I'm studying General Relativity and I'm getting a little confused with the relation between symmetries and conservation laws.

Indeed, in Classical Mechanics we prove from the variational principle that each symmetry of the lagrangian gives rise to a conservation law. This is Noether's theorem and is really a corolary from the Euler-Lagrange equations.

Now, in General Relativity, I've read that when the Lie derivative of the metric tensor with respect to a vector field $X$ vanishes, then the metric tensor has a symmetry under the transformation generated by the flow of the vector field, and this also gives rise to a symmetry.

For example: in the Schwarzschild metric

$$ds^2=\left(1-\dfrac{2GM}{r}\right)dt^2-\left(\dfrac{1}{1-\dfrac{2GM}{r}}\right)dr^2-r^2d\theta^2-r^2\sin^2\theta d\phi^2$$

we can easily show that $\mathfrak{L}_{\frac{\partial}{\partial t}}g=0$. In other words, the metric would be invariant under time translations. It is said this gives rise to conservation of energy. The same can be argued for spherical symmetry.

My question here is: it is said that if the metric tensor is invariant under a certain transformation given by the flow of a vector field $X$, that is, if the Lie derivative $\mathfrak{L}_X g =0$, then there is a conservation law.

But this conservation law is for what system? The quantity being conserved is for what system? I really didn't get here. So just for example, in the Schwarzschild metric energy is conserved. But for what system? I'm not getting for what system the conservation law deduced by symmetry of the metric applies.


2 Answers 2

  1. Consider an arbitrary matter action $S_m$ that is assumed to be general covariant under general coordinate transformations.

  2. Define the Hilbert stress-energy-momentum (SEM) tensor $$ T^{\mu\nu}~:=~\mp \frac{2}{\sqrt{|g|}}\frac{\delta S_m}{\delta g_{\mu\nu}}\tag{1} $$ in the Minkowski sign convention $(\pm,\mp,\mp,\mp)$.

  3. Diffeomorphism invariance leads (via Noether's 2nd theorem) to an off-shell identity. Using the matter eqs. of motion (eom) $$ \frac{\delta S_m}{\delta \phi}~\stackrel{m}{\approx}~0,\tag{2} $$ Noether's 2nd identity reads $$ \nabla_{\mu} T^{\mu\nu}~\stackrel{m}{\approx}~0 \tag{3}$$ 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.]

  4. Finally assume that the metric $g_{\mu\nu}$ has a Killing symmetry. The Noether's 2nd identity (3) together with a Killing vector field $K^{\mu}$ lead to an identity $$\frac{1}{\sqrt{|g|}} d_{\mu}\left(\sqrt{|g|}J^{\mu} \right)~=~ \nabla_{\mu} J^{\mu}~\stackrel{m}{\approx}~0, \tag{4}$$ where $$ J^{\mu}~:=~T^{\mu\nu} K_{\nu}.\tag{5} $$

  5. It is possible to extract an integrated on-shell conserved quantity from the identity (4) in the standard way via a 4-dimensional divergence theorem, cf. OP's title question.

  6. For more details, see my Phys.SE answer here and references therein.


Consider a geodesic with tangent vector $U^\mu$ $$U^\mu\nabla_\mu U^\nu=0,$$ and the metric has a Killing vector $$\nabla_{(\mu} X_{\nu)}=0,$$ then the quantity $U^\nu X_\nu $ is conserved along the geodesic. $$U^\mu\nabla_\mu (U^\nu X_\nu)=U^\mu U^\nu\nabla_\mu X_\nu =0.$$ The first equality follows from the geodesic equation and the second from the Killing equation.

Since $m U$ is the four-momentum of a particle in free fall on the geodesic, and $X$ selects a special coordinate, this is like saying the component of the four-momentum in this special coordinate is conserved.

Similarly suppose we are coupled to matter so there is a conserved symmetric energy momentum tensor $\nabla_\mu T^{\mu\nu}=0.$ Then the current $J^\nu\equiv X_\mu T^{\mu\nu}$ is conserved. $$\nabla_\nu J^\nu=T^{\mu\nu}\nabla_\nu X_\mu=0.$$


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