Thread: Killing vector View Single Post
 C. Spirit Sci Advisor Thanks P: 4,938 Hi there! The point is that the scalar quantity you formed is constant along the geodesic! Using your notation, $\triangledown _{U}(X_{i}U^{i}) = U^{j}\triangledown _{j}(X_{i}U^{i}) = U^{j}U^{i}\triangledown _{j}X_{i} + X_{i}U^{j}\triangledown _{j}U^{i}$. Note that $U^{j}U^{i}\triangledown _{j}X_{i}$ vanishes because $U^{j}U^{i}$ is symmetric in the two indices whereas, by definition of a killing vector, $\triangledown _{j}X_{i}$ is anti - symmetric in the two indices and it is very easy to show that the contraction of a symmetric tensor with an anti - symmetric one will vanish. The second term $X_{i}U^{j}\triangledown _{j}U^{i}$ vanishes simply because U is the tangent vector to a geodesic thus we have that $\triangledown _{U}(X_{i}U^{i}) = 0$. In particular note that if this geodesic is the worldline of some freely falling massive particle then its 4 - velocity is the tangent vector to the worldline and we can re - express the condition for the worldline being a geodesic in terms of the 4 - momentum of the particle (and for photons just define the geodesic condition like this) and we can have that if $X^{i}$ is a killing field on the space - time then $X_{i}P^{i}$ will be constant along this geodesic. It is a geometric way of expressing local conservation of components of the 4 - momentum; these killing fields are differentiable symmetries of the space - time and you might be able to see that more clearly by the fact that the lie derivative of the metric tensor along the killing field will vanish.