General relativity from a thermodynamic perspective

Abstract

I show that the gravitational dynamics in a bulk region of space can be connected to a thermodynamic description in the boundary of that region, thereby providing clear physical interpretations of several mathematical features of classical general relativity: (1) The Noether charge contained in a bulk region, associated with a specific time evolution vector field, has a direct thermodynamic interpretation as the gravitational heat content of the boundary surface. (2) This result, in turn, shows that all static spacetimes maintain holographic equipartition in the following sense: In these spacetimes, the number of degrees of freedom in the boundary is equal to the number of degrees of freedom in the bulk. (3) In a general, evolving spacetime, the rate of change of gravitational momentum is related to the difference between the number of bulk and boundary degrees of freedom. It is this departure from the holographic equipartition which drives the time evolution of the spacetime. (4) When the equations of motion hold, the (naturally defined) total energy of the gravity plus matter within a bulk region, will be equal to the boundary heat content. (5) After motivating the need for an alternate description of gravity (if we have to solve the cosmological constant problem), I describe a thermodynamic variational principle based on null surfaces to achieve this goal. The concept of gravitational heat density of the null surfaces arises naturally from the Noether charge associated with the null congruence. The variational principle, in fact, extremises the total heat content of the matter plus gravity system. Several variations on this theme and implications are described.

Appendices: Calculational details

Several calculational details and background results are collected together in these appendices in order not to distract the flow of ideas in the main paper. Many of them exist in the literature but are compiled together for the sake of completeness. I have given a fair amount of details (and sometimes alternative derivations of the results) in the hope that these will be useful. Some of these results are easier to obtain in the index-free differential form language but I have presented everything in the more familiar index language.

1.1 An identity relating the noether currents

It is obvious that $J^{ab}[q]=0$ identically, if $q_a=\nabla \phi$ is a pure gradient. Very often, we will need to find the Noether current for vectors of the form $v_a=f(x)\nabla _a\phi$. With this motivation, we will first prove a general relation between Noether currents for two vector fields $q^a$ and $v^a \equiv f(x) q^a$. Computing the expanded form of

$$\begin{aligned} J^a(v)=\nabla _b J^{ab}(v) = \nabla _b \left[ \nabla ^a (f q^b) - \nabla ^b (f q^a)\right] \end{aligned}$$

(130)

one can easily show that

$$\begin{aligned} J^a(v)= \nabla _b J^{ab} ( v) \!&= \! \nabla _b [f J^{ab} (q) {+} q^b f^a - q^a f^b] = f J^{a}(q)+J^{ab}(q) f_b + \nabla _b A^{ba}\nonumber \\&= 2 R^a_b f q^b + g^{ij} \pounds _v N^a_{ij} = g^{ij} \pounds _v N^a_{ij} + f ( J^{a}(q) - g^{ij} \pounds _q N^a_{ij})\nonumber \\ \end{aligned}$$

(131)

where we use the notation $f_a \equiv \nabla _a f$ and $A_{ij} \equiv q_i f_j - q_j f_i$. This allows us to obtain the relation we need, viz.,

$$\begin{aligned} J^a(fq)-f J^{a}(q)= g^{ij} \pounds _{fq} N^a_{ij} - f g^{ij} \pounds _q N^a_{ij} = f_b J^{ab}(q) + \nabla _b A^{ba} \end{aligned}$$

(132)

We take the dot product of this equation with $q_a$ to obtain

$$\begin{aligned} q_a J^a (fq) - f q_a J^a (q) = q_a f_b J^{ab} (q) + q_a \nabla _b A^{ba} \end{aligned}$$

(133)

Writing the last term as

$$\begin{aligned} q_a \nabla _b A^{ba}&= \nabla _b (q_a A^{ba}) - {\frac{1}{2}}A^{ba} J_{ba} (q) = \nabla _b (q_a A^{ba}) - {\frac{1}{2}}\, 2(q^b f^a) J_{ba} (q)\nonumber \\&= \nabla _b (q_a A^{ba}) - (q^b f^a) J_{ba} (q) \end{aligned}$$

(134)

we see that the last term in Eq. (134) cancels with first term in the right hand side of Eq. (133). Therefore, using the definition $A_{ij} \equiv q_i f_j - q_j f_i$ to simplify $\nabla _b (q_a A^{ba})$, we get:

$$\begin{aligned} q_a J^a (fq) - f q_a J^a (q) = \nabla _b (q_a A^{ba})= \nabla _b\left( \left[ q^a q^b - q^2 g^{ab} \right] \nabla _a f\right) \end{aligned}$$

(135)

In other words, we have the simple result

$$\begin{aligned} q_a J^a (fq) - fq_a J^a(q) = {\left\{ \begin{array}{ll} -\nabla _b(q^2 \mathcal {P}^{ab}(q)\nabla _a f) &{} \text {when}\,q^2\ne 0\\ +\nabla _b(q^b q^a \nabla _a f) &{} \text {when}\,q^2=0 \end{array}\right. } \end{aligned}$$

(136)

where $\mathcal {P}^{ab}(q) = g^{ab} - (q^a q^b / q^2)$ is the projection tensor orthogonal to $q^a$. This result provides one way of computing the Noether current for vector fields.

1.2 Noether currents for $u^a$ and $\xi ^a$

In the paper, we extensively used the Noether current associated with the vector $\xi _a=Nu_a=-N^2 \nabla _a t$. Here we will compute the Noether current for both the vector fields $u_a=- N \nabla _at$ and $\xi _a=-N^2 \nabla _a t$ because their comparison shows some structural differences which are of interest (and also because they can be computed without extra effort!). To calculate these currents, it is convenient to use the identity in Eq. (136). We start with $u_a = - N \nabla _at$ and use Eq. (136) with $q_a = - u_a/N,\ f= - N J^a [q]=0$ to get

$$\begin{aligned} - \frac{u_a}{N} J^a(u) = - \nabla _b \left( - \frac{1}{N^2} h^{ab} (-\nabla _a N)\right) = - \nabla _b (a^b/N) \end{aligned}$$

(137)

where $h^{ab} = g^{ab} + u^au^b$ is the spatial projection tensor and $h^{ab} \nabla _a N = N a^b$ where $a^b$ is the acceleration of $u^a$. To verify that this is indeed the acceleration, even in a general, time-dependent situation, let us compute it directly from the definition $a_k=u^l\nabla _lu_k$. We get:

$$\begin{aligned} a_k=u^l\nabla _l(-N\nabla _kt)= \left( u^l\nabla _lN\right) \frac{u_k}{N}+Nu^l\nabla _k\left( \frac{u_l}{N}\right) =u_ku^l\frac{\nabla _lN}{N}+\frac{\nabla _kN}{N} =h^l_k\frac{\nabla _lN}{N} \end{aligned}$$

