474 Screntific Proceedings, Royal Dublin Society. 
perpendicular to these two directions respectively. We are now 
in a position to determine the accelerations in these directions. 
To determine the acceleration along the radius vector we have, 
in the first place, from the radial component 7 an acceleration out- 
wards along 7 equal to 7. The second velocity component, namely 
the meridian component 76, gives an acceleration (v’/p) inwards 
along 7 equal to : 
is (76)? ee 62. 
ip 
and the third velocity component, 7 sin 00, gives an acceleration 
inwards equal to (v*/p) or 
Sas 1) : 
a2 a —--—/7 sin? O°. 
The whole acceleration along the radius vector is consequently 
+ — 7? — rsin? 04? = 1 — vr (8 + sin? 69). 
In the same manner the same velocity components give at 
once, remembering our lemma, accelerations along the meridian 
measured by 
i ae as) (r sin 06)? 
ei Sag) ees Ags ee ee 
oe at (r), rtan@ ’ 
and, therefore, the total acceleration along the meridian 1s 
1 
r 
In the same way the acceleration perpendicular to the meridian 
plane is | 
“ (776) — 2? sin 0 cos 06%; 
1 Gi amen 
r sin 0 dt ese), 
being the sum of the components 
trees 
sin O¢ 
> 9 cos 006, < (rsin Od): 
In the case of cylindrical coordinates we proceed in the same 
manner. 
In the same way if PP’ (fig. 4) be taken as an element of an 
electric current, it may be replaced by two elements of the same 
strength along PM and MP’, respectively, and the fundamental — 
formulee of Electrodynamics may be deduced at once. 
ee ee ee ee 
