Pass to globally conserved currents from locally conserved currents in curved spacetime

Question

Let us begin with a Lagrangian of the form

$$\mathscr L= \frac 12 \sqrt{-g}g^{\mu\nu}\partial_\mu\phi(x)\partial_\nu\phi(x)+\mathscr L_g,$$

where $$\mathscr L_g=\frac 1{16\pi k}\sqrt{-g}R.$$ Suppose as well that there are no Killing vectors associated to the metric $g^{\mu\nu}$ except for, say, a global timelike Killing vector if it helps the argument.

Associated to $\mathscr L$ is a locally conserved set of 10 currents from the Poincare group:

$$T^{\mu\nu}=\partial^\mu\phi(x)\partial^\nu\phi(x)-g^{\mu\nu}\mathscr L$$

for each spacetime translation, and $\epsilon_{\alpha\beta}x^\alpha T^{\mu\beta}$ for each spacetime rotation.

Locally we have $$\nabla_\mu T^{\mu\nu}=0$$ so these quantities are conserved only locally.

My question is, what is the obstacle to patching these locally conserved quantities together to make a globally conserved quantity:

$$Q=\int T^{0 \nu}f_\nu \;d^3x$$

with

$$dQ/dt=0$$

where $f_\nu$ might be a gluing function connecting the momentum flowing out of one patch of infinitesimal volume and into another?

(Edit: I realize there may not be a tensor associated to this conserved quantity but even a pseudo tensor involving only the fields would be satisfying, if it exists. So for example, to get the ball rolling, we can start with an object of the form

$M^{\lambda\mu\nu}=\frac 12\int_{a^\mu}^{x^\lambda}ds T^{\mu\nu}(s)-\frac 12\int_{a^\mu}^{x^\mu}ds T^{\lambda\nu}(s)$,

and then set $t^{\mu\nu}=\partial_\lambda M^{\lambda\mu\nu}$.

$t^{\mu\nu}$ is a psuedo tensor that is conserved $\partial_\mu t^{\mu\nu}=0$ generically by the antisymmetry in $\lambda,\ \mu$. Thus,

$t^{\mu\nu}=T^{\mu\nu}(x)+\frac 12\int^{x^\mu}ds\ \partial_\lambda T^{\lambda\nu}(s)+\frac 12\delta^{\mu}_\lambda T^{\lambda\nu}(x|_{x^\mu=a^\mu}).$

In flat space this quantity is almost the tensor we are looking for up to the boundary term $\frac 12\delta^{\mu}_\lambda T^{\lambda\nu}(x|_{x^\mu=a^\mu})$ where $\delta^\mu_\lambda$ is a Kronecker delta.

Of course, the boundary term ruins it from working in the limit. That, and the lack of symmetry in mu and nu, but this should give the idea of what could work with a better choice of starting point.)

Answer 1

1. Disclaimer: Let us here avoid the discussion of how to assign a stress-energy-momentum (SEM) pseudo-tensor $t^{\mu\nu}$ to the gravitational field. The word pseudo here refers to the fact that $t^{\mu\nu}$ is not a tensor wrt. general coordinate transformations; only a rigid subgroup thereof. In other words, the pseudo-tensor $t^{\mu\nu}\frac{\partial}{\partial x^{\mu}}\odot\frac{\partial}{\partial x^{\nu}}$ depends on coordinate systems. This makes it generically impossible to patch together a suitable global definition of e.g. gravitational energy and momentum in GR, cf. this Phys.SE post, this Physicsforum page, and links therein.

2. With pt. 1 in mind, the symmetric SEM tensor $$\tag{1} T^{\mu\nu}~:=~-\frac{2}{\sqrt{|g|}}\frac{\delta S_m}{\delta g_{\mu\nu}}$$ therefore contains only the matter part $S_m$. The matter action $S_m$ is assumed to be general covariant under general coordinate transformations. In other words, we are doing FT in curved space rather than GR. We stress in particular that the metric $g_{\mu\nu}$ is arbitrary and not assumed to satisfy the EFE. [In eq. (1) we have used the Minkowski sign convention $(+,-,-,-)$.]

3. Let us also for simplicity just consider scalar matter $$S_m~=~\int \!d^4x~{\cal L}(g,\phi,\partial\phi),$$ $$\tag{2}{\cal L}(g,\phi,\partial\phi)~:=~\sqrt{|g|} \left(\frac{1}{2}\phi_{,\mu} g^{\mu\nu} \phi_{,\nu} -V(\phi)\right),$$ Then we can ignore the Belinfante-Rosenfeld SEM tensor. For matter with non-zero spin, see my Phys.SE answer here.

4. Diffeomorphism invariance leads (via Noether's 2nd theorem) to an off-shell identity. cf. e.g. Ref. 1. Using the matter eqs. of motion (eom) $$\tag{3} \frac{\delta S_m}{\delta \phi}~\stackrel{m}{\approx}~0, $$ Noether's 2nd identity reads $$\tag{4} \nabla_{\mu} T^{\mu\nu}~\stackrel{m}{\approx}~0 $$ for an arbitrary metric $g_{\mu\nu}$. [Here the $\stackrel{m}{\approx}$ symbol means equality modulo matter eom. The connection $\nabla$ is the Levi-Civita connection.] Eq. (4) serves as an important consistency check. A matter source $T^{\mu\nu}$ to the EFEs should satisfy eq. (4), cf. the (differential) Bianchi identity.

