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)) Let g(θ1, θ2) = θ1x1 + θ2x2 − 1 2 From d dθj g(θ1, θ2) = 0, we have that (θ1 2 + θ1θ2 + θ2 2). x1 − θ1 − θ2 = 0, 1 2 x2 − θ2 − θ1 = 0, 1 2 2 from which we have θ1 = x1 − x2, θ2 = x2 − x1 4 3 2 3 4 3 2 3 Then So we need to find 2 I(x1, x2) = (x1 + x2 − x1x2). 2 2 3 2 (x1 + x 2 − x1x2) 2 inf x1,x2...
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3 3 0, then µ 0 and further x1 x2 = = = = min 2 3 2 x1 + 2 3 (5 − 2x1)2 − x1(5 − 2x1) which gives x1 = 10 11 , x2 = 35 11 and I(x1, x2) = 5.37. Thus lim sup n 1 n log P( Sn n ∈ F ) ≤ −5.37 Applying the lower bound part of the Cram´er’s Theorem we obtain lim inf n 1 n log P( Sn n ∈ F ) ≥ lim...
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exists a broad set of con­ ditions under which the large deviations bounds hold. Thus consider a general sequence of random variable Yn ∈ Rd which stands for (1/n)Sn in the i.i.d. case. Let φn(θ) = 1 log E[exp(n(θ, Yn))]. Note that for the i.i.d. case n φn(θ) = = 1 n 1 n log E[exp(n(θ, n−1Sn))] log M n(θ) =...
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0, we have 1P( Yn ≥ a) = P(exp(θYn) ≥ exp(θna)) n ≤ exp(−n(θa − φn(θ))) So we can get an upper bound sup(θa − φn(a)) θ≥0 In the i.i.d. case we used the fact that supθ≥0(θa − M (θ)) = supθ(θa − M (θ)) when a > µ = E[X]. But now we are dealing with the multidimensional case where such an identity does not make ...
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- Frobenious eigenvalue, which satisfies the following properties. 5 1. ρ > 0 is real. 2. For every e-value λ of B, |λ| ≤ ρ, where |λ| is the norm of (possibly complex) λ. 3. The left and right e-vectors of B denoted by µ and ν corresponding to ρ, are unique up to a constant multiple and have strictly positive co...
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ρnνi) The second identity is established similarly. Now, given a Markov chain Xn, a function f : Σ → Rd and vector θ ∈ Rd , consider a modified matrix Pθ = (e(θ,f (j))Pi,j , 1 ≤ i, j ≤ N ). Then Pθ is an irreducible non-negative matrix, since P is such a matrix. Let ρ(Pθ) denote its Perron-Frobenious eigenvalue. T...
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n 1≤j≤N where P n(i, j) denotes the i, j-th entry of the matrix Pθ applying Corollary 1, we obtain θ n . Letting φj = 1 and lim φn(θ) = log ρ(Pθ). n Thus the G¨artner-Ellis can be applied provided the differentiability of log ρ(Pθ) with respect to θ can be established. The Perron-Frobenious theory in fact can ...
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3.032 Mechanical Behavior of Materials Fall 2007 Using U(r): Measure parameters for U(r) in physical model to predict stresses that are high enough for elastic instabilities to occur (e.g., nucleation of defects in crystals). Images removed due to copyright restrictions. Please see: Fig. 1c in Gouldstone, Andrew, et a...
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ofibers: Do physical and mechanical properties differ from bulk polymers? Courtesy Gregory Rutledge. Used with permission. Images removed due to copyright restrictions. Please see Fig. 1 and 4 in Curgul, Sezen, et al. "Molecular Dynamics Simulation of Size-Dependent Structural and Thermal Properties of Polymer Nanof...
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18. Div grad curl and all that Theorem 18.1. Let A ⊂ Rn be open and let f : A −→ R be a differ­ entiable function. If �r : I −→ A is a flow line for �f : A −→ Rn, then the function f ◦ �r : I −→ R is increasing. Proof. By the chain rule, d(f �r) ◦ dt (t) = �f (�r(t)) · �r�(t) = �r�(t) �r�(t) ≥ 0. · � Corollary...
