A boundary term for the gravitational action with null boundaries

Abstract

Constructing a well-posed variational principle is a non-trivial issue in general relativity. For spacelike and timelike boundaries, one knows that the addition of the Gibbons–Hawking–York (GHY) counter-term will make the variational principle well-defined. This result, however, does not directly generalize to null boundaries on which the 3-metric becomes degenerate. In this work, we address the following question: What is the counter-term that may be added on a null boundary to make the variational principle well-defined? We propose the boundary integral of $2 \sqrt{-g} \left( \Theta +\kappa \right)$ as an appropriate counter-term for a null boundary. We also conduct a preliminary analysis of the variations of the metric on the null boundary and conclude that isolating the degrees of freedom that may be fixed for a well-posed variational principle requires a deeper investigation.

Appendices

Appendix 1: Decomposition of $\sqrt{-g}$ in terms of the determinant of metric on a 2-surface

One relevant question is whether there is a decomposition of $\sqrt{-g}$ in terms of $\sqrt{q}$, q being the determinant of the 2-metric $q_{AB}$ on the null surface, akin to the decomposition $\sqrt{-g}= N \sqrt{|h|}$ in the timelike and spacelike case. We shall prove in this appendix a preliminary result on the decomposition of the determinant of a $4\times 4$ metric in terms of the determinant of a $2\times 2$ sub-matrix, which will be later applied to a null surface.

We start by writing down a general result relating the determinant of a $2\times 2$ sub-matrix to the determinant of the whole matrix. We shall prove the result working with the metric written in the coordinates $\left( \phi ,x^1,x^2,x^3\right)$ with the components on the $\phi =$ constant surface being denoted by $h_{\alpha \beta }$:

$$\begin{aligned} g_{ab}= \left( \begin{array}{llll} g_{\phi \phi } &{} \quad g_{\phi 1} &{} \quad g_{\phi 2} &{} \quad g_{\phi 3}\\ g_{\phi 1} &{}\quad h_{11} &{}\quad h_{12} &{}\quad h_{13}\\ g_{\phi 2} &{}\quad h_{21} &{} \quad h_{22} &{} \quad h_{23}\\ g_{\phi 3} &{}\quad h_{31} &{} \quad h_{32} &{} \quad h_{33} \end{array}\right) . \end{aligned}$$

(49)

In this case, we can use the definition of an inverse matrix element applied to $g^{\phi \phi }$ to write

$$\begin{aligned} g= \frac{h}{g^{\phi \phi }}, \end{aligned}$$

(50)

where h is the determinant of $h_{\alpha \beta }$, the $3 \times 3$-matrix obtained by deleting the $\phi$-column and $\phi$-row from $g_{ab}$. Now, we can play the same game again with $h_{\alpha \beta }$. The determinant of the $2\times 2$ matrix $q_{AB}$, $A,B=2,3$, defined by $q_{AB}=h_{AB}$ satisfies an analogue of Eq. (50):

$$\begin{aligned} h= \frac{q}{h^{11}}, \end{aligned}$$

(51)

where $h^{11}$ is the 11-th component of the matrix $h^{\alpha \beta }$, the inverse of the matrix $h_{\alpha \beta }$. Substituting for h in Eq. (50), we obtain

$$\begin{aligned} g=\frac{q}{g^{\phi \phi }h^{11}}. \end{aligned}$$

(52)

Now, the denominator above can be expanded as follows:

$$\begin{aligned} g^{\phi \phi }h^{11} = g^{\phi \phi }g^{11} -(g^{1 \phi })^2, \end{aligned}$$

(53)

which is easiest to obtain by using the formula $h^{ab}=g^{ab}-\epsilon n^a n^b$. Thus, we obtain a relation relating the determinant of a $2\times 2$ sub-matrix with the determinant of the full $4\times 4$ matrix:

$$\begin{aligned} g=\frac{q}{g^{\phi \phi }g^{11} -(g^{1 \phi })^2}. \end{aligned}$$

(54)

Appendix 2: Working with a general $\ell _a=A\nabla _a \phi$

In this appendix, we shall consider the general case of our null normal being of the form $\ell _a = A \partial _a \phi =A v_{a}$, for an arbitrary scalar A which may depend on the metric. We no longer have the results $\nabla _a \ell _b=\nabla _b \ell _a$ everywhere and $\delta \ell _a=0$ everywhere that we had used profusely in Sect. 3.3. But we do have the result

$$\begin{aligned} \delta \ell _a = \delta A \nabla _a \phi =\delta \ln A ~\ell _a \end{aligned}$$

