u)_c \cr
& \qquad + [ \del(u)_c +a(u)_c] a(u)^c = -\four R^\top{}_\top \ ,\cr
}\eeq\message{check}%
its antisymmetric part vanishing due to the identity satisfied by the
spatial exterior derivative of the acceleration 1-form.
The quadratic term in the rotation has the alternate expression
\beq
[\omega(u)^2]^a{}_b = \omega(u)^a \omega(u)_b
- \delta^a{}_b \omega(u)^c \omega(u)_c\ .
\eeq
Similarly the magnetic part of the Riemann curvature is not manifestly
tracefree
\beq\eqalign{
\Hscr(u)^{ab} &= \del(u)_{[c} \theta(u)^a{}_{d]} \eta^{bcd}
-[\del(u)^a + 2a(u)^a ] \omega(u)^b + h^{ab} \del(u)_c\omega(u)^c\ ,\cr
\Hscr(u)^a{}_a &=
2 [ \del(u)_a - a(u)_a ] \omega(u)^a = 0\ ,\cr
}\eeq \message{check}
but its trace vanishes due to the divergence identity for the rotation vector.
Using these preliminary results, the
various spatial fields which represent the different independent pieces
of the curvature tensor may now be
expressed in terms of the spatial curvature
and the kinematic quantities of the congruence.
The electric and magnetic parts of the Weyl curvature tensor are
%
\Typeout{Give FW temporal derivative form also:}
\beq\eqalign{
E(u)_{ab} &
= -\half \{ \Lie(u)\sub{u} \theta(u)_{ab} -\Theta(u) \theta(u)_{ab}
+ 2\theta(u)_{c(a} \omega(u)^c{}_{b)}
\cr & \qquad
+ 4 \omega(u)^a \omega(u)_b - [ \del(u)_{(b} + a(u)_{(b}] a(u)_{a)} \cr
&\qquad - R(u)_{ab} \}\TF \ ,\cr
H(u)^{ab} &=\
\{ -\del(u)_{[c} \theta(u)^{(a}{}_{d]} \eta^{b)cd}
+[\del(u)^{(a} + 2a(u)^{(a}] \omega(u)^{b)} \}\TF \ .\cr
}\eeq \message{check}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\closeaux\bye
The scalar curvature and the Einstein tensor are
\beq\eqalign{
\four R &= 2 \{\Lie(u)\sub{u} + \Theta(u)\} \Theta(u)
-2[\del(u)_a +a(u)_a] a(u)^a \cr
& + \Tr \theta(u)^2 -\Theta(u)^2 + 2 \omega(u)^c \omega(u)_c
+ R\sym(u)^c{}_c \ ,\cr
2 \four G^\top{}_\top &= -2 \Fscr(u)^c{}_c =
\Tr \theta(u)^2 - \Theta(u)^2
- 6 \omega(u)^c \omega(u)_c - R\sym(u)^c{}_c \ ,\cr
2 \four G^\top{}_a &= 4 \del(u)_{[b} k(u)^b{}_{a]}
- 4 [ a(u) \times_u \vec\omega(u) ]_a\cr
&= - 2 \del(u)_b[ \theta(u)^{b}{}_a - \delta^b{}_a \Theta(u)]
- 2 \{[\vec \del(u) + 2a(u)] \times_u \vec\omega(u)\}_a\ ,\cr
\message{check?? symmetric part of acceleration derivative!!:}
\four G^a{}_b &= \{ \Lie(u)\sub{u} +\Theta(u) \}
[ \theta(u)^a{}_b -\delta^a{}_b \Theta(u)]
+ \half \delta^a{}_b [\Tr \theta(u)^2 - \Theta(u)^2] \cr
& \qquad - [\del(u)_{(b} + a(u)_{(b}]a(u)^{a)}
+ \delta^a{}_b [\del(u)_c + a(u)_c] a(u)^c \cr
& \qquad - 2 \omega(u)^a \omega(u)_b + \delta^a{}_b \omega(u)^c \omega(u)_c
+ G\sym(u)^a{}_b \ .\cr
}\eeq
Similarly the purely spatial part of the Ricci tensor is
\beq\eqalign{\Message{check symmetric part of acclereation derivative!!:??}
\four R^a{}_b &= -\{ \Lie(u)\sub{u} + \Theta(u) \} k(u)^a{}_b
- [\del(u)_b + a(u)_b]a(u)^a \cr
&\qquad -2 [\omega(u)^a \omega(u)_b - \delta^a{}_b \omega(u)^c \omega(u)_c]
+ R\sym(u)^a{}_b\ ,\cr
\four R_{ab} &= \{ \Lie(u)\sub{u} + \Theta(u) \} \theta(u)_{ab}
- 2 [\theta(u)^2]_{ab}
- [\del(u)_{(a} + a(u)_a]a(u)_{b)} \cr
&\qquad -2 [\omega(u)^a \omega(u)_b - \delta^a{}_b \omega(u)^c \omega(u)_c]
+ R\sym(u)_{ab}\ .??\cr
}\eeq
Recall that the spatial Lie derivative with
the expansion
scalar $\Theta(u) =-\Tr k(u)$ added as a multiplication operator has the
expression
\beq
[\Lie(u)\sub{u} +\Theta(u)] S^{a\ldots}_{\ \ b\ldots} =
(Mh^{1/2})^{-1} (h^{1/2} S^{a\ldots}_{\ \ b\ldots})_{,0}
\eeq
in a spatially-comoving observer-adapted frame.
Furthermore \ref{eq:sd}
shows that the acceleration term in the scalar curvature is a spacetime
divergence
\beq
[\del(u)_b + a(u)_b] a(u)^b = a(u)^\alpha{}_{;\alpha} = \four\div a(u)\ .
\eeq
Similarly the acceleration-augmented curl in the mixed part of the
Einstein tensor represents the spatial projection of the spacetime divergence
of the rotation 2-form.
Thus the Hilbert gravitational Lagrangian expressed in a spatially-comoving
observer-adapted frame has the form
\beq\eqalign{
\Lscr\H &= \four g^{1/2} \four R^\alpha{}_\alpha = \Lscr\adm
+2 \Lie(u)\sub{e_0} [ h^{1/2} \Theta(u) ]
-2 \four g^{1/2} a(u)^\alpha{}_{;\alpha}\ ,\cr
\Lscr\adm &= M h^{1/2} [ \Tr \theta(u)^2 - \Theta(u)^2
+ 2 \omega(u)^c \omega(u)_c + R\sym(u)^c{}_c]\ .\cr
}\eeq
In order to speak of the $e_0$ derivative term as a ``total time derivative"
in the usual sense, one must introduce a full splitting of spacetime.
Neglecting this term and the divergence defines the ADM Lagrangian
\Cite{Arnowit, Deser and Wheeler 1962, Misner, Thorne and Wheeler 1973}.
%%%%%%%%%%%%%%%%%%
\Subsection{Maxwell-like equations}
The Einstein equations in their Einstein or Ricci form
\beq
\four G^\alpha{}_\beta = \kappa \four T^\alpha{}_\beta\ ,\qquad
\four R^\alpha{}_\beta
= \kappa [ \four T^\alpha{}_\beta\
- \delta^\alpha{}_\beta \four T^\gamma{}_\gamma] \ ,\qquad
\eeq
may easily be split using the above splitting of the lefthand sides,
once the energy-momentum tensor is split, leading to a spatial scalar,
a spatial vector and spatial tensor equation.
