Analytic Mechanics PHYSICS 105B
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Five easy assignments for an Advanced Honors Seminar on analytic mechanics Spring 2005 Chun Wa Wong Department of Physics and Astronomy University of California Los Angeles CA 900954517 Copyright 2005 By Chun Wa Wong Seminar Title Physics 189 Honors seminar for Physics 105B Lecture 1 Seminar Description Advanced problem solving in analytic mechanics 1 unit PNP or letter grading Seminar Program of Study In this Honors Semi nar we developed problem solving skills in undergradu ate analytic mechanics Five topics were covered H A taste of geodesy Why is the Earth oblate to The art of estimation Why are we bigger than eas CA3 Lost horizon On a clear day you cannot see for ever he Probabilities Finding another Einstein o1 lnitial conditions Toppling pen falling toast and other issues in classical measurements The following notes are the corrected version of the handouts distributed to the class about a week before the discussion 1 gave in the class some missing steps in the handout and a sketch of the solution to the problems Assignment 1 A taste of geodesy Geodesy is the science of measurement or mapping of the properties of the Earth7s surface A good example of an interesting problem in geodesy is the gravitation acceleration g on earth as a function of the latitude 7r2 7 9 The angle 9 in spherical coordinates is sometimes called the co latitude angle For background also read Marion 107 and 1020 Newton already suggested in Piincipia 1687 that the Earth because of its rotation about its own axis is roughly an oblate spheroid with the polar radius RP 12 less than the equatorial radius Re a He therefore proposed that g m 981 fsin2 A ye1 160082 9 1 where f gt 0 Electronic address cwonnghysicsuc1a edu Problem 1 Show that on the surface of an oblate spheroid 1 1 mg12fcos29 2 where AR a 7 b 7 3 f Re a lt gt is called the attening parameter The Earth has an equatorial radius of 63784 km a polar radius of 63569 km and a mean radius of 63712 km radius of the sphere with the same volume as the actual volume of the Earth lts attening parameter is f m 1298 The Earth is also rotating about its own axis with angular velocity w The resulting centrifugal force contributes a term m to f Problem 2 Show that the centrifugal force due to the Earth7s spin about its own axis contributes a term of the form mge cos2 9 to Eq1 Show that w2a 1 This parameter m is a dimensionless number In geodesy it is de ned to be exactly 7 GM lt5 where G is the gravitational constant and M is the mass of the Earth Why is the Earth oblate If the Earth is a rotating liquid drop one can show that g on its surface has the form 1 with f m 52m7 f m 32m This simple expression reproduces the known values of g to within about 6 This neat result was rst found by Clairaut in 1743 in his book Theorie de la Figure de la Terre It is the background against which gravitational anomalies stand out To nd an almost complete derivation of Clairaut7s result we can start with Newton7s suggestion for the to tal potential energy on the surface of a spinning oblate spheroidal Earth at a distance r 7 19 from its cen ter T 19 7 ii l hp cos 19 7 cos2 1 6 where P2 cos 9 3 cos2 9 71 7 gives a convenient way of expressing Newton7s latitude dependence and w is the angular velocity of the Earth7s rotation The rst term describes the contribution of a spherical Earth of mass M while the third term is the centrifugal potential The justi cation of the second term will be given below in the Appendix on Legendre polynomials where we can derive some 80 of it leaving a loose end as an extra credit problemi It contains an empirical coef cient J2 called the dynamical form factor or the seconddegree zonal harmonic coef cient Problem 3 Consider the Earth as a spinning drop of liquid with an equipotential for its surface Use the equality 39i39p e of the polar and equatorial potentials to show that 1 J2 germ lt8 Hence show that Ag 917798 9 5 77m2 7711771171 9 98 98 f 2 2 2 f Appendix Legendre polynomials The function P2 used in Eq 9 is the Legendre polynomial of second degree The Legendre polynomials 1310086 appear in potential theory and more generally in the Taylor expansion of a function of cos 9 in powers of cos19 that have been suitably orthogonalizedi They can be used to solve a variety of problems involving 19 depen dence in the most economic way In this Appendix we give a short introduction to these polynomials and use them to a justify Eq16 b study what J2 should be for a rotating liquid drop of uniform mass density and c calculate Ag for our Earth with its original oblate shape if it is suddenly stopped from rotating 1 should add that a stopped Earth is unstable and is much more prone to earthquakes that will try to change its shape back to the sphericali This will of course take time but 1 would not want to live on such an unstable earth and de nitely not on Sumatra in Indonesia or near the North or South Polesi It is very likely that such a stopped earth will sooner or later become sphericali Let us begin by considering a point charge located at r T i on the zaxis The electrostatic ES potential on the zaxis at r T2 is proportional to 1 1 1pm 1PM 1 1 mzt7 10gt l0 where the Taylor series is convergent if t T T lt 11 If T gt 7 a convergent Taylor series must be expanded in a different way 1 1 1pm 1PM 1 1 WWZ 11gt l0 where t TT lt 11 Both cases are included by express ing the results as 1 00 7 2 7 7 12 Pr lt gt 7 lt 7 gt is the lesser greater of the two variables 7 and TA Thus the Coulomb potential everywhere due to a point charge located at r on the zaxis is proportional to the function Z 1 0 7 W Z p cosg 13 l0 gt where Plcos19 1 1 This expression can be simpli ed further by using the fact that Rlr7r lrgt172xtt2 14 where z cos19 and t 7 lt7 gt1 The function 