Energy Considerations. 
Potential Energy per unit area. 
ne 
PE, = PID (6.18 ) 
Potential Energy per unit area averaged over time 
cea 
—— | 2 
P.E. = t/osFe (6.19) 
+* 
tT rT 
Be cals 2 42x emt \\2 pgAS}._ 82x  4rt 
E.-= pa A? (sin( +3 ~ ))? dt (I-c s(- coy \)dt 
t* 
; 4 
_P9A Ti ye,.(872x  40(t*®+T))  .. (Br2x — 4nt* 
<a +4 = (sin(“T ot ) sin (“2 a )) (6.20) 
for T= kT or as T—>@ 5 
— A 
Pe.- (6.21 ) 
tok mat 
se 2 4r2m2x eum 4r2mx 2rmt\2 
1E.= T p> a any sin (Ze 2rmt\vat= Ups $a (Si aa = emmy)? dt 
athe Gr2m2x  2mrmty... ,47°UX  2nrqt 
+= Px > 2) &maqsin(=" TS Sea) S10 ema acta =—)\dlt (6.22) 
{* m=| q=l 
| Ee gra? 2-28" - 2h oe 
lim == Spo 2 pom TEpeT Wee 
poPE. “24g am Se ee (6.23) 
for fixed x= x, and t=t* 
gh? if -nT+E(x,,n Uses t*-T <nT-E(x,,9,T) 
to a 
” d 
PE. ‘| a if —nT +E(x,,n ae -t*<nT-E 
0 GTX, 4% 
Mi Tare = t*<-nT—E(x,n,T) 
or if mx, —t*+r<nT+ E(x,n,T) (6.24) 
aes 5 m=k+n 
re. oh 
m=k if m satisfies the inequalities 
-p(t+ Yo) + E(xn% pp ) Se — the p(t +14) - E(x, YonsP) 
a mx = 
and -p(t+%) + E(x, 90) <a aa = 8 T< p(t +4)-E (4 77,,) (6.25) 
Plate XWIL 
-II7- 