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∂f ∂y jˆ+ ˆı + on vector fields: Definition 18.5. Let A ⊂ R3 be an open subset and let F� : A −→ R3 be a vector field. The divergence of F� is the scalar function, which is defined by the rule div F� : A −→ R, div F� (x, y, z) = � · F� (x, y, z) = ∂f ∂x + ∂f ∂y + ∂f . ∂z 1 The curl of F� is the vector...
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and it is called incompressible if the divergence is zero, div F� = 0. : −→ � 3R Proposition 18.7. Let f be a scalar field and F� a vector field. (1) If f is C2, then curl(grad f ) = �0. Every conservative vector field is rotation free. (2) If F� is C2, then div(curl F� ) = 0. The curl of a vector field is incompres...
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� � � � F1 F2 F3 ∂2F2 ∂2F3 ∂x∂z ∂x∂y − = = − ∂2F3 ∂y∂x + ∂2F1 ∂y∂z + ∂2F2 ∂z∂x − ∂2F1 ∂z∂y = 0. This is (2). Example 18.8. The gravitational field � � F (x, y, z) = cx (x2 + y2 + z2)3/2 ˆı+ cy (x2 + y2 + z2)3/2 jˆ+ cz (x2 + y2 + z2)3/2 ˆ k, is a gradient vector field, so that the gravitat...
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3 A solution of the differential equation �2f = 0, is called a harmonic function. Example 18.11. The function f (x, y, z) = − c , (x2 + y2 + z2)1/2 is harmonic. 4 MIT OpenCourseWare http://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms ...
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Figure removed for copyright reasons. See Fig. 2 in Bockstaller et al. "Size-selective Organization of Enthalpic Compatibilized Nanocrystals in Ternary Copolymer/Particle Mixtures." J. Amer. Chem. Soc. 125 (2003): 5276-5277. Figure removed for copyright reasons. Figure removed for copyright reasons. See Fig. 2 in Bocks...
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Perfect Conductivity Lecture 2 Terry P. Orlando Dept. of Electrical Engineering MIT September 13, 2005 Massachusetts Institute of Technology 6.763 2005 Lecture 2 Outline 1. Persistent Currents 2. Parts of a Physical Theory 3. Circuits and Time Constants 4. Distributive Systems and Time constants A.Quasista...
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R Current conservation: i=i C + iL iL =iR Energy Conservation v = vC = vR + vL Massachusetts Institute of Technology 6.763 2005 Lecture 2 2. Constitutive Relations Massachusetts Institute of Technology 6.763 2005 Lecture 2 4 3. Summary Relation 1 jωC jωL R Massachusetts Institute of Technology 6.763 2005...
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Gauss Law Gauss’ Magnetic Law Conservations laws Charge conservation Also Poynting’s Massachusetts Institute of Technology 6.763 2005 Lecture 2 7 Distributied systems con’t 2. Constitutive Relations Local in space, linear time invariant Ohm’s Law 3. Summary relations Complex: Search first for first order i...
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QuasiStatics Solve first Image removed for copyright reasons. Please see: Figure 2.9, page 34, from Orlando, T., and K. Delin. Foundations of Applied Superconductivity. Reading, MA: Addison-Wesley, 1991. ISBN: 0201183234. Solve for E once B is found Boundary conditions: Massachusetts Institute of Technology 6....
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Electronics A Joel Voldman Massachusetts Institute of Technology Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/...
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devices > The electrical capacitor • What is the relation between Q and V? A I + - V ε E a d ⋅ = Q (cid:118) ∫ closed surface ε EA Q = ∇ × = ⇒ 0 E E r ( , ) t = −∇ V ( , ) r t ( ) V b V a − ( ) = − b ∫ a E l d ⋅ ( ) V b V a − ( ) = V Eg = ⇒ = E V g g Q = ε A V g = V CV = A ε g A ε g C = Cite as: Joel Voldman, course ...
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Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 6E - 5 Elements and element laws > Do this with all three basic elements > Resistor > Capacitor > Inductor I + R V V RI= - I + I + C V dV dt I C = L V dI dt V L = - - Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabricatio...
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7J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 6E - 8 Complex impedances > For LTI systems, use complex impedances instead • Impl...