7 1 7 Do CI7t W 7 gtlpdr 15 is called the generating function for the Legendre poly nomialsi It is relatively easy to use the generating function for P to derive the properties of Legendre polynomialsi Per haps the most interesting is the fact that these polyno mials are orthogonal in the sense that they satisfy the orthogonality condition 2 2l1 A PlzPlzdz 5m 16 We shall prove this as an example of how generating func tions can be used to nd properties wholesale leaving the explicit demonstration of a number of other properties to a problemi The orthogonality integral involves two Legendre poly nomials of arbitrary degrees Hence we shall need two generating functions Consider the integral of its square 1 1 d1 2 7 1Gztdz 7711721tt2 1 2 2 g 1nl1t lilnl1 7t l 2 2721 17 H 2z1 where we have expanded the result in a Taylor series Each generating function in the original integral can be expanded in its archtypical Legendre series to give 1 G2ztdz EM 1 PlzPlIdx 1s 1 ll 1 The two distinct in nite series obtained are equal to each other only if the orthogonality relation 16 holds Problem 4 Rodrigues7s formula Given the generating function Gzt Erma 19 n0 for a set of functions Fnzn 01oo show that l B Gzt FM nl dt 20 Use this result to verify that the rst three Legendre polynomials are Poltzgt1 P1ltzgtz7 P2ltzgt lt3z271gt 21gt Problem 5 Gravitational acceleration due to an equatorial bulge Consider a nonrotating object solid or liquid of uni form mass density with the oblate shape of our rotating Earth The polar sphere sphere of radius RP gives the same 9 everywhere on its surface 7 R We are inter ested in the change Ag at the pole and the equator under various circumstances a For the polar sphere show that A90 I Dye 72f b For the equatorial sphere polar sphere shell of thickness AR show that A9Rp9e 07 AgRege f the sum of 72f from part a and 3f from the mass shell R17 is now underground c Show that the equatorial bulge causes the changes AT AR17 cos2 9 AMoblate 23AMshelh AMprolate 13AMshe11 A9Rp9e 45f7 AgltRegtge 35 117 251 I shall explain in class how to nd the last equa tion It cannot be obtained exactly using Legendre polynomials alone You need to use the proper ties of more general functions called spherical har monics The last two equations show that a non rotating solid oblate earth of uniform density has Agye f5 d Show that the fourth equation of part c gives the value J2 Hence a rotating liquid of uni form density will give the resu ts f 54m7 Ayge 54m The observed result Agge m 32m means that J2 lt Use the actual value of Agge to nd an actual value of J2 for our Earth The density of the Earth is known to increase towards its center Does this explain why J2 lt 25f If you are interested in explaining this result more quantitatively you might need the following data The actual density increases from around 26 gcc in the crust to about 17 gcc at the very center of the Earth The mean mass density of the Earth is 552 gcc This coef cient J2 has to do with the second moment of the mass density distribution as compared to for a polar bulge Extra credit problem Use spherical harmonics you will need their orhtogo nality properties and the addition theorem or any other method to show that the equatorial bulge in an oblate spheroid of uniform density has the form given by the second term of Eq 6 with J2 25 Note that you may use b RP instead of a R8 in this term The equatorial radius a appears more frequently than the po lar radius 12 in geodesy because I is derived from a and another observed quantity and therefore has greater un certainty References Stacey 1 has very concise descriptions on many inter esting topics in earth physics Blakely 2 gives a more leisurely account of potential theory for geophysicists Torge 3 covers the subject of geodesy with Teutonic thoroughness These are books for graduate students but the advanced undergraduate can learn much from the introductory sections when suf ciently motivated 1 FD Sluey NW 0 m2 Em z onok eld Press Ens bane Australia 1992 Syd ed 2 R Blakely Pmmm mm m mamty am WWW amlzmtwm Cambridge Unwexsny Press Cambridge 1995 Sr 2 a W Toxge Ggadgsy de Gmym Berlin 2001 m1 ed Assignment 2 The art of estimation Radiated power of an accelerating charge EM waves like other waves must be powered by sources ile the driving terms in the inhomogeneous wave equations ltV2 7 E rt driving terms 22 It is obvious that chargefree systems do not emit EM waves because they have no EM elds Stationary charges too do not radiate because their electrostatic elds are timeindependent Even a charge moving with constant velocity cannot radiate because one can nd an inertial frame a Galilean frame for nonrelativistic parti cles and a Lorentz frame for relativistic particles where it is at rest Radiation is possible only if the charge is accelerating with a nonzero acceleration a waves carry energy the energy flux being uv where u is the energy density of the EM elds and V is the velocity of the wave The EM elds E and B of an EM wave serve the same mathematical functionality as the transverse displacement of the wave on a violin string This means that the EM energy density is quadratic in the elds The radiated power P is thus also quadratic in the EM elds Now the Maxwell equations are linear in both the elds and in the charges and currents producing them This is a consequence of the unspoken fact that the EM inter action is weak and does not give rise to nonlinear effects at the classical level Hence the radiated power must be proportional to q a where q is the charge whose ac celeration powers the EM radiation Thus positive and negative charges and positive and negative accelerations all radiate at the same rate These simple arguments based on symmetry and the linearity of the Maxwell equations thus show that q2a2 47m7 PA 23 where the squared charge has been