Mathematical Physics : Problems and Solutions of The Students Training Contest Olympiad in Mathematical and Theoretical Physics (May 21st - 24th, 2010)
G. S. Beloglazov, A. L. Bobrick, S. V. Chervon, B. V. Danilyuk, M. V. Dolgopolov, M. G. Ivanov, O. G. Panina, E. Yu. Petrova, I. N. Rodionova, E. N. Rykova, M. Y. Shalaginov, I. S. Tsirova, I. V. Volovich, A. P. Zubarev
aa r X i v : . [ m a t h - ph ] O c t MINISTRY OF EDUCATION AND SCIENCE OF RUSSIAN FEDERATIONSAMARA STATE UNIVERSITY
MATHEMATICAL PHYSICS
PROBLEMS AND SOLUTIONS
The Students Training Contest Olympiadin Mathematical and Theoretical Physics( on May 21st – 24th, 2010 )Special Issue № 3 of the Series«Modern Problems of Mathematical Physics»SamaraSamara University Press2010ДК 51-7+517.958ББК 22.311М 34Authors:G.S. Beloglazov, A.L. Bobrick, S.V. Chervon, B.V. Danilyuk, M.V. Dolgopolov,M.G. Ivanov, O.G. Panina, E.Yu. Petrova, I.N. Rodionova, E.N. Rykova,M.Y. Shalaginov, I.S. Tsirova, I.V. Volovich, A.P. ZubarevМ 34 Mathematical Physics : Problems and Solutions of The Students Training Con-test Olympiad in Mathematical and Theoretical Physics (May 21st – 24th, 2010) /[G.S. Beloglazov et al.]. – Ser. «Modern Problems of Mathematical Physics». – Spec.Iss. № 3. – Samara : Samara University Press, 2010. – 68 p.: il.ISBN 978-5-86465-494-1
The present issue of the series «Modern Problems in Mathematical Physics» represents theProceedings of the Students Training Contest Olympiad in Mathematical and Theoretical Physicsand includes the statements and solutions of the problems offered to the participants. The contestOlympiad was held on May 21st-24th, 2010 by Scientific Research Laboratory of Mathemati-cal Physics of Samara State University, Steklov Mathematical Institute of Russia’s Academy ofSciences, and Moscow Institute of Physics and Technology (State University) in cooperation.The subjects covered by the problems include classical mechanics, integrable nonlinear systems,probability, integral equations, PDE, quantum and particle physics, cosmology, and other areas ofmathematical and theoretical physics.The present Proceedings is intended to be used by the students of physical and mechanical-ma-thematical departments of the universities, who are interested in acquiring a deeper knowledge ofthe methods of mathematical and theoretical physics, and could be also useful for the personsinvolved in teaching mathematical and theoretical physics.
УДК 51-7+517.958ББК 22.311Editors: B.V. Danilyuk, M.V. Dolgopolov, M.G. Ivanov, I.S. Tsirova, I.V. VolovichInvited reviewers: M. N. Dubinin, Skobeltsyn Institute of Nuclear Physics ofMoscow State University, and Yu. N. Radayev, Institute for Problems in Mechan-ics of the Russian Academy of Sciences
The Olympiad and the given edition are supported by the grantsADTP № 3341, 10854 and FTP № 5163of the Ministry of Education and Science of the Russian Federation,and by Training and retrainings of specialist centerof Samara State University.
ISBN 978-5-86465-494-1 c (cid:13) Authors, 2010c (cid:13)
Samara State University, 2010c (cid:13)
Scientific Research Laboratory of Mathematical Physics, 2010c (cid:13)
Registration. Samara University Press, 2010 ontents
Introduction 4Problems and Solutions 91. VIRIAL FOR ANHARMONIC OSCILLATIONS . . . . . . . . . . . . . 92. METHOD OF SUCCESSIVE APPROXIMATIONS . . . . . . . . . . . 103. EVALUATION FOR ULTRAMETRIC DIFFUSION . . . . . . . . . . . 134. DOUBLE EFFORT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155. RANDOM WALK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166. THERMAL EVOLUTION EQUATIONS of the UNIVERSE . . . . . . . 207. ELECTRON THAT IS NOT GOING TO LEAVE . . . . . . . . . . . . 238. Х-SECTOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259. BY THE CRADLE OF LHC . . . . . . . . . . . . . . . . . . . . . . . . 2910. ‘WHIPPING Top-Toy’ from SAMARA . . . . . . . . . . . . . . . . . . 3111. LAPLACIAN ∆ SPECTRUM ON A DOUGHNUT . . . . . . . . . . . 4212. 3D DELTA FUNCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 4413. HEAT CONDUCTION EQUATION (HEAT SOURCE PRESENTS) . 4914. HEAT CONDUCTION EQUATION WITH NONLINEAR ADD-ON . 53Data on authors 57Annex: Statements of the Problems of the Second International Olympiad 58 ntroduction athematical Physics: Introduction
Modern Problems of Mathematical Physics. Special Issue № 3
PRESENCE OF AN INDIVIDUAL APPLICATION (REQUEST) from a partici-pant of The Olympiad (3 prize-winning places).3) Overall team topic scoring (1 – 3 prize-winning places on each subject).4) Separate team scoring among each of the years (second through sixth courses).5) Best team among the technical specialities of the institutes of higher education.6) Other nominations. Separate nomination is supported by the Center on Ad-vanced Training and Professional Development at Samara State University.The winners of The Olympiad held by correspondence participated in the daycompetition Olympiad held in Samara in September - 2010 (at the same time withthe Second International Conference and School on Mathematical Physics and itsApplications). For the above said winners, their travel and/or accommodation ex-penses were reimbursed.3 Contents of the problems for the Olympiad contestThe topic range of our ’Olympiad’ is related to mathematical methods in describ-ing physical phenomena based on the following units of mathematics and theoreticalphysics:theory of differential, integral equations, and boundary-value problems;theory of generalized functions, integral transform, theory of functions of complexvariable;functional analysis, operational calculus, spectral analysis;probability theory, theory of random processes;differential geometry and topology;theoretical mechanics, electrodynamics, relativity theory, quantum mechanics,and gravitation theory.New scientific methodological approach to composing the statements of the prob-lems for The Olympiad was first introduced in the sense that about a half of theproblems offered to the participants for the solution supposed that certain stage ofresearch (taken from original modern academic research in mathematical physicsand its applications) is involved. On the basis of the above mentioned approach, therecommendations on composing statements of the problems for The Olympiad aredeveloped.In the present issue, we quote the statements of problems offered to the par-ticipants of All-Russia Students Training Olympiad in Mathematical and Theoreti-cal Physics "Mathematical Physics" with International Participation (held on May21-24th, 2010).4 Results and resume of The OlympiadIn The Olympiad, the teams from the following institutes of higher educationand other organizations have participated:Belarusian State university, athematical Physics: Introduction
Modern Problems of Mathematical Physics. Special Issue № 3
Organizers of a series of the Mathematical Physics Olympiads: Alexander Andreev,George Beloglazov, Boris Danilyuk, Mikhail Dolgopolov, Vitaliy Garkin, Mikhail Ivanov,Yury Radaev, Irina Rodionova, Yury Samarsky, Irina Tsirova, Igor Volovich,Alexander Zubarev roblems and Solutions
Statements of the Problemsand Solutions at Students Training Olympiadon Mathematical & Theoretical Physics
MATHEMATICAL PHYSICS by Correspondencewith International ParticipationMay 21st – 24th, 2010
1. Virial for anharmonic oscillations
For a particle moving along the x axis with Hamiltonian H = p m + λx n , where λ is a positive constant, m is the mass of the particle, p is the momentum ofthe particle, n = 1 , , , . . . , obtain the relationship between the average valuesof kinetic h K i and potential energy h U i using two methods:(a) directly from the virial theorem (see explanation below);(b) from the condition (cid:28) ddt ( xp ) (cid:29) = 0 , which is true due to the fact that the motion of the particle is finite.Instruction: when a particle moves in a potential field, its Hamiltonian H andacting force ~F are defined by H = K + U, ~F = − grad U. Explanation to Problem 1. In classical mechanics time average values of kineticand potential energies of the systems performing finite motion are in rather simplerelationship.0
Modern Problems of Mathematical Physics. Special Issue № 3
The average value for a physical quantity G for a sufficiently large time interval τ is defined in a standard way: h G i = 1 τ τ ˆ G dt. If h K i is the average (for a rather long time interval) kinetic energy of the systemof point particles (radius-vectors of the particles given as ~r i ) subjected to forces ~F i ,then the following relation takes place: h K i = − *X i ~F i · ~r i + . ( (cid:7) ) The right hand side of equation ( (cid:7) ) is called Clausius virial, and the equation itselfexpresses the so called the virial Theorem. The proof of the theorem is given, forexample, in [1].SOLUTION(a) According to the given statement the particle is moving in the field of apotential force and possesses potential energy U ( x ) = λx n . The force equals F x = − dUdx = − λnx n − . Substitute it into the equation ( (cid:7) ) which expresses the virial theorem: h K i = − h ~F · ~r i = − h F x x i = 12 h λnx n − x i = n h λx n i = n h U i . (b) According to the given conditions, D ddt ( xp ) E = D dxdt p + x dpdt E = D pm p + xF x E = D p m + x ( − λ nx n − ) E == D p m E − n h λx n i = 2 h K i − n h U i , that is why h K i = n h U i .
