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-   -   Killing vector (http://www.physicsforums.com/showthread.php?t=666670)

c299792458 Jan24-13 06:37 PM

Killing vector
 
Let us denote by [itex]X^i=(1,\vec 0)[/itex] the Killing vector and by [itex]u^i(s)[/itex] a tangent vector of a geodesic, where [itex]s[/itex] is some affine parameter.

What physical significance do the scalar quantity [itex]X_iu^i[/itex] and its conservation hold? If any...? I have seen this in may books and exam questions. I wonder what it means...

WannabeNewton Jan24-13 06:57 PM

Re: Killing vector
 
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.

jfy4 Jan24-13 10:50 PM

Re: Killing vector
 
dat signature...

WannabeNewton Jan24-13 11:02 PM

Re: Killing vector
 
Quote:

Quote by jfy4 (Post 4242009)
dat signature...

:[ don't judge me T_T


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