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Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 6E - 10 Let’s analyze a circuit 1. Figure out what are you trying to determine 2. Replace eleme...
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i Z R L − i Z R R = 0 i R = V 0 Z L = + Z R Z C + V 0 Ls R + + 1 Cs i R = 2 LCs Cs + RCs V 0 + 1 Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded...
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. Figure out what are you trying to determine + -VC 2. Replace elements with complex impedances V0 + - 3. Assign node voltages & ground node 4. Write KCL at each node 5. Solve for node voltages 6. Use node voltages to find what you care about v1 V0 + - C C R R L L v2 Cite as: Joel Voldman, course materials for 6.77...
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777J Spring 2007, Lecture 6E - 16 Nodal analysis 1. Figure out what are you trying to determine v1 i2 i3 i1 C R v2 L V0 + - 2. Replace elements with complex impedances 3. Assign node voltages & ground node 4. Write KCL at each node 5. Solve for node voltages 6. Use node voltages to find what you care about v 2 ⎛ ⎜...
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Solve for node voltages v 2 = V 0 6. Use node voltages to find what you care about V C = v 1 − v 2 = V V − 0 0 2 LRCs 2 + LRCs Ls R + Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachuset...
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1 Z 2 1 Z 1 = Z Z 1 2 Z Z + 1 2 = Z 1 // Z 2 + 1 Z 2 Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Sprin...
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Z3 and Z4 in parallel Z1 and Z3 NOT in series Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007,...
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are you trying to determine 2. Replace elements with complex impedances = V 0 Z Z R L Z Z + R Z Z R L Z Z + R L + 1 ZZ C L 3. Collapse circuit in terms of series/parallel relations till circuit is trivial 4. Re-expand to find signal of = V 0 = V 0 interest Z Z R L + L Z Z R L ( Z + R RLs 1 ) Z Z Z C L L 1 Ls RLs +...
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one • Solve circuit • Repeat for all sources, then add responses > Turning off a voltage source gives a short circuit V0 + - short > Turning off a current source gives I0 open an open circuit Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring ...
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, then add v 2 = I 0 Z Z R C + Z Z + V 0 C Z R + Z C Z R responses v1 + - V0 v2 C R I0 Find v 2 v 2 = v 2 = + V R 0 R 1 I R 0 Cs 1 + R Cs I R V RCs 0 + RCs 0 + 1 Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http:...
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18.336 spring 2009 lecture 15 02/13/08 Finite Difference Methods for the One-Way Wave Equation � ut = cux u(x, 0) = u0(x) Solution: u(x, t) = u0(x + ct) Information travels to the left with velocity c. Three Approximations: ⎧ c ⎨ n n +1 − Uj Uj Δx U n − Uj n jc −1 Δx ⎪⎪⎪⎪⎪⎪⎪⎪ ⎪⎪⎪⎪⎪⎪⎪⎪ U n ⎩ c j+1 − U...
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2 2 uttΔt − cuxxΔx = c uxxΔt − cuxxΔx = cuxxΔx(r − 1) 1 2 1 2 = 0 if r = 1 r = cΔt Δx First order if r = 1 Courant number 1 � Downwind: Analogous: first order Centered: ut + 1 uttΔt − cux − 1 cuxxxΔx2 + O(Δt2) + O(Δx4) 6 2 Δt → 2 Δx First order in time → Second order in space Stability: Upwind: G − 1 Δt...
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Uj n n + Uj −1 + θ (Δx)2 n + λ � n 2 Uj −1 = c where λ = 2 Δt (Δx)2 θ How much diffusion? Lax-Friedrichs: Eliminate Uj n by λ = 1 θ = ⇒ (Δx)2 2Δt U n+1 j = 1 + r 2 � �� � n +1 + Uj 1 − r 2 � �� ≥0(for |r|≤1) � n Uj −1 ≥0(for |r|≤1) Monotone scheme n n +1 − Uj Uj −1 2Δx r = cΔt Δx Accuracy: ...
