Weiss variation for general boundaries

Abstract

The Weiss variation of the Einstein-Hilbert action with an appropriate boundary term has been studied for general boundary surfaces; the boundary surfaces can be spacelike, timelike, or null. To achieve this we introduce an auxiliary reference connection and find that the resulting Weiss variation yields the Einstein equations as expected, with additional boundary contributions. Among these boundary contributions, we obtain the dynamical variable and the associated conjugate momentum, irrespective of the spacelike, timelike or, null nature of the boundary surface. We also arrive at the generally non-vanishing covariant generalization of the Einstein energy-momentum pseudotensor. We study this tensor in the Schwarzschild geometry and find that the pseudotensorial ambiguities translate into ambiguities in the choice of coordinates on the reference geometry. Moreover, we show that from the Weiss variation, one can formally derive a gravitational Schrödinger equation, which may, despite ambiguities in the definition of the Hamiltonian, be useful as a tool for studying the problem of time in quantum general relativity. Implications have been discussed.

Appendices

Appendix A: Variation of bulk integrals

Here, we work out the variation of bulk integrals when boundary displacements are included. Consider a bulk integral of the form:

$$\begin{aligned} I = \int _U d^4 x \, F . \end{aligned}$$

(A1)

The variation of the integral takes the form:

$$\begin{aligned} \begin{aligned} \delta I&= \int _{U^\prime } d^4 x \, F^\prime - \int _U d^4 x \, F - {\mathcal {O}}(2) \\&= \int _U d^4 x \, \delta F + \int _{\partial U} d {\underline{\Sigma }}_\mu \, \delta x^\mu \, F . \end{aligned} \end{aligned}$$

(A2)

The second integral in the last equality describes the contribution from the boundary variation; one can see this by noting that $d {\underline{\Sigma }}_\mu \, \delta x^\mu$ corresponds to the volume swept out by the variation of the boundary induced by the displacement $\delta x^\mu$. The surface element $d {\bar{\Sigma }}_\mu$ is defined as

$$\begin{aligned} d {\underline{\Sigma }}_\mu := \frac{1}{3!} \epsilon _{\mu \alpha \beta \gamma } \, dx^\alpha \wedge dx^\beta \wedge dx^\gamma . \end{aligned}$$

(A3)

One can (by way of the divergence theorem) convert Eq. (A2) to a bulk integral of the form:

$$\begin{aligned} \delta I = \int _U d^4 x \left[ \delta F + \partial _\mu (F \, \delta x^\mu ) \right] . \end{aligned}$$

(A4)

Defining

$$\begin{aligned} f:=F/\sqrt{-g} , \end{aligned}$$

(A5)

one may write:

$$\begin{aligned} \delta I = \int _U d^4 x \left[ \sqrt{-g} \left( \delta f - \frac{1}{2} g_{\mu \nu } \delta g^{\mu \nu } f \right) + \partial _\mu (\sqrt{-g} f \delta x^\mu ) \right] . \end{aligned}$$

(A6)

If f is a scalar, one may write this as:

$$\begin{aligned} \begin{aligned} \delta I&= \int _U d^4 x \, \sqrt{-g} \left[ \delta f - \frac{1}{2} g_{\mu \nu } \delta g^{\mu \nu } f + {\bar{\nabla }}_\mu (f \delta x^\mu ) \right] \\&= \int _U d^4 x \, \sqrt{-g} \left[ \delta f - \frac{1}{2} g_{\mu \nu } \delta g^{\mu \nu } f \right] + \int _{\partial U} d \Sigma _\mu \, \delta x^\mu \, f . \end{aligned} \end{aligned}$$

(A7)

where $d \Sigma _\mu := \sqrt{-g} \, d {\underline{\Sigma }}_\mu$.