2. Method of successive approximations
Solve the integral Volterra equation of 2nd kind ϕ ( x ) = α ′ ( x )1 − α ( x ) x ˆ ϕ ( t ) dt + f ( x ) , (1) athematical Physics: Problems and Solutions x ∈ [0 , h ] , f ( x ) is a given (known) continuous on [0 , h ] function, α ( x ) ∈ C [0 , h ] (continuously differentiable function), and α ( x ) = 1 , α ′ ( x ) is the derivative.Perform your solution check.Vito Volterra (3 May 1860 — 11 October1940) was an Italian mathematician andphysicist, known for his contributions tomathematical biology and integral equa-tions.SOLUTIONUsing method of successive approximations, we find the solution of the equa-tion (1) through kernel resolvent K ( x, t ) = α ′ ( x )1 − α ( x ) : ϕ ( x ) = x ˆ R ( x, t ) f ( t ) dt + f ( x ) , (2) R ( x, t ) = ∞ X n =1 K n ( x, t ) , (3) K ( x, t ) = K ( x, t ) , (4) K n ( x, t ) = x ˆ t K ( x, s ) K n − ( s, t ) ds. (5)Using formula (5) we find the repeated kernel K ( x, t ) , K ( x, t ) K ( x, t ) = x ˆ t α ′ ( x )1 − α ( x ) α ′ ( s )1 − α ( s ) ds = (6) = α ′ ( x )1 − α ( x ) [ln(1 − α ( t )) − ln(1 − α ( x ))] ,K ( x, t ) = α ′ ( x )1 − α ( x ) x ˆ t α ′ ( s )1 − α ( s ) [ln(1 − α ( t )) − ln(1 − α ( s ))] ds = (7)2 Modern Problems of Mathematical Physics. Special Issue № 3 = α ′ ( x )1 − α ( x ) (cid:20) ln (1 − α ( t ))2 − ln(1 − α ( t )) ln(1 − α ( x )) + ln (1 − α ( x ))2 (cid:21) == α ′ ( x )2!(1 − α ( x )) [ln(1 − α ( t )) − ln(1 − α ( x ))] . Similarly, using the formula (5), K ( x, t ) = α ′ ( x )3!(1 − α ( x )) [ln(1 − α ( t )) − ln(1 − α ( x ))] , (8)we come to the conclusion that K n ( x, t ) = α ′ ( x )( n − − α ( x )) ln n − (cid:18) − α ( t )1 − α ( x ) (cid:19) . (9)Expression (9) should be substituted into the formula (3): R ( x, t ) = ∞ X n =1 α ′ ( x )1 − α ( x ) ln n − (cid:16) − α ( t )1 − α ( x ) (cid:17) ( n − . (10)Performing the transformation n − m in (10) and recalling the expansion e z = ∞ X m =0 z m m ! , (11)we obtain R ( x, t ) = α ′ ( x )(1 − α ( t ))(1 − α ( x )) . (12)Substituting (12) into formula (2), we obtain ϕ ( x ) = f ( x ) + α ′ ( x )(1 − α ( x )) x ˆ f ( t )[1 − α ( t )] dt. (13)Checking. Let us show that the function (13) is the solution of eq. (1). Designate J ( x ) = ϕ ( x ) − α ′ ( x )1 − α ( x ) x ˆ ϕ ( t ) dt, (14)and substitute the function (13) into the right side of equation (14). As a result, weshall obtain J ( x ) = f ( x ) + α ′ ( x )[1 − α ( x )] x ˆ f ( t )[1 − α ( t )] dt − (15) athematical Physics: Problems and Solutions − α ′ ( x )1 − α ( x ) x ˆ f ( t ) dt − α ′ ( x )1 − α ( x ) x ˆ dt t ˆ α ′ ( t )(1 − α ( s )) f ( s ) ds [1 − α ( t )] . In the last term of formula (15) let us change the integration order and calculatethe inner integral: x ˆ s α ′ ( t ) dt [1 − α ( t )] = 11 − α ( x ) − − α ( s ) . (16)The result should be substituted into the formula (15): J ( x ) = f ( x ) + α ′ ( x )[1 − α ( x )] x ˆ f ( t )[1 − α ( t )] dt − (17) − α ′ ( x )1 − α ( x ) x ˆ f ( t ) dt − α ′ ( x )(1 − α ( x )) x ˆ (1 − α ( s )) f ( s ) ds + α ′ ( x )1 − α ( x ) x ˆ f ( s ) ds ≡ f ( x ) . Our checking has shown that the function (13) is the correct solution of theequation (1).
3. Evaluation for ultrametric diffusion
When solving equations of the ultrametric diffusion type (that have a relation tothe description of conformational dynamics of complicated systems such as biomacro-molecules) the results can often be presented in the form of series of exponents. Twoof such series are represented below: R ( t ) = ∞ X n =0 a − n e − b − n t , S ( t ) = ∞ X n =1 n k a − n e − b − n t . Here t is time, R ( t ) and S ( t ) are probabilities that a system is in some definitegroups of states, k is some integer number, a > , b > are some parameters.Study the asymptotic behavior of functions R ( t ) and S ( t ) at t → ∞ and evalu-ate, if possible, their asymptotics using elementary functions depending on t .SOLUTIONLet us explore S ( t ) and R ( t ) = S ( t ) | k =0 + e − t . Note that the function x k a − x decreases, while the function e − b − x t increases with the growth of x . Then in theinterval x − ≤ n ≤ x the inequality takes place x k a − x e − b − ( x − t ≤ n k a − n e − b − n t ≤ x − k a − ( x − e − b − x t Modern Problems of Mathematical Physics. Special Issue № 3 takes place. Integrating it with respect to x from n to n + 1 gives (for n > ): a − ˆ n +1 n x k a − ( x − e − b − ( x − t dx ≤ n k a − n e − b − n t ≤ a ˆ n +1 n x − k a − x e − b − x t dx. Now, by summing over n from to ∞ , we obtain: ˜ S min ( t ) ˜ S ( t ) ˜ S max ( t ) where ˜ S ( t ) ≡ ∞ X n k a − n e − b − n t , ˜ S min ( t ) ≡ a − ˆ ∞ x k a − ( x − e − b − ( x − t dx, ˜ S max ( t ) ≡ a ˆ ∞ x − k a − x e − b − x t dx. By switching to new variables, we have: ˜ S min ( t ) = a − (ln b ) k − (ln t ) − k t − ln a ln b ˆ b − t (cid:18) − ln ( b − y )ln t (cid:19) − k y ln a ln b − e − y dy, ˜ S max ( t ) = a (ln b ) k − (ln t ) − k t − ln a ln b ˆ b − t (cid:18) − ln ( by )ln t (cid:19) − k y ln a ln b − e − y dy. Let us designate the function γ ( k ) ( z, t, α, β ) ≡ ˆ αt (cid:18) − ln βy ln t (cid:19) − k y z − e − y dy, considering αβ < for convergence.Note that the limit for this function is the Gamma-function (see the proof in [2]): lim t → + ∞ γ ( k ) ( z, t, α, β ) = ˆ ∞ y z − e − y dy = Γ( z ) . Then at t ≫ it is possible to write ˜ S min ( t ) = a − (ln b ) k − (ln t ) − k t − ln a ln b Γ (cid:18) ln a ln b (cid:19) (1 + o ( t )) , ˜ S max ( t ) = a (ln b ) k − (ln t ) − k t − ln a ln b Γ (cid:18) ln a ln b (cid:19) (1 + o ( t )) . Notation o ( t ) means that in the limit at t → ∞ the value of o ( t ) tends to zero.Since S ( t ) = ˜ S ( t ) + a − e − b − t , the final asymptotic evaluation is of the form (ln t ) − k t − ln a ln b : a − (ln b ) k − Γ (cid:18) ln a ln b (cid:19) (ln t ) − k t − ln a ln b (1 + o ( t )) ≤ S ( t ) ≤ athematical Physics: Problems and Solutions ≤ a (ln b ) k − Γ (cid:18) ln a ln b (cid:19) (ln t ) − k t − ln a ln b (1 + o ( t )) . Because R ( t ) = ˜ S ( t ) | k =0 + e − t + a − e − b − t , we have also a − (ln b ) − Γ (cid:18) ln a ln b (cid:19) t − ln a ln b (1 + o ( t )) ≤ R ( t ) ≤ a (ln b ) − Γ (cid:18) ln a ln b (cid:19) t − ln a ln b (1 + o ( t )) .
4. Double effort
Solve the Volterra integral equation ϕ ( x ) = x + ˆ x ( s − x ) ϕ ( s ) ds. (1)SOLUTIONThe considered equation is in fact Volterra integral equation of nd kind withcontinuous kernel. According to the theory of linear integral equations, it has aunique solution. To find its solution, we reduce it to Cauchy problem for ordinarydifferential equation. Let us assume that ϕ ( x ) is the solution of the equation (1). Double differentiationof the identity equation (1) by x gives: ϕ ′ ( x ) = 1 + ˆ x ( − ϕ ( s )) ds, (2) ϕ ′′ ( x ) = − ϕ ( x ) . (3) ϕ ( x ) = C sin x + C cos x. (4)Assuming x = 0 results in: ϕ (0) = 0 . (5) Special Issue No. 4 contains also another methods of solution (see [2]). Modern Problems of Mathematical Physics. Special Issue № 3 At x = 0 we have from eq. (2): ϕ ′ (0) = 1 . (6)Hence C = 1 , C = 0 , and ϕ ( x ) = sin x. (7)It is possible to perform a check: J ( x ) = x + ˆ x ( s − x ) sin s ds = (8) = x − ( s − x ) cos s | x + ˆ x cos s ds, (9) = x − x + sin( s ) | x = sin x. It is possible to propose other ways to solve this equation, such as with the helpof Laplace transformation, or by building a resolvent of the kernel by successiveapproximations method. However, the technique developed above is the simplest.
