The analysis in [

1] uses the fact that

\(u_aJ^a(\xi )\) is a 3-divergence, so that the spatial volume integral of

\(u_aJ^a(\xi )\) can be converted to a surface integral. In this case, it is natural to interpret

\(u_aJ^a(\xi )\) as a spatial density, viz., charge per unit volume of space. It turns out that similar results can be obtained even for the component of

\(J^a(\xi )\) in the direction of the normal to the equipotential surface along the following lines. It can be easily shown that

$$\begin{aligned} \hat{a}_{p}J^{p}(\xi )=-\left( g^{ij}-\hat{a}^{i}\hat{a}^{j}\right) \nabla _{i}\left( 2Nau_{j}\right) =-g^{ij}_{\perp }\nabla _{i}\left( 2Nau_{j}\right) \end{aligned}$$

(5.49)

where the tensor

\(g^{ij}_{\perp }\) acts as a projection tensor transverse to the unit vector

\(\hat{a}^i\). However in order to define a surface covariant derivative we need

\(\hat{a}^{i}\) to foliate the spacetime, which in turn implies

\(u^{i}\nabla _{i}N=0\). In this case Eq. (

5.49) can be written as

\(\hat{a}_{p}J^{p}(\xi )=-\mathcal {D}_{i}(2Nau^{i})\), where

\(\mathcal {D}_{i}\) is the covariant derivative operator corresponding to the induced metric

\(g^{ij}_{\perp }\) on the

\(N=\text {constant}\) surfaces with normal

\(\hat{a}_{i}\). (When

\(u^{i}\nabla _{i}N=0\), we have

\(a_{j}=\nabla _{j}\ln N\).) To obtain an integral version of this result, let us transform from the original

\((t,x^{\alpha })\) coordinates to a new coordinate system

\((t,N,x^{A})\) using

*N* itself as a “radial” coordinate. In this coordinate we have

\(a_{i}\propto \delta _{i}^{N}\) and thus

\(u^{N}=0\), thanks to the relation

\(u^{i}a_{i}=0\). Thus

\(\mathcal {D}_{i}(2Nau^{i})\) will transform into

\(\mathcal {D}_{\bar{\alpha }}(2Nau^{\bar{\alpha }})\), where

\(\bar{\alpha }\) stands for the set of coordinates (

\(t,x^{A}\)) on the

*N* = constant surface. Integrating both sides

\(\hat{a}_{p}J^{p}(\xi )=-\mathcal {D}_{\bar{\alpha }}(2Nau^{\bar{\alpha }})\), over the

\(N=\text {constant}\) surface will now lead to the result (with restoration of

\(1/16\pi \) factor):

$$\begin{aligned} \int d^{2}xdt\sqrt{-g_{\perp }}\,\hat{a}_{p}J^{p}(\xi )=-\int d^{2}x\sqrt{q}N\left( \frac{Na}{8\pi }\right) u^{t} =\int d^{2}x\left( \frac{Na}{2\pi }\right) \left( \frac{\sqrt{q}}{4}\right) \Bigg |^{t_1}_{t_2} \end{aligned}$$

(5.50)

where we have used the standard result

\(\sqrt{-g_{\perp }}=N\sqrt{q}\), with

*q* being the determinant of the two-dimensional hypersurface. The right hand side can be thought of as the difference in the heat content

\(Q(t_2)-Q(t_1)\) between the two surfaces

\(t=t_2\) and

\(t=t_1\) where:

$$\begin{aligned} Q(t)\equiv \int d^{2}x\left( \frac{Na}{2\pi }\right) \left( \frac{\sqrt{q}}{4}\right) =\int d^{2}x (Ts) \end{aligned}$$

(5.51)

This looks very similar to the result we obtained in the case of the integral over

\(u_i J^i\) earlier (see Eq. (

5.1) with

\(a^\alpha =r^\alpha \) on the

\(N=\) constant surface), but there is a difference in the interpretation of the left hand side. While

\(u_i J^i\) can be thought of as the charge density per unit

*spatial* volume, the quantity

\(\hat{a}_{p}J^{p}(\xi )\) represents the

*flux* of Noether current through a time-like surface; therefore,

\(\hat{a}_{p}J^{p}(\xi )\) should be thought of as a current per unit area per unit time. We will see later that the flux of Noether current through null surfaces leads to a very similar result.

We conclude this section with a discussion of the Newtonian limit of general relativity using the Noether current which has some amusing features. The Newtonian limit is obtained by setting

\(N^{2}=1+2\phi \),

\(g_{0\alpha }=0\) and

\(g_{\alpha \beta }=\delta _{\alpha \beta }\), where

\(\phi \) is the Newtonian potential [

3]. Then the acceleration of the fundamental observers turn out to be

\(a_{\alpha }=\partial _{\alpha }\phi \). Since the spatial section of the spacetime is flat, the extrinsic curvature identically vanishes and so does the Lie variation term. Also

\(2\bar{T}_{ab}u^{a}u^{b}=\rho _\mathrm{Komar}=\rho \), which immediately leads to (with

*G* inserted, see Eq. (

5.2)):

$$\begin{aligned} \nabla ^{2}\phi =4\pi G\rho \end{aligned}$$

(5.52)

the correct Newtonian limit. The same can also be obtained using the four velocity

\(u_{a}\). The Noether charge associated with

\(u_{a}\) turns out to have the following expression [

1]

$$\begin{aligned} D_{\alpha }a^{\alpha }=16\pi u_{a}J^{a}(u)=16\pi \bar{T}_{ab}u^{a}u^{b}+u_{a}g^{bc}\pounds _{u}N^{a}_{bc} \end{aligned}$$

(5.53)

In the Newtonian limit the following results hold

\(2\bar{T}_{ab}u^{a}u^{b}=\rho \) and

\(u_{a}g^{bc}\pounds _{u}N^{a}_{bc}=-D_{\alpha }a^{\alpha }\) (which follows from the Newtonian limit of the result

\(u_{a}g^{bc}\pounds _{\xi }N^{a}_{bc}=ND_{\alpha }a^{\alpha }+2a^{\alpha }D_{\alpha }N-Nu_{a}g^{ij}\pounds _{u}N^{a}_{ij}\) and the fact that in spacetime with flat spatial section the term

\(u_{a}g^{bc}\pounds _{\xi }N^{a}_{bc}\) identically vanishes). This immediately leads to Eq. (

5.52).

We also see that the Noether charge is positive as long as

\(\rho >0\) in the Newtonian limit. In fact, the Noether charge contained inside any equipotential surface is always a positive definite quantity as long as

\(r^\alpha \) and

\(a^\alpha \) point in the same direction (which happens when

\(\bar{T}_{ab}u^{a}u^{b}>0\)). To prove this we can integrate the Noether charge over a small region on a

\(t=\text {constant}\) hypersurface to obtain,

$$\begin{aligned} \int _{t=\mathrm {constant}}d^{3}x\sqrt{h}u_{a}J^{a}(u)&=\frac{1}{8\pi }\int _{N,t=\text {constant}}d^{2}x\sqrt{q}2Na^{\alpha }r_{\alpha } \nonumber \\&=\int _{N,t=\mathrm {constant}}d^{2}x\left( \frac{Na}{2\pi }\right) \left( \frac{\sqrt{q}}{4}\right) \end{aligned}$$

(5.54)

Since

\(\rho \) is positive definite in this case the fundamental observers are accelerating outwards and thus

\(r_{\alpha }a^{\alpha }=a\). The temperature as measured by these fundamental observers is a positive definite quantity and so is the entropy density and hence the positivity of Noether charge follows.