(138)

where we have used $\nabla _k\nabla _lt=\nabla _l\nabla _kt$ to arrive at the second term in the second equality. This shows that $Na_i=h^j_i\nabla _jN$ which is a useful result. Expanding out $\nabla _b (a^b/N)$ in Eq. (137), we get

$$\begin{aligned} \frac{u^a}{N} J^a(u) = \nabla _b \left( a^b/N\right) = \frac{1}{N} \nabla _b a^b - \frac{1}{N^2} \left( N a^2\right) = \frac{1}{N} \left( \nabla _i a^i - a^2\right) \end{aligned}$$

(139)

To proceed further, we note that, if $v^i$ is any spatial vector that satisfies the condition $u_iv^i=0$ and $D_i$ is the covariant derivative on the $t=$ constant surface, we have the result that

$$\begin{aligned} D_iv^i= D_\alpha v^\alpha \equiv \left( g^{ij} + u^i u^j\right) \nabla _i v_j = \nabla _i v^i - v_j a^j \end{aligned}$$

(140)

(Note that, in the adopted coordinates we must have $v^0=0$, since $u_i=-N\delta _i^0$.) Applying this to $v^i=a^i$, we have

$$\begin{aligned} D_ia^i= D_\alpha a^\alpha = \nabla _i a^i - a^2 \end{aligned}$$

(141)

Using Eq. (141) in Eq. (139), we get the Noether charge density to be:

$$\begin{aligned} u_aJ^a[u]=D_\alpha a^\alpha \end{aligned}$$

(142)

From the standard expression for the Noether current we also have:

$$\begin{aligned} D_ia^i = 2 R_{ab} u^a u^b + g^{ij}u_a \pounds _u N^a_{ij} \end{aligned}$$

(143)

Let us next consider the Noether current for $\xi ^a$. Using again Eq. (136) with, say, $q^a = u^a, \ f= N$, we find that

$$\begin{aligned} u_a J^a (Nu) - N u_a J^a(u) = \nabla _b (h^{ab} \nabla _a N) = \nabla _b (N a^b) \end{aligned}$$

(144)

Using Eq. (142) we get:

$$\begin{aligned} u_aJ^a(\xi )&= N u_aJ^{a}(u)+\nabla _j\left( Na^j\right) =N\left( D_\alpha a^\alpha +\nabla _ja^j+a^2\right) \nonumber \\&= 2N\nabla _ja^j=2D_\alpha \left( Na^\alpha \right) \end{aligned}$$

(145)

which is the final result we are after. Of course, in this particular case, one can also obtain this result by direct computation without using Eq. (136). The direct proof of this relation will proceed as follows:

$$\begin{aligned} u_b J^b(\xi )&= \nabla _i \left( u_b \left( \nabla ^b \xi ^i - \nabla ^i \xi ^b\right) \right) - \frac{1}{2} (\nabla _i u_b - \nabla _b u_i) J^{bi}\end{aligned}$$

(146)

$$\begin{aligned}&= \nabla _i\left[ u_b (N \nabla ^b u^i + u^i\nabla ^b N - (i\leftrightarrow b))\right] \nonumber \\&-\, (a_i u_b - a_b u_i) \left[ (\nabla ^b N) u^i + N \nabla ^b u^i)\right] \end{aligned}$$

(147)

$$\begin{aligned}&= \nabla _i \left[ N a^i + u^i u^b \nabla _b N + \nabla ^i N\right] - Na^2 - a_b \nabla ^b N \end{aligned}$$

(148)

$$\begin{aligned}&= \nabla _i \left[ N a^i + (g^{ib} + u^i u^b) \nabla _b N\right] - Na^2 - a_b \nabla ^b N \end{aligned}$$

(149)

$$\begin{aligned}&= \nabla _i ( 2 Na^i) - Na^2- a_b \nabla ^b N = 2 N \nabla _i a^i + a^i \nabla _i N - Na^2 \end{aligned}$$

(150)

where we have used $\nabla _{[i}\, u_{b]} = a_{[i}\, u_{b]}$ to get Eq. (147). Note that the last two terms, when combined as follows, will cancel each other:

$$\begin{aligned} Na^i \left[ \nabla _i \ln N - a_i\right]&= h^{ij}\nabla _j N \left[ \nabla _i \ln N - a_i\right] \nonumber \\&= (\nabla _j N) \left[ \frac{1}{N} h^{ij} \nabla _i N - h^{ij} a_i\right] =0 \end{aligned}$$

(151)

From Eq. (141), we also have

$$\begin{aligned} \frac{1}{2N}D_\alpha \left( 2N a^\alpha \right) =D_\alpha a^\alpha +a^\alpha D_\alpha (\ln N) =D_\alpha a^\alpha +a^2=\nabla _ia^i \end{aligned}$$

(152)

so that $D_\alpha (2N a^\alpha )=2N \nabla _ia^i$. This gives the result we are after.

Comparing with the Noether charge densities for $u^i$ (in Eq. (142)) and $\xi ^a$ (in Eq. (145)) we see that $a^\alpha$ has been replaced by $2Na^\alpha$ which turns out to be important for the thermodynamic interpretation for two reasons. (i) The factor 2 is crucial in giving the correct expression for the equipartition result through Eq. (55). (ii) More importantly, the combination $Na$ leads to a finite result on the horizon—leading to standard surface gravity—while the magnitude $a$ diverges on an $N=0$ surface on static spacetimes. This is also seen from the fact that, the correct local temperatures are always defined with the Tolman redshift factor $N$.

Finally, we give the steps involved in proving the expression for the Noether charge associated with $\zeta ^a$ which we needed in Sect. 7.3. It is slightly more convenient to expand the right hand side of Eq. (128) and show that it is equal to the left hand side. The computation is straightforward:

$$\begin{aligned} 2 r_a \zeta ^b \nabla _b u^a - ( r_a \pounds _\zeta u^a - u_a \pounds _\zeta r^a)&= 2r_a \zeta ^b \nabla _b u^a - r_a ( \zeta ^b \nabla _b u^a - u^b \nabla _b \zeta ^a)\nonumber \\&+ u_a (\zeta ^b \nabla _b r^a - r^b \nabla _b \zeta ^a) \nonumber \\&= r_a u_b \nabla ^b \zeta ^a - u_a r_b \nabla ^b \zeta ^a = r_a u_b ( \nabla ^b \zeta ^a - \nabla ^a \zeta ^b)\nonumber \\&= r_a u_b J^{ba} \end{aligned}$$

(153)

where we have used the result $r_au^a=0$ to obtain the second equality. This is the result quoted in the text.