converted effectively to the Gaussian units commonly used in atomic nuclear and particle physics The unknown proportionality con stant denoted A can only be a universal constant Problem 1 Use dimensional analysis to determine the dimension of the proportionality constant A Hence determine what it is in terms of a known physical con stant Any additional dimensionless proportionality constant cannot be determined from dimensional analysis The missing constant turns out to be 23 a result rst cal culated from electrodynamics by Larmor in 1895 Hence the formula we have found is called the Larmor radiation formula Why are we bigger than eas There were are and will be animals living on conti nents of earthlike planets around stars like our Sun in the Universe Suppose these animals breathe oxygen containing air with little hydrogen or helium gases in it even though hydrogen and helium are the most abundant elements in the Universe Other abundant elements in the Universe are nitrogen and oxygen because their atomic nuclei are the most sta ble ones after those of hydrogen and helium It is because of this that water H20 is so common in the Universe occurring not only on planets but also in many mete orites and comets We are interested in animals whose life depends on waterbased biochemistry Animals differ from plants in that they can move around rapidly They can also survive repeated falls from their heights Problem 2 Use the information given above and ad ditional information given below to answer the following questions a Estimate the radius and mass of the planets in units of the Earth radius and mass on which these animals are likely to be found in the past present or future b Estimate the size and mass of these animals them selves You may use the following physical constants Thermal energy at room temperature Troom 300 K kBTmom m 140 eV Binding or ionization energy of the electron in the hy drogen atom l Rydberg 14 eV The binding energy of the last least bound electron is m 01 Ry in large atoms 001 Ry in large molecules and 0003 Ry in biomolecules The mass of the hydrogen atom is 940 MeVCQ The speci c gravity of the Earth is 552 Hints According to Press Ami J Phys 488 597 1980 and available online free from UCLA computers at the web site of American Journal of Physics you need a planet small enough to allow H and He gases to escape at appropriate thermal temperatures in order to produce an oxygenrich atmosphere For part b assume that the injury suffered by animals falling from their height hence their size is proportional to their surface volume N biomolecules thick To hurt the animal you need to give him enough energy to ionize the least bound electrons from these surface biomolecules Assignment 3 Lost horizon On a clear day you cannot see forever Climb the tall mast of a ship in midocean and look around you You will see at the distance a circular bound ary between the wine dark sea and the bright blue sky This is the horizon Everybody knows that the higher up the mast you go the more the horizon recedes and the farther you can see A famous poet in ancient China had written with great poetic licence To see beyond a thousand miles Go up yet another floor What is not common knowledge is that the dark sea will become brighter as the horizon recedes until it be comes as bright as the day skyl Then the boundary that is the horizon will be lost to sight In this assignment we want to describe the physics of this lost horizon in simple termsll Problem 1 Observerls elevation Suppose the horizon becomes invisible beyond a dis tance of 300 kml How high above the Earth7s surface must you climb before you can verify the phenomenon How do you get to that elevation Lost horizon physics There are a number of amusing features in the physics of the lost horizonl An earth without any atmosphere will have a totally dark sky Horizons on this planet will stand out clear and stark because of re ected light from distant hills when the planet is illuminated by the Sun behind the observer Who can forget the picture of our lonely blue planet against the darkness of space brought back by our astronauts The situation on Earth is different It is the sky that is blue and bright while the sea or earth below the horizon is much darker So the rst issue is to understand Why the day sky is bright The day sky is bright because it re ects airtight ilel sunlight scattered by gas molecules in the atmosphere Suppose the differential scattering cross section dadQ of light by a gas molecule is spherically symmertic ilel independent of the angles 9 and The total cross section ist en da do a Ed 47r l 24 Let L0 be the radiance power per unit area per stera dian of the Sun meaning that the total solar power per unit area on Earth is L095 where 95 is the solid angle subtended by the Sun at the Earth7s locationl lnciden tally the product L095 is called the solar constant or total solar irradiancel Its value of 1370 Wm 2 is the starting point of the solar power industry The radiance from each air molecule the molecualr radiance is then Lmol Lonsjig L001 25 If the atmosphere has n air molecules per unit vol ume a thickness dz of the atmosphere contains ndz gas molecules er unit area This is the number of target particles that scatter the sunlight penetrating a distance dz into the atmosphere Hence this thickness of the at mosphere provide a source of airlight AL radiance of dLaLm Do dz 26 where L 7 CL C 7 7 27 oo 7 07 7 47 7 7 07 All the physics is contained in the parameter 6 called the scattering coefficient The total scattering cross section a of light by an air molecule is known from Rayleigh7s law to be proportional to A 4 where A is the wavelength of light From the known proportionality con stant in Rayleigh7s law and the density of air at sea level at standard temperature and pressure the sealevel value of