2 Probably used the wrong word "for" instead of "in". The former might confuse someone in that the answer refers to the energy of the spacetime. edited Nov 13 '15 at 4:56 auxsvr 1,80611 gold badge99 silver badges1313 bronze badges 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 forin the entire space-time. 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 for the entire space-time. 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. 1 answered Nov 12 '15 at 14:42 auxsvr 1,80611 gold badge99 silver badges1313 bronze badges 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 for the entire space-time.