1.3 Different expressions for $u_a g^{ij} \pounds _\xi N^a_{ij}$

We will obtain here the two other equivalent expressions for $u_a g^{ij} \pounds _\xi N^a_{ij}$ used in the text. Using $D_\alpha (2N a^\alpha )=2N \nabla _ia^i$ in Eq. (56) we get

$$\begin{aligned} u_a g^{ij} \pounds _\xi N^a_{ij} = 2 N (\nabla _i a^i - R_{ab} u^a u^b) \end{aligned}$$

(154)

On the other hand, from the standard identity for $R_{ab} u^a u^b$ (see e.g., page 541 of [44]) we have:

$$\begin{aligned} R_{ab} u^a u^b = u^a (\nabla _l \nabla _a u^l - \nabla _a \nabla _l u^l) = \nabla _l a^l - K^a_lK^l_a + \nabla _a ( K u^a) + K^2 \end{aligned}$$

(155)

where $K_{ij} = - \nabla _i u_j - u_i a_j$ is the extrinsic curvature. So

$$\begin{aligned} \nabla _i a^i - R_{ab} u^a u^b = K_{ij} K^{ij} - K^2 - \nabla _a (Ku^a) \end{aligned}$$

(156)

Substituting into Eq. (154) we get:

$$\begin{aligned} \frac{1}{2N} u_a g^{ij} \pounds _\xi N^a_{ij}&= \nabla _i a^i - R_{ab} u^a u^b = K_{ij} K^{ij} - K^2 - \nabla _a (Ku^a)\nonumber \\&= K_{ij} K^{ij} -u^a \nabla _a K \end{aligned}$$

(157)

This allows a simple physical interpretation for the combination $(K_{ij} K^{ij} - K^2)$ in terms of the Lie derivative of $N^a_{bc}$.

This result can also be obtained directly from the definitions of various quantities. To do this, we first note that, from the definition of extrinsic curvature, $K_{lm} = - \nabla _l u_m - a_m u_l$, we have the result

$$\begin{aligned} K_{lm} K^{ml} = \left( \nabla _l u_m + a_m u_l \right) \left( \nabla ^m u^l + a^l u^m \right) =(\nabla _l u_m) \, (\nabla ^m u^l) \end{aligned}$$

(158)

Further, using the notation $\alpha _i\equiv \nabla _i \ln N$, we can write:

$$\begin{aligned} \nabla _i ( N u_j)&= N \alpha _i u_j + N(-K_{ij} - u_i a_j) = - N K_{ij} + N (\alpha _i u_j - u_i a_j)\nonumber \\&= - N K_{ij} + N (\alpha _i u_j - h^k_j \alpha _k u_i)\nonumber \\&= - N K_{ij} + N (\alpha _i u_j - \alpha _j u_i) - u_iu_ju^k \nabla _k N \end{aligned}$$

(159)

Taking the symmetric and anti-symmetric parts of this expression we get

$$\begin{aligned} J_{ab} (\xi ) = 2 N ( \alpha _a u_b - \alpha _b u_a ) ; \qquad S_{ab}(\xi ) = - 2 N (K_{ab} + u_a u_b u^k \alpha _k) \end{aligned}$$

(160)

The rest of the calculation proceeds directly as follows:

$$\begin{aligned} u_b \left( \nabla ^b S - \nabla _i S^{bi}\right)&= \nabla _b \left( S u^b\right) - \nabla _i \left( u_b S^{bi}\right) - \left( \nabla _b u^b\right) S + \left( \nabla _i u_b \right) S^{bi}\nonumber \\&= \nabla _b \left( u^b \left( - 2 NK + 2N u^k \alpha _k\right) \right) - \nabla _i\left( 2N u^i u^k \alpha _k\right) \nonumber \\&+ K \left( -2NK +2N u^k \alpha _k\right) + S^{bi} \left( - K_{ib}-a_b u_i\right) \nonumber \\&= -2 \nabla _b \left( KN u^b\right) - 2 N \left( K^2 - Ku^k \alpha _k\right) \nonumber \\&+ 2 N \left( K_{ib} + a_b u_i\right) \left( K^{bi} + u^b u^i u^j \alpha _j\right) \nonumber \\&= - 2 \nabla _b \left( NK u^b \right) + 2 N \left( K_{ib} K^{bi} - K^2\right) + 2 N K u^k \alpha _k\nonumber \\&= 2 N \left( K_{ij} K^{ji} -K^2\right) {-} 2KN u^b \alpha _k {-} 2N \nabla _b \left( Ku^b\right) {+} 2 KN u^b \alpha _k\nonumber \\&= 2N\left[ K_{ij}K^{ji} - K^2 - \nabla _b (Ku^b)\right] = 2 N \left[ K_{ij} K^{ji} - u^b \nabla _b K\right] \nonumber \\ \end{aligned}$$

(161)

which agrees with Eq. (157) when we use the fact that the left hand side is $u_a g^{ij} \pounds _\xi N^a_{ij}$.

In the text we also related $u_a g^{ij} \pounds _\xi N^a_{ij}$ to $h_{ab}\, \pounds _\xi p^{ab}$ where $p^{ab} \equiv - \sqrt{h} (K^{ab} - h^{ab}K)$ is the momentum conjugate to $h_{ab}$. This result can be obtained as follows. We start with the standard result (see page 550 of [44]) used in the variation of the action functional with the $2K$ term:

$$\begin{aligned} - g^{ab} u_m \, \delta N^m_{ab} = 2 \delta K + K^{ab} \delta h_{ab}+D_a(h^a_b\delta u^b) \end{aligned}$$

(162)

which expresses the variational term that appears in the Einstein–Hilbert action in terms of the variation of the extrinsic curvature. The troublesome term is the one involving the spatial derivative $D_a(h^a_b\delta u^b)$, which, fortunately, vanishes when the variation is due to a diffeomorphism along $\xi ^a$. This is easy to see from:

$$\begin{aligned} D_a\left( h^a_b\pounds _\xi u^b\right) = D_a\left( h^a_b\pounds _\xi \left( \frac{\xi ^b}{N}\right) \right) = D_a\left( h^a_b\left[ \left( \frac{1}{N}\right) \pounds _\xi (\xi ^b)+ \xi ^b\pounds _\xi \left( \frac{1}{N}\right) \right] \right) =0 \end{aligned}$$

(163)

because $\pounds _\xi (\xi ^b)=0$ and $h^a_b\xi ^b=0$. As regards the rest of the terms in Eq. (162), an explicit calculation of $h_{ab} \delta p^{ab}$ gives

$$\begin{aligned} h_{ab} \delta \left[ (h^{ab} K - K^{ab})\sqrt{h}\right]&= K \sqrt{h}\, h^{ab}\, \delta h_{ab} + h_{ab} \sqrt{h} (- \delta K^{ab} + K \, \delta h^{ab} + h^{ab} \, \delta K)\nonumber \\&= 3 \sqrt{h}\, (h_{ab} \delta K^{ab} + K^{ab} \delta h_{ab}) - \sqrt{h} \, h_{ab}\delta K^{ab} \nonumber \\&= 2 \sqrt{h}\, h_{ab} \delta K^{ab} + 3 \sqrt{h}\,K^{ab} \delta h_{ab}\nonumber \\&= 2 \sqrt{h}\,\delta K + \sqrt{h}\,K^{ab} \delta h_{ab} \nonumber \\&= \sqrt{h}\,\left\{ 2\delta K + \,K^{ab} \delta h_{ab}\right\} \end{aligned}$$

(164)

It follows that:

$$\begin{aligned} \sqrt{h}\, u_c g^{ik} \pounds _\xi N_{ik}^c = - \, h_{ab}\, \pounds _\xi p^{ab} \end{aligned}$$

(165)

This result does not hold for the diffeomorphism along the vector $\zeta ^a$ and one will obtain an extra surface contribution in that case.