(55)

From Eq. (18), the boundary term on the null surface is

$$\begin{aligned} \sqrt{-g}Q[v_{a}]= & {} \frac{\sqrt{-g}}{A}Q[\ell _{a}]= \frac{1}{A}\left\{ \sqrt{-g} \nabla _c \left[ \delta \ell _{\perp }^{c}\right] - 2 \delta (\sqrt{-g} \nabla _a \ell ^{a}) + \sqrt{-g} (\nabla _a \ell _b \right. \nonumber \\&\left. -g_{ab} \nabla _c \ell ^{c}) \delta g^{ab} \right\} . \end{aligned}$$

(56)

Labelling the first term as $\sqrt{-g}R_1$, we have

$$\begin{aligned} \sqrt{-g}R_1\equiv \frac{\sqrt{-g}}{A} \nabla _a \left[ \delta \ell _{\perp }^{a}\right] =\frac{1}{A} \partial _{a}\left[ \sqrt{-g}\delta \ell _{\perp }^{a}\right] = \partial _{a}\left[ \frac{\sqrt{-g}}{A} \delta \ell _{\perp }^{a}\right] -\sqrt{-g}\delta \ell _{\perp }^{a}\partial _{a}\left( \frac{1}{A}\right) \nonumber \\ \end{aligned}$$

(57)

We shall now use the projector $\Pi ^a_{b}$ to separate out the surface derivatives in the first term.

$$\begin{aligned} \partial _{a}\left[ \frac{\sqrt{-g}}{A} \delta \ell _{\perp }^{a}\right]&=\partial _{a}\left[ \frac{\sqrt{-g}}{A} \Pi ^a_{b}\delta \ell _{\perp }^{b}\right] -\partial _{a}\left[ \frac{\sqrt{-g}}{A} k^a \ell _b \delta \ell _{\perp }^{b}\right] \nonumber \\&= \partial _{a}\left[ \frac{\sqrt{-g}}{A} \Pi ^a_{b}\delta \ell _{\perp }^{b}\right] - \delta \left( \ell _a \ell ^a\right) \partial _b\left( \frac{\sqrt{-g}}{A} k^b\right) \nonumber \\&\quad -\frac{\sqrt{-g}}{A} k^b \partial _b \left[ \delta \left( \ell _a \ell ^a\right) \right] \nonumber \\&= \partial _{a}\left[ \frac{\sqrt{-g}}{A} \Pi ^a_{b}\delta \ell _{\perp }^{b}\right] -\frac{\sqrt{-g}}{A} k^b \partial _b \left[ \delta \left( \ell _a \ell ^a\right) \right] , \end{aligned}$$

(58)

where the last step was obtained by using our assumption $\delta \left( \ell _a \ell ^a\right) =0$ on the null surface. Using this expression, we have

$$\begin{aligned} \sqrt{-g}R_1=\partial _{a}\left[ \frac{\sqrt{-g}}{A} \Pi ^a_{b}\delta \ell _{\perp }^{b}\right] -\frac{\sqrt{-g}}{A} k^b \partial _b \left[ \delta \left( \ell _a \ell ^a\right) \right] -\sqrt{-g}\delta \ell _{\perp }^{a} \partial _{a}\left( \frac{1}{A}\right) \quad \end{aligned}$$

(59)

The first term in Eq. (59) is a surface derivative on the null surface as $\Pi ^a_{b}\ell _a=0$. The second term in Eq. (59) has variations of the derivatives of the metric. We shall take out the $\delta$ to obtain

$$\begin{aligned} -k^b \partial _b \left[ \delta \left( \ell _a \ell ^a\right) \right] = -\delta \left[ k^b \partial _b \left( \ell _a \ell ^a\right) \right] + \delta k^b \partial _b \left( \ell _a \ell ^a\right) . \end{aligned}$$

(60)

