- #1
binbagsss
- 1,254
- 11
some questions I have seen tend to word as show that some quantity/tensor/scalar (e.g let this be ##K##) is constant along an affinely parameterised geodesic, others ask show covariantly constant.
the definiton of covariantly constant/ parallel transport is:
## V^a\nabla_u K = 0 ##for the quantity ##K## where ##V^a## is the tangent vector to the geodesicsimply constant is wr.t the affine paramter
##\frac{d}{ds} K =0 ##
but, it is often the case, to show the latter case, we use the chain rule , i.e. that ## \frac{d}{ds} = V^a \nabla_a## when showing covariantly constant
e.g for the proof that given a KVF ##K^u##, we make use of the chain rule (connections not needed since we are acting on a scalar) to show that along a geodesic ##V^uK^u## is conserved.
But are, simply being constant w.r.t the affine parameter, and being covariantly constant/parallel transported not different things physically?
thanks.
the definiton of covariantly constant/ parallel transport is:
## V^a\nabla_u K = 0 ##for the quantity ##K## where ##V^a## is the tangent vector to the geodesicsimply constant is wr.t the affine paramter
##\frac{d}{ds} K =0 ##
but, it is often the case, to show the latter case, we use the chain rule , i.e. that ## \frac{d}{ds} = V^a \nabla_a## when showing covariantly constant
e.g for the proof that given a KVF ##K^u##, we make use of the chain rule (connections not needed since we are acting on a scalar) to show that along a geodesic ##V^uK^u## is conserved.
But are, simply being constant w.r.t the affine parameter, and being covariantly constant/parallel transported not different things physically?
thanks.