For the variations arising from the diffeomorphism along $\xi ^a$, it is also possible to prove the result in Eq. (165) directly (without using Eq. (162)) as follows:

$$\begin{aligned} h_{ab} \pounds _\xi p^{ab}&= \pounds _\xi \left( h_{ab} \sqrt{h}\,\left( K h^{ab} - K^{ab}\right) \right) - p^{ab} \pounds _\xi h_{ab} \nonumber \\&= \pounds _\xi ( \sqrt{h}\, 2 K ) - p^{ab} ( - 2 N K_{ab})\nonumber \\&= 2 \sqrt{h}\, N u^a \partial _a K + 2 K \frac{1}{2} \sqrt{h}\, h^{ab} (- 2 N K_{ab}) + 2 N p^{ab} K_{ab}\nonumber \\&= 2 \sqrt{h}\, N u^a \partial _a K - 2 N \sqrt{h}\, K^2 + 2 N \sqrt{h}\, \left( K^2 - K_{ab} K^{ab}\right) \nonumber \\&= 2 N \sqrt{h}\, \left( u^a \partial _a K - K_{ab} K^{ab}\right) \end{aligned}$$

(166)

In the first equality, we have used the definition of $p_{ab} = \sqrt{h}(K h_{ab} - K_{ab})$; to get the second equality, we have used $\pounds _\xi h_{ab} = - 2N K_{ab}$. This result, when combined with Eq. (157), leads to the desired expression. We can therefore write the Noether current expression as

$$\begin{aligned} \sqrt{h}\, u_a g^{ij} \pounds _\xi N^a_{ij} = \sqrt{h}\, D_\alpha (2 N a^\alpha ) - 2 N \sqrt{h}\, R_{ab}\, u^a u^b = - h_{ab} (\pounds _\xi p^{ab}) \end{aligned}$$

(167)

This was the result used in the text.

1.4 Variation of the gravitational momentum flux $P^a_H$

We will derive the expression for $\delta ( \sqrt{-g}\,P^a_H)$, discussed in the text, from first principles. We start with the definition for $P^a$ based on the Einstein–Hilbert action for an arbitrary but fixed $q^a$:

$$\begin{aligned} P^a_H\equiv L_H q^a + g^{lm} \pounds _q N^a_{lm}=J^a_H-2G^a_bq^b \end{aligned}$$

(168)

We next introduce an arbitrary variation of the metric, keeping $q^a$ fixed. (The $q_a$ will change but we will not need it.). The variation of $\sqrt{-g}\,P^a_H$ gives

$$\begin{aligned} \delta ( \sqrt{-g}\,P^a_H)&= q^a \delta (R\sqrt{-g}\,) + \delta \left( f^{lm} \pounds _q N^a_{lm}\right) \nonumber \\&= q^a \left( R_{ij} \delta f^{ij} - \partial _c (f^{lm} \delta N^c_{lm})\right) + \delta \left( f^{lm} \pounds _q N^a_{lm}\right) + \pounds _q \left( f^{lm} \delta N^a_{lm}\right) \nonumber \\&-\, \pounds _q \left( f^{lm} \delta N^a_{lm}\right) \end{aligned}$$

(169)

where, in the second line, we have explicitly added and subtracted $\pounds _q (f^{lm} \delta N^a_{lm})$ which is a trick to get the symplectic structure. This gives

$$\begin{aligned} \delta \left( \sqrt{-g}\,P^a_H\right)&= \delta \left( f^{lm} \pounds _q N^a_{lm}\right) - \pounds _q \left( f^{lm} \delta N^a_{lm}\right) + q^a \left( R_{ij} \delta f^{ij}\right) + \pounds _q \left( f^{lm} \delta N^a_{lm}\right) \nonumber \\&-\,q^a \partial _c \left( f^{lm} \delta N^c_{lm}\right) \end{aligned}$$

(170)

The combination $Q^a\equiv f^{lm} \delta N^a_{lm}=\sqrt{-g}\,g^{lm} \delta N^a_{lm}$ which appears here is a tensor density of weight one and hence

$$\begin{aligned} \pounds _q \left( f^{lm} \delta N^a_{lm}\right) - q^a \partial _c \left( f^{lm} \delta N^c_{lm}\right)&= \pounds _q Q^a - q^a \partial _c Q^c = \partial _c\left( q^c Q^a - q^a Q^c\right) \nonumber \\&= \partial _c\left( f^{lm}\delta N^{[a}_{lm} q^{c]}\right) \end{aligned}$$

(171)

Therefore,

$$\begin{aligned} \delta (\sqrt{-g}\,P^a_H) - q^a R_{ij} \delta f^{ij}&= \delta \left( f^{lm} \pounds _q N^a_{lm}\right) - \pounds _q\left( f^{lm} \delta N^a_{lm}\right) + \partial _c\left( f^{lm}\delta N^{[a}_{lm} q^{c]}\right) \nonumber \\&= \delta f^{lm}\pounds _q N^a_{lm} - \left( \pounds _qf^{lm}\right) \delta N^a_{lm} + \partial _c\left( f^{lm}\delta N^{[a}_{lm} q^{c]}\right) \nonumber \\ \end{aligned}$$

(172)

where we have used the fact that $\delta$ and $\pounds _q$ operations on $N^a_{lm}$ commute when $\delta q^a=0$ (see Appendix 9.6). Defining the symplectic form

$$\begin{aligned} \sqrt{-g}\,\omega ^a (\delta , \pounds _q) \equiv \delta f^{lm} \pounds _q N^a_{lm} - (\pounds _q f^{lm}) \delta N^a_{lm} \end{aligned}$$

(173)

we can write Eq. (172) as:

$$\begin{aligned} \delta ( \sqrt{-g}\,P^a_H) - q^a R_{ij} \delta f^{ij}=\sqrt{-g}\,\omega ^a +\partial _c\left( f^{lm}\delta N^{[a}_{lm} q^{c]}\right) \end{aligned}$$

(174)

If we consider variations with the background being on-shell (i.e., $G_{ab}=0$ but $\delta G_{ab} \ne 0$), then the second term on the left hand side of Eq. (174) vanishes and we get

$$\begin{aligned} \delta ( \sqrt{-g}\,P^a_H) =\sqrt{-g}\,\omega ^a +\partial _c \left( f^{lm}\delta N^{[a}_{lm} q^{c]}\right) \end{aligned}$$

(175)

This is a completely covariant equation because $P_H^a$ is a covariant object. On using Eq. (168), we also find:

$$\begin{aligned} \delta \left( \sqrt{-g}\,[J^a_H- 2G^a_bq^b]\right) = \sqrt{-g}\,\omega ^a +\partial _c\left( f^{lm}\delta N^{[a}_{lm} q^{c]}\right) + q^aR_{ij}\delta f^{ij} \end{aligned}$$

(176)