5. Since the Lagrangian density (2) does not depend explicitly on spacetime, then the Noether's 1st theorem leads to an off-shell identity. Using the matter eom (3), Noether's 1st identity reads $$ \tag{5} d_{\mu}\Theta^{\mu}{}_{\lambda}~\stackrel{m}{\approx}~-\frac{\delta S_m}{\delta g_{\mu\nu}}g_{\mu\nu,\lambda}, $$ where $$\tag{6} \Theta^{\mu}{}_{\lambda}~=~\sqrt{|g|}\phi^{,\mu}\phi_{,\lambda} -\delta^{\mu}_{\lambda}{\cal L} ~=~ \sqrt{|g|} T^{\mu}{}_{\lambda} $$ is the canonical SEM tensor-density.

6. It is straightforward to check that Noether's 2nd identity (4) and Noether's 1st identity (5) are the same.

7. There is no conservation law associated with Noether's 2nd identity (4), cf. this Phys.SE post.

8. The Noether's 1st identity (5) would have lead to an on-shell conservation law if $\frac{\delta S_m}{\delta g_{\mu\nu}}$ were the true eom for the metric $g_{\mu\nu}$. But it is not. It is missing the Einstein tensor $G_{\mu\nu}$ from the Einstein-Hilbert action $S_{EH}$, which we did not include, cf. pts. 1-2.

9. Alternatively, we could demote the metric $g_{\mu\nu}$ from a field to a structure function, but then we would have explicit spacetime dependence in the Lagrangian density (2), and hence no symmetry to apply to Noether's 1st theorem to begin with. So there is no conservation law associated with Noether's 1st identity (5) as well.

10. Finally assume that the metric $g_{\mu\nu}$ has a Killing symmetry. The Noether's 2nd identity (4) together with a Killing vector field $K^{\mu}$ lead to an identity $$ \tag{7}\frac{1}{\sqrt{|g|}} d_{\mu}\left(\sqrt{|g|}J^{\mu} \right)~=~ \nabla_{\mu} J^{\mu}~\stackrel{m}{\approx}~0,$$ where $$\tag{8} J^{\mu}~:=~T^{\mu\nu} K_{\nu}, $$ cf. e.g. Ref. 2. It is possible to extract an integrated on-shell conserved quantity from the identity (7) in the standard way via a 4-dimensional divergence theorem.

References:

1. R.M. Wald, GR, Appendix E.1.

2. S.W. Hawking and G.F.R. Ellis, The Large Scale Structure of Space-Time, Section 3.2.

Answer 2

Gauss-Ostrogradsky theorem does not actually "care" for a "type" of space or functions that you've got, so it has the same form in curved space and in flat space: $$\int_D \mathrm{d}^4x \, \frac{\partial}{\partial x^\mu} [\text{Anything}] = \oint_{\partial D} \mathrm{d}\sigma_\mu [\text{Anything}]$$ where $\mathrm{d}^4 x$ and $\mathrm{d}\sigma_\mu$ are constructed from differentials of coordinates the same way as in flat space, e.g. $\mathrm{d}^4 x = \mathrm{d}x^0 \mathrm{d}x^1 \mathrm{d}x^2 \mathrm{d}x^3$ (not "real" 4-volume $\mathrm{d}\Omega = \sqrt{-g}\mathrm{d}x^0 \mathrm{d}x^1 \mathrm{d}x^2 \mathrm{d}x^3$).

For example, let's consider the electric current continuity equation $$\nabla_\mu j^\mu = 0.$$ Taking properties of Christoffel symbols into account, you can show that $$\nabla_\mu j^\mu = \frac{1}{\sqrt{-g}}\frac{\partial}{\partial x^\mu}\left(\sqrt{-g}j^\mu \right)$$ so in this case $$0 = \int \mathrm{d}\Omega \nabla_\mu j^\mu = \int \sqrt{-g} \mathrm{d}^4 x \frac{1}{\sqrt{-g}}\frac{\partial}{\partial x^\mu}\left(\sqrt{-g}j^\mu \right) = \int \mathrm{d}^4 x \frac{\partial}{\partial x^\mu}\left(\sqrt{-g}j^\mu \right) = (*)$$ and by O-G theorem $$(*) = \oint \mathrm{d} \sigma_\mu \sqrt{-g} j^\mu = \oint \mathrm{d}\Sigma_\mu j^\mu$$ where $\mathrm{d}\Sigma_\mu$ is "real" 3-surface. In other words you have $ \int j^0 \mathrm{d}V = \text{const}$.

However, for a symmetric 2-tensor $T^{\mu\nu}$, you have $$\nabla_\mu T^\mu_\nu = \frac{1}{\sqrt{-g}} \frac{\partial \left(\sqrt{-g}T^\mu_\nu \right)}{\partial x^\nu} - \frac12 \frac{\partial g_{\alpha\beta}}{\partial x^\nu} T^{\alpha\beta}$$ and "unfortunately" the second term can't be transformed to $$\frac{1}{\sqrt{-g}}\frac{\partial \left( \sqrt{-g} \,\mathrm{Something} \right)}{\partial x^\nu }.$$

Physically, this is because you didn't take the energy of the gravitational field into account. But as you can see from the above, you can't construct a proper tensor value of stress-energy tensor for gravitational field, you can only build a pseudo-tensor (this is compliant with equivalence principle).