Appendix B: Variation of covariant divergence

Here, we consider the variation of a boundary integral of the following form:

$$\begin{aligned} B = \int _U d^4 x \sqrt{-g} \left( {\nabla }_\mu V^\mu \right) , \end{aligned}$$

(B1)

where $V^\mu$ are the components of some vector field. First, we consider the variation of the divergence itself:

$$\begin{aligned} \begin{aligned} \delta [ {\nabla }_\mu V^\mu ]&= {\nabla }_\mu \delta V^\mu + V^\nu \delta {\Gamma }{^\mu }_{\nu \mu } \\&= {\nabla }_\mu \delta V^\mu + \frac{1}{2}V^\nu [ g^{\mu \sigma } {\nabla }_\nu \delta g_{\sigma \mu } ] \\&= {\nabla }_\sigma \delta V^\sigma + \frac{1}{2} g_{\mu \nu } [{\nabla }_\sigma V^\sigma ] \delta g^{\mu \nu } \\&\quad - \frac{1}{2} g_{\mu \nu } {\nabla }_\sigma [ V^\sigma \delta g^{\mu \nu }] . \end{aligned} \end{aligned}$$

(B2)

This result may be combined with the variation of the volume element to obtain:

$$\begin{aligned} \begin{aligned} \delta B =&\int _U d^4 x \sqrt{-g} \, {\nabla }_\sigma \left[ \delta V^\sigma - \frac{1}{2} \, V^\sigma \, g_{\mu \nu } \, \delta g^{\mu \nu } \right] \\&+ \int _{\partial U} \, d\Sigma _\mu \delta x^\mu \, {\nabla }_\nu V^\nu , \end{aligned} \end{aligned}$$

(B3)

which can be rewritten:

$$\begin{aligned} \begin{aligned} \delta B =&\int _U d^4 x \sqrt{-g} \, {\nabla }_\sigma \left[ \delta V^\sigma - \frac{1}{2} \, V^\sigma \, g_{\mu \nu } \, \delta g^{\mu \nu } \right] \\&+ \int _{\partial U} \, i_{\delta x} \left[ {\nabla }_\nu V^\nu \, \omega \right] , \end{aligned} \end{aligned}$$

(B4)

where the last term comes from the dispacement $\delta x^\mu =\Delta \lambda \, \xi ^\mu$ of the boundary, where the volume form $\omega$ is defined:

$$\begin{aligned} \omega := \frac{\sqrt{-g}}{4!} \, \epsilon _{\alpha \beta \mu \nu } \, dx^\alpha \wedge dx^\beta \wedge dx^\mu \wedge dx^\nu . \end{aligned}$$

(B5)

The interior product of the volume form is:

$$\begin{aligned} i_{V}\omega := \frac{\sqrt{-g}}{3!} \, \epsilon _{\alpha \beta \mu \nu } \, V^\alpha \, dx^\beta \wedge dx^\mu \wedge dx^\nu = d\Sigma _\sigma \, V^\sigma . \end{aligned}$$

(B6)

Now write:

$$\begin{aligned} \begin{aligned} \Delta V^\sigma&= \delta V^\sigma + \Delta \lambda \, \pounds _\xi V^\sigma \\ \Delta g^{\mu \nu }&= \delta g^{\mu \nu } + \Delta \lambda \, \pounds _\xi g^{\mu \nu } , \end{aligned} \end{aligned}$$

(B7)

which yields:

$$\begin{aligned} \begin{aligned} \delta B =&\int _{\partial U} d\Sigma _\sigma \left[ \Delta V^\sigma - \frac{1}{2} \, V^\sigma \, g_{\mu \nu } \, \Delta g^{\mu \nu } \right] \\&- \Delta \lambda \int _{\partial U} \, d\Sigma _\sigma \left[ \pounds _\xi V^\sigma - \frac{1}{2} \, V^\sigma \, g_{\mu \nu } \, \pounds _\xi g^{\mu \nu } \right] \\&+ \int _{\partial U} \, i_{\delta x} \left[ {\nabla }_\nu V^\nu \, \omega \right] . \end{aligned} \end{aligned}$$

(B8)

This can be written as (using the identity $d i_V \omega =\nabla _\mu V^\mu \omega$):

$$\begin{aligned} \begin{aligned} \delta B =&\int _{\partial U} d\Sigma _\sigma \left[ \Delta V^\sigma - \frac{1}{2} \, V^\sigma \, g_{\mu \nu } \, \Delta g^{\mu \nu } \right] \\&- \Delta \lambda \int _{\partial U} \, \pounds _\xi i_V \omega + \Delta \lambda \int _{\partial U} \, i_{\xi } d i_V \omega . \end{aligned} \end{aligned}$$