Substituting in Eq. (59), we obtain

$$\begin{aligned} \sqrt{-g}R_1 =\,&\partial _{a}\left[ \frac{\sqrt{-g}}{A} \Pi ^a_{b}\delta \ell _{\perp }^{b}\right] -\frac{\sqrt{-g}}{A} \delta \left[ k^b \partial _b \left( \ell _a \ell ^a\right) \right] +\frac{\sqrt{-g}}{A}\delta k^{b} \partial _b \left( \ell _a \ell ^a\right) \nonumber \\&\quad -\sqrt{-g}\delta \ell _{\perp }^{a}\partial _{a}\left( \frac{1}{A}\right) \nonumber \\ =\,&\partial _{a}\left[ \frac{\sqrt{-g}}{A} \Pi ^a_{b}\delta \ell _{\perp }^{b}\right] - \delta \left[ \frac{\sqrt{-g}}{A} k^b \partial _b \left( \ell _a \ell ^a\right) \right] - \frac{\sqrt{-g}}{2A} \left[ k^b \partial _b \left( \ell _a \ell ^a\right) \right] g_{ij} \delta g^{ij}\nonumber \\&\quad +\sqrt{-g}k^b \partial _b \left( \ell _a \ell ^a\right) \delta \left( \frac{1}{A}\right) +\frac{\sqrt{-g}}{A}\delta k^{b} \partial _b \left( \ell _a \ell ^a\right) -\sqrt{-g}\delta \ell _{\perp }^{a}\partial _{a}\left( \frac{1}{A}\right) \nonumber \\ \end{aligned}$$

(61)

Here, all the variations of the derivatives of the metric are in the first two terms, assuming A does not depend on the derivatives of the metric. The second term in Eq. (56) is

$$\begin{aligned} \sqrt{-g}R_2\equiv - \frac{2}{A} \delta \left( \sqrt{-g} \nabla _a \ell ^{a}\right) =-2 \delta \left( \frac{\sqrt{-g}}{A} \nabla _a \ell ^{a}\right) +2\sqrt{-g} \nabla _a \ell ^{a}\delta \left( \frac{1}{A}\right) \end{aligned}$$

(62)

Substituting Eqs. (61) and (62) back in Eq. (56), and using the relation

$$\begin{aligned} \nabla _a \ell ^a+\frac{k^a}{2} \partial _a \left( \ell _b \ell ^b\right) = \delta ^{a}_{b} \nabla _a \ell ^b + k^a \ell _b \nabla _a \ell ^b = \Pi ^a_{b}\nabla _a \ell ^b, \end{aligned}$$

(63)

in three places, the boundary term on the null surface reduces to

$$\begin{aligned} \sqrt{-g}Q[\ell _{a}]&=\partial _{a}\left[ \frac{\sqrt{-g}}{A} \Pi ^a_{b}\delta \ell _{\perp }^{b}\right] - 2 \delta \left( \frac{\sqrt{-g}}{A} \Pi ^a_{b}\nabla _a \ell ^b\right) \nonumber \\&\quad + \frac{\sqrt{-g}}{A} \left( \nabla _a \ell _b -g_{ab} \Pi ^c_{d}\nabla _c \ell ^d\right) \delta g^{ab}\nonumber \\&\quad +2\sqrt{-g} \Pi ^a_{b}\nabla _a \ell ^b \delta \left( \frac{1}{A}\right) +\frac{\sqrt{-g}}{A}\delta k^{b} \partial _b \left( \ell _a \ell ^a\right) -\sqrt{-g}\delta \ell _{\perp }^{a}\partial _{a}\left( \frac{1}{A}\right) , \end{aligned}$$

(64)

When $A=1$, the last three terms vanish and this result reduces to the result in Eq. (29). To see why the second-to-last term did not appear for $A=1$, note that this term will only have normal derivatives as $\ell _a \ell ^a$ is fixed to zero everywhere on the null surface. Hence, only $\delta k^{\phi }$ contributes in our $\left( \phi ,y^1,y^2,y^3\right)$ coordinate system. When $A=1$, $\delta k^{\phi }=\delta \left( \ell _ak^{a}\right) =0$. In the general case, $k^a\ell _a=-1$ means $k^{\phi }A=-1$, which gives

$$\begin{aligned} k^{\phi }=-1/A; \quad \delta k^{\phi }=\frac{\delta \ln A}{A}. \end{aligned}$$

(65)

In Eq. (64), we have succeeded in separating out a total derivative on the surface and a total variation to remove all derivatives of the metric. The counter-term to be added in this case is the integral over the null surface of

$$\begin{aligned} 2 \frac{\sqrt{-g}}{A} \Pi ^a_{b}\nabla _a \ell ^b =2 \frac{\sqrt{-g}}{A} \left( q^a_{b}\nabla _a \ell ^b-\ell ^a k_{b}\nabla _a \ell ^b\right) =2 \frac{\sqrt{-g}}{A} \left( \Theta +\kappa \right) . \end{aligned}$$