can readily be computed It is 116 X 109 0 7A4 km71 28 when A is given in nml Problem 2 Use Rayleigh7s law to explain why the sky is blue but sunsets are red Extra credit problem Find the formula for Rayleigh7s law and verify the numerical proportionality constant given in the last equation The human eye can see light in the optical window where A is between 385 nm and 760 nml lts greatest sensitivity appears at Abest m 560 nm which is yellow ish greenl Outside the optical window light is absorbed very strongly by water in the eye through which light must pass to get the photosensitive cells at the inner back surface of our eyeballsl Problem 3 a Calculate the numerical values of G and Bo Abest b Let Lz be the AL radiance emitted by a certain atmospheric source at distance z from an observer Show that this AL radiance is diminished by the amount dLz i Lzdz 29 after it has traversed a distance dz on its way to the observer along the line of sight Note that the dz in this expression is the thickness of the intervening atmosphere not the thickness of the source even though we use the same symbol dz for it Hence show that the AL on arrival is exponentially atten uated Lz L0e m 30 where L0 is the radiance of the atmospheric source We have now reverted back to a more gen eral value for We are now ready to calculate the radiance of the day sky due to scattered sunlight L Lm 5e43de Loo 31 0 where L00 is a fraction C m 5 X 10 6 of the solar radiance Loi What can we see The human eye can only see light in its narrow optical windowi This means that the perceived brightness of an object can be de ned as B KLd 32 where K is the luminous ef ciency of the eye at wave length A K is nonzero only in the optical windowi To distinguish two objects of dfferent brightness B1 and B2 lt B1 their contrast 731732 0 B1 33 must exceed a certain value 6 Experience suggests that e m 002 The sea or earth just below the horizon at distance d from the observer appears darker than the day sky be cause there is no contribution to its AL radiance beyond 0 Lsea Loo 5e43de 0 L000 7 a 34 For the earth just below the horizon there is a small additional contribution due to sunlight re ected from it This can be allowed for by suitably increasing the effec tive distance d Hence the contrast between sea and sky at the horizon is K oddA CH f feKd ei oOxbes 35 Problem 4 De ne the limit of visibility of the hori zon by the condition C e m 002 Find the distance d beyond which the horizon is no longer visible using the value 60 Abe for the standard atmosphere We have describe a simple model 1 of how the hori zon disappears from sight due to the loss of contrast with distance from the day sky in a uniform standard atmo sphere The effects of nonuniform air density of air borne particles aerosols of atmospheric refraction and other effects must be added in a more realistic treatmenti 1 CF Bohren and AB Fraser Am J Phys 54 222 1986 Assignment 4 Probabilities Finding another Einstein ls there another Einstein living somewhere now on an other planet around another star in another galaxy in our big big Universe For that matter are there intelligent beings elsewhere even if none is as good a physicist as Einstein Consider rst the question How big is our Universe First imagine going up the mountain on a cloudless night You will see a bright band of light with its center in the constellation Scorpius This is the Milky Way our galaxy It contains aome 400 billion stars like our Sunl Remove the entire Milky Way from the night sky What is left is not total darkness but a very very faint galaxy shine There are probably 100 billion galaxies in the Universe more or less uniformly distributed in all directions cov ering a universe that is 14 billion light years in radius This is because the Universe was born in a Big Bang 14 billion years ago Galaxies were rst seen with good resolution in 1845 by William Parsons the third Earl of Rosse of Ireland using his own 72inch re ector telescope newly built for that purpose That telescope would remain the worlds biggest for the next 75 years By the early 1920 a big controversy raged as to whether galaxies were as close as stars in the Milky Way or island universes77 far from the Milky Way This con troversy was settled by Hubble in his 1925 paper describ ing distance measurements done with the 100inch re ector on Mt Wilsonl He used the periodluminosity relation for pulsating Cepheid variable stars in these dis tant objects to show that one of these Cepheids in the Andromeda galaxy was around a million light years ly away1 Hubble went on to classi y galaxies as spiral elliptical lenticular or irregular in shape Our Milky Way is a spiral galaxy about 100000 ly in diameter Our Sun is located around 28000 ly from the galactic center Our nearest galactic neighbor is Andromeda or M31 number 31 in the Messier Catalog at a distance of 29 million ly Our Milky Way is the most massive but second largest after Andromeda galaxy of our neighborhood galaxies called the Local Group which is made up of 3 large and 30 plus small galaxies2 By 1929 Hubble had deduced with the help of previous observations of Doppler red shift of galactic spectral lines the fact that galaxies at distance E from us were receding from us at the constant fractional rate H vZ where v is its recessional velocity along the line of sight This constant H is now called the Hubble constantl lts mea surement can be said to mark the beginning of modern observational cosmology Given this background on our Universe the answers to the questions we pose at the beginning of this note are too vast to visualize comfortably In fact it is absolutely mind boggling that there could be so many galaxies on which so many Einsteins could be living So let us cut the questions down to size and ask them only for our galaxy the Milky Way These are