(B9)

From Cartan’s magic formula $di+id=\pounds$, one can write:

$$\begin{aligned} \pounds _\xi i_V \omega = d i_\xi i_V \omega + i_\xi d i_V \omega . \end{aligned}$$

(B10)

After a cancellation, one finds that:

$$\begin{aligned} \begin{aligned} \delta B =&\int _{\partial U} d\Sigma _\sigma \left[ \Delta V^\sigma - \frac{1}{2} V^\sigma g_{\mu \nu } \Delta g^{\mu \nu } \right] - \Delta \lambda \int _{\partial U} d i_\xi i_V \omega . \end{aligned} \end{aligned}$$

(B11)

The term proportional to $\Delta \lambda$ on the right-hand side of Eq. (B11), being the exterior derivative of a (two-)form, vanishes when integrated over $\partial U$ via the boundary of a boundary principle, but we choose to keep it for now. The second interior product takes the form:

$$\begin{aligned} i_{\xi }i_{V}\omega := \frac{\sqrt{-g}}{2!} \, \epsilon _{\alpha \beta \mu \nu } \, V^\alpha \, \xi ^\beta \wedge dx^\mu \wedge dx^\nu . \end{aligned}$$

(B12)

The exterior derivative takes the form:

$$\begin{aligned} \begin{aligned} di_{\xi }i_{V}\omega&:= \frac{1}{3!} \partial _\sigma \left( \sqrt{-g} \, \epsilon _{\alpha \beta \mu \nu } \, V^\alpha \, \xi ^\beta \right) dx^\sigma \wedge dx^\mu \wedge dx^\nu \\&\,= \frac{1}{3!} \, \partial _\sigma \left( \sqrt{-g} \, \epsilon _{\alpha \beta \mu \nu } \, V^{[\alpha } \, \xi ^{\beta ]}\right) dx^\sigma \wedge dx^\mu \wedge dx^\nu . \end{aligned} \end{aligned}$$

(B13)

We recognize the quantity in the bracket as the Hodge star of the antisymmetric tensor $V^{[\alpha } \, \xi ^{\beta ]}$. Now consider the following expression for the antisymmetric tensor $A^{\mu \nu }$:

$$\begin{aligned} \begin{aligned} d{^\star }A&:= \frac{1}{3!} \partial _{[\sigma } \left( {^\star }A_{\mu \nu ]} \right) dx^\sigma \wedge dx^\mu \wedge dx^\nu , \end{aligned} \end{aligned}$$

(B14)

where ${^\star }A_{\mu \nu } = \sqrt{-g} \epsilon {_{\mu \nu \alpha \beta }} \, A^{\alpha \beta }$ are the components of the Hodge dual of the tensor $A^{\mu \nu }$. In a four-dimensional Lorentzian manifold, the double dual operator of a 3-form is the identity, so one may write:

$$\begin{aligned} \begin{aligned} d{^\star }A&= \frac{1}{3!} \epsilon _{\sigma \mu \nu \tau } \, \epsilon ^{\tau \alpha \beta \gamma } \partial _{[\alpha } \left( {^\star }A_{\beta \gamma ]} \right) dx^\sigma \wedge dx^\mu \wedge dx^\nu \\&= - \epsilon ^{\tau \alpha \beta \gamma } \partial _{\alpha } \left( {^\star }A_{\beta \gamma } \right) d{\underline{\Sigma }}_\tau \\&= 2 \partial _{\alpha } \left( \sqrt{-g} A^{\tau \alpha } \right) d{\underline{\Sigma }}_\tau \end{aligned} \end{aligned}$$

(B15)

It follows that the expression in Eq. (B13) can be rewritten:

$$\begin{aligned} \begin{aligned} di_{\xi }i_{V}\omega&:= 2 \partial _{\alpha } \left( \sqrt{-g} V^{[\sigma } \, \xi ^{\alpha ]}\right) d{\underline{\Sigma }}_\sigma , \end{aligned} \end{aligned}$$

