Conservation of energy and Killing-field

Question

In general relativity we have no general conservation of energy and momentum. But if there exists a Killing-field we can show that this leads to a symmetry in spacetime and so to a conserved quantity. Thats what the mathematic tells us. But I don't understand what's the meaning of an timelike/spacelike/lightlike Killing-field? For conservation of energy is it important to have the field $\partial_t $ or must it be an arbitrary timelike Killing-field?

If we look at the Kerr solution we see that for some spacetime region the field $\partial_t $ will be timelike and for another it will be spacelike Killing-field. What consequences has this for the conserved quantities, since its still a Killing-field?

Answer 1

A physical system in GR is never isolated, in general, as it interacts with the curved metric, i.e., the gravitational background. (However a notion of isolated system can be given in the particular case of an asymptotically flat spacetime as discussed in auxsvr's answer.)

Apparently this fact prevents the existence of conserved quantities because the "external" system may furnish contributions to every quantity obtained by integrating the components of the stress energy tensor $T_{ab}$ over any notion of 3D rest space and this contribution may change in time. The formal conservation law $$\nabla^a T_{ab}=0$$ does not produce any true conservation law as instead it happens in flat spacetime.

However, if there is a timelike Killing vector field $K$, an observer evolving along the tangent lines of $K$ viewes the gravitational background as stationary (see below).

The current $J_b := K^aT_{ab}$ turns out to be properly conserved in view of the Killing equation for $K$ and the formal conservation law $\nabla^a T_{ab}=0$, $$\nabla^b J_b=0 \tag{1}\:.$$ Indeed, if $\Sigma$ is any spacelike smooth 3-surface transverse to $K$ (it generally does not exist if $K$ is not timelike), it can be viewed as a rest space of an observer evolving along $K$ (moving $\Sigma$ itself with the flow generated by $K$, obtaining surfaces $\Sigma_t$, where $t$ is the parameter along the integral curves of $K$).

At rest with that 3D space, the background is stationary: Using the notion of time $t$ parametrising the curves tangent to $K$ as the time coordinate together with other three spacelike coordinates on $\Sigma$, it turns out that $\partial_t g_{ab}=0$. This is nothing but the Killing equation written in the said coordinates.

An elementary use of the theorem of the divergence proves that (1) implies $$\int_{\Sigma_{t_1}} J^b n_b d\Sigma = \int_{\Sigma_{t_3}} J^b n_b d\Sigma $$ where $d\Sigma$ is the natural notion of volume measure induced by the metric on $\Sigma_t$ and we have assumed that $J$ vanishes sufficiently fast in the spatial directions. We conclude that there is a conserved (in time) total quantity $$Q = \int_{\Sigma_{t}} J^b n_b d\Sigma$$ with respect to the given notion of time.

ADDENDUM. Regarding Kerr metric there is an interesting phenomenon discovered by Penrose and related with the fact that the external timelike Killing vector (the one that approaches the standard Minkowski Killing time far away from the black hole) becomes spacelike inside the ergosphere of the black hole.

Generally speaking, if $K$ is timelike and you have a particle with four momentum $p$ evolving along a geodesics, $$p^a \nabla_a (K^bp_b)=0\tag{2}$$ as a consequence of both the Killing equation and the geodesic equation. Identity (2) says that

the energy of the particle, $E := -K^bp_b$, referred to the notion of time associated to $K$ is conserved in time.

If the particle breaks into two particles, the same conservation law leads to the identity $$-K^bp_b = -K^bp^{(1)}_b - K^bp^{(2)}_b\quad \mbox{i.e.}\quad E= E_1 + E_2\tag{3}$$ Since $K$ and $p, p^{(1)}, p^{(2)}$ are future-oriented, the energies $E,E_1,E_2$ are all positive and $0 <E_i \leq E$.

Everything I wrote is valid also if $K$ is not timelike, in this case $-K^bp_b$ is conserved but it has not the meaning of energy and its sign can be arbitrary.

Suppose that the initial particle breaks just inside the ergosphere of a Kerr black hole. Assume that the a particle entered the ergosphere arriving from a region far away from the black hole (so that $E>0$). Suppose also that part $1$ remains inside the ergosphere whereas part $2$ comes out and reaches the initial asymptotic region.