The first two are analogous to the source-driven Maxwell equations
for the electromagnetic field, while the
the spatial scalar and spatial vector identities satisfied by the
gravitoelectric and gravitomagnetic fields are analogous to the
homogeneous Maxwell equations.
The first pair of Maxwell-like equations analogous to the sourcefree
Maxwell equations are equations (\ref{eq:gemeqsourcefree}).
The spatial scalar and spatial vector equations which result from the
measurement of the Ricci form of the Einstein equations
provide the much more complicated analogs of the source driven pair
of Maxwell equations
\beq\eqalign{\label{eq:gemeqsources}
\four R^\top{}_\top
&= [ \del(u)_c - g(u)_c] g(u)^c - \half H(u)^c H(u)_c \cr & \qquad
+ [\Lie(u)\sub{u} +\Theta(u)] \Theta(u) +\Tr \theta(u)^2 -\Theta(u)^2 \cr
&= \kappa [ \four T^\top{}_\top - \half\four T^\alpha{}_\alpha ]\ ,\cr
2 \four R^\top{}_a
&= - \{[\vec \del(u) - 2 \vec g(u)] \times_u \vec H(u)\}_a \cr &\qquad
- 2\del(u)_b[ \theta(u)^{b}{}_a - \delta^b{}_a \Theta(u)] \cr
&= 16\pi \four T^\top{}_a \ .\cr
}\eeq
The complication arises from the new ingredient described by the spatial metric
which has no analog in linear electrodynamics.
In the linearized Einstein equations, the analogy of these four equations
with Maxwell's equations
becomes not only very close but even useful.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\Newpage
\Section{Mixed commutation formulas}
The commutation formula for two spatial covariant derivatives acting on
a spatial tensor field leads to spatial curvature via the spatial
Ricci identities. This formula represents the purely spatial part of the
splitting of the corresponding spacetime commutation formula, i.e., of
the corresponding spacetime Ricci identity.
The remaining members of the family of spatial identities which result from
splitting the latter identity are also useful identities.
All of the these identities together
are equivalent to commutation formulas among various combinations
of the spatial covariant derivative and the Fermi-Walker temporal derivative
acting on the members of the family of spatial fields which represent a
given spacetime tensor field. This provides a more straightforward
method of deriving the splitting identities and a more useful packaging
of their content.
In practice the Ricci identities are used to interchange the order of
two successive covariant derivatives in an expression. The spatial
identities obtained from the spacetime Ricci identities allow one to
interchange the order of two spatial covariant derivatives or of
a spatial covariant derivative and a Fermi-Walker temporal derivative.
It is equally
convenient to be able to interchange the order of a Lie temporal
derivative and a spatial covariant derivative. For the Lie temporal
derivative this leads to the Lie derivative of the spatial connection,
a spatial tensor field
which appears in the spatial Bianchi identities of the second kind.
\Subsection{Splitting the Ricci identities}
The Ricci identity for a vector field $Z$ is
\beq
[\four\del_\gamma, \four\del_\delta] Z^\alpha
= \four R^\alpha{}_{\beta\gamma\delta} Z^\delta \ .
\eeq
Since the operator $[\four\del,\four\del]$ itself has an electric and
a magnetic part as a 2-form operator, while the vector field $Z$ has
both a temporal and a spatial part, four different spatial identities
result from splitting this identity. These are equivalent to commutation
formulas involving $[\del(u),\del(u)]$ and $[\del\fw(u),\del(u)]$ acting
on either a spatial vector field or on $u$ itself.
For an arbitrary vector field $Z$ one has
\beq\eqalign{
[\del(u)_\gamma,\del(u)_\delta] Z^\alpha &=
2 P(u)^\mu{}_{[\gamma} P(u)^\nu{}_{\delta]} P(u)^\alpha{}_\sigma
\four\del_\mu [ P(u)^\epsilon{}_\nu P(u)^\sigma{}_\rho
\four\del_\epsilon Z^\rho ] \cr
&= P(u)^\mu{}_\gamma P(u)^\nu{}_\delta P(u)^\alpha{}_\rho
[\four\del_\mu, \four\del_\nu] Z^\rho \cr
&\qquad + 2 \omega(u)_{\gamma\delta} \del\fw(u) Z^\alpha
- 2 k(u)^\alpha{}_{[\gamma} P(u)^\epsilon{}_{\delta]}
u_\rho Z^\rho{}_{;\epsilon} \cr
}\eeq
and
\beq\eqalign{
[\del\fw (u), \del(u)_\beta] Z^\alpha &= \cr
&= P(u)^\alpha{}_\gamma P(u)^\delta{}_\beta u^\epsilon
[\four\del_\epsilon,\four\del_\delta] Z^\gamma \cr
&\qquad + a(u)_\beta \del\fw (u) Z^\alpha + a(u)^\alpha u_\gamma
Z^\gamma{}_{;\delta} P(u)^\delta{}_\beta \cr
&\qquad + k(u)^\alpha{}_\beta u_\gamma Z^\gamma{}_{;\delta} u^\delta
+ P(u)^\alpha{}_\gamma Z^\gamma{}_{;\delta} k(u)^\delta{}_\beta\ .\cr
}\eeq
Restricting each of these to a spatial vector field $Z$ and using the
spacetime Ricci identity for the spacetime commutator term leads to the
two identities
\beq\eqalign{
[\del(u)_\gamma,\del(u)_\beta] Z^\alpha &=
( [P(u)\four R]^{\alpha\beta}{}_{\gamma\delta}
- 2 k(u)^\alpha{}_{[\gamma} k(u)^\beta{}_{\delta]} )Z_\beta \cr
&\qquad + 2 \omega(u)_{\gamma\delta} \del\fw(u) Z^\alpha \cr
}\eeq
and
\beq\eqalign{
[\del\fw (u), \del(u)_\beta] X^\alpha &=
(-\Hscr_{\beta\gamma} \eta(u)^{\gamma\alpha}{}_\delta
+ a(u)^\alpha k(u)_{\delta\beta} - k(u)^\alpha{}_\beta a(u)_\gamma)
Z^\delta \cr
&\qquad + a(u)_\beta \del\fw(u) Z^\alpha
+ k(u)^\delta{}_\beta \del(u)_\delta Z^\alpha \ .\cr
}\eeq
On the other hand letting $Z=u$ one obtains
\beq
[\del(u)_\gamma,\del(u)_\delta] u^\alpha =
\eta(u)_{\gamma\delta\beta} \Hscr(u)^{\alpha\beta}
+ 2\omega(u)_{\gamma\delta} a(u)^\alpha \ ,
\eeq
and
\beq
[\del\fw (u), \del(u)_\beta] u^\alpha = -\Escr(u)^\alpha{}_\beta
- [k(u)^2]^\alpha{}_\beta + a(u)^\alpha a(u)_\beta \ .
\eeq
From first of these four identities, re-expressing the left hand side with
the spatial Ricci identity, one recovers the
relationship between the magnetic-magnetic electric part of the spacetime
curvature ($\sim \Fscr_{\alpha\beta}$) and the spatial curvature tensor.