(B16)

so that:

$$\begin{aligned} \begin{aligned} \delta B =&\int _{\partial U} d\Sigma _\sigma \left[ \Delta V^\sigma - \frac{1}{2} V^\sigma g_{\mu \nu } \Delta g^{\mu \nu } \right] \\&- 2 \Delta \lambda \int _{\partial U} \nabla _{\alpha } \left( V^{[\sigma } \, \xi ^{\alpha ]}\right) d{\Sigma }_\sigma . \end{aligned} \end{aligned}$$

(B17)

Appendix C: Lie derivative of connection coefficients

The Lie derivative measures the deviation from formal invariance under a diffeomorphism. We consider here a one-parameter diffeomorphism $x^\prime (\epsilon ,x)$, which to first order in $\epsilon$, takes the form:

$$\begin{aligned} x^\prime (\epsilon ,x) = x + \epsilon \, \xi (x) + O(\epsilon ^2). \end{aligned}$$

(C1)

To first order in $\epsilon$, this can be easily inverted to obtain (noting that $\xi (x^\prime )=\xi (x) + O(\epsilon )$):

$$\begin{aligned} x(\epsilon ,x^\prime ) = x^\prime - \epsilon \, \xi (x^\prime ) + O(\epsilon ^2), \end{aligned}$$

(C2)

and it follows that:

$$\begin{aligned} \begin{aligned} \frac{\partial {x^\prime }^\mu }{\partial x^\nu }&= \delta {^\mu }{_\nu } + \epsilon \, \frac{\partial \xi ^\mu }{\partial x^\nu } + O(\epsilon ^2) \\ \frac{\partial x^\mu }{\partial {x^\prime }^\nu }&= \delta {^\mu }{_\nu } - \epsilon \, \frac{\partial \xi ^\mu }{\partial {x^\prime }^\nu } + O(\epsilon ^2) . \end{aligned} \end{aligned}$$

(C3)

For a tensor field $T^{\mu ...}_{\nu ...}(x)$, one can characterize the deviation from formal invariance under the diffeomorphism $x^\prime (\epsilon ,x)$ in the following way:

$$\begin{aligned} (\pounds _\xi T)^{\mu ...}_{\nu ...}(x):=\lim _{\epsilon \rightarrow 0} \frac{1}{\epsilon }\left\{ T^{\mu ...}_{\nu ...}(x^\prime (\epsilon ,x))-{T^\prime }^{\mu ...}_{\nu ...} (x^\prime (\epsilon ,x))\right\} , \end{aligned}$$

(C4)

where ${T^\prime }^{\mu ...}_{\nu ...}$ represents the components of the coordinate transformed tensor. The above formula defines the Lie derivative and may be applied directly to the connection coefficients.

To work out the explicit expression for the Lie derivative, first recall the transformation law for connection coefficients:

$$\begin{aligned} {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }}=\left[ \frac{\partial {x^\prime }^\gamma }{\partial x^\sigma }\frac{\partial x^\alpha }{\partial {x^\prime }^\mu }\frac{\partial x^\beta }{\partial {x^\prime }^\nu }\right] {\bar{\Gamma }}^\sigma {_{\alpha \beta }} - \frac{\partial x^\alpha }{\partial {x^\prime }^\mu }\frac{\partial x^\beta }{\partial {x^\prime }^\nu } \left[ \frac{\partial ^2 {x^\prime }^\gamma }{\partial x^\alpha \partial x^\beta }\right] . \end{aligned}$$

(C5)

Expanding ${{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }}(x^\prime (\epsilon ,x))$ to lowest order in $\epsilon$ (quantities without arguments are implicitly functions of x),

$$\begin{aligned} \begin{aligned} {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }}(x^\prime (\epsilon ,x)) =&~ {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} + \epsilon \, \xi ^\sigma \frac{\partial {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }}}{\partial x^\sigma } + O(\epsilon ^2), \end{aligned} \end{aligned}$$