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uxx− Δt c2uxxxxΔx2 1 utttΔt2 − 6 = 6 ut − cux = 0 Δt 2 1 uttΔt − 2 2 Stability: λ = r2 c uxx = 0 − 2 24 G = r2 + r 2 e ikΔx + (1 − r 2) + r2 − r 2 e−ik(Δx) = (1 − r 2) + r 2 cos(kΔx) + ir sin(kΔx) Worst case: kΔx = π ⇒ G = 1 − 2r2 Stable if |r| ≤ 1 Image by MIT OpenCourseWare. 4 MIT OpenCourseWare http://...
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1 I-campus pro ject School-wide Program on Fluid Mechanics Modules on Waves in fuids T. R. Akylas & C. C. Mei CHAPTER SEVEN INTERNAL WAVES IN A STRATIFIED FLUID 1 Introduction. The atmosphere and o c e a n are continuously stratifed due to change in temperature, composition and pressure. These changes in the ocean and ...
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denoted as u, v and w . The fuid particle has to satisfy the continuity equation 1 Dp @u @ v @w + + + 0 (2.1) p Dt @x @ y @ z and the momentum equations ...
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q . Then for fxed e and q (or s), p is : independent of pressure p p(e , q ): (2.5) The motion that takes place is assumed to be isentropic and without change of phase, so that e and q are constant for a material element. Therefore Dp @ p De @ p Dq + : (2.6) 0 Dt @ e Dt @ q Dt In other words, p is constant for a mater...
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p p , (2.9) @ t @x ; @ v @ p p , (2.10) @ t @ y ; @w @ p p gp: (2.11) @ t @ z ; ; Next, we consider that the wave motion results from the perturbation of a state of equilibrium, which is the state of rest. So the distribution of density and pressure is the hydrostatic equilibrium distribution given by When the motion d...
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that the density perturbation at a point is generated by a v ertical advection of the background density distribution. The continuity equation (2.7) for incompressible fuid stays the same, but the momentum equations (2.9) to (2.11) assume the form ...
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, respectively, of the equations (2.17) to (2.19), and we obtain 2 2 0 @ u @ p pp , (2.21) @x@ t @x ; 2 2 2 0 @ v @ p pp , (2.22) @ y@ t @ y ; 2 2 2 0 0 @ w @ p @ p pp g : (2.23) 2 @ t @ t@ z @ t ; ; If we substitute equations (2.21) and (2.22) into equation (2.20), we obtain 2 0 2 0 2 1 @ p @ p @ w ; pp @x @ y @ t@ z ...
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Next, we use equation (2.24) to eliminate p from equation (2.26), which gives the 0 following partial diferential equation for w : 5 2 2 2 2 2 @ @ w @ w 1 @ @w @ w @ w 2 2 2 2 2 2 @ t @x @ y pp @ z @ z @x @ y + + pp + N + , (2.27) 0 where we defne 2 g @ pp N (z ) , (2.28) ; p @ z which has the units of frequency (rad...
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z @x @ y + + + N + : (2.30) 0 The assumption above is equivalent to the Boussinesq approximation, which applies when the motion has vertical scale small compared with the scale of the background density. It consists in taking the density to b e constant in computing rates of change of momentum from accelerations, but ...
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6 g [ pp(z + ( ) pp(z )] g ( , (3.31) d pp ; � dz and it is negative. Applying Newton's law t o the fuid parcel of unit volume, we have 2 @ ( d pp pp g ( (3.32) 2 @ t dz or where 2 @ ( 2 @ t 2 + N ( , (3.33) 0 2 g d pp N (z ) , (3.34) ; pp dz which is called the buoyancy frequency or the Brunt V iasialia frequency. T...
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is the frequency. In order for (4.35) to satisfy the governing equation (2.30) for the vertical perturbation velocity, ! and k must be related by the dispersion ;! relation ...
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37) ;! j j l k cos(<) sin (e) (4.38) ;! j j m k sin(<) (4.39) ;! j j The coordinate system in the wavenumb e r space is given in the fgure 1. The dispersion relation given by equation (4.36) reduces to 2 ! N cos(<): (4.40) Now w e can write expressions for the quantities p , p , u and v . From equation (2.20) we 0 0 ca...
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8 m 2 2 2 1/2 (k + l + m ) φ 2 (k 2 + l ) 1/2 θ m l k Figure 1: Coordinate system in the wavenumb e r space. 9 0 2 N p p w sin(kx + ly + mz ! t ): (4.42) 0 0 ; !g ; The horizontal velocity components can b e found from equations (2.17) and (2.18), which give (u, v ) (k , l )(k + l )...