The last of these identities involving the electric-electric part of
the curvature is necessary to finish the discussion of
relative acceleration.
A relative acceleration vector is the
second spatial Fermi-Walker derivative of a spatial
``relative position vector"
$Y$ which undergoes spatial Lie transport along $u$.
Its expression
involves the commutator of the spatial Fermi-Walker derivative and the
spatial covariant derivative acting on $u$ itself, leading to the appearance
of the electric-electric electric part of the spacetime curvature.
Using the appropriate commutation formula in eq. \ref{eq:ra}
leads to
\beq\eqalign{
\del\fw (u)^2 Y^\alpha &= \{ -\Escr(u)^\alpha{}_\beta
+[\del(u)_\beta + a(u)_\beta] a(u)^\alpha \} Y^\beta\ ,\cr
{\rm or} \qquad
\del\fw (u)^2 Y &= \{ -\Escr(u)
+\del(u) a(u) + a(u) \otimes a(u)^\flat \}\rightcontract Y\ ,\cr
}\eeq
For a geodesic congruence, the acceleration vanishes and this reduces to the
geodesic deviation equation involving only the electric-electric electric part
of the Riemann tensor, which is interpreted as describing the tidal
gravitational field.
\Subsection{Commuting $\Lie(u)\sub{u}$ and $\del(u)$}
\message{warning Ca0b terms?}
The commutator of the temporal Lie derivative and the spatial covariant
derivative of a spatial tensor field is easiest to evaluate in an
observer-adapted frame. For a spatial vector field one easily finds
\beq
[\Lie(u)\sub{u}, \del(u)_b] X^a = a(u)_b \Lie(u)\sub{u} X^a
+ \Lie(u)\sub{u} \Gamma(u)^a{}_{bc} X^c \ ,
\eeq
or in any frame
\beq
[\Lie(u)\sub{u}, \del(u)_\beta] X^\alpha = a(u)_\beta \Lie(u)\sub{u} X^\alpha
+ L(u)^\alpha{}_{\beta\gamma} X^\gamma\ ,
\eeq
where the spatial tensor $L(u)$ is the temporal Lie derivative of
the spatial connection, given the overused kernel symbol $H$ by Ferrarese
\Cite{1987, 1988, 1989}.
This tensor is most easily evaluated using the
commutation formula for $[\partial_0,\partial_a]$ in a spatially-comoving
spatially-holonomic partially-normalized observer-adapted frame.
The result
\beq\eqalign{
L(u)^a{}_{bc} &= \Lie(u)\sub{u} \Gamma(u)^a{}_{bc} \cr
&= h^{ad} [ \theta(u)_{\{\delta\beta||\gamma\}_-}
+ \theta(u)_{\{db} a(u)_{c\}_-} ] \cr
}\eeq
easily generalizes to the frame-independent formula
\beq\eqalign{
\Lie(u)\sub{u} \Gamma(u)^\alpha{}_{\beta\gamma}
&= P(u)^{\alpha\delta} [\theta(u)_{\{\delta\beta||\gamma\}_-}
+ \theta(u)_{\{\delta\beta} a(u)_{\gamma\}_-} ] \cr
&= P(u)^{\alpha\delta} \theta(u)_{\{\delta\mu;\nu\}_-}
P(u)^\mu{}_\beta P(u)^\nu{}_\gamma \ .
}\eeq
This formula is a closely related to the more familiar result for the
variation of the Christoffel symbols induced by the variation of a
metric.
\Newpage
\Section{Splitting the Bianchi identities of the second kind}
The next level of splitting addresses the derivatives of curvature, the most
important of which occur in the Bianchi identities which hold for both
the spacetime curvature tensor and the various spatial curvature tensors.
\Subsection{Spacetime identities}
The Bianchi identity for the curvature 2-form has the simple index-free form
\beq
\four D \four \bfOmega =
d \four \bfOmega + \four\bfomega\wedge \four\bfOmega
-\four\bfOmega\wedge \four\bfomega = 0
\eeq
in terms of the spacetime covariant exterior derivative $\four D$,
where right contractions are implied between the adjacent tensor-valued
indices of the curvature and connection forms,
or in index form
\beq
\four R^\alpha{}_{\beta[\gamma\delta;\epsilon]} =0\ .
\eeq
Taking the two-sided dual of this equation leads to
\beq
\four \Gb^{\alpha\beta\gamma\delta}{}_{;\delta} = 0\ ,
\eeq
which has the contracted form
\beq
\four G_{\alpha}{}^\beta{}_{;\beta} =0
\eeq
called the contracted Bianchi identity.
These identities simplify the expression for the divergence of the
Weyl tensor
\Typeout{how does divergence free weyl follow from Bianchi
identities? more here:}
\beq\eqalign{
\four C^{\alpha\beta}{}_{\gamma\delta}{}^{;\delta} &=
-\four \Gb^{\alpha\beta}{}_{\gamma\delta}{}^{;\delta}
+ 2 \delta^{[\alpha}{}_{[\gamma}
\four G^{\beta]}{}_{\delta]} {}^{;\delta}
+ \fraction16 \delta^{\alpha\beta}{}_{\gamma\delta} \four R^{;\delta}\ .\cr
&= \four G^{[\alpha}{}_\gamma{}^{;\beta]}
+ \fraction16 \delta^{\alpha\beta}{}_{\gamma\delta} \four R^{;\delta}\ ,\cr
}\eeq
enabling it to be expressed in terms of the covariant derivatives of the
Einstein tensor and the scalar curvature.
The identities for the divergence of the Weyl and Einstein tensors can each
be split to yield useful information about the curvature.
The spacetime split of the Weyl divergence identity yields the analog of
Maxwell's equations for its electric and magnetic parts as discussed by
Tr\"umper \Cite{1967}, but the results are more accessible in work by
Hawking \Cite{1966} and Ellis \Cite{1971},
while the spacetime split of the Einstein tensor divergence identity
provides the framework for the discussion of the preservation
of the constraints in the evolution of solutions of the
``initial value problem".
Introduce the Einstein field equation tensor
\beq
\four E^\alpha{}_\beta =
\four G^\alpha{}_\beta - \kappa \four T^\alpha{}_\beta
\eeq
which vanishes when the Einstein equations are satisfied.
The Einstein tensor has identically vanishing divergence. The -momentum
tensor of the source of the gravitational field must therefore have
vanishing divergence. This is either a consequence of the source field
equations or, in the case of a perfect fluid with an explicit equation of
state, is equivalent to them. When this divergence is zero, the Einstein
field equation tensor has vanishing divergence
\beq
\four E^\alpha{}_{\beta;\alpha}=0\ .