questions close to the heart of SETl Search for Extraterrestrial Intelligence enthusiasts who hope to detect radio signals soon from other intelligent civilizations in our galaxy After all they have been searching for these signals only for 50 years In 1961 Frank Drake then president of the SETl Insti tute and now Emeritus Professor of Astronomy and As trophysics at the University of California at Santa Cruz wrote an equation that estimates the number N of de tectable civilizations in our galaxy based on a number of relevant inputs We shall not go into the speci cs of the Drake equation because one cannot yet estimate many of the necessary inputs reliably 3 Suf ce it to say that SETl enthusiasts typically think that N is 105 or 105 but there are sensible people who believe on scienti c or religious grounds that we on Earth are quite uniquel4 In any case the number of galactic civilizations that have existed in the galactic lifetime must have been vastly larger than Ni Let us just take a stab in the dark by going to a middling position between the extremists of all stripes and just assume that there are M million such civilizations in the history of our galaxy We shall carry our fudge factor M along but when push comes to shove we shall set it to 1 How long might each of these civilizations last First look at our own civilizationl5 Agriculture was discov ered rst in the Fertile Crescent in modern lraqSyria before 8000 BC and then independently in China in the Americas and elsewhere including Southeast Asia and New Guinea So we can claim some 10000 years of farming civilization on good scienti c evidence On the religous right where science is bad and Darwin is a dirty word one could rely instead on the estimate given by Bishop James Ussher who lived in lreland in the 17th century He interpreted the Bible literally and came to the conclusion that God created the Earth in the year 4004 BC6 The human civilization could go on for thousands of years more It is likely however that it will sooner or later disappear with or without the help of a biblical apoca lypsel We could easily mess up by ourselves in nuclear warfare or in a population explosion if we have already learned the lesson that world wars bring nothing but suf feringl It is more likely however that Nature will give us a hand with climate changes leading to a deep drought from which even the mighty American empire will not survive Many past empires had thus succumbed to the cruel whim of Mother Nature and found their glorious monuments turned back to dus To make the optimists happy let us suppose that each of these civilizations lasts 100000f years A fudge factor f has been inserted here to keep the pessimists sadl Sup pose an Einstein lives on Earth once every two centuries This too is very hypothetical but at least we have had two giants of physics Newton and Einstein within the last four centuries So then there might be some 500f Einsteins in each of these civilizations Assuming all these what are the chances of our nd ing a new Einstein living now on another planet in our galaxy Probabilities We need a little bit of probability theory to proceed furtherim Consider an event with only two outcomes success or failure iiei nding or not nding a new Ein steini Let p be the probability of success per event or time period Therefore 4 1 7 p is the probability of failure per event The binomial theorem states that 10 4 1 1 Zn 7pmqn m7 36 m0 where the binomial coef cient is 37 gives the number of combinations of n objects taken m at a time irrespective of their orders The term M 7 7pmqn m 38gt is called the binomial distribution or probability of nd ing m successes and n7m failures in n independent trials The binomial distribution has an important limiting case when n 7 00 an p 7 0 in such a way that np a remains nitei Then 1 TL m7 qm 7 pap A 6711 n7ml an The resulting distribution am 7 Pm me 40 is called the Poisson distribution This is a properly nor malization distribution because 2 Pm remains 1 in the limit This result can be checked explicitly by using the Taylor expansion m e Z a i 41 The Poisson distribution gives the probability of no success as P 0 i 6 one success as P1 ae and of exactly m successes as P m i For applications to real problems the derivation given above of the Poisson distribution might have appeared too elegant to be practically usefuli That this is not the case can be appreciated after we have examined a more downtoearth derivation 8 to give us a better un derstanding of the idea of successes and failures in the binomial trials described above Suppose particles are emitted from a radioactive source at the probability rate M per second meaning that the probability of emitting one particle during a short time At is MAti Let R t be the probability that n particles are emitted during the time t Then P1At MAti The probability that no particle is emitted during the same time At must be P0At m 1 7 MAt if the emission of 2 or more particles during the short time At is neglected This will be exact in the limit At 7 0 The probability that n particles are emitted during the time t At can then be related to the probabilities at time t as Pnt At PntP0A Pn71tP1At7 42 provided that At is so short that the probability of emit ting 2 or more particles during At is negligible In this way we can derive the rstorder differential equation DE dPnt 7 Pn t At 7 Pnt dt 7 A30 At ManO Pn71tlA 43 For n 71 P71 0 because such a quantity does not exist Then 7 7MP0t P0t e mi 44 The general solution turns out to be 7 W 7 z Pnt 7 Te 1 i 45 This is to be proved in the following problemi Problem 1 The Poisson distribution Pn Let Pn LN 46 Find the DE satis ed by fn Show that the solution is Mt MOt fn1t dt i 47 Hence obtain the solution 45 sequentially starting with n Note that the two Poisson distributions are related as