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along wave crests, and there is no pressure gradient in this direction. The restoring force on a particle is therefore due solely to the component g cos < of gravity in the direction of motion. The restoring force is also proportional to the component of density change in this direction, which is cos < per unit displac...
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an important form of motion that is often observed. For instance, it is quite common on airplane journeys to see thick layers of cloud that are remarkably fat and extensive. Each cloud layer is moving in its own horizontal plane, but diferent layers are moving relative t o e a c h other. ...
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y a n t F l u i s i n g B u o y y d a a n c y n c y G r o u p V e l o c i t y c it y e l o e V s a h P H i g h P r e s s u r e ( u , w , P = 0 ) L o w ( u , w , P P H i g h r e s s u r e = 0 ) P r e s s u r e X - axis Figure 2: The instantaneous distribution of velocity, pressure, and buoyancy perturba- tions in an i...
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11 4.1 Dispersion Efects. In practice, internal gravity waves never have the form of the exact plane wave given by equation (4.35), so it is necessary to consider superposition of such waves. As a consequence, dispersion efects become evident, since waves with diferent frequencies have diferent phase and group velociti...
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C sin <(sin < cos e , sin < sin e , cos <): (4.46) g Therefore, the magnitude of the group velocity is ( ) sin <, and its direction is at j k j k ; ! j j ; N ; ! an angle < to the vertical, as illustrated in the fgure 3. To illustrate the efects of dispersion, we consider the case of two dimensional mo- tions. We consi...
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12 m θ 2 2 2 1/2 (k + l + m ) C g φ φ 2 (k 2 + l ) 1/2 m l k Figure 3: Wavenumb e r vector and group velocity vector. 13 wave packet. The phase velocity is given by the equation (4.45), where the wavenumb e r vector k makes an angle < with the horizontal direction (see fgure 1, but...
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ed is 1 1 u^(k , m ) dx dz exp( ikx imz )u(x, z ) (4.49) ;1 ;1 f ; ; g Z Z and 1 1 1 u(x, z ) dk dm exp( ikx imz ) u^(k , m ) : (4.50) 2 4 ;1 ;1 f ; ; g Z Z The Fourier transform of the equation (2.30) is given by the equation 2 2 2 @ w^ N k + w^ , (4.51) 0 2 2 2 @ t k + m which has solution of the form w^(k , m, t)...
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0 ^ 2 @ p p N 0 @ t g w ^ : (4.54) 0 ^ 2 p N 0 p (k , m, t) iA(k , m ) ex p ( i! t) + iB (k , m ) exp(i! t) , (4.55) g ! (k , m ) f; g where the constants A and B are determined from the Fourier transform of the initial conditions for p , given by the equations 0 0 p (x, z , 0) f (x, z ), (4.56) 0 @ p @ t (x, z , 0) 0...
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by the equations ^ f (x, z ) exp x < z T cos(k x + mz~ ), (4.61) 1 1 1 2 2 2 2 ~ 2 2 2 ; ; 1 1 (k k ) 1 (m m~ ) 1 (k + k ) 1 (m + m~ ) 2 2 2 2 ~ ~ ^ f (k , m ) exp + exp : ; ; 2 2 2 2 2<T 2 < 2 T 2 < 2 T ; ; ; ; ( � ! � !) (4.62) ...
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2 ~ � parts as time increases. This two parts propagate in opposite directions from each other. Since the x and z components of the main wavenumb e r are equal and positive and the wave p a c ket has the same modulation along the x and z directions (< T ), the two parts of the initial wave packet travel towards the mid...
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initial wave packet looks almost without variation in the x direction. The wave packet splits in two parts as time increases. These two parts propagate in opposite directions from each other, in a way similar to the previous example. The interference efect b e t ween these two wave packets for early times is more inten...
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16 4.2 Saint Andrew's Cross. Here we discuss the wave pattern for internal waves produced by a localized source on a sinusoidal oscillation, like an oscillating cylinder for example, in a fuid with constant density gradient (the buoyancy frequency is constant). For sinusoidal internal waves, the wave energy fux I p u (...
