X_iu^i: Physical Significance & Conservation

[c299792458]

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

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

 

[WannabeNewton]

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.

 

[jfy4]

dat signature...

 

[WannabeNewton]

dat signature...

:[ don't judge me T_T