A slightly different, but equivalent, expression is:

$$\begin{aligned} \partial _b \left\{ \delta \left( \sqrt{-g}\,J^{ab}_H\right) - \sqrt{-g}\,g^{lm}\delta N^{[a}_{lm} q^{b]}\right\}&= \sqrt{-g}\,\omega ^a + 2 \delta \left( G^a_b q^b \sqrt{-g}\,\right) \nonumber \\&+ q^a \sqrt{-g}\,G_{lb}\delta g^{lb} \end{aligned}$$

(177)

which is valid off-shell. If we now assume that the original spacetime is on-shell with $G_{ab}=0$ but $\delta G_{ab}\ne 0$, then:

$$\begin{aligned} \partial _b \left\{ \delta \left( \sqrt{-g}\,J^{ab}_H\right) - \sqrt{-g}\,g^{lm}\delta N^{[a}_{lm} q^{b]}\right\} = \sqrt{-g}\,\omega ^a + 2 \delta \left( G^a_b q^b \sqrt{-g}\,\right) \end{aligned}$$

(178)

We next compute the Lie derivative of the gravitational energy density. To do this from first principles, we again start from the definition of $P^a_H$ in Eq. (168) (for $q^a = \xi ^a$) and obtain from it the result

$$\begin{aligned} \sqrt{h} u_a P^a_H[\xi ] = - t_a f^{lm} \pounds _\xi N^a_{lm} - \sqrt{-g}\,L_H \end{aligned}$$

(179)

where $t_a \equiv - (u_a/N) = \delta _a^0$ in the adopted coordinates. The variation of this expression, on using the standard result for $\delta (\sqrt{-g}\,L_H)$, gives

$$\begin{aligned} \delta \left( \sqrt{h}\ u_a P^a_H\right) = - t_a \delta \left( f^{lm} \pounds _\xi N^a_{lm}\right) - \sqrt{-g}\,G_{ab} \delta {g^{ab}} + \partial _c \left( f^{lm} \delta N^c_{lm}\right) \end{aligned}$$

(180)

Using the usual trick for obtaining the symplectic structure and introducing a factor $t_a \xi ^a =1$ in one of the terms, we get

$$\begin{aligned} \delta \left( \sqrt{h}\ u_a P^a_H\right)&= - t_a\left[ \delta \left( f^{lm} \pounds _\xi N^a_{lm}\right) - \pounds _\xi \left( f^{lm} \delta N^a_{lm}\right) \right] - t_a \pounds _\xi \left( f^{lm} \delta N^a_{lm}\right) \nonumber \\&+\, (t_a \xi ^a)\partial _c \left( f^{lm} \delta N^c_{lm}\right) - \sqrt{-g}\,G_{ab} \delta {g^{ab}} \end{aligned}$$

(181)

Defining the symplectic form $\omega ^a$ as in Eq. (173) and using the fact that, for any tensor density $Q^a$ we have the result $\pounds _\xi Q^a - \xi ^a\partial _c Q^c = \partial _c ( \xi ^{[c} Q^{a]})$, we get

$$\begin{aligned} \delta \left( \sqrt{h}\ u_a P^a_H\right) + R_{ab} \delta f^{ab} = \sqrt{h}\, u_a \omega ^a - \partial _c \left[ u_a f^{lm} \delta N^{[c}_{lm} u^{a]} \right] \end{aligned}$$

(182)

Simplifying the second term on the right hand side, we find that $u_a f^{lm} \delta N_{lm}^{[c}u^{a]} = -h^c_a f^{lm} \delta N^a_{lm}$ leading to the result:

$$\begin{aligned} \delta \left( \sqrt{h}\ u_a P^a_H\right) + R_{ab} \delta f^{ab} = \sqrt{h}\, u_a \omega ^a + \partial _c \left[ h^c_a f^{lm} \delta N^a_{lm} \right] \end{aligned}$$

(183)

Further, using $R_{ab} \delta f^{ab}= \sqrt{-g}\,G_{ab} \delta {g^{ab}} = - \sqrt{-g}\,G_{ab} \pounds _\xi {g^{ab}}$ and the Bianchi identities, we can simplify the second term on the left. These together lead to the result

$$\begin{aligned} \pounds _\xi \left( \sqrt{h}\, u_a P^a_H(\xi )\right) = \partial _c \left[ \left( 2 G^c_b \xi ^b + h^c_a {g^{lm}} \pounds _\xi N^a_{lm}\right) \sqrt{-g}\,\right] \end{aligned}$$

(184)

This is the result used earlier.

1.5 Variational formulation based on the null surfaces

We used in the text the following result:

$$\begin{aligned} R_{ab} \ell ^a\ell ^b&= \ell ^j (\nabla _i\nabla _j - \nabla _j\nabla _i)\ell ^i = \nabla _i(\kappa \ell ^i) - \nabla _i\ell ^j\nabla _j\ell ^i - \nabla _j (\ell ^j (\kappa + \theta )) + ( \nabla _i\ell ^i)^2\nonumber \\&= - \nabla _i(\theta \ell ^i)- [\nabla _i \ell ^j \nabla _j\ell ^i - (\nabla _i\ell ^i)^2 ] \equiv - \nabla _i (\theta \ell ^i) - \mathcal {S} \end{aligned}$$

(185)

where we have used the standard result $\nabla _i \ell ^i = \theta + \kappa$ (with $\kappa$ defined through $\ell ^i \nabla _i \ell ^j = \kappa \ell ^j$) and defined $\mathcal {S}$ by,

$$\begin{aligned} \mathcal {S}\equiv [\nabla _i \ell ^j \nabla _j\ell ^i -(\nabla _i\ell ^i)^2] \end{aligned}$$

(186)

When we integrate expressions over a null surface with the measure $d\lambda \ d^2 x \sqrt{\sigma }$, we can ignore terms of the kind $\nabla _i(\phi \ell ^i )$ for any scalar $\phi$ since they produce only boundary contributions. To see this we only need to note that

$$\begin{aligned} \nabla _i (\phi \ell ^i) = \frac{d\phi }{d\lambda }+\phi \frac{d}{d\lambda }(\ln \sqrt{\sigma }) = \frac{1}{\sqrt{\sigma }} \frac{d}{d\lambda } (\sqrt{\sigma }\,\phi ) \end{aligned}$$

(187)

where we have used $\ell ^i\nabla _i=d/d\lambda$ and $\nabla _i \ell ^i=d( \ln \sqrt{\sigma })/d\lambda$. So the integral over the null surface of the first term in Eq. (185) is given by

$$\begin{aligned} \int \limits _{\lambda _1}^{\lambda _2} d\lambda \, d^2x\, \sqrt{\sigma }\, \nabla _i ( \theta \ell ^i) = \int d^2x \sqrt{\sigma }\, \theta \bigg |_{\lambda _1}^{\lambda _2} \end{aligned}$$

(188)

which is just a boundary contribution which can be ignored. We therefore conclude that

$$\begin{aligned} \int \limits _{\lambda _1}^{\lambda _2} d\lambda \ d^2x\, \sqrt{\sigma }\, \left[ -2R_{ab} + T_{ab}\right] \ell ^a\ell ^b = \int \limits _{\lambda _1}^{\lambda _2} d\lambda \ d^2x\, \sqrt{\sigma }\, [2\mathcal {S} + T_{ab}\ell ^a\ell ^b ] \end{aligned}$$

(189)

when we consider variations of the null vectors $\ell ^a$ which vanish at the boundaries ($\lambda = \lambda _1, \lambda _2$). It follows from the arguments given in the text that the field equations can be obtained by varying $\ell ^a$ in the expression:

$$\begin{aligned} Q\equiv \int \limits _{\lambda _1}^{\lambda _2} \frac{d\lambda \ d^2x}{16\pi }\, \sqrt{\sigma }\, [2\mathcal {S} + 16\pi T_{ab}\ell ^a\ell ^b ] \end{aligned}$$

(190)

where we have reintroduced $16\pi G$ with $G=1$.