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k + l f ; g which is parallel to the group velocity, according to equation (4.46). Therefore, for internal waves the energy propagates in the direction of the group velocity, which is parallel to the surfaces of constant phase. This fact means that internal waves generated by a localized source could never have the fam...
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of the group velocity and by the fact that the phase velocity N cos < C cos < cos e , cos < sin e , sin < (4.65) ; ! f g k ; ! j j ...
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isotropy has been verifed in dramatic experiments by Mowbray and Stevenson. By oscillating a long cylinder at various frequencies vertically in a stratifed fuid, equal phase lines are only found along four beams forming \St. Andrew's Cross", see fgure 5 for ! /N :7 and ! /N :9. It can be verifed that the 0 0 angles a...
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state as the state of rest, fuid properties are constant on horizontal surfaces and, furthermore, the boundaries are horizontal. Solutions of the perturbation equation (2.27) can b e found in the form ...
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67) ; The equation for w^ (z ) can be found by substitution of equation (5.67) into the governing equation (2.27). We obtain 1 @ @w^ (N ! ) 2 2 2 2 pp @ z @ z ! ; 2 pp + (k + l ) w^(z ) 0 (5.68) The boundary conditions for this equation are the bottom condition of no fux across it, given by the equation a n d a t t h e...
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2 2 + (k + l ) w^(z ) 0 at z : (5.72) 0 @ z ! 2 To simplify the governing equation for w^(z ), we make the Boussinesq approximation, such that equation (5.68) simplifes to ...
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of equation (5.73) under the boundary condi- tions (5.72) and (5.69). We frst consider the case where ! > N . For this case the 2 2 general solution has the form sinh[m(z + H )] (! N ) 2 2 2 2 2 w^ (z ) with m (k + l ), (5.74) sinh(mH ) ! ; 2 which already satisfes the bottom boundary condition. The free-surface bound...
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22 2 2 N ! gm tanh(mH ): (5.77) ; For a given value of the frequency ! , this dispersion relation gives a countable set of values for the modulus of the horizontal component ( k + l ) of the wavenumber, or for 2 2 a given value of the modulus of the horizontal component of the wavenumber, we have a c o u n table set of...
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! , n , 2, 3, : : : , (5.80) 1 2 2 2 2 2 n + ( k + l )H which is close to the result given by the dispersion relation given by the free-surface boundary condition (5.77). The value of m for the case with a free-surface is slightly larger than the case with the rigid lid approximation. If the ocean is perturbed with a ...
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23 5.2 Free Waves in a semi-infnite region. The atmosphere does not have a defnite upper boundary as does the ocean, so solutions of equation (5.73) will now b e considered for the case of a semi-infnite domain z > 0. In this case there are two types of solutions, the frst b e i n g typifed by the case N constant. The ...
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solve initial-value problems, and have the from of Fourier integrals. When N varies with z , there is another type of solution possible, namely, one that satisfes the condition at the ground yet decays as z . These are waveguide modes, ! 1 and there are, in general, only a fnite numb e r possible. A simple example is p...
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and ! can assume any value b e t ween 0 and N . This is not true for the case when 2 N ! N , when the frequency ! can assume only a fnite set of values in the range 2 1 N ! N . In this case, the solution of equation (5.73) for the frst layer is given by 2 1 the equation ...
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2 2 2 (! N ) At t h e intersection z H b e t ween the two l a yers, the perturbation pressure p and the 0 vertical velocity w should be continuous. Alternatively, this condition can be expressed in terms of the ratio 0 p Z , (5.85) p w 0 which must b e the same on both sides of the boundary. It is convenient to refer t...
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discrete waveguide modes and the continuous spectrum of sinusoidal modes. The relative amplitude of the diferent modes depends on the initial state. 6 Energetics of Internal Waves. The energy equation for internal waves, under the assumption of small perturbations, incompressible and inviscid fuid and irrotational fow,...
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7), (2.18) and (2.19), respectively, b y u, v and w , b y m ultiplying equation (2.16) by g p pN / p , and then adding the result and by taking into account the equation 2 0 2 of continuity (2.7) and the defnition of the buoyancy frequency. We obtain @ 1 1 gp @ (p u) @ (p v ) @ (p w) 2 2 2 0 0 0 0 pp u + v + w + + + + ...