(66)

We shall now do the analysis of what is to be fixed on the null boundary in this case. Since Eq. (17) are taken to be valid even on variation, the relations $q^{ab}\ell _a=0$ and $q^{ab}k_a=0$ are respected by the variations and terms of the form $\ell _a \ell _b\delta q^{ab}$, $\ell _a k_b\delta q^{ab}$ and $k_a k_b\delta q^{ab}$ would reduce to zero, just as in Sect. 3.3. Thus, we can simplify $g_{ab}\delta g^{ab}$ as follows:

$$\begin{aligned} g_{ab} \delta g^{ab}&=g_{ab}\left[ \delta q^{ab}-\delta \left( \ell ^{a} k^{b}\right) -\delta \left( \ell ^{b} k^{a}\right) \right] = q_{ab}\delta q^{ab}+2 \left( \ell _{a}k_{b} + \ell _{b}k_{a}\right) \delta \left( \ell ^{a} k^{b}\right) \nonumber \\&= q_{ab}\delta q^{ab}+2 \ell _{b}k_{a} \delta \left( \ell ^{a} k^{b}\right) \nonumber \\&= q_{ab}\delta q^{ab} - 2 k_{a}\delta \ell ^{a}- 2 \ell _{a}\delta k^{a}=q_{ab}\delta q^{ab} - 2 k_{a}\delta \ell ^{a}- 2 \delta \ln A, \end{aligned}$$

(67)

where we have used Eq. (65) in the last step.

Next, we shall simplify $\left( \nabla _{a} \ell _b \right) \delta g^{ab}$. We have

$$\begin{aligned} \left( \nabla _{a} \ell _b \right) \delta g^{ab}&= \left( \nabla _{a} \ell _b \right) \delta q^{ab} - \delta \left( \ell ^a k^b \right) \nabla _a \ell _b- \delta \left( \ell ^b k^a \right) \nabla _a \ell _b \nonumber \\&= \left( \nabla _{a} \ell _b \right) \delta q^{ab} -\delta {\ell ^a} k^b \nabla _a \ell _b - \delta {k^b }\ell ^a \nabla _a \ell _b-\delta {\ell ^b} k^a \nabla _a \ell _b - \delta {k^a }\ell ^b \nabla _a \ell _b\nonumber \\&= \left( \nabla _{a} \ell _b \right) \delta q^{ab} - \delta {\ell ^a} k^b \left( \nabla _a \ell _b+\nabla _b \ell _a\right) - \kappa \delta {k^b } \ell _b-\frac{\delta k^a}{2}\partial _a \ell ^2 \nonumber \\&= \left( \nabla _{a} \ell _b \right) \delta q^{ab} - \delta {\ell ^a} k^b \left( \nabla _a \ell _b+\nabla _b \ell _a\right) - \kappa \delta \ln A-\frac{\delta k^a}{2}\partial _a \ell ^2, \end{aligned}$$

(68)

where we have used $\ell ^a \nabla _a \ell _b=\kappa \ell _b$ (see Eq. 13) and $\ell _a\delta k^a=\delta \ln A$ (see Eq. 65). The expression $\left( \nabla _{a} \ell _b \right) \delta q^{ab}$ can be simplified as follows:

$$\begin{aligned} \left( \nabla _{a} \ell _b \right) \delta q^{ab}&= \delta ^m_a \delta ^n_b \left( \nabla _{m} \ell _n \right) \delta q^{ab} \nonumber \\&= \left( q^m_a -\ell ^m k_a-k^m \ell _a\right) \left( q^n_b-\ell ^n k_b-k^n \ell _b\right) \left( \nabla _{m} \ell _n \right) \delta q^{ab} \nonumber \\&= \left( q^m_a -\ell ^m k_a \right) \left( q^n_b-\ell ^n k_b \right) \left( \nabla _m \ell _n \right) \delta q^{ab} \nonumber \\&=\left( q^m_a q^n_b \nabla _m \ell _n - q^m_a \ell ^n k_b \nabla _m \ell _n - q^n_b \ell ^m k_a \nabla _m \ell _n \right) \delta q^{ab} \nonumber \\&=\left( \Theta _{ab}- \frac{k_b q^m_a\partial _m \ell ^2}{2}-\kappa q^n_b k_a \ell _n \right) \delta q^{ab} \nonumber \\&=\Theta _{ab} \delta q^{ab} \end{aligned}$$