\eeq
The splitting of this equation takes the following form when expressed
in a spatially-comoving observer-adapted frame
\beq\eqalign{
0 &= \four E^\alpha{}_{\top;\alpha} =
( L h^{1/2})^{-1} \Lie(u)\sub{e_0} ( h^{1/2} \four E^\top{}_\top) \cr
& \qquad-\L^{-2} \del(u)^c ( \L^2 \four E^\top{}_c )
+ \theta(u)^d{}_c \four E^c{}_d\ ,\cr
0 &= \four E^\alpha{}_{a;\alpha} =
( L h^{1/2})^{-1} \Lie(u)\sub{e_0} ( h^{1/2} \four E^\top{}_a) \cr
& \qquad 2 \four E^\top{}_c k(u)^c{}_a - a(u)_a \four E^\top{}_\top
+ [\del(u)_c +a(u)_c] \four E^c{}_a\ .\cr
}\eeq
One can divide the Einstein field equations into ``{\it dynamical equations}"
\beq
\four E^a{}_b =0 \qquad {\rm or} \qquad P(u) \four E =0
\eeq
and ``{\it constraint equations}"
\beq\eqalign{
\four E^\top{}_\top &=0\cr
\four E^\top{}_a &=0 \qquad {\rm or} \qquad
\four E^\top{}_\beta P(u)^\beta{}_\alpha =0\ .
}\eeq
The significance of the above divergence equations is then the following.
Suppose one considers a ``{\it tubular region}" of spacetime with respect
to the congruence, namely a part of spacetime obtained by starting with
an ``{\it initial hypersurface}" transversal to the congruence with or
without boundary and then Lie dragging it into the future a fixed
proper time interval along the congruence.
If the ``dynamical equations" are satisfied within the tubular region
swept out in this manner, then the above equations clearly show that
the ``constraint equations", if satisfied at the initial hypersurface,
remain satisfied along the congruence within this tubular region.
By the same process used above to evaluate divergences with a
spatially-comoving observer-adapted frame, one may split the divergence
of the Weyl tensor and then split the Bianchi identities involving the
covariant derivatives of the Einstein tensor and the scalar curvature.
If the Einstein equations are satisfied, these latter terms in the Bianchi
identities may be replaced by the corresponding derivatives of the
energy-momentum tensor which then provide source terms for the Maxwell-like
equations for the electric and magnetic parts of the Weyl tensor.
This was first done by Tr\"umper \Cite{unpublished, see also Tr\"umper 1967}
and discussed by Ellis \Cite{1971}.
To emphasize the similarity with Maxwell's equations,
one considers the spatial fields $E^\alpha{}_\beta$ and $H^\alpha{}_\beta$
as vector-valued 1-forms, but since they are symmetric upon index
shifting, the operations on the vector-valued 1-forms must be symmetrized
to respect that symmetry.
A divergence and curl (the $\Scurl$ operator of section \ref{sec:Scurl})
can then be defined for
these fields in a natural way
using the spatial covariant exterior derivative $D(u)$
\beq\eqalign{
[ \div(u) \bfE(u)]^\alpha &=
\del(u)^\beta E^\alpha{}_\beta = -[\delta(u) \bfE(u)]^\alpha \ ,\cr
[ \Scurl(u) \bfE(u)]^{\delta\alpha} &=
\eta(u)^{\beta\gamma(\alpha} \del(u)_\beta E^{\delta)}{}_\gamma
= [\dualp{u} D(u) \bfE(u)]^{(\delta\alpha)} \ .\cr\message{index level??}
}\eeq
Similarly the spatial
cross product operation can be applied to their 1-form indices
to reflect taking the spatial dual of the wedge product of a spatial
1-form with the vector-valued 1-form in a way that makes the curl operation
just ``del cross"
\beq
[X \times_u \bfE(u)]^{\delta\alpha} =
X_\beta E(u)^{(\delta}{}_\gamma \eta(u)^{\alpha)\beta\gamma}\ .
\eeq
In the splitting of the Weyl divergence
$\four C_{\alpha\beta\gamma\delta}{}^{;\delta}$ one must recall the tracefree
condition on the Weyl tensor and the first Bianchi identity for the Weyl tensor,
which imply
\beq
\four C^{\alpha\gamma}{}_{\gamma\delta}{}^{;\delta} =0\ ,\qquad
\four C_{[\alpha\beta\gamma]\delta}{}^{;\delta} =0\ .
\eeq
Taking these into account, the splitting of the Weyl divergence
$C^\delta_{\alpha\beta\gamma;\delta}$
may be represented by the following independent spatial fields
\beq
\{ \four C^\delta{}_{\top a \top;\delta} ,
\four C^\delta{}_{(ab)\top ;\delta} ,
[\dual{}\four C]^\delta{}_{\top a \top ;\delta} ,
[\dual{}\four C]^\delta{}_{(ab)\top ;\delta} \}\ ,
\eeq
where the traces of the second and fourth fields vanish, i.e., by two
spatial vectors and two tracefree symmetric tensors. These are explicitly
%
\beq\eqalign{
\four C^\delta{}_{\top a \top;\delta}
&= -[\div(u) E(u)]^a + \eta(u)^a{}_{bc} H(u)^b{}_d \sigma(u)^{cd}
+ 3H(u)^a{}_c \omega(u)^c \ ,\cr
\four C^\delta{}_{(ab)\top ;\delta}
&= - \Lie(u)\sub{u} E(u)^{ab}
[ \sigma(u)^{(a}{}_c +\omega(u)^{(a}{}_c] E(u)^{b)c}
- \fraction23 ??\Theta(u) E(u)^{ab}
- h^{ab}\sigma_{cd} E(u)^{cd} \cr
&\qquad + \{ [ \curl(u) + 2 a(u) \times_u ] H(u)\}^{ab} \ ,\cr
[\dual{}\four C]^\delta{}_{\top a \top ;\delta}
&= -[\div(u) H(u)]^a + \eta(u)^a{}_{bc} E(u)^b{}_d \sigma(u)^{cd}
+ 3E(u)^a{}_c \omega(u)^c \ ,\cr
[\dual{}\four C]^\delta{}_{(ab)\top ;\delta}
&= -\Lie(u)\sub{u} H(u)^{ab}
[+ \sigma(u)^{(a}{}_c +\omega(u)^{(a}{}_c] H(u)^{b)c}
- \fraction23 ??\Theta(u) H(u)^{ab}
- h^{ab}\sigma_{cd} H(u)^{cd} \cr
&\qquad + ??\{ [ \curl(u) + 2 a(u) \times_u ] E(u)\}^{ab} \ ,\cr
& {\bf CHECK DETAILS ; h??} \cr
}\eeq
The form of these equations given by Tr\"umper or Ehlers \Cite{1961,1993}
and later by Hawking \Cite{1966} and Ellis \Cite{1971}
uses the Fermi-Walker temporal derivative
instead of the Lie temporal derivative. This change is easily made
resulting in the equations
\beq\eqalign{
\four C^\delta{}_{\top a \top;\delta}
&= \del\fw(u)\sub{u} E(u)^{ab}
+ [-3 \sigma(u)^{(a}{}_c +\omega(u)^{(a}{}_c] E(u)^{b)c}
- ??\Theta(u) E(u)^{ab}
+ h^{ab}\sigma_{cd} E(u)^{cd} \cr
&\qquad + \{ [ \curl(u) + 2 a(u) \times_u ] H(u)\}^{ab} \ ,\cr
\four C^\delta{}_{(ab)\top ;\delta}
&= \del\fw(u)\sub{u} H(u)^{ab}
+[-3 \sigma(u)^{(a}{}_c + \omega(u)^{(a}{}_c ] H(u)^{b)c}
- ??\Theta(u) H(u)^{ab}
+ h^{ab}\sigma_{cd} H(u)^{cd} \cr
&\qquad - \{ [ \curl(u) + 2 a(u) \times_u ] E(u)\}^{ab} \ ,\cr
& still to correct\cr
}\eeq
the form given by Ellis and Bruni \Cite{1989}.