(C6)

and expanding ${{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }}(x^\prime (\epsilon ,x))$ to lowest order in $\epsilon$, we obtain the following expression:

$$\begin{aligned} \begin{aligned} {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }}(x^\prime (\epsilon ,x)) =&~ {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} + \epsilon \frac{\partial \xi ^\gamma }{\partial x^\sigma } {\bar{\Gamma }}^\sigma {_{\mu \nu }} - \epsilon \frac{\partial \xi ^\sigma }{\partial x^\mu } {{\bar{\Gamma }}^\prime }{^\gamma }{_{\sigma \nu }} \\&- \epsilon \frac{\partial \xi ^\sigma }{\partial x^\nu } {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \sigma }} - \epsilon \frac{\partial ^2 \xi ^\gamma }{\partial x^\mu \partial x^\nu } + O(\epsilon ^2). \end{aligned} \end{aligned}$$

(C7)

The Lie derivative of the connection coefficients then takes the form:

$$\begin{aligned} \begin{aligned} \pounds _\xi {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} =&~ \xi ^\sigma \frac{\partial {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }}}{\partial x^\sigma } - \frac{\partial \xi ^\gamma }{\partial x^\sigma } {\bar{\Gamma }}^\sigma {_{\mu \nu }} + \frac{\partial \xi ^\sigma }{\partial x^\mu } {{\bar{\Gamma }}^\prime }{^\gamma }{_{\sigma \nu }} \\&+ \frac{\partial \xi ^\sigma }{\partial x^\nu } {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \sigma }} + \frac{\partial ^2 \xi ^\gamma }{\partial x^\mu \partial x^\nu }. \end{aligned} \end{aligned}$$

(C8)

Now one might expect $\pounds _\xi {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }}$ to be covariant. To see that it is in fact covariant, consider the following expression:

$$\begin{aligned} \frac{\partial \xi ^\mu }{\partial x^\nu } = {\bar{\nabla }}_\nu \xi ^\mu - {\bar{\Gamma }}^\mu {_{\nu \tau }} \xi ^\tau , \end{aligned}$$

(C9)

which is just a rewriting of the formula for the covariant derivative ${\bar{\nabla }}_\nu \xi ^\mu$.

One can show that:

$$\begin{aligned} \begin{aligned} \pounds _\xi {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} =&~ \xi ^\sigma [ \partial _\sigma {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} - \partial _\mu {{\bar{\Gamma }}^\prime }{^\gamma }{_{\nu \sigma }} + {\bar{\Gamma }}^\tau {_{\mu \nu }} {{\bar{\Gamma }}^\prime }{^\gamma }{_{\tau \sigma }} \\&+ {{\bar{\Gamma }}^\prime }{^\gamma }{_{\nu \tau }} {\bar{\Gamma }}^\tau {_{\mu \sigma }} - {{\bar{\Gamma }}^\prime }{^\gamma }{_{\tau \nu }} {\bar{\Gamma }}^\tau {_{\mu \sigma }} - {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \tau }} {\bar{\Gamma }}^\tau {_{\nu \sigma }} ]\\&+ {\bar{\nabla }}_\mu ({\bar{\nabla }}_\nu \xi ^\gamma ) + {\bar{\nabla }}_\mu \xi ^\sigma T^\gamma {_{\sigma \nu }}, \end{aligned} \end{aligned}$$

(C10)

where the torsion tensor is $T^\gamma {_{\mu \nu }}:= {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} - {{\bar{\Gamma }}^\prime }{^\gamma }{_{\nu \mu }}$. One may write:

$$\begin{aligned} {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} = {{\bar{\Gamma }}^\prime }{^\gamma }{_{\nu \mu }} + T^\gamma {_{\mu \nu }}. \end{aligned}$$

(C11)