5. Random walk
A particle performs random walk on one-dimensional lattice situated on the OX axis, the nodes of the lattice have the coordinates m = 0 , ± , ± , ... . At theinitial time moment t = 0 the particle is at the origin of the coordinates. At randomtime moments t , t , t ,. . . the particle performs the jumps into adjacent lattice nodeswith the probabilities of a jump leftwards and rightwards equal to α , the probabilityof remaining still being β = 1 − α . The time intervals between the jumps t i +1 − t i , i = 0 , , , ... are independent random quantities which have the same exponentialdistribution Φ( t ) = 1 τ exp( − t/τ ) with expectation τ . Find:(a) dispersion of the location of the particle as time function t ;(b) probability that the particle is in m -th node at time moment t .SOLUTION1. Let us find the probability p ( t, n ) that the particle during time interval (0 , t ] would perform exactly n jumps, taking into account that this probability is thedistribution function of the Poisson process. Let t , t , ... be time instants of thejumps, so that t = 0 < t < t < ... < t n − < t n < t < t n +1 . (1)Probability p ( t, n ) that the particle during the time interval (0 , t ] would performexactly n jumps, can be represented as p ( t, n ) = M [ I ( t n < t < t n +1 )] , (2) athematical Physics: Problems and Solutions M [ ... ] is expectation and I ( t n < t < t n +1 ) = (cid:26) , if t n < t < t n +1 , , if t n ≥ t or t ≥ t n +1 . (3)Let us perform Laplace transformation of the function p ( t, n ) : ˆ p ( s, n ) = ˆ ∞ dte − st p ( t, n ) = M (cid:20) ˆ ∞ dte − st I ( t n < t < t n +1 ) (cid:21) == M (cid:20) e − st n − e − st n +1 s (cid:21) . (4)Due to the fact that t n = n X i =1 τ i is the sum of independent random variables, then M (cid:2) e − st n (cid:3) = M " exp − s n X i =1 τ i ! = n Y i =1 M [exp ( − sτ i )] = (5) = n Y i =1 ˆ ∞ dτ i exp ( − sτ i ) Φ( τ i ) = ˆΦ n ( s ) , where ˆΦ( s ) = ˆ ∞ dτ exp ( − sτ ) Φ( τ ) is Laplace image of Φ( τ ) . From this, ˆ p ( s, n ) = M (cid:20) e − st n − e − st n +1 s (cid:21) = ˆΦ n ( s ) 1 − ˆΦ( s ) s . (6)Due to the fact that Φ( t ) = 1 τ exp( − t/τ ) , it follows that ˆΦ( s ) = 1 /τs + 1 /τ , (7)and ˆ p ( s, n ) = (1 /τ ) n ( s + 1 /τ ) n +1 . (8)Transforming from Laplace image to the original, we obtain the distribution functionfor the Poisson process p ( t, n ) = t n e − t/τ τ n n ! . (9)2. Let us find the dispersion D ( t ) of the location of the particle as a functionof time t . Location of the particle after n jumps is defined by the random variable8 Modern Problems of Mathematical Physics. Special Issue № 3 X n ( t ) ≡ ξ + ξ + ... + ξ n , where ξ i , i = 1 , ..., n are independent random variables,possessing the values ± with the probability α and 0 with the probability β == 1 − α . Due to the fact that M (cid:2) ( ξ i ) (cid:3) = α , M( ξ i ) = 0 and M [ ξ i ξ j ] = 0 for i = j ,the dispersion equals to D ( t ) = ∞ X n=0 p ( t, n )M (cid:2) ( X n ( t )) (cid:3) = ∞ X n=0 p ( t, n ) αn. (10)Substituting into this formula the expression for p ( t, n ) , we obtain D ( t ) = ∞ X n=1 t n e − t/τ τ n n ! αn = αe − t/τ tτ ∞ X n=1 t n − τ n − ( n − α tτ . (11)3. Let us find the probability that the particle is located at node m at timemoment t . Let h n ( m ) designate the probability that the particle is located at thepoint with coordinate m after n jumps (transitions). Then the probability f ( m, t ) that the particle is at location m at time moment t will be given by formula f ( m, t ) = ∞ X n =0 p ( t, n ) h n ( m ) . (12)Function h n ( m ) equals to h n ( m ) = M [ δ m,ξ + ξ + ... + ξ n ] , (13)where δ m,n is the Kronecker delta. Due to the fact that δ m,n = 12 π ˆ π − π e i ( m − n ) ϕ dϕ , itis possible to write h n ( m ) = 12 π M (cid:20) ˆ π − π e i ( m − ξ − ξ − ... − ξ n ) ϕ dϕ (cid:21) = 12 π ˆ π − π e imϕ n Y j =1 M (cid:2) e − iξ j ϕ (cid:3) dϕ. (14)Note that M (cid:2) e − iξ j ϕ (cid:3) = α (cid:0) e − iϕ + e iϕ (cid:1) + 1 − α = α cos ϕ + 1 − α. (15)Therefore h n ( m ) = 12 π ˆ π − π e imϕ ( α cos ϕ + 1 − α ) n dϕ. (16)Substituting this equation for h n ( m ) , and earlier found expression for p ( t, n ) intothe formula for f ( m, t ) , we obtain f ( m, t ) = ∞ X n =0 t n e − t/τ τ n n ! 12 π ˆ π − π e imϕ ( α cos ϕ + 1 − α ) n dϕ = (17) athematical Physics: Problems and Solutions = 12 π e − t/τ ˆ π − π e imϕ dϕ ∞ X n =0 ( α cos ϕ + 1 − α ) n t n τ n n ! == 12 π e − t/τ ˆ π − π e imϕ e t ( α cos ϕ +1 − α ) /τ dϕ == 12 π e − t/τ (cid:18) ˆ π e imϕ e t ( α cos ϕ +1 − α ) /τ dϕ + ˆ π e − imϕ e t ( α cos ϕ +1 − α ) /τ dϕ (cid:19) == 1 π e − t/τ ˆ π cos( mϕ ) e t ( α cos ϕ +1 − α ) /τ dϕ = 1 π e − αt/τ ˆ π cos( mϕ ) e tα cos ϕ/τ dϕ. The last equation can be reproduced in a different form, using integral representationof Bessel function J m ( z ) : J m ( z ) = i − m π ˆ π cos ( mϕ ) e iz cos ϕ dϕ. (18)As a result, we obtain f ( m, t ) = i m e − αt/τ J m (cid:18) − i αtτ (cid:19) . (19)Taking into consideration that J m ( iz ) ≡ i m I m ( z ) , (20)where I m ( z ) = ∞ X k =0 (cid:0) z (cid:1) k + m k ! ( k + m )! (21)are the modified Bessel functions, we write the final result for the probability oflocation of the particle at node m at time moment t : f ( m, t ) = ( − m e − αt/τ I m (cid:18) − αtτ (cid:19) = e − αt/τ I m (cid:18) αtτ (cid:19) . (22)Asymptotics of f ( m, t ) at t → ∞ with the asymptotic behavior of modifiedBessel functions I m ( z ) = e z √ πz (cid:0) O (cid:0) z − (cid:1)(cid:1) at z → ∞ (23)considered, is of the form: f ( m, t ) = e − αt/τ e αt/τ p παt/τ (cid:0) O (cid:0) t − (cid:1)(cid:1) = 1 p παt/τ (cid:0) O (cid:0) t − (cid:1)(cid:1) . (24)0 Modern Problems of Mathematical Physics. Special Issue № 3
6. Thermal equations of the Universe evolution
It is assumed that at high temperature (at early stage of the evolution of theUniverse) it is possible to describe matter using field theory. Equation of state withgood approximation corresponds to ideal quantum gas of massless particles (in thegeneral case, it can be a mixture of ideal Bose- and Fermi-gases). In this theory,under the condition that the temperature T is far from mass threshold yet (radiation dominance, ρ = 3 p ), thermodynamic functions are given by the formulae: ρ = 3 p = π N ( T ) T , (1) s = 2 π N ( T ) T , (2)where N ( T ) is the function related to the number of bosonic and fermionic degreesof freedom ( N ( T ) = N b ( T ) + 78 N f ( T ) ), ρ and p are equilibrium energy density andpressure of the matter, s is specific entropy. All expressions are written in the unifiedsystem of units c = ~ = 1 .Formulate the dynamic equations of the evolution of the Universe in terms oftemperature.Note. The required equations are not Einstein’s equations in the standard form(see Einstein equations, for example, in [3, 4]). It is proposed to write the equationsof the evolution of the Universe in Friedmann model using thermodynamic func-tions and temperature as function of time. It is possible to do this with the use ofenergy conservation law and condition of adiabatic expansion of the Universe in theframework of the standard cosmologic model.Instruction. Introduce auxiliary function ǫ ( T ) = ka T , (3)where k = 0 , ± respectively for flat, open and closed models of the Universe withtime-dependent scale factor of a ≡ a ( t ) . By radiation here, we mean any relativistic object, including relativistic matter as well asphotons.athematical Physics: Problems and Solutions (cid:18) ˙ aa (cid:19) + ka = 8 π Gρ (4)in terms of temperature [4] (here G is the gravitation constant). It is consideredthat stress energy tensor of the Universe takes the form of energy momentum tensorfor ideal liquid [3]. We shall use the energy conservation law ddt ( ρa ) = − p ddt ( a ) (5)(the change in energy in a comoving volume element, d ( ρa ) , is equal to minus thepressure times the change in volume, − p d ( a ) ) and the fact that in the standardcosmological model it is supposed that the Universe undergoes adiabatic expansion ddt ( sa ) = 0 (6)(the entropy per comoving volume element remains constant).Let us write Einstein–Friedmann equation (4) in terms of temperature supposingthat temperature value is far from the mass threshold (see e. g. [3]). We deal withmatter which is found in thermodynamical equilibrium at almost all the time stagesduring cosmological expansion, so the chemical potential is considered to be zero.Taking into consideration (1) and auxiliary function (3) let us represent eq. (4)in the form (cid:18) ˙ aa (cid:19) + ǫ ( T ) T = 4 π GN ( T ) T . (7)2 Modern Problems of Mathematical Physics. Special Issue № 3
Taking into account equation of state ρ = 3 p , from the energy conservation law(5) we obtain the relationship ˙ aa = −
14 ˙ ρρ , (8)which in agreement with (1) would take the form ˙ aa = − ˙ TT −
