Register to reply

Killing vector

by c299792458
Tags: conservation, general relativity
Share this thread:
c299792458
#1
Jan24-13, 06:37 PM
P: 71
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...
Phys.Org News Partner Physics news on Phys.org
Engineers develop new sensor to detect tiny individual nanoparticles
Tiny particles have big potential in debate over nuclear proliferation
Ray tracing and beyond
WannabeNewton
#2
Jan24-13, 06:57 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,638
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
#3
Jan24-13, 10:50 PM
jfy4's Avatar
P: 647
dat signature...

WannabeNewton
#4
Jan24-13, 11:02 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,638
Killing vector

Quote Quote by jfy4 View Post
dat signature...
:[ don't judge me T_T


Register to reply

Related Discussions
Killing vector in Stephani Advanced Physics Homework 3
Killing Vector Equations Special & General Relativity 19
Killing vector on S^2 Special & General Relativity 11
Killing Vector Advanced Physics Homework 3
Killing vector help Differential Geometry 0