We next compute the Noether current for the null vector $\ell ^a$ used in the text. Let $\ell _a$ be a null congruence defining a null surface which may not be affinely parametrized. If we take $\ell _a=A(x)\nabla _a B(x)$, then $\ell ^i\nabla _i\ell _j=\kappa \ell _j$ where $\kappa =\nabla _iA\nabla ^iB=\ell ^a\nabla _a\ln A$. We can compute the Noether current for $\ell _a$ using our Eq. (136) and noting that the Noether current for $q_a = \ell _a/A$ is zero. This gives

$$\begin{aligned} \frac{\ell _a }{A} J^a (\ell )&= \nabla _b \left( \frac{\ell ^b }{A}\ \frac{\ell ^a}{A} \nabla _a A\right) = \nabla _b \left( \frac{\ell ^b}{A}\,\kappa \right) = \nabla _b( \kappa \ell ^b)\frac{1}{A} - \kappa \ell ^b\frac{1}{A^2}\nabla _b A\nonumber \\&= \frac{1}{A}\left\{ \nabla _b( \kappa \ell ^b) - \kappa ^2\right\} \end{aligned}$$

(191)

It follows that

$$\begin{aligned} \ell _a J^a (\ell ) = \nabla _b( \kappa \ell ^b) - \kappa ^2= \ell ^b\nabla _b\kappa + \kappa ^2 + \kappa \theta - \kappa ^2 = \frac{d\kappa }{d\lambda } + \theta \kappa \end{aligned}$$

(192)

where we have used $\nabla _a \ell ^a = \theta + \kappa$ and $\ell ^a \nabla _a = d/d\lambda$. The quantity $\nabla _b( \kappa \ell ^b) - \kappa ^2$ is analogous to the right hand side of Eq. (141) in the case of a null vector. But the projection to a surface ‘orthogonal’ to a null vector is not well defined for us to introduce a covariant derivative. Instead, we have to work with a co-null vector $k_a$ defined such that $k^2=0$ and $k_a \ell ^b =-1$. Then the projection to the 2-surface is provided by the projection vector $q^a_b = \delta ^a_b + k^a\ell _b + k_b\ell ^a$ and we can relate $\mathcal {D}_a (\kappa \ell ^a) \equiv q^{ab} \nabla _a (\kappa \ell _b)$ to $\nabla _a(\kappa \ell ^a)$. In this computation, it is useful to note the identity $\ell _b \nabla _a(\phi \ell ^b)=0$ for any scalar $\phi$. This leads to the result

$$\begin{aligned} \mathcal {D}_a (\kappa \ell ^a) = \nabla _a (\kappa \ell ^a) + k^b[\ell _b (\ell ^a \nabla _a \kappa )+\kappa \ell ^a \nabla _a \ell _b] =\nabla _a (\kappa \ell ^a) - \left( \frac{d\kappa }{d\lambda } +\kappa ^2\right) \end{aligned}$$

(193)

Therefore,

$$\begin{aligned} \nabla _a (\kappa \ell ^a) - \kappa ^2 = \mathcal {D}_a (\kappa \ell ^a) + \frac{d\kappa }{d\lambda } \end{aligned}$$

(194)

These results lead to the expressions used in the text.

1.6 Lie derivatives of expressions involving connections

We derive a series of results related to Lie derivatives, which will be useful in several derivations in the paper. The Lie derivative $\pounds _q T$ of some tensor (or tensor density) $T$ has the structure of terms involving $q\partial T$ plus a series of $T\partial q$. So, a variation $\delta (\pounds _q T)$ will involve the sum of terms each having either $T\delta q$ or $q\delta T$. The sum of all terms involving $q \delta T$ will lead to $\pounds _q (\delta T)$ while the sum of terms with $T\delta q$ will be $\pounds _{\delta q} T$. Therefore,

$$\begin{aligned} \delta (\pounds _q T) = \pounds _q (\delta T) + \pounds _{\delta q} T. \end{aligned}$$

(195)

So, for any tensorial object $T$, the operations $\delta$ and $\pounds _q$ commute if $\delta q =0$.

This result, however, is not true for non-tensorial objects involving connections, etc. Since we need to handle them, we will derive a series of results for a particular set of them. To begin with, we have the expression for the Lie derivative of the connection

$$\begin{aligned} \pounds _v\Gamma ^a_{bc}=\nabla _b \nabla _c v^a+R^a_{cmb}v^m \end{aligned}$$

(196)

which can be obtained from first principles, knowing the transformation properties of the connection or from its definition in terms of the metric. As an aside, we point out the following curious consequence. In flat spacetime, we get a non-zero result:

$$\begin{aligned} \pounds _v[\Gamma ^a_{bc}]_\mathrm{flat} = \nabla _b \nabla _c v^a \end{aligned}$$

(197)

(In fact, if we are using Cartesian coordinates, this result tells us that the Lie derivative of zero is non-zero; i.e, $\pounds _v 0 = \partial _b \partial _c v^a$!). This fact is sometimes used to define a background subtraction scheme in order to make expressions involving connections covariant. The essential idea is to work with the difference $\Gamma ^a_{bc}-\Gamma ^a_{bc}|_{flat}$, which, of course, is a tensor. The result in Eq. (196) can be re-written as:

$$\begin{aligned} \pounds _q \Gamma ^a_{bc} = {\left[ \pounds _{q} {\Gamma ^a_{bc}}\right] _\mathrm{std}} + \partial _b \partial _c q^a \end{aligned}$$

(198)

where ${\left[ \pounds _{q} {....}\right] _\mathrm{std}}$ stands for the Lie derivative computed treating the indexed object $(....)$ as though it was a tensor or tensor density. From this, we get the results:

$$\begin{aligned} \pounds _q N^a_{bc}&= {\left[ \pounds _{q} {N^a_{bc}}\right] _\mathrm{std}} - \partial _b \partial _c q^a + {\frac{1}{2}}\left( \delta ^a_c \partial _b \partial _l q^l + \delta ^a_b \partial _c \partial _l q^l\right) \end{aligned}$$

