Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Your last question in the first paragraph supposes a notion of energy other than "that which is conserved by a time-like Killing field". Which notion would that be? – ACuriousMind Nov 12 '15 at 14:16
I'm not sure if I understand what you mean? – Alpha001 Nov 17 '15 at 17:16
The "correct" definition of "energy" is mostly circular: It's the quantity conserved under time translation symmetry. The question "For conservation of energy is it important to have the field $\partial t$ or must it be an arbitrary timelike Killing-field?" becomes trivial if "energy" is defined as "that which is conserved under a time-like Killing field" (or, equally, it is trivial if energy is that which is conserved under $\partial_t$). So for that questions to make sense, you must have a definition of energy that is neither of those. – ACuriousMind Nov 17 '15 at 17:27

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.

share|cite|improve this answer
Isn't the reason asymptotically flat space-times are preferable to model physical systems that they can be considered as isolated? – auxsvr Nov 12 '15 at 19:29
Indeed, this is not an isolated system. It is a system interacting with a stationary environment. – Valter Moretti Nov 12 '15 at 20:57
What do you mean by "isolated system"? – auxsvr Nov 12 '15 at 21:06
"Physically, asymptotically flat space-times represent isolated systems", cf. Robert Geroch and Jeffrey Winicour. Linkages in general relativity. Journal of Mathematical Physics, 22(4):803-812, 1981. – auxsvr Nov 13 '15 at 5:11
Also, it is not necessary that the Killing vector is timelike, it is necessary that it is asymptotically timelike, and this is for the reason described in my answer below, namely that the scalar defined using it has the appropriate asymptotic behaviour. In Kerr space-time, $K^a$ may be space-like! – auxsvr Nov 13 '15 at 5:13

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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