14 ˙ N ( T ) N ( T ) . (9)Using the condition of adiabatic expansion of the Universe (6), we can find ˙ aa = −
13 ˙ ss . (10)Substituting into (10) the expression for specific entropy (2), we obtain ˙ aa = − ˙ TT −
13 ˙ N ( T ) N ( T ) . (11)Comparing (9) and (11), we come to the conclusion that ˙ N ( T ) = 0 . So therelation between the scaling factor and temperature should have the form ˙ aa = − ˙ TT . (12)Substituting this equation into eq. (7), we obtain one of the dynamic equations ofthe evolution of the Universe: ˙ TT ! + ǫ ( T ) T = 4 π GN ( T ) T . (13)To write the second required equation, let us multiply both parts of the eq. (2) by a ,express a T and substituting it into the auxiliary eq. (3), we obtain the equation ǫ ( T ) = k (cid:20) π N ( T ) S (cid:21) / , (14)where S ≡ sa is the total entropy in the volume defined by the radius of curvature a . We need to note that N , S hence ǫ are constant in the considered temperature(or time) range (however a is not constant) due to the particle counting by allowedparticles degrees of freedom thresholds.As the result, the dynamic equations of the evolution of the Universe in termsof temperature and entropy are the equations (13) and (14). athematical Physics: Problems and Solutions
7. Trapped electron
Consider an isolated conducting sphere of radius R carrying the total charge Q .At the distance a > R from its center, there is a point charge q ( qQ > ). Findpotential of the system ϕ ( ~r ) and the force −→ F ( a ) acting on the point charge. Analyzethe limit lim a → R +0 F ( a ) , explain the obtained result.SOLUTIONThe present problem can be solved using method of image charges.It is known that for any two point electric charges of opposite sign, it is alwayspossible to find such a spherical surface that the resulting potential on it wouldbe zero. Radius of the sphere and the distance from its center to the charges isdetermined uniquely if the values of the charges and the distance between themare known. So, the system under discussion (the point charge and the conductingsphere) is equivalent to the set of point charges. Thus let us place a charge q on aline connecting the center of the sphere O with the q charge at the distance of d inthe direction of the charge q . One more charge, q = Q − q , we shall place at the O point, see fig. 1.Let us place the origin of the reference frame also in O point and direct OX axisto the point charge q (leftwards). Let us write the condition that the total potential(due to the charges q and q ) is zero at points ( R, , ) and ( − R, , ).Hence we obtain that d = R /a , q = − qR/a , thus q = Q + qR/a .The total potential inside the full sphere equals q /R and outside is given by theexpression: ϕ ( ~r ) = qr + q r + q r . (1)The same in the Cartesian coordinates: ϕ ( ~r ) = q p ( x − a ) + y + z − Ra q q ( x − R a ) + y + z + Q + q Ra p x + y + z , (2) q r RaP d Ox q r r F(a)
Fig. 1 Modern Problems of Mathematical Physics. Special Issue № 3
The force on q acts along axis OX . Its projection is given by F = qQa + q Ra − (cid:18) − R a (cid:19) . (3)It is easy to check that if we represent a as R + ∆ , in the limit at ∆ → the expression for the force would become F ∼ − q / (2∆) which corresponds to theinteraction force of a point charge with non-charged conducting plane.At a → ∞ , F → qQa (Coulomb force, as expected).At a → R , no matter how large Q is, even if Qq > , we have F → −∞ ,i. e. the force becomes very strong and attractive!Let us find the distance at which the effect of attraction starts. Let us write theinteraction force (3) in a dimensionless form: f ( s ) = αs − s − s ( s − , (4)where α = Qq > , s = aR > , f ( s ) = R q F x ( a ) . (5)The force occurs to be zero at a distance a = s R , where s > satisfying theequation s − s ( s − = α. (6)Function f ( s ) graphs are shown in fig. 2 at α = {
2; 1; 0 . } . In these cases s == { .
43; 1 .
62; 1 . } , and force f ( s ) reaches its maximum { . , . , . } when s = s max = { . , . , . } .This explains why electrons can’t escape from metals, even though they arerepelled by the other electrons. If an electron manages to escape and gets to a smalldistance from the surface of metal, the other electrons conspire to bring it back byrearranging themselves in such a way to create a huge image charge which attractsthe electron back to the metal with a strong electric force! athematical Physics: Problems and Solutions s max s f max sf(s) α=2α=1α=0.5 Fig. 2
8. Х-sector Consider the following model of Higgs sector with two doublet scalar fields φ and φ transformable as SU (2) doublets, with the weak hypercharge generator of Y W = 1 , and with each component of the doublet being a complex scalar field.Suppose both fields acquire parallel vacuum averages (vevs) of the type h φ i i = 1 √ (cid:18) v i (cid:19) ( i = 1 , , (1)with the values v , v (these vacuum averages lead to gauge bosons mass matrix asin the Standard Model with the replacement v = v + v ). The most general formof the potential energy function (potential) for a model with two Higgs doublets israther complicated . However, the model hermitian potential possessing the mainproperties can be written in the following form: V ( φ , φ ) = − µ ( φ † φ ) − µ ( φ † φ ) − µ ( φ † φ ) − (cid:0) µ (cid:1) ∗ ( φ † φ ) ++ λ ( φ † φ ) + λ ( φ † φ ) + λ ( φ † φ )( φ † φ )+ λ ( φ † φ )( φ † φ )+ λ φ † φ ) + λ ∗ φ † φ ) , where µ and λ may be the complex numbers. Some extension of Peskin & Schroeder [6] problem 20.5. Recall that in the Standard Model (SM), the gauge bosons masses come from the term | Dφ | in the Lagrangian, where we set φ equal to its vacuum expectation value v . It is only for reasons of simplicity that the SM contains just a single Higgs doublet. Supersym-metric extensions of the SM typically contain two or more Higgs doublets, and singlets. Modern Problems of Mathematical Physics. Special Issue № 3 (a) Obtain the conditions that for the direction in field space at given configura-tion of vevs (1) the potential is bounded below at large field values. (The analogueof λ > for the theory with a single Higgs doublet.)(b) Find the conditions to impose on the parameters µ and λ , so that the config-uration of vacuum averages (1) gives strictly local (locally stable) minimum of thispotential.(c) In the unitary gauge (rotation to canonical form), one linear combination ofthe upper components φ and φ nulls, while another becomes a physical field. Showthat charged physical Higgs field is of the form: H + = φ +1 sin β − φ +2 cos β, (2)where β is defined by the relation tg β = v v . (3)(d) Investigate if the CP invariance breaks in the given potential. Substantiatethe obtained results.SOLUTION(a) First of all, note that µ i =1 , and λ i =1 ... are all real due to the fact thatLagrangian is Hermitian. But µ and λ may be the complex numbers.Second, find the condition under which both φ and φ have parallel non-zerovevs (recall that in the case of a single Higgs doublet the gauge symmetry allowsthe vacuum expectation value (vev) to be taken in the form (0 , v ) , with v real). Nowuse a rotation to make h φ i = 1 √ (cid:18) v (cid:19) , where v is real. Note that after we havedone this we have used all of our gauge rotation freedom, so the vevs of φ are stillcompletely general, i. e. h φ i = 1 √ (cid:18) v ′ v ′′ (cid:19) . So we rewrite the potential in terms ofthe two (complex) vevs of φ , and of the v . Letting v = v ′ + v ′′ , we find V = F ( v , v ) − µ v v ′′ − ( µ ) ∗ v v ′′ ∗ + λ v ( v ′′ ∗ v ′′ ) + λ v v ′′ + λ ∗ v v ′′ ∗ , (4)where we denote F ( v , v ) the function dependent on values v and v only.First we want to answer two questions: (i) What is the condition that v ′′ is real?(ii) What is the condition that v ′ is zero? If we enforce these conditions, it ensuresthat the vevs take the form of (1), where both v and v are real. This is what itmeans for the vevs to be ’aligned’. So, how to find these conditions? It becomes clearif we rewrite v ′′ ≡ ae iθ . The potential written in terms of a and θ is given by: V ( a, θ ) = ( Stuff not dependent on θ or a ) − Re µ v a cos θ + Im µ v a sin θ ++ 14 λ v a + 14 Re λ v a cos 2 θ − Im λ v a sin 2 θ. (5) athematical Physics: Problems and Solutions v ′′ is ensured by forcing θ = 0 , π . We can see that θ = 0 or π willbe a stable minimum of the potential if the second derivative of expression (5) ispositive at these values of phase θ , i. e.Re λ − Re µ v a < , (6)and in the limit of large field valuesRe λ < . (7)In this case the Eqn. (5) will be minimized for cos 2 θ = 1 . We must show that v ′ = 0 .Recall that the sum v ′ + v ′′ is fixed to be v . So if we can arrange the potentialso that it is energetically advantageous for all the vevs to go into v ′′ , we are done.This is equivalent to saying that we want V ( a, θ ) to be minimized for a → ∞ . Thisis accomplished if λ + λ < . (8)So, together, eqns. (7), (8) guarantee that the vacuum expectation values can bealigned.(b) Now what is still required, is to show that this is a stable minimum. Usingaligned forms for the vevs, rewrite the potential in terms of v and v . What addi-tional conditions on the parameters are necessary to guarantee a stable minimum?Stability is equivalent to saying that there is a positive mass squared for fluctuationsabout the minimum. In other words, we examine the mass matrix Hessian H = ∂ V ( v , v ) ∂v ∂ V ( v , v ) ∂v ∂v ∂ V ( v , v ) ∂v ∂v ∂ V ( v , v ) ∂v in minimum. (9)In order to have a stable minimum the matrix of second derivatives needs to bepositively definite. This does not mean that all second derivatives need to be positive.Both eigenvalues, e i , need to be positive to ensure the stability of the minimum.The above matrix is to be evaluated at the minimum, setting to zero the derivativesevaluated at the vevs (i. e. where ∂V∂v = ∂V∂v = 0 ). Since invariant Tr( H ) = e + e and Det( H ) = e e , we can simply require ∂ V ( v , v ) ∂v + ∂ V ( v , v ) ∂v > , and Det( H ) > . Straight-forward algebra gives: H = λ v + Re µ v v − Re µ + λ v v − Re µ + λ v v λ v + Re µ v v . (10)8 Modern Problems of Mathematical Physics. Special Issue № 3
