511] 



of three veolors dx, u, v, we conclude that the vai-iables Jiiust 

 vanish. Thus 



Similarly 



ft ^■'^""* 1 / I ƒ 



0«' 





ft ^•^'"' -A 17 



Ó.rf 



By adding the first two, and snhti'acting the last, and considei-ing 



that kl^nm = I^Lma GtC. WC find 



I in 



Ki.lm 



Now we know 



/;/„, — 2 p"^ ha, I in = — :S p^l 



Im 

 a 



Im 1 

 b ' 



and so from (4') we see that .onr values for (bt^ -. 



' Im ' 



du^' = — :^{hn) 



dxl u"' 



(4) 



constitiite the only solution consistent with ail conditions. 



10. To explain the applications of sections 3 and 4 we j)i'Ocee(l 

 as follows. Suppose a vector V, in point P, be marked in the 

 compass-rigid. After a geodesical displacement to Q the marked 

 vector will have got the components 



i hn I 



yc, 



ihi'l K"« 



Now if we have in Q a vector with components ]^" -^ dV", 

 where dV" now represents some increment of the component V", 

 then obviously the components of the geodesic differential are 



<i?F" -f 2:{lm) \d.rl F"'. 



This geodesic differential will be a vector itself, being the difference 

 of two vectors, while d V" are no vector-components. 



If the line-element PQ itself is drawn in the compass-i-igid as a 

 vector with components (/.c«, and displaced geodesically, then in Q 

 the arrow will have got components 



\lin \ 



This arrow we have called the geodesic pro/ongation of the 



