<Dd> <Dl> <Dd> Thank you for pointing out the diagonal aspect for my expression . I'm a little confused though, I did not have that g = g, I had that they were inverses of each other . Is g not the inverse of g? Second, are you saying p and p are not both covariant? In my case momentum was a covariant vector, following the well know chant "co is low". Are you saying g u does not equal u? Finally, I am also unfamiliar with writing gamma and way you have in the formula for the stationary observer, can you explain why you wrote it the way you did? Thank you .--Preceding unsigned comment added by Jfy4 (talk contribs) 05: 16, 11 April 2011 (UTC) </Dd> </Dl> </Dd> <Dl> <Dd> Thank you for pointing out the diagonal aspect for my expression . I'm a little confused though, I did not have that g = g, I had that they were inverses of each other . Is g not the inverse of g? Second, are you saying p and p are not both covariant? In my case momentum was a covariant vector, following the well know chant "co is low". Are you saying g u does not equal u? Finally, I am also unfamiliar with writing gamma and way you have in the formula for the stationary observer, can you explain why you wrote it the way you did? Thank you .--Preceding unsigned comment added by Jfy4 (talk contribs) 05: 16, 11 April 2011 (UTC) </Dd> </Dl> <Dd> Thank you for pointing out the diagonal aspect for my expression . I'm a little confused though, I did not have that g = g, I had that they were inverses of each other . Is g not the inverse of g? Second, are you saying p and p are not both covariant? In my case momentum was a covariant vector, following the well know chant "co is low". Are you saying g u does not equal u? Finally, I am also unfamiliar with writing gamma and way you have in the formula for the stationary observer, can you explain why you wrote it the way you did? Thank you .--Preceding unsigned comment added by Jfy4 (talk contribs) 05: 16, 11 April 2011 (UTC) </Dd> <Dl> <Dd> <Dl> <Dd> <Dl> <Dd> Notice that g is the inverse matrix of g, that is, <Dl> <Dd> g α β g β γ = δ γ α (\ displaystyle g ^ (\ alpha \ beta) g_ (\ beta \ gamma) = \ delta _ (\ gamma) ^ (\ alpha) \,) </Dd> </Dl> </Dd> <Dd> where δ is the Kronecker delta . So the corresponding element on the diagonal of the inverse will not be the inverse of the diagonal generally, but usually only when the off - diagonal elements are zero . </Dd> <Dd> Unlike some people, I prefer to show all raising and lowering of indices explicitly . To answer your questions: <Ul> <Li> No, it is not . </Li> <Li> No, I am not saying that . </Li> <Li> I would not use u at all . </Li> <Li> See the following calculation: </Li> </Ul> <Dl> <Dd> u α = d x α d t d t d τ = v α u t . (\ displaystyle u ^ (\ alpha) = (\ frac (dx ^ (\ alpha)) (dt)) (\ frac (dt) (d \ tau)) = v ^ (\ alpha) u ^ (t) \, .) </Dd> </Dl> </Dd> <Dd> Thus for a diagonal and spatially isotropic metric <Dl> <Dd> − c 2 = g α β u α u β = g t t (u t) 2 + g s s v 2 (u t) 2 . (\ displaystyle - c ^ (2) = g_ (\ alpha \ beta) u ^ (\ alpha) u ^ (\ beta) = g_ (tt) (u ^ (t)) ^ (2) + g_ (ss) v ^ (2) (u ^ (t)) ^ (2) \, .) </Dd> </Dl> </Dd> <Dd> Solving for u gives <Dl> <Dd> u t = c − 1 g t t + g s s v 2 . (\ displaystyle u ^ (t) = c (\ sqrt (\ frac (- 1) (g_ (tt) + g_ (ss) v ^ (2)))) \, .) </Dd> </Dl> </Dd> <Dd> Is that clear? JRSpriggs (talk) 10: 44, 11 April 2011 (UTC) </Dd> </Dl> </Dd> </Dl> </Dd> </Dl>

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