In this case $E_1\geq 0$, because the geodesic of this particle is again future oriented as $K$ is. However it is now possible that $E_2<0$, because $K$ is spacelike in the ergosphere even if $p_2$ is still timelike and future directed therein. As $E= E_1+E_2$, it could happen that $$E_1>E>0\:.$$ As a matter of fact we have extracted energy from the black hole. This phenomenon is possible because the Killing vector $K$ becomes spacelike inside the ergosphere.

Answer 2

Valter Moretti's answer is very nice, and I learned some things from reading it. This answer is meant to be a lower-level explanation of the basics of this topic, giving a treatment in the same style that is found when most GR texts introduce this topic.

Definition of a Killing vector

A Killing vector $\xi$ is vector field describing a symmetry of a spacetime. If we move every point in the spacetime by an infinitesimal amount, the direction and amount being determined by the Killing vector, then the metric gives the same results. A Killing vector can be defined as a solution to Killing's equation,

$$ \nabla_a \xi_b + \nabla_b \xi_a = 0,$$

i.e., the covariant derivative is asymmetric on the two indices.

Now suppose that $p$ is a tangent vector along a geodesic. By this we mean that it satisfies the geodesic equation $p^a \nabla_a p^b = 0$. This equation states not just that $p$ stays tangent to a geodesic, i.e., is transported parallelly to itself, but also that it's parallel-transported along this geodesic in an affine manner, so that it doesn't "change length" as we go along. This is an affine notion of "not changing length," not a metrical one, so it applies equally well if $p$ is null rather than timelike. For a massive or massless particle moving inertially, the momentum is a tangent vector to the geodesic in this sense.

The conserved quantity associated with a Killing vector

Theorem: Under the assumptions given above, $p_b \xi^b$ is a conserved quantity along the geodesic. That is, it is constant for a test particle.

Proof: We prove this by showing that

$$p^a\nabla_a (p_b \xi^b)=0.$$

Application of the product rule gives

$$p^a \xi^b \nabla_a p_b+p^a p_b \nabla_a \xi^b.$$

The first term vanishes by the geodesic equation, and the second term by the antisymmetry expressed by the Killing equation, combined with the symmetry of $p^a p_b$.

None of the above is changed if we scale $p$ by some factor. In the case of a massive particle, it may be more convenient to let $p$ be the momentum per unit mass, so that all expressions depend only on the geodesic.

Nowhere in this argument was it necessary to make any assumptions about whether $\xi$ was timelike, null, or spacelike.

Some special cases

Some special cases are of interest. Suppose that the metric is independent of one coordinate $x^\mu$. Then $\partial_\mu$ is a killing vector , and $p_\mu$ is conserved.

A series of further restrictions to more and more special cases: ---

If the metric is independent of $t$, where $t$ is a timelike coordinate, it's $p_t$ (a component of the covariant momentum vector) that's conserved, but it's usually $p^t$ that we call "the" energy.

When the metric is also diagonal, it's $g_{tt} p^t$ that is conserved. For the Schwarzschild metric, this is $(1-2M/r)E$, where E is the energy measured by a static observer, i.e., an observer whose velocity vector is parallel to the Killing vector. (Note that this sequence of interpretations doesn't work when the Killing vector isn't timelike.)

Null or spacelike Killing vectors: an example

To see what happens when $\xi$ isn't timelike, it's helpful to look at the special case of a photon infalling from $r=+\infty$ into a Schwarzschild black hole. Then in Schwarzschild coordinates, $ds^2=0$ gives $dr/dt=\pm A$, where $A=1-2M/r=g_{tt}$.

Outside the horizon, $dr/dt=-A$, $p^t$ is the energy measured by a static observer hovering at $r$, and the conserved quantity $p_t=p^tA$ is the redshifted energy seen by an observer at infinity.

As the same particle passes into the interior of the horizon, its trajectory now has $dr/dt=+A$, which is still negative. There are no static observers here, so $p^t$, which is negative, is not the energy seen by any observer.

Answer 3

If the metric is asymptotically flat, it is straightforward to assign meaning to a quantity, such that it resembles the energy we know from special relativity. In particular, for the Kerr metric we may regard $(\partial_t)^a$ as a vector representing the stationary observer at infinity, where space-time is Minkowski, and the rest appears to said observer as an isolated system. If the definition of a scalar is such that it matches the energy in the asymptotic region, say $-p_a(\partial_t)^a$ with $p^a$ the momentum of a test particle, then we may regard it as energy in the entire space-time.