(199)

$$\begin{aligned} \pounds _q(f^{bc} N^a_{bc})&= {\left[ \pounds _{q} {(f^{bc} N^a_{bc})}\right] _\mathrm{std}} - f^{bc} \partial _b \partial _c q^a + f^{ab} \partial _b \partial _l q^l\nonumber \\&= {\left[ \pounds _{q} {(f^{bc} N^a_{bc})}\right] _\mathrm{std}} - \sqrt{-g}\,K^a \end{aligned}$$

(200)

where

$$\begin{aligned} K^a = g^{lm} \partial _l\partial _m q^a - g^{al} \partial _l\partial _m q^m \end{aligned}$$

(201)

For $V^a \equiv - g^{bc} N^a_{bc}$, we have:

$$\begin{aligned} \pounds _q ( \sqrt{-g}\,V^a) = {\left[ \pounds _{q} {\sqrt{-g}\,V^a)}\right] _\mathrm{std}} + \sqrt{-g}\,K^a \end{aligned}$$

(202)

Further, using $L_\mathrm{sur} \equiv \partial _a (\sqrt{-g}\,\, V^a)$, we obtain the result:

$$\begin{aligned} \pounds _q L_\mathrm{sur} = \pounds _q \partial _a(\sqrt{-g}\,V^a) = \pounds _q \partial _a (\sqrt{-g}\,V^a)\Big |_\mathrm{std} + \partial _a (\sqrt{-g}\,K^a) \end{aligned}$$

(203)

We are now in a position to work out the Lie derivative of $\sqrt{-g}\,\, L_\mathrm{quad} \equiv \sqrt{-g}\,\, L_H - L_\mathrm{sur}$ needed in the paper. We have

$$\begin{aligned} \pounds _q(\sqrt{-g}\,L_\mathrm{quad})&= \pounds _q(\sqrt{-g}\,L_H - L_\mathrm{sur})\Big |_\mathrm{std} - \partial _a (\sqrt{-g}\,K^a)\nonumber \\&= \partial _a \left[ \sqrt{-g}\,L_\mathrm{quad} q^a \right] - \partial _a (\sqrt{-g}\,K^a) \end{aligned}$$

(204)

Further, let us note the following two useful facts: (i) Because of the specific structure of Eqs. (198) and (199), the operations $\delta$ and $\pounds$ still commute on $\Gamma ^a_{bc}$ and $N^a_{bc}$ even though they are non-tensorial. (ii) Using the notation $\mathcal {V}^a\equiv \sqrt{-g}\,\, V^a$ and Eq. (202), we see that $\sqrt{-g}\,\, K^a$ disappears in the following combination (which occurs frequently):

$$\begin{aligned} \pounds _q ( \mathcal {V}^a) - q^a \partial _c (\mathcal {V}^c) - \sqrt{-g}\,K^a&= {\left[ \pounds _{q} {\mathcal {V}^a}\right] _\mathrm{std}} - q^a \partial _c ( \mathcal {V}^c) \nonumber \\&= q^c\partial _c \mathcal {V}^a - \mathcal {V}^c\partial _c q^a + \mathcal {V}^a\partial _c q^c - q^a\partial _c\mathcal {V}^c \nonumber \\&= \partial _c( q^c \mathcal {V}^a - q^a \mathcal {V}^c) \end{aligned}$$

(205)

These results allow us to relate the Noether currents for the two Lagrangians $L_H$ and $L_\mathrm{quad}$. Using the standard result for the Noether current for the Einstein–Hilbert action $J^a_H$, we can re-write the result in Eq. (112) in the form

$$\begin{aligned} \sqrt{-g}\,J^a_\mathrm{quad} = \sqrt{-g}\,J^a_H- \left[ \pounds _q ( - \sqrt{-g}\,V^a ) + q^a \partial _c ( \sqrt{-g}\,V^c) + \sqrt{-g}\,K^a\right] \qquad \end{aligned}$$

(206)

Using the notation $\mathcal {V}^a\equiv \sqrt{-g}\,\, V^a$, we see that $\sqrt{-g}\,\, K^a$ disappears in the expression in square brackets because of Eq. (205), giving

$$\begin{aligned} J^a_\mathrm{quad} = J^a_H + \nabla _b \left( V^{[a} q^{b]} \right) = J^a_H + \nabla _b \left( - \frac{f^{lm}}{\sqrt{-g}\,} N^{[a}_{lm} q^{b]}\right) = J^a_H + \nabla _b \left( g^{lm} N^{[b}_{lm} q^{a]}\right) \end{aligned}$$

(207)

where we have used $\mathcal {V}^a = \sqrt{-g}\,\, V^a = - f^{lm} N^a_{lm}$. If we write $J^a_H\equiv J^a_\mathrm{quad}+J^a_\mathrm{sur}$, thereby defining a Noether current associated with $L_\mathrm{sur}$ then:

$$\begin{aligned} J^{a}_\mathrm{sur} = - \nabla _b \left( V^{[a} q^{b]}\right) = \nabla _b \left( V^{[b} q^{a]}\right) =\nabla _bJ^{ab}_\mathrm{sur}; \quad J^{ab}_\mathrm{sur}\equiv V^{[b} q^{a]}=q^aV^b-q^bV^a \end{aligned}$$

(208)

1.7 Variation of the gravitational momentum flux $P^a_\mathrm{quad}$

Here we give some of the algebraic details related to the discussion in Sect. 7.2. We begin by relating $P^a_H$ and $P^a_\mathrm{quad}$ by

$$\begin{aligned} \sqrt{-g}\,P^a_H&= f^{lm} \pounds _q N^a_{lm} + q^a L_H \sqrt{-g}\,\nonumber \\&= \pounds _q \left( f^{lm} N^a_{lm}\right) - N^a_{lm}\pounds _q f^{lm} + q^a \left( \sqrt{-g}\,L_q -\partial _c \left( f^{lm} N^c_{lm}\right) \right) \nonumber \\&= - P^a_\mathrm{quad} +\pounds _q \left( f^{lm}N^a_{lm}\right) - q^a \partial _c \left( f^{lm} N^c_{lm}\right) \end{aligned}$$

(209)

Using the notation $\mathcal {V}^a=\sqrt{-g}\,V^a = - f^{lm} N^a_{lm}$ and using Eq. (205), we get

$$\begin{aligned} \sqrt{-g}\,P^a_H = - P^a_\mathrm{quad} - \left[ \pounds _q (\mathcal {V}^a) - q^a \partial _c \mathcal {V}^c\right] = - P^a_\mathrm{quad} - \left[ \sqrt{-g}\,K^a + \partial _c ( q^{[c} \mathcal {V}^{a]})\right] \end{aligned}$$

(210)

This gives, on combining with the on-shell result in Eq. (175), the result:

$$\begin{aligned} \delta P^a_\mathrm{quad}&= - \delta \left( \sqrt{-g}\,P^a_H\right) - \delta \left( \sqrt{-g}\,K^a\right) + \partial _c \left( q^{[c} \delta N^{a]}_{lm} f^{lm} + q^{[c} N^{a]}_{lm} \delta f^{lm}\right) \nonumber \\&= -\sqrt{-g}\,\omega ^a - \delta \left( \sqrt{-g}\,K^a\right) + \partial _c \left( q^{[c} N^{a]}_{lm} \delta f^{lm}\right) \end{aligned}$$

(211)

The extra term $\delta (\sqrt{-g}\,K^a)$ arises because of the non-tensorial character of $P^a_\mathrm{quad}$.