The divergence of the Weyl tensor is specified by the once contractec
Bianchi identities of the second kind as explained in the appendix.
\centerline{FIX UP}
\Typeout{Also give source terms.}
\Newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\Subsection{Spatial identities}
\message{warning: Ca0b terms?}
The form taken by the Bianchi identities of the second kind for each of
the three spatial curvature tensors is most easily obtained by first evaluating
the spatial covariant exterior derivative of the Lie spatial curvature 2-form
\beq\eqalign{
D(u) \bfOmega\lie(u) &=
d(u) \bfOmega\lie(u) + \bfomega(u) \rightcontract \wedge \bfOmega\lie(u)
- \bfOmega\lie(u) \rightcontract \wedge \bfomega(u) = \ldots \cr
&= d(u)^2 \bfomega(u)
= 2 \omega(u)^\flat \wedge \Lie(u)\sub{u} \bfomega(u)\ ,\cr
}\eeq
or in index form
\beq
R\lie(u)^{\alpha\beta}{}_{[\gamma\delta||\epsilon]}
= 2 L(u)^\alpha{}_{[\gamma}{}^{||\beta}
\omega(u)_{\delta\epsilon]}\ ,
\eeq
yielding the result of Ferrarese \Cite{1965}.
Defining the nonsymmetric Lie
spatial Einstein tensor as the sign-reversed two-sided spatial dual of the
spatial curvature tensor
\beq\eqalign{
G\lie(u)^\alpha{}_\beta &=
-\fraction14 \eta(u)^{\alpha\gamma\delta} \eta(u)_{\beta\mu\nu}
R\lie(u)_{\gamma\delta}{}^{\mu\nu}\ ,\cr
&= R\lie(u)^{[\alpha\delta]}{}_{\beta\delta}
-\half P(u)^\alpha{}_\beta R\lie(u)^{\gamma\delta}{}{\gamma\delta} \cr
&= R\lie(u)^\alpha{}_\beta
-\half P(u)^\alpha{}_\beta R\lie(u)^{\gamma}{}_\gamma \ ,\cr
}\eeq
leads to the usual formula in terms of the Ricci tensor.
Then taking the two-sided spatial dual of the previous identity
on each group of antisymmetric indices results in the divergence of the
Lie spatial Einstein tensor
\beq
G\lie(u)^{\alpha\beta}{}_{||\beta} =
- \eta(u)^{\alpha\beta\gamma} \omega(u)^\delta
\theta(u)_{\delta\beta||\gamma} \ .
\eeq
The symmetric part of the identity is instead
\beq
R\lie(u)^{(\alpha\beta)}{}_{[\gamma\delta||\epsilon]}
= 2 \theta(u)^{\alpha\beta}{}_{||[\gamma} \omega(u)_{\delta\epsilon]}\ .
\eeq
More complicated versions of these formulas may be obtained for the
remaining two spatial curvature tensors by considering the derivatives of
the difference tensors relative to the Lie spatial curvature.
\Newpage
\Section{``Time without space defines space without time" and vice versa}
\message{POORLY WRITTEN SECTION??}
The congruence point of view defines ``time" as the proper time measured
by the test observers but establishes no correlation between the
proper times of different observers, which is instead a property of ``space".
One can define a useful space, however, namely the quotient
$\four M/\flow(u)$ of the spacetime by the 1-parameter group of diffeomorphisms
generated by $u$, i.e., the space of observer worldlines.
The term ``{\it observer space}" is a convenient label for this space, which
is without any Riemannian structure or any notion of time, precisely because
of the lack of correlation between different observers.
For example, in order to integrate the Einstein equations in this point of view
one needs an initial value hypersurface to get started, where contraints must
be satisfied by certain initial data. This initial hypersurface then leads
to a slicing of the spacetime under dragging along by the observer congruence,
yielding a nonlinear reference frame, in terms of which the initial data
may then be evolved. Without such additional structure one can only
consider ``{\it propagation equations}" describing the evolution of the
gravitational and source fields along each individual observer worldline,
and certain ``{\it constraint}" equations, as
described by Ellis \Cite{1971}.
Similarly in the hypersurface point of view, one can take the quotient of the
spacetime by the slicing to obtain a time line but there is no explicit
notion of ``points of space" which may be associated with the time line.
Of course there is implicitly because of the existence of the normal
congruence, but explicitly introducing it leads to a full splitting.
In order to convert the split tensor equations involving spatial
Lie derivatives along $u$ and spatial covariant derivatives into explicit
partial differential equations, one must introduce local coordinates.
In the interest of simplification it seems clear that these coordinates
should be adapted to the congruence in some way.
This essentially
means introducing a full splitting of spacetime.
For a rotating congruence this means introducing a slicing
and an associated time coordinate function $t$ (i.e., a parametrized slicing)
and the equivalence class of comoving
spatial coordinate systems for the congruence.
For a nonrotating congruence it is clear that the integrable family of
orthogonal hypersurfaces should be taken as the slicing, allowing the
freedom to use spatial coordinates which are not comoving along the
normal congruence.
The spatial Lie derivative by $u$ can then be represented
as a rescaled time coordinate derivative, plus a spatial Lie derivative
if the spatial coordinates are not comoving.
However, the spatial covariant
derivative in the case of a nonintegrable connection, i.e., associated
with a rotating congruence, cannot be represented in terms of
spatial coordinate derivatives alone and is necessarily
accompanied by a time derivative. This means that the ``constraint"
equations involving spatial derivatives of the kinematical
quantities and only first order spatial Lie derivatives
along $u$ necessarily involve second coordinate time derivatives,
as does the spatial curvature.
Only in the case of a nonrotating congruence, with the slicing taken to be
the orthogonal slicing to the congruence, is it possible to
represent the constraint equations in terms of only first order
coordinate time derivatives. Only then is it meaningful to speak
of the contraints as defining an ``initial value problem" in the usual
sense.
\endinput
This can be eliminated
by instead considering the Lorentz boost $B(u,U)S$ into the local rest
space of the observer congruence.
This vector has the same length as $S$, a quantity
which remains constant, and hence must also undergo a rotation in that
local rest space, which can therefore be described completely by an
angular velocity $\zeta$ along the worldline of the gyro.
To quantify this rotation one must choose a spatial orthonormal frame
$\{e_a\}$ along the worldline in order to measure the relative rotation
of the boosted spin vector.
Since boosts preserve inner products, the components of the boosted
spin vector with respect to an orthonormal frame in the observer
local rest space are the same as those of the original spin vector
with respect to the inverse boost of the spatial frame into the local
rest space of the gyro
\beq
e_a \cdot_u B(u,U) S = B(U,u) e_a \cdot_U S \ .
\eeq
By choosing such a spatial frame which is transported
along the worldline by co-rotating Fermi-Walker transport, the frame
remains parallel to itself except for the additional rotation that it
needs to remain nonrotating with respect to the observer congruence.
