13
$\begingroup$

In special relativity, a conserved current is one that satisfies $$\partial_\mu T^{\mu \nu \rho \ldots} = 0.$$ In this case, one can show that we have a tensorial conserved quantity $$Q^{\nu \rho \ldots} = \int d^3x \, T^{0 \nu \rho \ldots}, \quad \frac{dQ}{dt} = 0.$$ In general relativity, a conserved current instead satisfies $$\nabla_\mu T^{\mu \nu \rho \ldots} = 0.$$ Under what circumstances can we construct a tensorial conserved quantity? Is it always possible when $T$ has rank one? What conditions do we need when $T$ has rank greater than one?

$\endgroup$
10
$\begingroup$

If $J$ is a four vector satisfying $\nabla_a J^a =0$ and $\Sigma$, $\Sigma'$ are smooth spacelike (Cauchy) hypersurfaces with the latter in the future of the former, it holds $$Q := \int_\Sigma J^a n_a dS = \int_{\Sigma'} J^a n'_a dS'\tag{1}$$ where $dS$ is the measure induced by the metric and $n$ the future-pointing unit co-vector normal to $\Sigma$ and similar definitions are valid for $\Sigma'$, provided $J$ rapidly vanishes at spacetime infinity. The proof is based on the divergence theorem referring to a cylindric surface $C$ with bases on $\Sigma$ and $\Sigma'$ respectively and lateral surface larger and larger towards spacelike infinity. Indeed, if $V$ is the solid cylinder whose frontiers is $\partial V = C $ $$0 = \int_V \nabla_aJ^a dv = \int_{\partial V} J^a N^a dS = -\int_\Sigma J^a n_a dS + \int_{\Sigma'} J^a n'_a dS' + 0\tag{2}$$ where $N$ is the outward normal unit vector to $C$ and $0$ stays for the vanishing lateral contribution and we used the fact that $N = -n$ on $\Sigma$. (2) immediately implies (1).

Now suppose that $T^{abcd...}$ is symmetric and $\nabla_a T^{abcd...}=0$. If $K,K',K'',...$ are Killing vectors (possibly equal) and we define $$J^a := T^{abcd...}K_bK'_cK''_d...$$ we have $$\nabla_aJ^a = \nabla_a (T^{abcd...}K_b K'_c K''_d...)$$ $$= (\nabla_a T^{abcd...})K_bK'_cK''_d...+ T^{abcd..} [(\nabla_aK_b) K'_c K''_d... + K_b (\nabla_aK'_c) K''_d... + K_b K'_c (\nabla_aK''_d)...]\:.$$ The right-hand side vanishes because $\nabla_a T^{abcd...}=0$ and, for instance, using $T^{abcd...}=T^{bacd...}$ $$ T^{abcd..} (\nabla_aK_b) K'_c K''_d... = \frac{1}{2}T^{abcd..} (\nabla_aK_b + \nabla_bK_a) K'_c K''_d... =0$$ in view of $$\nabla_aK_b + \nabla_bK_a=0$$ because $K$ is Killing.

In summary $$J^a := T^{abcd...}K_bK'_cK''_d...$$ defines a conserved charge if

(a) $\nabla_a T^{abcd...}=0$,

(b) $K,K',K'',...$ are Killing vectors (possibly equal)

(c) $J\to 0$ rapidly at spacelike infinity.

Condition (b) can be weakened to the requirement that some of the $K$s is a conformal Killing vector provided the corresponding trace of $T$ vanishes. E.g., if $K'$ is a conformal Killing vector it must also hold $T_{a}\:^{bad...}=0$.

$\endgroup$
4
$\begingroup$

Any conserved $T^{ab\ldots c}{}_{;c} = 0$ and symmetric tensor $T^{ab\ldots c} = T^{(ab\ldots c)}$ gives rise to a conserved 4-current $J^a$ by contraction with Killing vectors, see e.g. this answer for the argument (the argument is essentially the same regardless of the tensor rank, as long as all but one index is contracted with a Killing vector). Alternatively, see Valter Moretti's answer. I will instead focus on the statement that $Q$ is constant.

By taking $Q = \int d^3x (J^an_a)$ for a time-like, unit normal field $n_a$, we find by the generalization of the Leibniz integral rule that $$ Q_{;a}n^a \equiv \dot{Q} = \int d^3x (J^bn_b)_{;a}n^a = \int d^3x\left(J^a\dot{n}_a + J_{a;b}n^an^b\right). $$ By the same argument as in special relativity, we can replace the second term in the integrand by the 3-divergence, and convert it to a surface term via Stokes' Theorem on smooth manifolds (possibly by considering a limit). Thus if $J^a \to 0$ sufficiently fast as we approach infinity, so that the total flux goes to zero, and $\dot{n}_a = 0$ (or indeed if it is orthogonal to $J^a$) we find that $\dot{Q} = 0$.

Note that to the best of my knowledge the requirement that $J^a\dot{n}_a = 0$ cannot be circumvented, which becomes important if we consider the rest frame of an accelerated observer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.