Les courants moyens – quel rôle jouent
Transcription
Les courants moyens – quel rôle jouent
Netherlands Institute for Sea Research, Texel, Pays-Bas Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Theo Gerkema, Leo Maas & Hans van Haren 24 mai 2013 OGOA Lyon 2013 1/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Internal waves and mean flows Common situation: internal waves in presence of (meso-scale) background flows. How does that flow affect the wave’s frequency? In various forms, one finds the following assertion in the literature: • "Advection leads to a Doppler shift." (Kunze, 1985) • "A background flow (. . . ) provides a way to increase the apparent frequency of near-inertial waves through Doppler shifting." (Zhai et al., 2005) • ". . . a Doppler shift in the frequency of the inertial wave brought about by the mean flow past the mooring." (White, 1972) • . . . observations of horizontal velocity are computed from data recorded on instrumented moored buoys. The mean drift causes non negligible Doppler shifts in the observed frequencies (given by kU)." (Frankignoul, 1970) • "a constant current (. . . ) implies a Doppler shift." (Olbers, 1981) OGOA Lyon 2013 2/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Internal waves and mean flows Internal waves and mean flows Where does this idea come from? With background flow U, the first terms of the linearized equations of motion read ∂ ∂ +U +··· = ··· ∂t ∂x Substituting a wave solution sin(ωt − kx), with wavenumber k and wave frequency ω, we find the combination of terms ω − Uk + · · · = · · · One can then define ω 0 = ω − Uk, the intrinsic frequency. In the oceanographic literature, it is common to refer to Uk as a "Doppler shift" (Bretherton & Garrett 1968, Kunze 1985, etc.) OGOA Lyon 2013 3/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? "Doppler shifts" – a confusion of tongues? "Doppler shifts" – a confusion of tongues? In physics, "Doppler shifts" refer to a difference in frequency between source and observer; here mean flows never create a "Doppler shift". In oceanographic parlance, "Doppler shifts" Uk refer to a difference in frequency between two observers; here mean flows always (by definition!) create a "Doppler shift". . . . and sometimes the two are mixed up. . . In the oceanographic context, one often compares frequencies of sources (e.g. tidal) with observed ones (e.g. in moorings) – one should then not apply the expression Uk ! To avoid confusion I shall call the former "Doppler shift"; the latter, "quasi-Doppler shift". OGOA Lyon 2013 4/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Doppler shifts revisited Doppler shifts revisited Doppler (1842, 1846): how do movements of source and observer affect the frequency? He found that the frequency of the source, ωs , is shifted to ω = ωs c − vo c − vs c wave speed vs velocity of source vo velocity of observer There is a Doppler shift only if source and observer move relative to each other (vs 6= vo ) OGOA Lyon 2013 5/13 The theory was put to the test by Buys Ballot in 1845. He had musicians play trumpets on a steam train. Result: change in tone agrees with theory. Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Doppler shifts revisited Doppler (1847): the effect of a steady uniform flow Source and observer both at rest. Wavelength changes (see figure). Successive wave crests propagate under identical circumstances and thus need equal times to bridge distance from source to observer. Hence observed frequency equal to frequency of emittance → no Doppler shift. All cases collected in general formula by Bateman (1917): ω = ωs c + U − vo c + U − vs So if vs = vo , we have no Doppler shift – irrespective of mean flow U. OGOA Lyon 2013 6/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Analytical example Analytical example Linear long interfacial waves in presence of mean flow U. NB: no wave dispersion included at this point! Governing equations: ∂u ∂u ∂η +U = − g0 ∂t ∂x ∂x ∂η ∂η h1 h2 ∂u +U + = F(t, x) ∂t ∂x h ∂x with shear u, reduced gravity g0 and total water depth h: g0 = g ρ2 − ρ1 ; ρ∗ h = h1 + h2 . A moving source explicitly included: forcing term F. OGOA Lyon 2013 7/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Analytical example Specify source: oscillating at frequency ωs and moving at speed vs : F(t, x) = sin(ωs t) G0 (x − vs t) for an arbitrary function G (expressing F as its derivative). Transform equations to frame of reference moving with the flow, and combine the resulting equations: 2 ∂F ∂2 η 2∂ η − c = , ∂t2 ∂x2 ∂t with c = (g0 h1 h2 /h)1/2 , the phase speed of proper wave propagation. For a system starting from rest, the solution reads 1 η= 2c OGOA Lyon 2013 Z t dτ 0 8/13 Z x+c(t− τ ) x−c(t− τ ) dξ ∂F (τ, ξ ) . ∂τ Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Analytical example The inner integral can be solved, and after a transformation back to the original system, we obtain 1 η (t, x) = 2 Z t 0 n o dτ sin(ωs τ ) G0 (X− ) + G0 (X+ ) , with X± = x − (±c + U )t + (±c + U − vs )τ. The remaining integral can be solved analytically if we choose a (moving) point source, described by a delta-distribution: G0 = δ. The end result is: η = η− + η+ with η± = h (±c + U )t − x i 1 sin ωs 2 | ± c + U − vs | ±c + U − vs for all x satisfying (±c + U )t < x < vs t or vs t < x < (±c + U )t; for all other x, η± = 0. OGOA Lyon 2013 9/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Analytical example Setting: two fixed observers, left and right to initial position of source: Solution for several cases with mean flow or/and moving source: → mean flow creates no Doppler shift OGOA Lyon 2013 10/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Analytical example The solution can be easily modified to include a movement of the observer, at velocity vo , via x = xo + vo t. Then: ω= ± c + U − vo ωs , ± c + U − vs k= ωs . ±c + U − vs Returning now to quasi-Doppler shifts: From the above, it follows that a fixed observer (vo = 0) measures ±c + U ω= ωs ± c + U − vs and an observer moving with the mean flow (vo = U): ±c ω0 = ωs ±c + U − vs Combining these with the expression for k: ω 0 = ω − Uk (quasi-Doppler shift) Expression is invariant for movement of source (vs ), thus hiding the fact that ω 0 , ω and k – each individually – depend on it ! OGOA Lyon 2013 11/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Dispersion Dispersion So far wave dispersion was not considered. How does this change the Doppler relations? Example: source moving toward an observer at rest. One can still argue that C ω= ωs C − vs but the phase speed C is no longer known a priori ! C depends on ω → dispersion relation needed to close the problem. E.g., for ω = Uk + (f 2 + c2 k2 )1/2 , i.e. Coriolis disperion & mean flow: C± = −(vs f 2 − Uωs2 ) ± ωs [c2 (ωs2 − f 2 ) + (U − vs )2 f 2 ]1/2 ωs2 − f 2 featuring movement of source vs and mean flow U. OGOA Lyon 2013 12/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ? Conclusions Conclusions 1 It is important to distinguish Doppler shifts – difference in frequency between source and observer – from quasi-Doppler shifts ("Uk") – difference in frequency between two observers. 2 A Doppler shift in frequency occurs only if source and observer move relative to each other. 3 The presence of a steady mean flow does not change this fact ! 4 This remains so if wave dispersion is included. 5 A mean flow does affect wavelength and amplitude! Reference: Gerkema et al. (2013), J. Phys. Oceanogr. 43, 432-441. OGOA Lyon 2013 13/13 Les courants moyens – quel rôle jouent-ils dans l’effet Doppler ?