We denote λ = λ + λ + Re λ .This shows that stability conditions are equivalent to: λ v + 2 λ v + (cid:18) v v + v v (cid:19) Re µ > , (11)and λ λ + 2 λ v v Re µ + 2 λ v v Re µ > ( λ ) − λ Re µ , (12)by taking first derivatives with respect to all the scalar fields, and setting them equalto zero (i. e. in minimum).In particular case Re µ = 0 we see simply λ λ > ( λ ) , and λ > , λ > . (13)In addition, we will requite concavity, which is implied to be positive, whichassures us that we’re at a minimum and not at a saddle point. The tricky part ofthe problem is to show that indeed it is possible to have the vevs parallel, i. e. ofthe form of (1). To do this, the best way is to use the SU (2) rotation to force thevev of φ to have the right form. Then all we need to show is that the potentialis minimized (i. e. the appropriate derivatives satisfy the conditions stated above)when φ takes the right form.Thus the conditions for a local stable minimum of the potential in this problemare (7), (8), (11), (12), see also [7].(c) In the SM, there are 4 degrees of freedom in the Higgs doublet, three ofwhich are consumed by the W + , W − and Z . When there are 2 Higgs doublets,they contain in total 8 degrees of freedom, so the 5 remaining after goldstones areconsumed. Now, the vev given breaks the SU (2) × U (1) Y symmetry to U (1) em ,therefore in general there would be 3 goldstone bosons, and 5 physical Higgs fields.Two of these remaining degrees of freedom are charged, and three are neutral. Thetask in this problem is to determine which two charged degrees of freedom are eaten,and which two charged degrees of freedom remain. The simplest way to do this isto consider the components of the two Higgs doublets as being part of a largervector. An orthogonal transformation will rotate the different components amongstthemselves. In particular, we can find the basis where the vev is entirely in oneneutral component. The charged piece associated with this neutral component isthe would-be goldstone boson that is eaten. The charged Higgs is the piece whichis orthogonal to this goldstone boson that is eaten. So, considering the neutralcomponents (which have the vevs), we have: (cid:18) φ ′ φ ′ (cid:19) = (cid:18) cos β sin β − sin β cos β (cid:19) (cid:18) φ φ (cid:19) . (14) Thus, in fact, condition (6) is equivalent to positive sign of squared mass for third Higgs(pseudoscalar) boson m = − v Re λ + Re µ v v , and condition (8) at large field values is equivalentto positive sign of squared mass for charge Higgs boson m H ± = − v λ + Re λ ) + Re µ v v .athematical Physics: Problems and Solutions (say φ ′ ), we get (cid:18) cos β sin β − sin β cos β (cid:19) (cid:18) v v (cid:19) = (cid:18) v (cid:19) . (15)It follows from (15) that tg β = v v . So, the whole vev lives in the φ ′ field, whichmeans that the charged component of φ ′ is the Goldstone Boson eaten by the W .That means φ ′ = φ cos β + φ sin β is the Goldstone, and φ ′ = H + = φ +1 sin β − φ +2 cos β (16)is the physical charged Higgs field.(d) In the models with two doublets of scalar fields CP invariance can be vio-lated by the terms of the potential containing ( φ † φ ) or ( φ † φ ) with the complexparameters µ and λ . In the case of real parameters CP invariance is not broken.
9. By the cradle of LHC Generally, it is possible to describe a scattering experiment in the following way(see fig. 3):1) sufficiently wide uniform beam of particles is prepared sothat it is possible to assume the momentum of each particle beequal to ~p = ~ ~k , where ~k is wave vector, ~ – Planck constant;2) this beam of particles is directed to stationary target con-sisting of identical particles;3) at certain distance from the target, products of reactionof the particles from the beam with the particles forming thetarget, are registered at different angles. Fig. 3
Thus at sufficiently large distances from the target the wave function of theparticles is the superposition of plane incident wave ψ |||| = exp { i~k ~r } and sphericalscattered wave ψ J = exp { ikr } /r . Here, ~r = x~ i + y~ j + z~ k, r = | ~r | .(a) Calculate flux density of the probability ~j = ~ m i ( ψ ∗ ~ ∇ ψ − ψ ~ ∇ ψ ∗ ) for thewave functions ψ = e i~k · ~x and ψ J . Here, m is the mass of the particle, and ~x = ~ix .(b) Illustrate the obtained result with the help of the graph: draw the pattern ofthe vector field ~j (lines of the ~j vector) in both cases. To build such graph, use anyavailable computer software suitable for building graphics (plots).(c) Prove that ~ ∇ · ~j = 0 , ~ ∇ · ~j J ∼ δ ( ~r ) . Once one goes to the proper basis for describing the Higgs mechanism, there is really only onedoublet (in the above case it is φ ′ ) that acts as the Higgs and has three of four degrees of freedomthat are eaten. So, we can say that it is a model with "two-complex scalars". Large Hadron Collider. Modern Problems of Mathematical Physics. Special Issue № 3
SOLUTION(a) ~j = ~ mi ( ψ ∗ ~ ∇ ψ − ψ ~ ∇ ψ ∗ ) = ~ mi ~ i [ e − i~k~x ik x e i~k~x − e i~k~x ( − ik x ) e − i~k~x ] = ~ i ~ k x m ,~j J = ~ mi ( ψ ∗ J ~ ∇ ψ J − ψ J ~ ∇ ψ ∗ J ) == ~ mi ~n r h e − ikr r ddr e ikr r − (cid:16) ddr e − ikr r (cid:17) e ikr r i = ~ km ~n r r . Here, ~n r = ~r/r. (b)The field in both cases is represented by a graph in x – y plane at z = 0 (inMathematica realization, see fig. 4). In[1]:=
Needs @ "VectorFieldPlots‘" D ; In[2]:=
VectorFieldPlot @8
1, 0 < , x, -
1, 1 < , y, -
1, 1 In[3]:= VectorFieldPlot B 8 x, y
1, 1 < , y, - 1, 1 < ,ScaleFunction ® H LF Out[3]= a bFig. 4 It is possible to see that the spherical wave ψ J is diverging from the origin ofcoordinates.(c)Vector field ~j is uniform, ~j = ~ i ~ k x m = ~ const , so that ~ ∇ · ~j = 0 .It is possible to represent vector ~j J in the form: ~j J = ~ km ~n r r = ~ km ~rr = − ~ km ~ ∇ r . Hence ~ ∇ · ~j J = − ~ km ∆ 1 r = ~ km πδ ( ~r ) . athematical Physics: Problems and Solutions 10. ‘Whipping Top-Toy’ from Samara Rigid ball of the mass m with radius R rests on smooth rigid horizontal surface.Center of mass C of the ball is at the distance of l from its geometric center O . Mass ofthe material is symmetrically distributed along the volume of the ball relatively to OC axis, and also any plane containing that axis. Moments of inertia of the ballrelatively to OC axis and any axis passing the center of mass and perpendicular to OC axis, are equal respectively to J and J . During certain time period, the ballis accelerated around the static vertical axis passing its center O . The moment oftime when the action of the "accelerating" forces is finished, is chosen as the timeorigin. At this moment the ball has the angular velocity −→ ω , directed vertically up,and OC axis makes some angle ε with the rotation axis of the ball. The center ofmass of the ball is lower than its mechanical center (see fig. 5). O l C t=0 w e Fig. 5. Rotating ball at time moment t = 0 Study the motion of the ball at t > . Obtain the equation of motion of the balland find the integrals of motion (conserved values). Show that the center of massof the ball is lifting up (at a certain relationship between the parameters), and findout, whether it can approach its top position at which the −→ OC vector is directedvertically up.SOLUTIONIn order to describe the motion of the ball we introduce an inertial frame ofreference S (with axes X , Y , Z ), that is at rest relatively to the horizontal surface,and an noninertial frame S ′ (with axes X ′ , Y ′ , Z ′ ), that is rigidly bound