In such a frame the co-rotating Fermi-Walker total spatial covariant
derivative of the boosted spin vector reduces to the
ordinary derivative of its orthonormal components and therefore defines
the {\it relative angular velocity} of the boosted spin vector with respect
to the observer congruence, related to the derivative of the relative
rotation in the usual way.
Except for the time dilation, this angular
velocity will agree with the angular velocity of the original spin
vector of the gyro in its local rest space with respect to the boosted
spatial frame.
In order to calculate the co-rotating Fermi-Walker
total spatial covariant derivative of the
boosted spin vector,
one only needs an expression for the part of the boost
which maps
between the local rest spaces of $U$ and $u$. This was discussed in
section (\ref{sec:splitU}).
Expressing this map in terms of the spatial momentum rather than the
relative velocity
\beq
P(u) \rightcontract B(u,U) \rightcontract P(U) =
[ P(u)
- \gammauu ^{-1} (\gammauu + 1 )^{-1} \Vec p \otimes
p^\flat ]
\rightcontract P(u) \rightcontract P(U)
\eeq
leads to the following expression for the boosted spin vector
\beq
\vec \Sscr = B(u,U) \rightcontract S =
[ P(u) - \gammauu ^{-1} (\gammauu + 1 )^{-1}
\Vec p \otimes p^\flat ] \rightcontract \vec S\ .
\eeq
It coincides with $\vec S$ at points of the
worldline where the observed spin is perpendicular to the relative
velocity of the worldline.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
pig
However, this expression for $\zeta\cfw$
is still not the classic gyro precession
formula, since in the nonrelativistic limit $\gammauu \to 1$, there
is an additional space curvature precession term
\Cite{Thorne 1989} which has twice the value of
the spin orbit term leading to a net coefficient of $\fraction32$ instead
of $\half$.
The classic precession formula in fact measures the angular
velocity of the spin vector with respect to a preferred spatial
orthonormal frame defined at all points of space and
which undergoes co-rotating Fermi-Walker transport
along the observer congruence. The spatial frame along the worldline
of the gyro which undergoes spatial co-rotating Fermi-Walker transport
along that worldline itself rotates with an angular velocity relative
to the ``fixed spatial frame" by an amount which is twice the
spin orbit value
in the traditional scenario, leading to the factor of $\fraction32$.
The orthonormal
spatial frame $\{ e_a\}$ has already been assumed to be transported
along the observer
congruence by co-rotating Fermi-Walker transport.
Since the connection 1-form in an orthonormal frame is antisymmetric, one
can define a space curvature angular velocity by
\beq\eqalign{
D\cfw(U,u) \Sscr^a / d\tau_u
&= d \Sscr^a / d\tau_u + \Gamma(u)^a{}_{bc} \nu^b \Sscr^c \cr
&= d \Sscr^a / d\tau_u -
\eta(u)^a{}_{bc} \zeta\sc{}^b \Sscr^c \ ,\cr
}\eeq
where
\beq
\zeta\sc{}^a = -\half \eta(u)^{abc} \Gamma(u)_{bdc} \nu^d \ .
\eeq
This expression for the total spatial covariant derivative is
justified below in section \ref{sec:extscd}.
One then has the angular velocity of the boosted spin relative to
this ``fixed frame" as the sum of the two
\beq\eqalign{
d \Sscr^a / d\tau_u &=
\eta(u)^a{}_{bc} [\zeta\cfw + \zeta\sc]{}^b \Sscr^c \ .\cr
&\equiv \eta(u)^a{}_{bc} \zeta^b \Sscr^c \ .\cr
}\eeq
Of course now that ordinary derivatives are being used one has to
choose the ``fixed frame" carefully in order that this make some
physical sense. The space curvature term comes from the spatial curvature
and leads to a nontrivial rotation of any vector which is parallelly
transported around in a loop in ``space". An adequate discussion
of these matters requires the structure of a nonlinear reference frame.
%and conditions of stationarity or quasi-stationarity.
EDIT:
By choosing such a spatial frame which is transported
along the worldline by co-rotating Fermi-Walker transport, the frame
remains parallel to itself except for the additional rotation that it
needs to remain nonrotating with respect to the observer congruence.
In such a frame the co-rotating Fermi-Walker total spatial covariant
derivative of the boosted spin vector reduces to the
ordinary derivative of its orthonormal components and therefore defines
the {\it relative angular velocity} of the boosted spin vector with respect
to the observer congruence, related to the derivative of the relative
rotation in the usual way.
Except for the time dilation, this angular
velocity will agree with the angular velocity of the original spin
vector of the gyro in its local rest space with respect to the boosted
spatial frame.
\beq\eqalign{
D\fw (U,u)/d\tau_U
&= \gammauu [\del\fw (u) + \del(u)\sub{\Vec\nu}]\ ,\cr
&= \gammauu \del\fw (u) + \del(u)\sub{\Vec p} \ ,\cr
D\cfw (U,u)/d\tau_U
&= \gammauu [\del\cfw (u) + \del(u)\sub{\Vec\nu}]\ ,\cr
&= \gammauu \del\cfw (u) + \del(u)\sub{\Vec p} \ ,\cr
D\lie (U,u)/d\tau_U
&= \gammauu [ \Lie(u)\sub{u} + \del(u)\sub{\Vec\nu}]\ ,\cr
&= \gammauu \Lie(u)\sub{u} + \del(u)\sub{\Vec p} \ ,\cr
}\eeq
which may then be rescaled by the factor $d\tau_U/d\tau_u = \gammauu ^{-1}$
to correspond to a derivative with respect to the observer proper time
\beq\eqalign{
D\fw (U,u)/d\tau_u
&= \del\fw (u) + \del(u)\sub{\Vec\nu} \cr
&= \del\fw (u) + \gammauu ^{-1}\del(u)\sub{\Vec p} \ ,\cr
% & \equiv \del\fw (U,u) \ ,\cr
D\cfw (U,u)/d\tau_u
&= \del\cfw (u) + \del(u)\sub{\Vec\nu} \cr
&= \del\cfw (u) + \gammauu ^{-1}\del(u)\sub{\Vec p} \ ,\cr
% & \equiv \del\fw (U,u) \ ,\cr
D\lie (U,u)/d\tau_u
&= \Lie(u)\sub{u} + \del(u)\sub{\Vec\nu}\ ,\cr
&= \Lie(u)\sub{u}
+ \gammauu ^{-1}\del(u)\sub{\Vec p} \ .\cr
% & \equiv \del\lie (U,u) \ .\cr
}\eeq
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The definition of the electric and magnetic fields is such that the Lorentz
4-force on a particle of charge $q$, mass $m$ and 4-velocity $U$ takes the form
\beq\eqalign{
m \four D U / d \tau_U &=
q F \rightcontract U = q \gamma [ E(u) + \nu(U,u) \times_u B(u) ]
\qquad {\rm or} \cr
m \four D U / d \tau_u &=
q F \rightcontract U = q \gamma [ E(u) + \nu(U,u) \times_u B(u) ] \ ,\cr
}\eeq
where the derivative is the absolute derivative along the
proper-time-parametrized curve in the first case, and along the re-parametrized
curve in the second. This derivative is
called the total covariant derivative in this text as described in Appendix A.