We can also obtain corresponding expressions for $\delta ( \sqrt{-g}\,J^a_\mathrm{quad})$ by relating it to $\delta (\sqrt{-g}\,J^a_H)$ using Eq. (113). Then we get:

$$\begin{aligned} \delta \left( \sqrt{-g}\,J^a_\mathrm{quad}\right)&= \delta \left( \sqrt{-g}\,J^a_H\right) + \partial _b \left( \delta f^{lm} N^{[b}_{lm} q^{a]} + f^{lm}\delta N^{[b}_{lm} q^{a]}\right) \nonumber \\&= \delta f^{lm}\pounds _q N^a_{lm} - \pounds _qf^{lm} \delta N^a_{lm} + \partial _c\left( f^{lm}\delta N^{[a}_{lm} q^{c]}\right) + 2 q^b \delta \left( \sqrt{-g}\,G^a_b \right) \nonumber \\&+\, q^a\sqrt{-g}\,G_{ij} \delta g^{ij} + \partial _b \left( \delta f^{lm} N^{[b}_{lm} q^{a]} + f^{lm}\delta N^{[b}_{lm} q^{a]}\right) \nonumber \\&= \sqrt{-g}\,\omega ^a + \partial _b \left( \delta f^{lm} N^{[b}_{lm} q^{a]}\right) + E^a \end{aligned}$$

(212)

where the equations of motion terms are:

$$\begin{aligned} E^a \equiv 2 q^b \delta \left( \sqrt{-g}\,G^a_b \right) + q^a\sqrt{-g}\,G_{ij} \delta g^{ij} \end{aligned}$$

(213)

It is possible to obtain the above result by direct computation as well. This derivation is instructive in clarifying the origin of the $K^a$ term. Let us start with the definition:

$$\begin{aligned} \sqrt{-g}\,\, P^a_\mathrm{quad} =(N^a_{ij} \pounds _q f^{ij} - L_\mathrm{quad} q^a\sqrt{-g}\,) \end{aligned}$$

(214)

and compute the on-shell variation $\delta P^a_\mathrm{quad}$. To do this, we will first obtain a preliminary result for the on-shell variation of $L_\mathrm{quad}$ in a specific form. We begin with the standard on-shell result

$$\begin{aligned} q^d\delta ( L_\mathrm{quad} \sqrt{-g}\,) = q^d \partial _c ( N^c_{ab}\ \delta f^{ab}) \end{aligned}$$

(215)

and re-write it in the following way:

$$\begin{aligned}&q^d\delta (L_\mathrm{quad}\sqrt{-g}\,)=q^d \partial _c (N^c_{ab}\, \delta f^{ab}) =\partial _c \left( q^d N^c_{ab} \, \delta f^{ab}\right) -\left( \partial _c q^d\right) N^c_{ab} \, \delta f^{ab}\nonumber \\&\quad = \left[ \partial _c \left( q^c N^d_{ab} \delta f^{ab}\right) -N^c_{ab} \left( \delta f^{ab}\right) \partial _c q^d \right] +\partial _c \left[ \left( q^d N^c_{ab} - q^c N^d_{ab}\right) \delta f^{ab}\right] \nonumber \\ \end{aligned}$$

(216)

To arrive at the second line, we have performed the usual symplectic trick of adding and subtracting the quantity $\partial _c (q^c N^d_{ab}\delta f^{ab})$. Let $v^d \equiv N^d_{ab} \delta f^{ab}$, so that

$$\begin{aligned} q^d \delta \left( L_\mathrm{quad} \sqrt{-g}\,\right) =\left[ \partial _c \left( q^c v^d\right) - v^c \partial _c q^d\right] + \partial _c \left[ q^{[d} N^{c]}_{ab} \delta f^{ab}\right] \end{aligned}$$

(217)

Then it is easy to see that (with $\mathcal {V}^a = - f^{lm} N^a_{lm}$)

$$\begin{aligned} \pounds _q ( v^d)&= \pounds _q \left[ -\delta \mathcal {V}^a- f^{ab}\delta N^d_{ab} \right] =-\pounds _q \left[ \delta \mathcal {V}^a\right] -\pounds _q\left[ f^{ab}\delta N^d_{ab} \right] \nonumber \\&= {\left[ \pounds _{q} {v^d}\right] _\mathrm{std}}-\delta \left( \sqrt{-g}\,K^a\right) =\partial _c \left( q^c v^d\right) - v^c \partial _c q^d-\delta \left( \sqrt{-g}\,K^a\right) \nonumber \\&= \delta f^{ab} \pounds _q N^d_{ab} + N^d_{ab} \delta \left( \pounds _q f^{ab}\right) \end{aligned}$$

(218)

where we have used Eq. (202) to obtain the third equality. Therefore

$$\begin{aligned} \partial _c \left( q^c v^d\right) - v^c \partial _c q^d=\left[ \delta f^{ab} \pounds _q N^d_{ab} + N^d_{ab} \delta \left( \pounds _q f^{ab}\right) \right] +\delta \left( \sqrt{-g}\,K^a\right) \end{aligned}$$

(219)

where we have used the fact that when $\delta q^a =0$ the Lie differentiation and variation commute giving $\pounds _q \delta (f^{ab}) = \delta \pounds _q(f^{ab})$. So

$$\begin{aligned} \delta ( q^d L_\mathrm{quad}\sqrt{-g}\,) = \left\{ \delta f^{ab} \pounds _q N^d_{ab} + N^d_{ab}\delta \pounds _q f^{ab}\right\} -\partial _c \left[ q^{[c} N^{d]}_{ab} \ \delta f^{ab}\right] + \delta \left( \sqrt{-g}\,K^d\right) \end{aligned}$$

(220)

Once we have the variation of $L_\mathrm{quad}$ in place, we can compute the variation $\delta P^a_\mathrm{quad}$ defined with $L_\mathrm{quad}$. We get

$$\begin{aligned} \delta P^d_\mathrm{quad}&= \left\{ \delta N^d_{ab} \pounds _q f^{ab} + N^d_{ab} \delta ( \pounds _q f^{ab}) -\delta f^{ab} \pounds _q N^d_{ab} - N^d_{ab}\delta \pounds _q f^{ab} \right\} \nonumber \\&- \delta (\sqrt{-g}\,K^a)+\partial _c \left[ q^{[c} N^{d]}_{ab} \delta f^{ab}\right] \nonumber \\&= \left\{ \delta N^d_{ab} \pounds _q f^{ab} - \delta f^{ab} \pounds _q N^d_{ab}\right\} + \partial _c\left[ q^{[c} N^{d]}_{ab} \delta f^{ab}\right] -\delta (\sqrt{-g}\,K^a)\qquad \qquad \end{aligned}$$

(221)

This again has a symplectic structure in the first term and an explicit surface term in the second.