to the ball.The origin of the frame S coincides with the initial position of the center of the ball,the axis Z is directed vertically and the plane Y OZ is chosen so that it contains theaxis OC at the initial moment t = 0 . The origin of the frame S ′ coincides with themass center of the ball and the axis Z ′ – with the axis OC , so that at the moment t = 0 the axis Z ′ makes an angle ε with the axis Z . Additionally, the axis X ′ at this Place in Russia where this competition is held and assessed. Modern Problems of Mathematical Physics. Special Issue № 3 moment has the direction similar to that of the horizontal axis X , and the axis Y ′ belongs to the plane Y OZ (see fig. 6). O C t=0 w e e z ’ zx x ’ y ’ y Fig. 6. Coordinate axes and the origins of S , S ′ at the moment t = 0 The reference frames S and S ′ being chosen, the Euler angles ϕ , θ , ψ that definethe orientation of the ball (and the frame S ′ ) relatively to the frame S , are given bythe following values at the initial moment: ϕ (0) = 0 , θ (0) = ε, ψ (0) = 0 . (1)Components of the angular velocity ~ω of the ball in the frames S and S ′ at anyinstant of time are given by the cinematic Euler’s formulas: ω x = ˙ θ cos ϕ + ˙ ψ sin θ sin ϕ,ω y = ˙ θ sin ϕ − ˙ ψ sin θ cos ϕ,ω z = ˙ ϕ + ˙ ψ cos θ, (2) ω x ′ = ˙ ϕ sin θ sin ψ + ˙ θ cos ψ,ω y ′ = ˙ ϕ sin θ cos ψ − ˙ θ sin ψ,ω z ′ = ˙ ϕ cos θ + ˙ ψ. (3)For the initial values of the components of the angular velocity in S the equa-tions (2), with the account of (1), give ω x (0) = ˙ θ (0) ,ω y (0) = − ˙ ψ (0) sin ε,ω z (0) = ˙ ϕ (0) + ˙ ψ (0) cos ε. (4)On the other hand, it is known that the initial angular velocity ~ω (0) = ~ω hasthe same direction as the axis Z and, hence, ω x (0) = ω y (0) = 0 , ω z (0) = ω . (5) athematical Physics: Problems and Solutions ˙ ϕ (0) = ω , ˙ θ (0) = 0 , ˙ ψ (0) = 0 . (6)As seen from the pictures, the cartesian coordinates x m , y m , z m of the masscenter of the ball in S at the initial moment of time have the following values: x m (0) = 0 , y m (0) = l sin ε, z m (0) = − l cos ε. (7)For t < (until the rotating forces action has stopped) the center of mass of theball moves in the horizontal plane along the circle of the radius l sin ε with the axis Z passing through its center. Therefore, the velocity ~v m of the mass center of theball in the frame S at the initial instant of time is given by ~v m (0) = [ ~ω ~r m (0)] , (8)where ~r m (0) is the initial value of the radius-vector ~r of the mass center of the ballin S (that is the vector with the components x m (0) , y m (0) , z m (0) ). From (5), (7)and (8) it follows that the initial velocity of the mass center of the ball in S isdirected opposite to the axis X and is equal to v m (0) = ω l sin ε, (9)or ˙ x m (0) = − ω l sin ε, ˙ y m (0) = ˙ z m (0) = 0 . (10)The equations of motion of the ball for t > can be derived as the Lagrangeequations of the 2-nd kind. The ball is a mechanical system with five degrees offreedom. One can use the Euler angles ϕ , θ , ψ and coordinates x m , y m of the masscenter in S as the independent generalized coordinates of the ball, the coordinate z m at any instant of time being defined by z m = − l cos θ. (11)The Lagrange function has the form L = T − U, (12)where T is the kinetic energy of the ball and U is its gravitational potential energy(these values are defined in the frame S ).The kinetic energy of the ball may be expressed in the form T = mv m τ, (13)4 Modern Problems of Mathematical Physics. Special Issue № 3 where τ is its kinetic energy of rotation. As the axes x ′ , y ′ , z ′ are directed along theprincipal axes of inertia of the ball, then it follows τ = 12 ( J x ′ ω x ′ + J y ′ ω y ′ + J z ′ ω z ′ ) , (14)where J x ′ , J y ′ , J z ′ are the moments of inertia of the ball relatively to the axes x ′ , y ′ , z ′ , with, according to the problem statement, J x ′ = J y ′ = J, J z ′ = J . (15)Inserting the expressions (3) and (15) into (14) one gets τ = J ϕ sin θ + ˙ θ ) + J ϕ cos θ + ˙ ψ ) . (16)The expression for the velocity’s square of the mass center of the ball with theaccount of (11) gets the form v m = ˙ x m + ˙ y m + l sin θ ˙ θ . (17)From (13), (16) and (17) it follows that T = m x m + ˙ y m ) + m l sin θ ˙ θ + J ϕ sin θ + ˙ θ )++ J (cid:16) ˙ ϕ cos θ + 2 ˙ ϕ ˙ ψ cos θ + ˙ ψ (cid:17) . (18)The potential energy of the ball with the account of (11) is given by the expression U = − mgl cos θ. (19)Inserting the equations (18) and (19) into (12) one gets the following explicitexpression for the Lagrange function of the ball: L = m x m + ˙ y m ) + 12 ( ml sin θ + J ) ˙ θ + 12 ( J sin θ + J cos θ ) ˙ ϕ ++ J ψ + J cos θ ˙ ϕ ˙ ψ + mgl cos θ. (20)It is known that the Lagrange equations of the 2-nd kind for a system with idealholonomic constraints and S degrees of freedom under the absence of dissipativeforces have the form ddt (cid:18) ∂L∂ ˙ q j (cid:19) − ∂L∂q j = 0 ( j = 1 , , ..., S ) , (21) athematical Physics: Problems and Solutions q j ( j = 1 , , ..., S ) are independent generalized coordinates of the system.If q j is a cyclic coordinate of the system (that is, the Lagrange function doesn’tdepend on it), then the corresponding generalized momentum p j = ∂L∂ ˙ q j is the inte-gral of motion.According to (20), x m , y m , ϕ , ψ are cyclic coordinates of the ball. Therefore, thegeneralized momentums ∂L∂ ˙ x m = m ˙ x m , (22) ∂L∂ ˙ y m = m ˙ y m , (23) ∂L∂ ˙ ϕ = J sin θ ˙ ϕ + J cos θ (cos θ ˙ ϕ + ˙ ψ ) , (24) ∂L∂ ˙ ψ = J (cos θ ˙ ϕ + ˙ ψ ) (25)are the integrals of motion of the ball that are constant and are defined by the initialconditions (1), (6), (7) and (10). Writing down the corresponding conservation laws,one gets the following system of the 1-st order differential equations: ˙ x m = − ω l sin ε, (26) ˙ y m = 0 , (27) J ˙ ϕ sin θ + J ω cos ε cos θ = ω ( J sin ε + J cos ε ) , (28) ˙ ϕ cos θ + ˙ ψ = ω cos ε (29)(when deriving the equation (28), that presents the conservation law for the gener-alized momentum p ϕ = ∂L∂ ˙ ϕ , the equation (29) was used).From (26) and (27) and the initial conditions (7) it follows that x m = − ω tl sin ε, (30) y m = l sin ε. (31)6 Modern Problems of Mathematical Physics. Special Issue № 3 Hence, the mass center of the ball moves in the vertical plane that is normalto the axis y and that passes through the initial position of the mass center. Thehorizontal component of the ball’s velocity stays constant and equal to the initialvelocity ~v m (0) .Note that the equations (26) – (28) may be also derived from the conservationlaws for the components of momentum and angular momentum of the mechanicalsystem in an inertial frame. In order to do this one has to consider the directionsof external forces, applied to the ball. These forces are the reaction force ~N that isdirected vertically up and the gravitational forces that are applied to every elementof the ball and are directed vertically down, their resultant force m~g being appliedto the mass center of the ball.As the sum of the external forces that are applied to the ball, is directed vertically,the X and Y components of the ball’s momentum ~P = m~v m (32)are conserved for t ≥ . The corresponding conservation laws, with the account ofthe initial conditions (10), give the equations (26) and (27).The angular momentum ~M of the ball in the frame S may be presented in theform ~M = [ ~r ~P ] + ~µ, (33)where ~µ is the intrinsic angular momentum momentum of the ball (that is the angu-lar momentum of the ball in the mass center reference frame that moves translatoryto S ). The components of ~µ in S ′ are defined by µ x ′ = J ω x ′ , µ y ′ = J ω y ′ , µ z ′ = J ω z ′ . (34)As the moments of all external forces in S that are acting on the ball are directedhorizontally for t ≥ , the projection M z of the angular momentum on the axis Z is the integral of motion. Taking into account (30), (31), (11), (26) and (27), iteasy to check that Z -component of the vector [ ~r ~P ] has the constant value equal to mω l sin ε . Hence, with the account of (33), it follows that M z = mω l sin ε + µ z , (35)or that Z -component of the ball’s rotate momentum µ z is an integral of motion.Using the well-known linear transformation law for the components of an arbi-trary vector for a frame rotation and expressing all the coefficients of such transfor-mation as the functions of Euler angles, one may show that µ z = sin θ sin ψµ x ′ + sin θ cos ψµ y ′ + cos θµ z ′ . (36)Inserting the equations (34), where the angular velocity components are givenby the formulas (3), into (36), one gets µ z = ( J sin θ + J cos θ ) ˙ ϕ + J cos θ ˙ ψ. (37) athematical Physics: Problems and Solutions µ z coincides with the gen-eralized momentum p ϕ . Hence the conservation law for µ z , written with the accountof the initial conditions (1) and (6) and of the equation (29), is the equation (28).Inserting the equation (20) into (21) and letting q j = θ , the Lagrange equationfor the ball that corresponds to the generalized coordinate θ is obtained: ( ml sin θ + J )¨ θ + ml sin θ cos θ ˙ θ + ( J − J ) sin θ cos θ ˙ ϕ + (38) + J sin θ ˙ ϕ ˙ ψ + mgl sin θ = 0 . Note that the equations (28) and (29) form a system of linear algebraic equationsrelatively to ˙ ϕ and ˙ ψ with the coefficients that depend only on the angle θ . Havingsolved this system, one gets the dependencies of these generalized velocities on θ : ˙ ϕ = ω J sin θ ( J sin ε + J cos ε − J cos ε cos θ ) , (39) ˙ ψ = ω J sin θ [cos ε ( J sin θ + J cos θ ) − cos θ ( J sin ε + J cos ε )] . (40)After having inserted the expressions (39) and (40) into (38) one may get thenonlinear 2-nd order differential equation that defines the dependency θ ( t ) . In orderto get this dependency one may also use the conservation law for the total mechanicalenergy of the ball E = T + U. (41)This quantity coincides with the generalized energy of the ball and is conserved dueto the absence of both dissipative forces and explicit dependency of the Lagrangefunction on time.Setting the explicit expression for the energy of the ball that follows from (18),(19) and (41), equal to its initial value, that is found from (1), (6) and (10), andtaking into account the conservation laws (26) and (27), one gets ( ml sin θ + J ) ˙ θ + ( J sin θ + J cos θ ) ˙ ϕ + J ˙ ψ + 2 J cos θ ˙ ϕ ˙ ψ − mgl cos θ == ( J sin ε + J cos ε ) ω − mgl cos ε. (42)Substituting the expressions (39) and (40) into the last equation, one obtainsthe following nonlinear 1-st order differential equation that defines the dependency θ ( t ) : sin θ (1 + β sin θ ) ˙ θ = ω (cos ε − cos θ )Θ( θ ) , (43)where Θ( θ ) is quadratic in cos θ and has the form: Θ( θ ) = a + 2 a cos θ + a cos θ, (44)8 Modern Problems of Mathematical Physics. Special Issue № 3 where a − a are given by: a = − βγ − (cid:0) ( α − cos 3 ε + ( α + 1)(3 α − 1) cos ε (cid:1) , (45) a = 14 (cid:0) ( α − 1) cos 2 ε + α + 1 (cid:1) , (46) a = 2 βγ, (47)Dimensionless parameters α , β , γ in equations (45) – (47) and equation (43), arerelated to the known parameters in the following way: α = J J , (48) β = ml J , (49) γ = glω = ω m ω (50)(here ω m = p g/l is the cyclic frequency of a flat simple pendulum of length l ).The region of motion of the ball top is defined by the sign and the roots ofthe function Θ( θ ) . In order to investigate them one has to specify the range of theparameters that are present in the function. As J, J ≥ and as for a particular caseof the ball being a rotator the moments of inertia have the values J = 0 , J = 0 , weconclude that α is not negative. On the other hand, it is known that none of theprinciple moments of inertia is bigger than the sum of the other two, which gives J ≤ J . Therefore, we conclude that ≤ α ≤ . Further, as it is easy to notice, ≤ β < ∞ and < γ < ∞ .Taking into account the equations (45) – (47), one finds that: ≤ a ≤ , (51) ≤ a < ∞ . (52)Concerning a , one may notice that this parameter is not bound from below andapproaches its maximum value with respect to βγ at βγ = 0 . Further analysis showsthat for ≤ α ≤ , ≤ ε ≤ π the parameter a reaches its maximum value, that isequal to a = 4 , at α = 2 , ε = π . For small ε the parameter a is maximized withrespect to α at α = 0 and increases monotonically with ε . Hence, we finally concludethat − ∞ < a ≤ . (53) athematical Physics: Problems and Solutions Θ( ε ) > themotion of the top will always go on in the area where cos ε ≥ cos θ or, that is, where θ ≥ ε . Correspondingly, for Θ( ε ) < the mass center will always be lower than inits initial position during the motion. Finally, for Θ( ε ) = 0 the motion will be goingon at constant θ , that is, the mass center of the top will be moving at constantheight (however, this regime is unstable with respect to small variations in α, β, γ ).The last result follows from the fact that for Θ( ε ) = 0 , as we shall see further, theright-hand side of (43) is negative everywhere except for at θ = ε .From the equations (44) and (45) – (47) it follows: Θ( ε ) = − ε (( α − 1) cos ε + βγ ) (54)Hence, the condition that the mass center lifts up during the initial stages ofmotion is: Θ( ε ) > ⇐⇒ ( α − 1) cos ε + βγ < (55)Note, in particular, that for the tops with α > the mass center will alwaysbe moving downwards under the condition ε < π irrespectively to the kinematicparameters of the problem.Remark. Note that in [2] the case for small ε is considered in details and analyticalsolutions are obtained, and the analysis of trajectories is carried out also.Let us now study the behavior of the roots of the parabola, defined by thedependency Θ(cos θ ) . As a ≥ , the parabola is open upwards (the case a = 0 willbe understood as the limiting one). The minimum of the parabola corresponds to cos θ equal to z m ≡ − a a . As from the equations (45) – (47), (48) – (50) it followsthat a , a contain independent parameters, one concludes that − ∞ < z m ≤ . (56)Further, from (44) one gets that: Θ m ≡ Θ( z m ) = − a a + a . (57)Substitution of (45) – (47) leads to the expression: Θ m = − βγ − ( α cos ε + sin ε ) βγ − cos ε ( α − sin ε (1 − α ) ) . (58)As the parameters α, βγ may be considered as independent, one can say that Θ m ≤ − ( α cos ε + sin ε ) − cos ε ( α − sin ε (1 − α ) ) . (59)The last expression becomes the equality when βγ = 14 ( α cos ε + sin ε ) . Afterelementary transformations one arrives at Θ m ≤ − ε α − 1) cos ε ) . (60)0 Modern Problems of Mathematical Physics. Special Issue № 3 Therefore, it is seen that the minimum of the parabola defined by the dependency Θ(cos θ ) is a strictly nonpositive quantity that achieves its maximum value equalto zero, under ε = π (and, simultaneously, βγ = 14 α ). Hence, the binomial Θ(cos θ ) does always have two real roots with respect to cos θ . Let us introduce the notation z + , z − for the bigger and the smaller root correspondingly: z ± = − a a ± r ( a a ) − a a (61)Further, notice that: dΘ(cos θ )d cos θ (cid:12)(cid:12)(cid:12)(cid:12) θ = ε = 4 βγ cos ε + α cos ε + sin ε. (62)The last equation is strictly nonnegative under ε ≤ π . Hence, for a motion thatstarts from the position where ε ≤ π , one can state that the region of motion willbe defined by max[ − , z + ] ≤ cos θ ≤ cos ε (63)when the condition (55) is satisfied and cos ε ≤ cos θ ≤ min[ z + , (64)when it is not satisfied.Considering the equations (63) – (64) and the equation (61), it is easy to get thecondition that at its highest position the mass center belongs to the same horizontalplane as the center of the ball. In order for this to happen the condition (55) andthe equation z + = 0 must hold, which is equivalent to a = 0 or βγ + (cid:0) ( α − cos 3 ε + ( α + 1)(3 α − 1) cos ε (cid:1) = 0 . (65)Note that the last condition can be satisfied only if it admits real roots for α (the other parameters being fixed) and only if one of the roots is in the range ≤ α ≤ . The roots for α may be transformed to the form: α , = 1cos ε ( − sin ε ± q sin ε − βγ cos ε ) (66)The condition of the existence of two real roots implies the inequality sin ε − βγ cos ε ≥ (67) Further in the text, the condition ε ≤ π will always be considered true.athematical Physics: Problems and Solutions α ≥ implies than only the bigger root may be a physicallymeaningful value and that the following inequality holds: sin ε cos ε − βγ cos ε ≥ , (68)Note that it already includes the inequality (67). The condition α ≤ leads to theinequality cos ε + 2 βγ cos ε ≥ , (69)which is satisfied automatically.We can finally conclude that for a top that was initially at the position where ε <π , the highest position for its mass center will be θ = π if the conditions (66), (55)hold, the necessary condition for the first one being (69).Let us now study the values of the root z + . When (55) holds (the top lifts up),it is always true that z + ≤ cos ε . The lower bound for z + may be derived afterinserting the equations (45) – (47) into (61), which gives, after transformations: z + = − λ + p ( µ + τ ) + λ − τ µ , (70)where λ = sin ε + α cos ε, (71) τ = cos ε ( α cos ε + ( − α ) sin ε ) , (72) µ = 4 βγ. (73)As in (70) the parameter µ is independent with respect to the other ones, onecan minimize this expression with the account of µ ≥ . It is easy to show that theexpression for z + is minimal in the limit µ → , converging to the value z + min = τλ (74)Now, considering α as a variable, it is easy to see that the equation z + min isminimal for α = 0 and in this case is equivalent to z + min | α =0 = − cos ε . Therefore,we conclude that, irrespectively to the parameters of the problem, if the mass centerlifts up at the beginning of motion, it is always in the range [ ε, π − ε ] .Similarly, for the downwards motion