Proof that the first curvature vector P of a curve C is at right angles to C
First, some notation: write T_{,a} for the (ordinary) partial derivative of the tensor T with respect to x^{a}. This will avoid a lot of messy notation (as if there is not enough already!).
To show the result, we compute
P, T  = 


= 

(Definition of the inner product)  
= 

(Definition of P^{i})  
= 


= 


(since g^{li}g_{ij} = _{j}^{l}) 
Looking at the term on the right, we distribute to get 3 terms of the form
g_{qj,p}  ds 
ds 
ds 
=  ds 
ds 
ds 
by the chain rule, 
and we notice that the other two terms are identical to this (except for dummy variables). Thus, we are left with


= 

(by symmetry of the g_{ij})  
= 

(definition of the inner product)  
= 

(refer back to the Proof of 5.5 to check this)  
=  0, 
as asserted.
Just close this window to return to the lecture.