\endinput
%pig
REWRITE:
First consider the spatial projection of the acceleration
\beq\eqalign{
P(u) \four D U / d\tau_U
&= P(u)[\gammauu \four\del\sub{u} + \four\del\sub{\Vec p}]
[\gammauu u + \Vec p\,] \cr
&= [ \gammauu \del\fw (u) +\del(u)\sub{\Vec p}\,]
[\gammauu u + \Vec p\,] \cr
&= \cdots = D\fw (U,u) \Vec p /d\tau_U + \gammauu ^2 a(u)
-\gammauu k(u)\rightcontract \Vec p\ .\cr
}\eeq
Rescaling to observer proper time
\beq
P(u) \four D U / d\tau_u
= D\fw (U,u) \Vec p /d\tau_u - \vec F{}\g\fw
\eeq
leads to the definition of
the Fermi-Walker {\it spatial gravitational force}
\beq
\vec F{}\g\fw = \gammauu [ - a(u) + k(u) \rightcontract \Vec\nu \,] \ .
\eeq
The spatial projection of the acceleration equation then takes the suggestive
form
\beq
D\fw (U,u) \Vec p /d\tau_u
= \vec F{}\g\fw + \vec F \ ,
\eeq
in which the spatial force is accompanied by an apparent gravitational force
due to the motion of the observers.
This same equation can be expressed in terms of the other total spatial
covariant derivatives leading to corresponding spatial gravitational forces
which differ from each other by the same terms as the corresponding derivatives.
For the Lie total spatial covariant derivative the correction term also
depends on whether the covariant or contravariant spatial momentum is
differentiated, leading to two different force terms.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The temporal part of the acceleration equation is similarly obtained.
First projecting the acceleration along $u$ yields
\beq\eqalign{
- u \cdot \four D U / d\tau_U
&=-u \cdot [\gammauu \four\del\sub{u}
+ \four\del\sub{\Vec p}\,] [\gammauu u + \Vec p\,] \cr
&= D\fw (U,u) E /d\tau_U -[ \gammauu u\cdot \four\del\sub{u} \Vec p
+ u \cdot \four\del\sub{\Vec p} \,\Vec p\,]\ .\cr
}\eeq
Since the various total spatial covariant derivatives agree on functions,
the identifying subscript may be dropped, while
the term in square brackets may be ``integrated by parts" using
the relation $u\cdot \Vec p=0$ to obtain
\beq\eqalign{
- u \cdot \four D U / d\tau_U
&= D (U,u) E /d\tau_U
+[ \gammauu \Vec p \cdot_u \four\del\sub{u} u
+ \Vec p \cdot_u \four\del\sub{\Vec p} u ]\ ,\cr
&= D (U,u) E /d\tau_U
+[ \gammauu a(u) \cdot_u \Vec p
+ \Vec p \cdot_u \bivec \theta(u) \cdot_u \Vec p ]\ ,\cr
}\eeq
where $\bivec \theta(u) = \theta(u)^\sharp$ is the dyad notation.
Solving for the derivative of $E$, rescaling the proper time
and using the temporal component of the force equation leads to
\beq\eqalign{\label{eq:derenergyF}
D (U,u) E /d\tau_u
&= - \gammauu [ a(u) \cdot \Vec\nu
+ \Vec\nu \cdot_u \bivec \theta(u) \cdot_u \Vec\nu ]
+ \wp \ ,\cr
&= [ \vec F\g\fw + \vec F ] \cdot_u \Vec\nu \cr
&= [ \vec F\g\cfw + \vec F ] \cdot_u \Vec\nu \cr
&= [ \vec F\g\lies + \vec F ] \cdot_u \Vec\nu
+ \gammauu \Vec\nu \cdot_u \bivec \theta(u) \cdot_u \Vec\nu \cr
&= [ \vec F\g\lief + \vec F ] \cdot_u \Vec\nu
- \gammauu \Vec\nu \cdot_u \bivec \theta(u) \cdot_u \Vec\nu \ .\cr
}\eeq
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The best one can do to eliminate this spatial dependence of the orientation
of the static frame along the gyro worldline in a spatially covariant
manner is to transport an orthonormal spatial frame $\{\Escr_a\}$ along
the gyro worldline by spatial co-rotating Fermi-Walker transport
\beq
D\cfw(U,u) \Escr_a /d\tau_u = 0 \ .
\eeq
Such a spatial frame remains parallel to itself except for the additional
rotation it needs to remain non-rotating with respect to the observer
congruence (as well as the additional boost it needs to remain spatial
with respect to $u$). In such a frame the co-rotating Fermi-Walker
total spatial covariant derivative of $\Sscr$ reduces to the ordinary
derivative of its components $\Sscr^a$
\beq
D\cfw(U,u) \Sscr^a /d\tau_u = d \Sscr^a /d\tau_u \ .
\eeq
This provides the physical interpretation of this total spatial covariant
derivative.
Since the spatial vector $\Sscr$ has the same constant magnitude as the
spin vector $S$, one can define the {\it co-rotating relative angular velocity}
of $\Sscr$ with respect to the observer congruence by the equation
\beq\eqalign{
D\cfw(U,u) \Sscr^a/ d\tau_u &= \eta(u)^a{}_{bc} \zeta\cfw{}^b \Sscr^c
\ , \qquad {\rm or} \cr
D\cfw(U,u) \Sscr / d\tau_u &= \zeta\cfw \times_u \Sscr \ .\cr
}\eeq
The previous discussion shows that $\zeta\cfw$ directly describes the rotation
of $\Sscr$ with respect to an orthonormal spatial frame which undergoes
spatial co-rotating Fermi-Walker transport along the gyro worldline.
However, because of the presence of the spatial covariant derivative in
the total spatial covariant derivative, the spatial curvature leads to
a finite rotation of this transported frame with respect to the static
frame for a worldline segment which begins and ends on the same static
observer worldline, i.e., corresponding to a closed loop in the space of
observers. Only the noncovariant result describes the net rotation of
$\Sscr$ relative to the static axes for such a curve segment.
The Schiff formula corresponds to this noncovariant result.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
For an orthonormal static frame, the expression for the spatial co-rotating
Fermi-Walker derivative decomposes into the ordinary derivative minus
a term from the relative angular velocity of a co-rotating Fermi-Walker
frame relative to the static frame
\beq\eqalign{
D\cfw(U,u) \Sscr^a / d\tau_u
&= d \Sscr^a / d\tau_u + \Gamma(u)^a{}_{bc} \nu^b \Sscr^c \cr
&= d \Sscr^a / d\tau_u -
\eta(u)^a{}_{bc} \zeta\sc{}^b \Sscr^c \ .\cr
}\eeq
One then has the angular velocity of the boosted spin vector relative to
the static frame as the sum of the relative angular velocity of the transport
condition with respect to the static frame
and the relative angular velocity of the spin vector
with respect to the observer congruence (i.e., with respect to the
co-rotating Fermi-Walker transport condition)
\beq\eqalign{
d \Sscr^a / d\tau_u &=
\epsilon_{abc} [\zeta\cfw + \zeta\sc]{}^b \Sscr^c \ .\cr
& \rightarrow \zeta = \zeta\cfw + \zeta\sc \ .
