The Kelvin Wave

13. The Kelvin Wave Let us take the linearized shallow water equation on a f-plane with D = constant. A solid wall makes possible a rather special wave trapped at the wall, a gravity wave modified by rotation. Consider the following configuration: z y Ocean x Figure 13-1 Figure by MIT OpenCourseWare. The peculiar property of the wave is that the velocity normal to the wall is identically zero everywhere v≡0 Then the equations of motion are: 1. ut = -gηx The pressure gradient along the wall determines the acceleration the velocity along the wall is in 2. fu = -gηy geostrophic balance with the pressure gradient perpendicular to the wall 3. ηt+Dux = 0 the sea surface adjusts itself to the velocity gradient along the wall Find one equation for η From 1 gηx = -ut From 3 ηt = -Dux ⇒ ηtt = -D ∂ of 1 ∂x gηxx = - ∂ (ut) ∂x ∂ 1 (ut) = ηtt D ∂x ηtt-gDηxx = 0 Assume the usual dependence e-iωt -ω2η-gDηxx = 0 ⇒ ηxx + 2 ω η=0 gD If we now assume a wave traveling along the wall e ikx so the traveling wave is ei(kx-ωt) we obtain the dispersion relationship 2 2 ω = (gD)k ω = ±cok co = gD This is the shallow water gravity wave dispersion relationship without rotation!! So, how does rotation affect the wave? Put η = a(y) ei(kx-ωt) Combine the momentum equations of 1 and 2 ∂ of 2 → fut = -gηyt ∂x From 1: ut = -gηx or fηx - ηyt = 0 With the given shape of η: fka+ω da =a dy da fk = − dy a ω or as ⇒ a(y) = ηoexp ( -fk y) ω ω = co k - f a(y) = ηoe co y Full solution η = ηo exp − (f/coy)cos(kx-ωt) Notice that if the wave is on the y>0 side as in the above configuration lim y → +∞ η = finite = 0 co = + gD = ω >0 k and k>0 The wave travels in the +x direction If the wave is on the y<0 side co = - gD < 0 lim y → −∞ η = finite = 0 and k<0 The wave travels in the – x direction In either case, in the northern hemisphere f>0. The Kelvin wave always travels with the wall on its right side. The wave amplitude decreases exponentially away from the wall. The wave is trapped along the wall by rotation. Rotation does not affect the particle motion and wave propagation; only traps the wave to the coastline. Synopsis of Kelvin wave v≡0 1. ut = -gηx 2. fu = -gηy 3. ηt+Dux= 0 ηxx +(ω2 /gD) η = 0 η = ei(kx-ωt) ⇒ ω = ± cok co = gD η = a(y)ei(kx-ωt) a(y) = ηoe -f k y ω η = ηoexp(-f/coy) cos (kx-ωt) The decay scale is co = Rossby deformation radius f Rotation does not affect the particle motion or wave propagation, only traps the wave at the coast. The Kelvin wave always travels with the coast on its right side in the Northern Hemisphere Di rec ti o f no ss gre pro Di rec t o ion s res rog fp Figure. 13-2 Northern hemisphere Kelvin waves on opposite sides of a channel that is wide compared with the Rossby radius. In each vertical plane parallel to the coast, the currents (shown by arrows) are entirely within the plane and are exactly the same as those for a long gravity wave in a nonrotating channel. However, the surface elevation varies exponentially with distance from the coast in order to give a geostrophic balance. This means Kelvin waves move with the coast on their right in the northern hemisphere and on their left in the southern hemisphere. Figure by MIT OpenCourseWare. ω f Poincare modes n=3 Kelvin wave n=2 n=1 ω = Co k ωmin kL Figure 13-3. Summary dispersion relation for the right moving waves. Figure by MIT OpenCourseWare.