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H ] dxdy 2 2 1 2 1 1 2 2 2 2 Z Z Z Z Z � � 2 2 ; ; ; ; 1 1 2 2 2 2 p g [ (H h) ] + p g [(H h) H ] dxdy (6.88) 2 2 1 2 2 2 ; ; ; ; Z Z � � If we skip the constant terms in the equation above, we end up only with the potential energy associated with the energy due to the perturbation, which is equal to 1 1 2 2 p g + g ...
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pNp h dxdydz , (6.91) ; 2 @ z 2 Z Z Z Z Z Z where h is the displacement of a fuid element from its equilibrium position. Since the density of a fuid element at its p ertu rb ed level z + h is equal to the density pp(z ) at its equilibrium position, the perturbation density p is given by 0 0 @ pp p pp(z ) pp(z + h) h , ...
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0 2 E pp (u + v + w ) > + (6.94) 2 2 pN p 2 When integrated over a large volume, equation (6.87) shows that the rate of change of energy over that volume is equal to the fux of energy across the sides. since this fux is also periodic, the average over a large plane area is approximately the same as ...
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ariety o f m e c hanisms. Often the source region is approximately horizontal, so the vertical velocity component can efectively b e specifed on some horizontal surface, and the motion away from the source can b e calculated from the equations of motion. We frst consider the case in which air or water is moving with un...
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and the momentum equations 28 @ u @ u @ u @ p p + u + w , (7.98) @ t @ x @ z @ x ; � � @ w @ w @ w @ p p + u + w pg : (7.99) @ t @ x @ z @ z ; ; � � The fuid density has to satisfy the equation Now we write the horizontal velocity i n t h e form 1 D p p D t (7.100) 0 u(x, z ) U (z ) + u (x, z ), (7.101) 0 We substitut...
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equation, its linearized form is 0 @u @u @U @ p p + U + w , (7.103) @ t @x @ z @x ; � � @w @w @ p p + U pg : (7.104) @ t @x @ z ; ; � � For the density equation (7.100) we obtain 1 @ p @ p @ p @ p 0 + U + u + w (7.105) 0 p @ t @x @x @ z � � We consider the wave motion as a result from the perturbation of the state of e...
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um equation assume the form 29 0 0 @ u @ u @ U @ p pp + U + w , (7.106) @ t @ x @ z @ x ; � � @ w @ w @ p 0 0 pp + U p g , (7.107) @ t @ x @ z ; ; � � and the density equation assumes the form 1 @ p @ p @ p 0 0 0 + U + w : (7.108) 0 pp @ t @ x @ z � � As we did in section 2, we would like to reduce the system of equati...
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110) @x@ t @x @x @ z @x ; 2 2 � � 0 Third, we eliminate the u variable from the equation (7.110) above. To do so, we use equation (7.109) and the x derivative of the continuity equation (7.102). After the u 0 variable is eliminated, equation (7.110) assumes the form 2 2 2 0 @ w @ w @w @U @ p pp U + (7.111) ; ; ; @ z@ t...
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t @x@ t @x @ t@ z @x@ z @ t @x ; ; ; � � � � and with equation (7.105) we can eliminate p from equation (7.112). The result is the 0 equation 2 2 2 2 0 2 0 @ w @ w @ w @ p @ p @ pp 2 pp + 2 U + U U gw , (7.113) 2 2 @ t @x@ t @x @ t@ z @x@ z @ z ; ; ; � � Next, we apply the operator to equation (7.113). Then, we elimina...
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1 @ @w @ @ 1 @ @U @w @ w 2 2 2 2 + U + pp + U pp + N (z ) , 0 @ t @x @x pp @ z @ z @ t @x pp @ z @ z @x @x ; 2 2 � � � � (7.115) where N (z ) is the buoyancy frequency defned according to equation (2.28). If we assume that w varies with z much more rapidly than pp(z ), then we can write 1 @ @w @ w 2 pp (7.116) pp @ ...
