Thread: Killing vector
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WannabeNewton
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#2
Jan24-13, 06:57 PM
C. Spirit
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Hi there! The point is that the scalar quantity you formed is constant along the geodesic! Using your notation, [itex]\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}[/itex]. Note that [itex]U^{j}U^{i}\triangledown _{j}X_{i}[/itex] vanishes because [itex]U^{j}U^{i}[/itex] is symmetric in the two indices whereas, by definition of a killing vector, [itex]\triangledown _{j}X_{i}[/itex] 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 [itex]X_{i}U^{j}\triangledown _{j}U^{i}[/itex] vanishes simply because U is the tangent vector to a geodesic thus we have that [itex]\triangledown _{U}(X_{i}U^{i}) = 0[/itex]. 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 [itex]X^{i}[/itex] is a killing field on the space - time then [itex]X_{i}P^{i}[/itex] 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.