follows Pm Pmt aM 48 So a success in nding a new Einstein in a given time period is just like the success of having a particle emitted from a radioactive source during a time perio i We are now ready to do interesting problems with probabilities Problem 2 Finding new Einsteins on Earth Assume that a new Einstein appears on Earth once every two centuries a What is the probability that in the next twelve months you will not nd a new Einstein ii nd just one new Einstein and iii nding two new Einsteins b Repeat the answers for the next 20 years the aver age time of a generation c Repeat the answers for the next 75 years the aver age human lifetime It is comforting to know that Einsteins are not as rare as one might suppose If you live to a hundred years the chances of being a contemporary with an Einstein are about 03 Why not 05 The grandfather Alexander Moszkowski of one of my colleagues even wrote the rst biography of the real Einstein What about other civilizations in our galaxy Before considering the problem let us rst calculate a few interesting properties of the Poisson distribution 40 Problem 3 Mean and variance of the Poisson distri bution Show that for the Poisson distribution 40 a ltmgt E E b 02 ltmi ltmgt2gt aA f200 mPm a7 We are nally in a position to calculate some interest ing probabilities Problem 4 The number of simultaneous civilizations in out galaxy Assume that there have been M million civilizations on different planets in different solar systems in our galaxy in the 14 billions year life of the Universe For simplic ity we shall even approximate the galaxy lifetime by the 10 universe lifetime even though it is highly unlikely that there was any life in the Universe during its rst two billion years a Find the mean number of civilizations in the galaxy for the following time periods the next 12 months ii in a generation 20 y iii in a human lifetime 75 y and iv in the average lifetime of 100 OOOf years of the civilization itself b We ourselves are already living in such a civiliza tion For M l calculate the probability P2 a that there are two other civilizations in the galaxy within the time periods speci ed in part a using f l in part iv What about new Einsteins anywhere in the galaxy The probabilities are not as hard to nd as one might imagine Problem 5 The number of living Einsteins in our galaxy Assume that there are fM billion Einsteins living in our galaxy on different planets in different solar systems in the 14 billions year life of the Universe This may seem like too many Einsteins but Bill Gates has more money in dollars than the galaxy has Einsteinsl a Find the mean number of Einsteins living in our galaxy in the following time periods the next 12 months ii in a generation 20 y iii in a human lifetime 75 y and iv in the average lifetime of 100 OOOf years of the civilization itself b Calculate the probability P2a that there are two Einsteins in our galaxy within the time periods speci ed in part a using fM l in part iv 1 httpwwwastrucaedukeelgalaxies 2 httpsedslplarizonaedumessiermoremwhtml 3 httpwwwstation1netD0uglasJ0nesdrakehtm 4 httpwwwastrobionetnewsarticle610html 5 J Haywood Atlas of world history Barnes amp Noble New York 1997 6 httpwwwreligioustolerance0rgevdatehtm 7 J Mathews and KL Walker Mathematical methods in the physical sciences Wiley New York 1983 8 ML Boas Mathematical methods of physics Benjamin New York 1970 Assignment 5 Initial conditions Events and laws of nature Eugen Wigner spoke with great perception on the laws of nature in his Nobel Lecture given on the occasion of his winning the Nobel Prize in physics in 1963 He said The surprising discovery of Newton7s age is just the clear separation of laws of nature on the one hand and initial conditions on the other The former are precise beyond any thing reasonablei We know virtually nothing about the latter In other words the laws of physics when realized with the same initial conditions for any causal physical sys tem always give the same outcome It is our inability to control all initial conditions that causes classical mea surements to give different results under the same deter ministic physical laws Consider the example of a tumbling toast that has fallen from a tablet Given an initial angular velocity w as it clears the table its subsequent behavior is to tally deterministic with alternating segments of falling distances where the buttered side is up then down then up and so on alternatinglyi Since the falling distance d 12gt2 is quadratic in the time t while its rota tional angle 9 wt increases linearly in time it is clear that there is a simple correlation between the falling dis tance and the rotational angle d 12g9w21 Hence the segment lengths scale in a deterministic way as 1w2 for different w values Furthermore the angular velocity w of the tumbling toast can be determined from energy conservation by equating the rotational kinetic energy to the change in the gravitational potential energy as the toast leaves the table after clipping through an angle of a without slip ping lf there is slipping the resulting kinetic energy of translation must also be included One thus expects that 0J2 is proportional to zsina where z is the initial overhang of the center of mass of the toast from the edge of the table In other words the scale of the segments can be expected to be inversely proportional to the over hang z A detailed analysis of the tumbling toast can be found in Ref 2 Of course when a buttered toast is accidently pushed off a tea table its overhang from the tables edge can take on a range of values As a result its angular veloc ity when it leaves the table is correspondingly uncertaini This is why not all falling toasts will end up with the buttered side down on the sanded oori These uncon trollable changes in the initial conditions are responsible for some of