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@ z @ t @x @ z @ z @x @x ; 2 2 2 � � � � (7.118) We can simplify this equation further by assuming that the velocity U is constant. In this case we end up with an equation of the form @ @ @ w @ w @ w 2 2 2 2 2 + U + + N (z ) , (7.119) 0 @ t @x @x @ z @x 2 2 2 � � Next, we discuss boundary conditions for the equatio...
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, a n d w e obtain the linear boundary 0 condition w(x, z , t) U on z (7.121) 0 @ ( @ x For an infnite atmosphere we need a radiation condition, which ensures that the energy fux is away from the ground. In other words, energy is radiated away from the ground by t h e i n ternal waves generated by the topography. For...
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) 0 where k is the wavenumb e r of the topography. The boundary condition (7.121) at the ground for this case assumes the form w(x, z , t) A U k cos(kx ) at z (7.123) 0 0 In this example, we assume a constant b u o yancy frequency N for the entire atmosphere. Under such condition, we can assume a solution for the stea...
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to the equation above, if > k , them m is real and we obtain waves which U N propagates through the atmosphere. If k , them m is imaginary and we have a U solution which decays exponentially away from the ground. ...
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velocity given by equation (7.124), we have 0 m p pU A p cos(kx + mz ) (7.128) k Now, we compute the average vertical energy fux, and we obtain the expression 0 2 1 m F pU A p , (7.129) z 2 k and since we need a positive vertical average energy fux for the energy to b e radiated away from the ground, we chose m as give...
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For the vertical wavenumb e r m equal to zero, we h a ve the horizontal cut-of wavenumber k , given by the equation c 34 N k : (7.132) c U This wavenumb e r divides the two types of solutions (N /U > k and N/U k), and corresponds to a wavelength 2 k equal / to the horizontal distance traveled by a fuid c particle in o...
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7.134) if ; g ...
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a horizontal constant speed U passing over a localized topography z ( (x). We assume that the buoyancy frequency N is constant. For a reference frame fxed to the ground, the governing equation (7.119) assumes the form 2 2 2 2 @ @ w @ w @ w 2 2 U + + N : (7.137) 0 2 2 2 2 @x @x @ z @x In this reference frame we will ob...
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case. The governing equation in the moving reference frame is given by equation (2.30) without the y component. In other words, 36 2 2 2 2 @ @ w @ w @ w 2 2 0 2 2 0 2 @ t @ (x ) @ z @ (x ) + + N , (7.138) 0 where x is the horizontal axis in the moving reference frame. The horizontal axis in the 0 fxed and moving frame ...
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^ 0 0 f (k) f (x) ex p ( ikx )dx (7.141) ;1 ; Z and +1 1 ^ 0 f (x) f (k) exp(ikx )dk : (7.142) 2 ;1 Z We apply the Fourier transform to the governing equation (7.138) and to the boundary condition (7.140). The governing equation (7.138) in the wavenumb e r domain assumes the form 2 2 @ @ ^w 2 2 2 2 2 @ t @ z ; ; k + ...
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(7.143) we end up with the governing equation for w^ , which follows: 0 2 0 @ w^ 2 2 2 2 2 0 ! + ( N k ! k ) w^ (7.146) 0 2 @ z ; Solutions of the equation (7.146) are in the form 0 w^ (k , z ) A exp( im(k)z ), (7.147) ; where m(k) can b e obtained in terms of k and ! by substituting equation (7.147) into equation (7....
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e wave frequency with respect to the wavenumber, then ...
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: For k 0 w e have that • 2 m(k , N /U ) , (7.152) ; ; j j : 2 1/2 U N 2 N U k if k 8 2 N N 1/2 h i i k if k > ; ) U U { : ; ; j j [ ; )J For k 0 we have that • m(k , N /U ) , (7.153) ; j j : 2 1/2 U N 2 N U k if k where we used in equation (7.148) the fact that 8 2 N N 1/2 h i i k if k > ; ) U U { : ; ; j j [ ; )J !...
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39 2 m C U , (7.155) gx ; k ; ! 2 j j which implies a negative v alue for the horizontal component of the group velocity. From this equation we also realize that the horizontal component of the group velocity is in the same direction as the horizontal component of the phase speed, but smaller magnitude. This implies th...
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