One can use the above to obtain:

$$\begin{aligned} \begin{aligned} \pounds _\xi {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} =&~ \xi ^\sigma [ \partial _\sigma {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} - \partial _\mu {{\bar{\Gamma }}^\prime }{^\gamma }{_{\sigma \nu }} + {\bar{\Gamma }}^\tau {_{\mu \nu }} {{\bar{\Gamma }}^\prime }{^\gamma }{_{\sigma \tau }} - {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \tau }} {\bar{\Gamma }}^\tau {_{\sigma \nu }}\\&- \partial _\mu T^\gamma {_{\nu \sigma }} + {\bar{\Gamma }}^\tau {_{\mu \nu }} T^\gamma {_{\tau \sigma }} + {\bar{\Gamma }}^\tau {_{\mu \sigma }} T^\gamma {_{\nu \tau }} \\&- {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \tau }} T^\tau {_{\sigma \nu }}] + {\bar{\nabla }}_\mu ({\bar{\nabla }}_\nu \xi ^\gamma ) + {\bar{\nabla }}_\mu \xi ^\sigma T^\gamma {_{\sigma \nu }}, \end{aligned} \end{aligned}$$

(C12)

and upon identifying the curvature and covariant derivative of the torsion tensor, one obtains the result:

$$\begin{aligned} \begin{aligned} \pounds _\xi {{\bar{\Gamma }}^\prime }{^\gamma }{_{\mu \nu }} =&~ {\bar{\nabla }}_\mu {\bar{\nabla }}_\nu \xi ^\gamma + \xi ^\sigma {\bar{R}}{^\gamma }_{\nu \sigma \mu } + {\bar{\nabla }}_\mu \left( \xi ^\sigma T^\gamma {_{\sigma \nu }} \right) . \end{aligned} \end{aligned}$$

(C13)

Appendix D: Divergence expression

Here, we compare the divergences of a tensor with respect to two different metric-compatible connections. First, a useful result is derived for two metric tensors $g_{\mu \nu }$ and $\eta _{\mu \nu }$:

$$\begin{aligned} \begin{aligned} \partial _\sigma \left[ \frac{\sqrt{-\eta }}{\sqrt{-g}}\right]&= \frac{\sqrt{-\eta }}{\sqrt{-g}} \left[ \frac{\partial _\sigma \sqrt{-\eta }}{\sqrt{-\eta }} - \frac{\partial _\sigma \sqrt{-g}}{\sqrt{-g}} \right] \\&= - \frac{\sqrt{-\eta }}{\sqrt{-g}} \left[ \Gamma {^\tau }_{\tau \sigma } - {\bar{\Gamma }}{^\tau }_{\tau \sigma } \right] = - \frac{\sqrt{-\eta }}{\sqrt{-g}} \, W{^\tau }_{\tau \sigma } . \end{aligned} \end{aligned}$$

(D1)

For a tensor $T{^\mu }{_\nu }$ and a scalar field $\varphi$, one may obtain the following expression (where ${\bar{\nabla }}_\alpha \eta _{\mu \nu } = 0$):

$$\begin{aligned} \begin{aligned} \nabla _\mu (\varphi \, T{^\mu }{_\nu }) = \,&{\bar{\nabla }}_\mu (\varphi \, T{^\mu }{_\nu }) + \varphi \, W{^\mu }_{\mu \sigma } \, T{^\sigma }{_\nu } - \varphi \, W{^\sigma }_{\mu \nu } \, T{^\mu }{_\sigma } \\ = \,&\varphi \, {\bar{\nabla }}_\mu T{^\mu }{_\nu } + \varphi \, W{^\mu }_{\mu \sigma } \, T{^\sigma }{_\nu } - \varphi \, W{^\sigma }_{\mu \nu } \, T{^\mu }{_\sigma } \\&+ T{^\mu }{_\nu } \, \partial _\mu \varphi . \end{aligned} \end{aligned}$$

(D2)

Now if $\varphi ={\sqrt{-\eta }}/{\sqrt{-g}}$, one can show that:

$$\begin{aligned} \begin{aligned} {\sqrt{-g}} \, \nabla _\mu \left[ \frac{\sqrt{-\eta }}{\sqrt{-g}} \, T{^\mu }{_\nu }\right] = \,&\sqrt{-\eta } \left( {\bar{\nabla }}_\mu T{^\mu }{_\nu } - W{^\sigma }_{\mu \nu } \, T{^\mu }{_\sigma } \right) . \end{aligned} \end{aligned}$$