the uncertainties of classical measurements It might appear that these initial conditions in classical measurements are uncontrollable only because of techno logical or nancial constraints One is then tempted to believe that when given suf cient resources such classi cal measurements can always be taken to the precision 11 allowed by the state of the art i This is indeed true of classical deterministic systems that are in some sense orderly or wellbehavedi It is not true for badly behaved or chaotic systems that show extreme sensitivity to ini tial conditions Chaotic systems can be deterministic if they always evolve in time in the same way for the same initial conditions However they behave very differently as time goes on even for very small changes in the ini tial conditions They are then said to show deterministic chaos Deterministic chaos The study of deterministic chaos began at least as early as 1890 when Poincar found that two planets gravitat ing around a star could have very complicated orbits that could be analyzed readily in phase space His work was continued by others notably the Russian Kolmogorov and the American Smalei When digital computers be came available numerical solutions of nonlinear differen tial equations were readily obtainedi They in turn stim ulated experimental studies of nonlinear dynamics espe cially the sensitivity to initial conditions and the char acterization of the resulting chaotic motion of dynamical systems poised at the brink of mayhemi Peitgen J rgens and Saupe 3 have written a very readable book on the subject A simple illustration of the sensitivity to initial condi tions can be made by standing a pen with a round top on its rounded headi Since the upright position is unstable the pen is bound to fall down but it would be hard to predict along which direction it will fall if it was origi nally poised very close to its potential maximum at the perfectly vertical position A little unseen breath of air could easily have made the pen topple in one direction instead of another This extreme sensitivity to initial conditions is often called a butterfly e cecti The term comes from the title of a talk given by EiNi Lorenz in December 1972 to AAAS American Associa tion for the Advancement of Science Does the ap of a butter y7s wings in Brazil set off a tornado in Texas The title was actually made up by two organizers of the meeting Lorenz himself had used a sea gull metaphor in 1963 when referring to the unpredictatbility of the weather one ap of a sea gull s wings would be enough to alter the course of the weather forever Reference to sensitive initial conditions has been traced back to at least 1898 when WiSi Franklin wrote in the review of a book in mechanics that the ight of a grasshopper in Montana may turn a storm aside from Philadelphia to New York 4 Catastrophy and hysteresis Let us nd a problem to do that will illustrate the difference between stable and unstable solutions of non linear differential equations NLDE Problem 1 Catastrophy and hysteresis Consider the rstorder NLDE W p7 fr7 Here b gt 0 and p are real parameters with 13 7 bzl 49 a On the 117 plane sketch the curve p that gives the position of static equilibrium of the system where z39 0 Mark the minimum position of A and its maximum position B Show between A and B as a dashed curve and its values outside this region as two solid curves Mark as C the point on one solid curve where has the same value as the point B Mark as D the point on the other solid curve where has the same value as the point A b The gure you have drawn is called a stability dia gram for the NLDEl lt separates the zp plane into two regions Above the dashsolid curve i gt 0 This means that all motion there has positive ve locity and therefore moves the system to the right with time Denote this result with a few small rightpointing arrows placed above and near the curve to denote the direction of the velocity Below the stability curve i lt 0 so that the system moves to the left Denote this result by short left pointing arrrows just below the curve Show that the system moves towards the solid parts of the curve but away from the dashed part For this reason points on the solid parts of the curve are called attractors or xed points while points on the dashed part of the curve are called repellorsl c If the system at equilibrium is made to move from a large and negative value of I by gradually increas ing the value of the parameter p determine what happens as it passes the point B Also determine what happens at A when one is moving down the right solid curve by gradually decreasing the value of p The sudden changes you will nd are called catastrophies while the historydependent behavior is called hysteresis For help on this problem read Wong 5 Sectl 414 A common example of both catastrophy and hysteresis can be found in ferromagnetism when driven by an in creasing or a decreasing external magnetic eld B or Hi In astronomy the Great Attractor in the sky is a region of space in the constellation Centaurus towards which all our nearest galaxies are moving The Great Attractor has a mass of 1016 solar massesl Bifurcation At p 0 the stability curve you have drawn in Prob lem 1 gives two stable solutions at Is ib and one unstable solution at rum The problem is studied further in the next probleml Problem 2 Pitchfork bifurcation a On the bz plane sketch the two p 0 stable solu tions Is as two solid curves and the single unstable solution rum 0 as a dashed line lndicate with short uppointing or downpointing arrows the di rection of the velocity when one is slightly off an equilibrium position b Show that for b lt 0 the equilibrium position where z39 0 is re 0 Show that the equi librium is a stable onel Draw the b lt 0 stable solutions Is 0 as a solid line in your bz plotl You have just construct a