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by: Kylie Bartoletti DVM
Kylie Bartoletti DVM

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This 0 page Class Notes was uploaded by Kylie Bartoletti DVM on Monday November 2, 2015. The Class Notes belongs to PHYS 4421 at Georgia Institute of Technology - Main Campus taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/234285/phys-4421-georgia-institute-of-technology-main-campus in Physics 2 at Georgia Institute of Technology - Main Campus.


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Date Created: 11/02/15
Chapter 15 Waves lExercise 153 Ship Waves a Consider a plane wave with frequency 000 and wave vector go as measured in the water s frame and a and 7 as measured in the boat s frame For an observer with position in the water s frame and 73 in the boat s frame the phase he measures is er 0 TO39 a0t in terms of water s frame variables and k Y not in terms of boat s frame variables Since the phase is invariant under the change of reference frame we can equate the above two expressions and then differentiate both sides with respect to t noting that for an observer moving together with the boat ix Jdt 7 dYdt 0 we get a 000 E 7 By looking at Fig 153 and use 7 to denote the wave vector as measured in the water s frameas the text does we get a 000 ukcos 5 b 0 is the angle between V go 7 and 7 elementary trigonometry then givessee the sketch on the right tan0 Vgo sin u Vgo cos For stationary wave pattern a 0 using the ak we got in part a we see that 000k ukcos c For capillary waves 000 m 1 7pk3 Vgo BangBk 32 ypk Plugging these and u aokcos into the expression for tan 0 we get tan0 3 tan l 2tan2 Capillary wave pattern for a given 0 exists only when we can find some 5 E Z 7r ie only forward waves can contribute to the pattern satisfying the above equation And it s easy to show that indeed for any 0 we can nd such a 5 given by tan 3 J9 8tan20 4tan0 when 0 lt 7r2 and tan 3 J9 8tan20 4tan0 when 0 gt 7r2 For gravity waves 000 m Jge k and Vgo 12 W and we get tan0 tan l 2tan2 Only when 0 lt arcsinl3 can we find some 5 E 7I2 7r satisfying this equation which is tan l i 41 8tan20 4tan0 both solutions are valid This means that the gravitywave pattern is con ned to a trailing wedge with an opening angle 0W 2arcsinl3 QED 2Tsunamis from Japan We can treat this as a 2dimensional problem ie only consider the horizontal components of the velocity which are almost independent of Z In what follows V is the 2dimensional derivative operator a The mass per unit area is pD g and the mass ux per unit length is pD g m pDV to the first order in perturbation Then by mass conservation BpD 5Bt V pDV 0 2 agat V D7 0 i assuming constant p The Navier Stokes equation in this case is WBt VP p g whose vertical component tells us P pgc Z and whose horizontal components then tell us JV8t ch ii Applying at to both sides of eqn i and then plugging in eqn ii we get 925th gV DVg b By plugging in a plane wave solution to the wave equation we nd the dispersion relation 9 r a ngTJr iVDD kkz m ngTn i2VDD kk2 The imaginary part of a only affects the decayingor growing of the the wave amplitude but not it s propagation direction and furthermore it s smaller than the real part by a factor of 11 ltlt 1 where l is the wave length and l is the scale over which D varies Thus we take a m klgj Using the Hamilton s equations ofr notion introduced in Chapter 6 Geometrical Optics we get ii Yak V7coa kk and dkdt Vya k21g DVYD Thus we see the direction of wave propagation is always de ected towards the shollower part of the ocean c Create in the bottom of the Paci c Ocean a ridge going from Japan to LAwith equidepth contours being elliptical curves and LA being at the focus This ridge will act as a lens focusing those Tsunamis towards LA Note that only very slight de ectionver slight difference in ocean depth is suf cient Assume Japan extends 500km long and the distance between Japan and LA is about 10000km Then only about 500km10000km 5 change in the ocean depth is needed QED 3Solitary Waves in a Deformable Conduit a Assuming that a vary slowly with height the solution for this part can be found in Section 1245 Blood Flow BampT The only difference between there and here is that here we no longer neglect gravity By adding a plg term to the driving force we get see eqn1272 Poiseuille s Law BampT Q 7W43P132 Plg8771 b The force per unit area on the conduit wall applied by uid2 consists of two parts the pressure contribution p2 and the viscous contribution fmmm Since uid2 has no vertical motion we have p2 constant p2gz Now let s calculate fmwm in cylindrical coordinates The velocity field due to the change in a is T v with v arBaBt fmwm Tquot 2n20 2n2238vBr l3vr 2n2ar2BaBt where we have used formulas analogous to those in Box 102 BampT fmwmh a 2n2laBaBt Thus the force per unit area applied by uid2 on the wall is constant ngz 2n2laBaBt and similarly the force per unit area applied by uidl on the wall is p1 2n1laBaBt Equating these two forces and noting that 172 gtgt 171 we get p1 p2gz 2n2laBaBt constant Thus the PDE relating Q and a is Q a48n1p2g plg a2n2agtaaargtazgt The other PDE relating Q and a is given by mass conservationie volume conservation assuming constant p for uidl as follows 27ra2dz Q21 Q22 2 27raBaBt aQaz 21 c 27raBaBt 87Ia2Bt BQBZ let Q Q0 Q1fZ at then az Q10fz ct where we ve set the additive constant to zero without loss of generality Chapter 22 Nonlinear Dynamics of Plasmas 1Ex222 Particle Energy in a Wave original solution by Chris Hirata dskdt ddt j12mv2Fovdv j12mv23FOVatdv I12mev2BDVBF0vBvBv VBFoBzdv by equation 2216 The second term in is odd in v making its contribution to the integral vanish After integrating by part we get risk1t JDvaFovBvmevdv Now look at the exporession 2217 for Dv Dv ezeomg 0 61mm For the nonresonant piece of Dv kv ltlt 00 so we can replace the denominator of the integrand with a m 00 nezeomeThen we get Dv 1nme I dkskmi which is actually vindependent Plugging the above expression into equation and integrate by part we get risk1t I DmeFovdv Dmen I dkskmi Writing this as dskdt 12 I deaJisk 12 I deskdt we see this implies half the wave energy is kinetic 2Ex 224Cerenkov Powerby Alexander Putilin The emission rate of plasmons is given by 2237 9 W ezmyeokzh w k 7239 Each plasmon has energy hwy so the radiated power per unit time is gt P 12703 d3kWhm 8287r260 Id3ka3k26a k v The integration is over the region lt kmaxoutside this region waves are strongly Landau damped A good estimate of kmax is inverse Debye length kmaXN12D see the discussion at the end of Chapter 2135 Since kip lt 1 we can approximate ak by a constant mpg Choosing T to point along Zaxis we have 2 2 2 3 2 P e mpg87 empldmd k1k 5a k v ezmgesnzeo Wm dszdkz1ki k 5m k 4 kiltkr2m4m EV2gtdzk1lkf cageV5 2 2 2 ernaX quot EV2gt 2 2 2 e mpg87 60v 0 anrdkiki mpgv 82mge47reovlnkmaxvape Note that P depends on kmax logarithmically So if v is sufficiently large it doesn t make much difference what particular definition we use for km ezm eS zeov 3 Nonlinear Excitation of Particle Motion original solution by Chris Hirata Power expand 7 W W W higher order terms where W is of order Equot The equation ofmotion is JFKalt2 emEcosE E 0011 Ecosk Y 0021


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