The pressure of water waves upon a fixed obstacle 417 
the pressure conditions at z = €, given by (25), are satisfied to the second 
order, namely p = 0 and 
Op opdp odd op 
Gt) Oa Ox Oz dz 
To the first order, (24) and (25) represent plane waves of amplitude h 
reflected at the plane x = 0. We may now evaluate the additional pressure 
upon this plane per unit width. We put x = 0 in (27) and integrate with 
respect to z from —o to €. The first order term is the periodic force 
(2gph/«) cos ot; for the additional quadratic terms we obtain 
— hgpC? + 2gph€ cos ot — gph" sin? ot, (28) 
the second term in (28) coming from the expansion of e*¢. We put in the 
value of € from (25), noting that we only need this to the first order; and we 
obtain for the additional steady force P per unit width of the plane, taking 
mean values, 
P = 39ph?, (29) 
where h is the amplitude of the incident waves. 
It may be remarked that instead of using the fact that the second order 
term in the expansion of ¢ is zero, it would have sufficed for the present 
purpose to assume ¢ to be purely periodic, an assumption made by Larmor 
(1920) in the corresponding calculation for sound waves. It is well known 
that waves of finite amplitude possess linear momentum in the direction 
of propagation; the average amount, to second order terms, is mph2V per 
wave length, V being the wave velocity. On the other hand, if we calculate 
the rate of transfer of linear momentum across a vertical plane, we obtain 
an average rate of 4gph?; this gives in one period one-half the average 
momentum in one wave-length. The average pressure P given by (29) may 
be regarded as due to the reversal of this flow of momentum. We notice 
also that P is equal to one-half the average density of energy in the standing 
oscillations, and this may again be connected with the fact that the group 
velocity for water waves is one-half the wave velocity. For plane waves of 
amplitude / incident upon the plane x = 0 at an angle « to the plane, we 
may take 
@ = (2gh/c) e cos (ka sin a) sin (ot — Ky cos ig an 
€ = 2h cos (kx sin a) cos (ot—Ky Cos @). 
We obtain now, instead of (28), the quadratic terms for the additional 
pressure as 
— dgpl? + 2gph" cos (ot — Ky cos a) 
—gph?{ cos? a cos? (at — Ky cos %) + sin® (ot — Ky COS &)}. (31) 
478 