bifurcation diagram where the single stable equilibrium solution Is 0 for b lt 0 separates into two branches as b becomes positive This diagram is said to describe a pitch fork bifurcation We shall next indicate how successive bifurcations give rise to deterministic chaosl Deterministic chaos in nonlinear maps Many adult insect populations do not overlap from season to season e population rn1 in one season depends only on the population In of the last season An interesting example of this type is described by the rstorder nonlinear difference equation zn1 bznl 7 50 Since 117 1 has a maximum of 14 at z 12 it is convenient to restrict the parameter b to l lt b lt 4 so that the population value expressed in units of a certain maximum population will always remains in the interval 1 01 A mapping equation such as Eq 50 is called a nonlinear map Equation 50 itself is called a logistic or population map from the French word logis that means lod in or house and here used to refer to the population according to Ref 3 pl 45 A detailed description of this equation and the com plexities of its solutions has been given by May The complexity arises from the nonlinear term 7bzi describ ing a simple damping mechanism or environmental con straint that prevents the population from exploding Problem 3 Show that the equilibrium solutions of the logistic map are stable for l lt b lt 3 but unstable for b gt 3 For help read my book Section 414 or Thornton 7 Sectl 47 For 3 lt b lt 345 one can show by explicit calculation that the unstable solution becomes periodic with a period of two This means that the system oscillates between two populations in succeeding seasonsl Another famous example of a double period comes from the biblical ac count of the seven years of plenty followed by seven years of famine that happened in ancient Egypt This periodic ity in turn might have been caused by the periodic popu lation explosions of locusts The perioddoubling bifur cation of the logistic map is shown explicitly in the bifur cation diagrams that can be found in Wong or Thorntoni As 12 increases further the system undergoes further perioddoubling bifurcations at well de ned values of 121 This occurs more and more frequently with an accumu lation point at bu 3157 where the number of periods becomes in nite Beyond the accumulation point bu the system exhibits a behavior so complicated that periodicity is lost and the system becomes aperiodici Its time evolution is now very sensitive to the value of the initial population Two slightly different initial populations no matter how close will eventually give rise to very different population evo lutions after a suf ciently large number of seasons The system is then said to be unpredictable and chaotic even though it is still completely deterministici This is deter ministic chaosi Exponential growth in unstable systems We have heard so much about population explosions that every one of us know about the danger to the en vironment posed by the exponent growth of bacteria or people including people in Heaven As we all know ex ponential growth results when the population increase dP is proportional both to the present population Pt and to the time period dt under consideration dP MPtdt a Pt Poem 51 where M is the growth constanti Alternatively one can separate t into n equal periods of length to tni Then Pn Pofn where f e A Mtoi 52 In population growth the most interesting time period to use is the generation time ie the average birthto birth time between successive generations The danger of population explosion is often called a population bombi This is an apt metaphori The grand daddy of all destructive bombs is the atomic bomb where the ssion reaction runs away77 exponentiallyi A few details will make the situation much more graphici In a ssile nucleus such as U235 each neutron induced ssion produces 1 25 neutrons on the aver a e Each produced neutron has a geometrical nonescape probability 73 lt 1 of remaining in the bomb to cause an other ssion reaction and a capture ssion probability a of being captured by another nucleus without produc ing a ssion reactioni One can thus de ne a criticality constant w C1a 53 for a ssion bomb or reactor The growth constant can be written in the form M 54gt where To is the mean lifetime of the reaction ie the av erage time between two successive ssion reactions or the generation time The dif culty of controlling nuclear s sion comes from that fact that To is only 10 8 s compared to a generation time of 20 y for the human population As a result even when C 7 1 can be reduced to 10 4 there are 104 successive generations of ssion neutrons in just 1 secon 1 An analogous situation occurs in an unstable mechan ical systemi Here a small difference 50 in an initial con dition will cause a runaway exponential increase 5 age 55 in the difference between the two resulting motionsi Very much the same situation can occur for the general onedimensional map In1 56 If we start with a very small difference 50 then the dif ference in the next generation is 51 50f zo age 57 where f dfdzi Whether the system is stable will depend on the nature of f which is studied in our last problemi Problem 4 Lyapunov exponent a De ne An by the equation an 50 geekquot 58 Show that 1 n71 7 I A 7 n 12111110 11 59 after n generationsi Lyapunov has proved that in the limit n A 00 An takes on a wellde ned value A that is independent of the starting value 10 of the mapping This limiting value is called the Lya punov exponen i b Examine Fig 425 on p 177 of Thornton and Marion 7 that gives the Lyapunov exponent of the logistic map as a function of a b of this note Explain why A changes with at Describe the impli cations of the fact that A gt 0 for certain ranges of a values For help read Thornton Secti 4181
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