(D3)

We now consider the divergence of a current $k^\mu = \xi ^\tau \, {\tilde{T}}{^\mu }{_\tau }$, where $\xi ^\mu$ is the dual of a vector field:

$$\begin{aligned} \begin{aligned} \nabla _\sigma k^\sigma = \,&\nabla _\sigma \left( \xi ^\tau \, {\tilde{T}}{^\sigma }{_\tau }\right) \\ = \,&\xi ^\tau \nabla _\sigma {\tilde{T}}{^\sigma }{_\tau } + {\tilde{T}}{^\sigma }{_\tau } \, \nabla _\sigma \xi ^\tau \\ = \,&\xi ^\tau \nabla _\sigma {\tilde{T}}{^\sigma }{_\tau } + {\tilde{T}}{^\sigma }{_\tau } \, {\bar{\nabla }}_\sigma \xi ^\tau + {\tilde{T}}{^\sigma }{_\nu } \, W{^\nu }_{\sigma \tau } \, \xi ^\tau \\ = \,&\left( \nabla _\sigma {\tilde{T}}{^\sigma }{_\tau } + {\tilde{T}}{^\sigma }{_\nu } \, W{^\nu }_{\sigma \tau } \right) \xi ^\tau \ + {\tilde{T}}{^\sigma }{_\tau } \, {\bar{\nabla }}_\sigma \xi ^\tau . \end{aligned} \end{aligned}$$

(D4)

Upon setting ${\tilde{T}}^{\mu \nu } = \varphi \, T^{\mu \nu }$, and substituting Eq. (D2):

$$\begin{aligned} \begin{aligned} \nabla _\sigma (\varphi \, \xi ^\tau \, {T}{^\sigma }{_\tau }) = \,&\xi ^\tau \left[ \nabla _\sigma (\varphi \, {T}{^\sigma }{_\tau } ) + \varphi \, {T}{^\sigma }{_\nu } \, W{^\nu }_{\sigma \tau } \right] \\&+ \varphi \, {T}{^\sigma }{_\tau } \, {\bar{\nabla }}_\sigma \xi ^\tau \\ = \,&\xi ^\tau \biggl [ \varphi \, {\bar{\nabla }}_\sigma {T}{^\sigma }{_\tau } + T{^\mu }{_\tau } \, \partial _\mu \varphi + \varphi \, W{^\sigma }_{\sigma \mu } \, T{^\mu }{_\tau } \biggr ]\\&+ \varphi \, {T}{^\sigma }{_\tau } \, {\bar{\nabla }}_\sigma \xi ^\tau , \end{aligned} \end{aligned}$$

(D5)

and setting $\varphi ={\sqrt{-\eta }}/{\sqrt{-g}}$, we obtain the following result:

$$\begin{aligned} \begin{aligned} \nabla _\sigma \left[ \frac{\sqrt{-\eta }}{\sqrt{-g}} \, \xi ^\tau \, {T}{^\sigma }{_\tau } \right] = \frac{\sqrt{-\eta }}{\sqrt{-g}} \biggl [&{\bar{\nabla }}_\sigma {T}{^\sigma }{_\tau } \, \xi ^\tau + {T}{^\sigma }{_\tau } \, {\bar{\nabla }}_\sigma \xi ^\tau \biggr ]. \end{aligned} \end{aligned}$$

(D6)

Alternatively, defining the tensor $S^{\sigma \mu } = \varphi \, T^{\sigma \mu }$, the above may be rewritten:

$$\begin{aligned} \begin{aligned} \nabla _\sigma \left[ \xi ^\tau \, {S}{^\sigma }{_\tau } \right] = \,&\frac{\sqrt{-\eta }}{\sqrt{-g}} \, \xi ^\tau \, {\bar{\nabla }}_\sigma \left( \frac{\sqrt{-g}}{\sqrt{-\eta }} {S}{^\sigma }{_\tau } \right) + {S}{^\sigma }{_\tau } \, {\bar{\nabla }}_\sigma \xi ^\tau . \end{aligned} \end{aligned}$$

(D7)