}\eeq
Thorne \Cite{Thorne et al 1986} had referred to the relative
angular velocity $\zeta\sc$ as the space curvature term.
Of course now that ordinary derivatives are being used one has to
choose the static frame carefully in order that this make some
physical sense.
It remains simply to evaluate the relative angular velocity $\zeta\cfw$,
which one does by computing the total spatial covariant derivative
of the boosted spin vector $\Sscr$ along the gyro worldline.
Using the tensor contraction notation and formula (\ref{eq:boostback})
for the boost, this vector is expressed in terms of the projected
spin vector $\vec S = P(u) S$ by the formula
\beq
\Sscr = B(u,U) \rightcontract S =
[ P(u) - \gammauu ^{-1} (\gammauu + 1 )^{-1}
\Vec p \otimes p^\flat ] \rightcontract \vec S\ .
\eeq
It coincides with $\vec S$ at points of the
worldline where the observed spin is perpendicular to the relative
velocity of the worldline.
One can now compute the co-rotating Fermi-Walker
total spatial covariant derivative of the boosted spin vector
in terms of the derivatives of the spatial quantities
$p$, $E = \gammauu $, and $\vec S$.
The result gives an explicit formula for
relative angular velocity $\zeta\cfw(U,u)$ when expressed in the complete
notation which
has the expression
\beq\eqalign{
\zeta\cfw &= -\half \vec H(u)
- \gammauu (\gammauu +1)^{-1} \Vec\nu \times_u \vec F
+ (\gammauu + 1 )^{-1} \, \Vec\nu \times_u \vec F\g\fw \cr
&= \zeta\gmag + \zeta\thom + \zeta\geo \cr
}\eeq
in terms of the spatial projection $\vec F$ of the applied force or
equivalently
\beq\eqalign{
\zeta\cfw &= - \half\vec H(u)
- \gammauu ^2 (\gammauu + 1 )^{-1} \Vec\nu \times_u \vec a\cfw(U,u) \cr
&\qquad + \gammauu \Vec\nu \times_u
[ \vec F\g\cfw - (\gammauu + 1 )^{-1} \Vec\nu \times_u \vec\omega(u) ]
\cr
}\eeq
in terms of the relative acceleration of the worldline.
This formula was given by Massa and Zordan \Cite{1975}.
\message{Thomas term which formula: comment??}
The first term
is the gravitomagnetic precession due
to the rotation of the observer congruence which is present even for
a gyro at rest with respect to the observer congruence.
When $\Vec\nu=0$,
the formula reduces to \Eq (\ref{eq:restspinp}) which
describes a gyro carried by an observer of the observer congruence.
In the context of weak fields and slow motion, this effect was
first studied by Thirring and Lense \Cite{1918}.
In flat spacetime with $u$ a unit timelike Killing vector field,
the spatial gravitational force is zero and
the second term in either formula
\beq
\zeta\thom =
- \gammauu ^2 (\gammauu + 1 )^{-1} \Vec\nu \times_u \vec a\cfw(U)
= - \nu^{-2} ( \gammauu - 1) \Vec\nu \times_u \vec a\cfw(U)
\eeq
is the Thomas precession due to the acceleration of the gyro
\Cite{Thomas 1927}. For circular motion with angular velocity $\vec\Omega$,
the precession angular velocity is $[\gammauu -1]\vec\Omega$, as described in
exercise 6.9 of Misner, Thorne and Wheeler \Cite{1973}.
The first term in their \Eq (6.28) is exactly the boosted
spin vector $\Sscr$. In the nonrelativistic limit the Thomas
precession reduces to
\beq
\zeta\thom \rightarrow -\half \nu \times_u \vec a(U) =
- \half \nu \times_u \vec F \ ,
\eeq
where no distinction remains among the four possible spatial acceleration
expressions under these conditions.\Message{??}
For a geodesic in an arbitrary spacetime, the Thomas
precession term expressed in terms of $\vec F$
vanishes leaving the last term which
describes the geodetic precession, also known as the
DeSitter \Cite {1916} or Fokker \Cite {1920} precession
\beq\eqalign{
\zeta\geo &= ( \gammauu + 1 )^{-1} \, \Vec\nu \times_u \vec F\g\fw(U,u) \cr
&= \gammauu ( \gammauu + 1 )^{-1} \, \Vec\nu \times_u
[ \vec g(u) +\half \Vec\nu \times_u \vec H(u)
- \theta(u)\rightcontract \Vec\nu ] \ .\cr
}\eeq
In the nonrelativistic limit $\gammauu \to 1$, neglecting terms
of second order in the velocity,
this term has the limiting expression
\beq
\zeta\so = \half \Vec\nu \times_u \vec g(u) \ .
\eeq
Thorne \Cite{1989} describes this nonrelativistic term as an
``induced gravitomagnetic precession" or ``spin-orbit" precession
since it corresponds to the gravitomagnetic precession due to
an additional ``induced" gravitomagnetic field
$\vec H(u)\ind = - \Vec\nu \times_u \vec g(u)$ induced by the motion
of the gyro in the gravitoelectric field in analogy with the induced
magnetic field due to motion in an electric field.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In order to see this result,
consider the Schwarzschild
line element in the isotropic cartesian coordinate form
\beq
ds^2 = - (1-2U) dt^2 + (1+2U) \delta_{ab} dx^a dx^b \ ,
\eeq
where $U = M/r$ and $r= (\delta_{ab} x^a x^b)^{1/2}$.
Let $u= (1-2U)^{-1/2} \partial/\partial t$ be the usual Schwarzschild
observer 4-velocity, and let $ e_a = (1+2U)^{-1/2} \partial/\partial x^a$
be the static spatial orthonormal frame.
Then the spatial structure functions for this observer-adapted orthonormal
frame, the spatial connection components, and the space curvature
precession, in the limit $U\ll 1$ are
\beq\eqalign{
C^a{}_{bc} &= 2\delta^a{}_{[b} g(u)_{c]} \ ,\cr
\Gamma(u)_{abc} &= 2\delta_{b[a} g(u)_{c]} \ ,\cr
\zeta\sc{}^a &= \eta(u)^{abc} \nu_b g(u)_c
= [\Vec\nu \times_u \vec g(u) ]^a \ ,\cr
}\eeq
where the frame components of the gravitoelectric field are
\beq
g(u)_a = e_a U \ .
\eeq
In this limit the space curvature precession, which has a
completely different
origin from the spin-orbit precession, has twice the magnitude of the latter
precession, leading to the famous factor $\fraction32$ of the
Schiff formula \Cite{Schiff 1960}
\beq
\zeta\so + \zeta\sc \rightarrow \frac{3}{2} \nu \times_u \vec g(u) \ .
\eeq
This total precession term is referred to as the {\it geodetic precession}.
(Even Schiff
omits this discussion for a rotating body, quoting only the result.)
================== RFC 822 Headers ==================
Date: Wed, 4 Jun 1997 05:16 EST
From: JANTZEN@UCIS.VILL.EDU (Bob Jantzen, drbob, or ms_ani)
Message-Id: <009B542EF28CE360.88EF@UCIS.VILL.EDU>
To: jantzen@vxrmg9.icra.it
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