Killing vector


by c299792458
Tags: conservation, general relativity
c299792458
c299792458 is offline
#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
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
WannabeNewton
WannabeNewton is offline
#2
Jan24-13, 06:57 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 4,933
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
jfy4 is offline
#3
Jan24-13, 10:50 PM
jfy4's Avatar
P: 647
dat signature...

WannabeNewton
WannabeNewton is offline
#4
Jan24-13, 11:02 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 4,933

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