(69)

where we used $\ell _a\delta q^{ab}=\delta \left( q^{ab} \ell _a \right) -q^{ab}\delta \ell _a=- q^{ab}\ell _a \delta \ln A=0$ to get to the second line, $k_a k_b \delta q^{ab}=0$ to get to the third line and the definition of $\Theta _{ab}$ and $\kappa$ (see Appendix A.2.5 in [32] and Eq. (13) in this paper) to get to the fourth line. The final result is obtained using $\ell _a q^{ab}=0$ and the fact that $q^m_a\partial _m \ell ^2$, a derivative on the null surface, is zero since $\ell ^2$ is zero all over the null surface. Next, we shall simplify the last two terms in Eq. (68) as follows:

$$\begin{aligned} - \kappa \delta \ln A-\frac{\delta k^a}{2}\partial _a \ell ^2 =&- \kappa \delta \ln A-\frac{\delta k^\phi }{2}\partial _\phi \ell ^2 =- \kappa \delta \ln A-\frac{\delta \ln A}{2A}\partial _\phi \ell ^2 \nonumber \\ =&-\delta \ln A \left( \kappa + \frac{\partial _{\phi }\ell ^2}{2A}\right) =-\delta \ln A \left( \kappa - \frac{k^a \partial _{a}\ell ^2}{2}\right) \nonumber \\ =&-\delta \ln A \left( \kappa + \tilde{\kappa }\right) . \end{aligned}$$

(70)

Here, the first step made use of the fact that $\ell ^2=0$ all over the null surface and hence only the derivative along $\phi$ contributes and the second step as well as the second-to-last step used Eq. (65). The last line used the definition $2\tilde{\kappa }=-k^{a}\nabla _{a}\ell ^{2}$ (see Appendix A.2.6 in [32]). The remaining terms in Eq. (64) are

$$\begin{aligned} 2\sqrt{-g} \Pi ^a_{b}\nabla _a \ell ^b \delta \left( \frac{1}{A}\right)&=-2 \frac{\sqrt{-g}}{A}\left( \Theta +\kappa \right) \delta \ln A \end{aligned}$$

(71)

$$\begin{aligned} \frac{\sqrt{-g}}{A}\delta k^{b} \partial _b \left( \ell _a \ell ^a\right)&=2\frac{\sqrt{-g}}{A} \tilde{\kappa } \delta \ln A \end{aligned}$$

(72)

$$\begin{aligned} -\sqrt{-g}\delta \ell _{\perp }^{a}\partial _{a}\left( \frac{1}{A}\right)&=\frac{\sqrt{-g}}{A} \left( \delta \ell ^a\partial _a \ln A+g^{ab}\partial _a \ln A\delta \ell _b\right) \nonumber \\&=\frac{\sqrt{-g}}{A} \left( \delta \ell ^a\partial _a \ln A+\ell ^{a}\partial _a \ln A\delta \ln A\right) \nonumber \\&=\frac{\sqrt{-g}}{A} \left( \delta \ell ^a\partial _a \ln A+(\kappa -\tilde{\kappa })\delta \ln A\right) \end{aligned}$$

(73)

Eq. (72) can be obtained using the same manipulations that we performed in Eq. (70). Combining all the above results, the boundary term in Eq. (64) becomes

$$\begin{aligned} \sqrt{-g}Q[\ell _{a}]&=\partial _{a}[\frac{\sqrt{-g}}{A} \Pi ^a_{b}\delta \ell _{\perp }^{b}] - 2 \delta \left[ \frac{\sqrt{-g}}{A} \left( \Theta +\kappa \right) \right] + \frac{\sqrt{-g}}{A} \left[ \left( \Theta _{ab}-\left( \Theta \right. \right. \right. \\&\quad \left. \left. \left. +\,\kappa \right) q_{ab}\right) \delta q^{ab}\right] +\frac{\sqrt{-g}}{A} \left( 2 k_{a}\left( \Theta +\kappa \right) - k^b \left( \nabla _a \ell _b+\nabla _b \ell _a\right) \right. \\&\quad \left. +\,\partial _a \ln A\right) \delta \ell ^{a}, \end{aligned}$$