the minimal value for z + is cos ε . The maxi-mum value may be derived from the equation (70). Maximizing this expression withrespect to µ , one obtains that z + reaches its maximum value that is equal to unityin the limit µ → ∞ . Therefore, one can conclude that for any initial position of thetop the parameters of the problem may be chosen in such a way that the top may ap-proach arbitrarily close to the position θ = 0 . This result, in particular, correspondsto a simple case ω = 0 .In conclusion, notice that the equation (43) may be easily integrated in elemen-tary functions by the use of the variable change τ = cos θ .2 Modern Problems of Mathematical Physics. Special Issue № 3 11. Laplacian ∆ spectrum on a doughnut b ay x Fig. 7 Consider a torus made of a rectangular block ≤ x ≤ a, ≤ y ≤ b with glued opposite sides (see fig. 7) where theidentical arrows mark the sides to be glued to-gether. For sufficiently big ratio a/b , it is pos-sible to implement such torus nearly withoutdeformations, as the surface of a doughnut in3D space.If inside the doughnut magnetic field with the flux Φ is cre-ated, and also magnetic flux Φ is passing through the hole ofthe doughnut (see fig. 8), then the wave function of the station-ary state of the charged particle with the charge e and mass m , on the surface of the torus, is the eigenfunction Ψ( x, y ) ofthe operator Fig. 8 ˆ H = − ~ m "(cid:18) ∂∂x − i ec ~ A x ( x, y ) (cid:19) + (cid:18) ∂∂y − i ec ~ A y ( x, y ) (cid:19) , (1)where A ( x, y ) is a vector potential, ~ is Planck constant, and c is speed of light.Function Ψ( x, y ) satisfies to periodic boundary conditions: Ψ( x, 0) = Ψ( x, b ) , Ψ(0 , y ) = Ψ( a, y ) , (2) Ψ ′ y ( x, 0) = Ψ ′ y ( x, b ) , Ψ ′ x (0 , y ) = Ψ ′ x ( a, y ) . (3)Suppose the magnetic field turns to zero on the surface, we can make cuts ofthe surface and, using gauge transformation, nullify A on the surface of the torus.Then the wave function would be discontinuous at the cuts (such cuts are possibleto make exactly on the border of the rectangle). It is also clear that | Ψ( x, y ) | shouldnot change.After the gauge transformation, the task to find stationary states is modified andrequires to find eigenfunctions ψ ( x, y ) of the operator ˆ H = − ~ m "(cid:18) ∂∂x (cid:19) + (cid:18) ∂∂y (cid:19) , (4)with phase shift (lagging) boundary conditions: ψ ( x, 0) = e iϕ ψ ( x, b ) , ψ (0 , y ) = e iϕ ψ ( a, y ) , (5) athematical Physics: Problems and Solutions ψ ′ y ( x, 0) = e iϕ ψ ′ y ( x, b ) , ψ ′ x (0 , y ) = e iϕ ψ ′ x ( a, y ) , (6)(a) Find eigenfunctions and eigenvalues of the operator ˆ H .(b) What is the relationship between the fluxes Φ , Φ and phase displacements ϕ , ϕ ?SOLUTIONLet’s make mentioned above cuts of doughnut surface and nullify vector poten-tial A ′ with the help of gauge transformation A ′ ( x, y ) = A − ▽ f ( x, y ) = 0 . (7)Thus the Hamiltonian ˆ H is simplified. In spite of the fact that the wave functionwould undergo certain changes, it is clear that the probability | Ψ( x, y ) | of particlelocation should not be changed by gauge (gradient) transformation, i. e. the newwave function ψ ( x, y ) differs from the old one Ψ( x, y ) only in phase factor [8]: ψ ( x, y ) = Ψ( x, y ) exp (cid:18) − ie ~ c f ( x, y ) (cid:19) , (8)where f ( x, y ) is the function of gauge transformation, ψ ( x, y ) is the eigenfunctionof the new modified Hamiltonian ˆ H , that is in fact Laplace operator.a) Let us find eigenfunctions ψ ( x, y ) of the transformed Hamiltonian ˆ H . On R , eigenfunctions of Laplace operator can be chosen in the form of plane waves exp( i kr ) : ψ ( x, y ) = exp( ik x x ) exp( ik y y ) . (9)It is clear that any linear combination of eigenfunctions, for which the value k x + k y is the same, will also be an eigenfunction. Let us find k x and k y from the boundaryconditions (5), (6): k x = − ϕ a + 2 πn a , k y = − ϕ b + 2 πn b , (10)where n , n ∈ Z .Eigenvalues of ˆ H are found from the equation ˆ H ψ ( x, y ) = Eψ ( x, y ) and havethe form: E = ~ m "(cid:18) − ϕ b + 2 πn b (cid:19) + (cid:18) − ϕ a + 2 πn a (cid:19) . (11)Let us prove that we have found the complete basis of the ˆ H eigenfunctions.If the torus is without phase lagging, then we obtain the regular Fourier series.With phase lagging, not quite regular Fourier series is obtained, however it is re-ducible to the regular one: ψ ( x, y ) = exp h − i (cid:16) ϕ xa + ϕ yb (cid:17)i Ψ( x, y ) . Modern Problems of Mathematical Physics. Special Issue № 3 Here, Ψ is a regular periodic function for which the normal Fourier series is writtenas an expansion using the basis in L for which the completeness has been alreadyproved. Multiplication by exp h − i (cid:16) ϕ xa + ϕ yb (cid:17)i in terms of space L is unitarytransformation.b) Let us find the relationship between the fluxes of magnetic field Φ , Φ andphase lagging ϕ , ϕ . The flux of magnetic field is: Φ = ˛ S H d S = ˛ S rot A d S or,after a transformation using Stokes theorem: Φ = ˛ l A d l . Then the fluxes of magneticfield inside the torus and through the hole of the doughnut are respectively equalto: Φ = ˆ b A y ( x, y ) dy, (12) Φ = ˆ a A x ( x, y ) dx. (13)Recalling expression (7) and substituting it into (12) and (13): Φ = f ( x, b ) − f ( x, , (14) Φ = f ( a, y ) − f (0 , y ) . (15)It is possible to find the differences in values of f ( x, y ) in these points using expres-sion (8) and boundary conditions (2), (3), (5), (6). As a result we obtain: Φ i = ~ ce ϕ i = Φ ϕ i π , (16)where i = 1 , ; Φ is the quantum of magnetic flux. It is necessary to note that inthis case the magnetic flux is not quantized. 12. 3D Delta function Coulomb wave function of the ground state has the form ϕ c ( ~p ) = 8 παµ | ϕ c ( r = 0) | ϕ p , | ϕ c ( r = 0) | = α µ π , ϕ p = ( ~p + µ α ) − and satisfies Schroedinger equation in the momentum representation.The values of hyperfine splitting of the ground level of hydrogen-like atom withaccuracy up to α found on the basis of quasipotential built from the diagrams ofthe order of α and higher, are found most easily if to assume ϕ c ( ~p ) ≈ (2 π ) δ ( ~p ) | ϕ c ( r = 0) | . ( ⋆ ) athematical Physics: Problems and Solutions ⋆ ) by proving the following statements:(a) δ ( x ) = lim a → π aa + x , x ∈ R ; (b) πδ ( x )2 x = lim a → a ( a + x ) , x ∈ R ;(c) δ ( ~p ) = δ ( p )2 πp , p = | ~p | , p > ; (d) δ ( ~p ) = lim a → aπ ( a + p ) , p = | ~p | , p > .Direction: equations (a) – (d). containing generalized functions, of the class D ′ should be proved on the space of the basic functions of the class D .SOLUTIONThe above statement ( ⋆ ) is based on the fact that the value of the square ofmodule of the wave function (present in a matrix element) in coordinate space at r = 0 has the order of α .Let us prove the (a) statement.Let x ∈ R , ϕ ( x ) ∈ D ( R ) is the test function. Then (cid:18) lim a → aπ x + a , ϕ ( x ) (cid:19) = ∞ ˆ − ∞ dx lim a → aπ x + a ϕ ( x ) == lim a → aπ ∞ ˆ − ∞ dxϕ ( x ) x + a = lim a → aπ πi Res x = ia ϕ ( x ) x + a == lim a → aπ πi ϕ ( ia )2 ia = ϕ (0) = ( δ ( x ) , ϕ ( x )) . The first and the last parts are underlined to focus on equality of the functionalshence the generalized functions themselves.Derivation of the (с) relationship: δ ( ~p ) = δ ( p x ) δ ( p y ) δ ( p z ) = (2 π ) − ˆ e i~k~p d~k == (2 π ) − π ˆ π ˆ ∞ ˆ e ikp cos θ k dk sin θdϕdθ = (2 π ) − ∞ ˆ ˆ − e ikpx dxk dk == (2 π ) − ∞ ˆ e ikpx ikp (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − k dk = 1(2 π ) ip ∞ ˆ (cid:0) e ikp − e − ikp (cid:1) kdk = 1(2 π ) ip ∞ ˆ − ∞ e ikp kdk = Description of hyperfine splitting is based on computation of quantum distributions beingsquaresd by the wave function. From the explicit form of the wave function in the momentumrepresentation it turns that the terms of the lowest order in the expansion of a matrix element are α . Modern Problems of Mathematical Physics. Special Issue № 3 = 1(2 π ) ip ∂i∂p ∞ ˆ − ∞ e ikp dk = − πp δ ′ ( p ) = δ ( p )2 πp . where the known relation pδ ′ ( p ) = − δ ( p ) is utilized (the proof is partial integration).To check (d), we consider the test function Φ( ~p ) ∈ D ( R ) and perform thefollowing transformations: (cid:18) lim a → aπ p + a ) , Φ( ~p ) (cid:19) = ˆ d~p lim a → aπ p + a ) Φ( ~p ) == ˆ d Ω lim a → aπ ∞ ˆ p Φ( p, Θ , ϕ )( p + a ) dp == ˆ d Ω lim a → aπ ∞ ˆ p Φ( p, Θ , ϕ )( p + a ) d p − − ∞ ˆ p ′ Φ( − p ′ , Θ , ϕ )( p ′ + a ) dp ′ = { let Φ( p, Θ , ϕ ) = Φ( − p, Θ , ϕ ) } = ˆ d Ω lim a → a π ∞ ˆ − ∞ p Φ( p, Θ , ϕ )( p + a ) dp == ˆ d Ω lim a → a π πi Res p = ia p Φ( p, Θ , ϕ )( p + a ) ( residue in the pole of the 2nd order ) == ˆ d Ω lim a → aiπ lim p → ia ddp p Φ( p, Θ , ϕ )( p + ia ) == ˆ d Ω lim a → aiπ lim p → ia (cid:26) p Φ + p Φ ′ ( p + ia ) − p Φ( p + ia ) (cid:27) == ˆ d Ω lim a → aiπ (cid:26) ia Φ( ia, Θ , ϕ )(2 ia ) + ( ia ) Φ ′ ( ia, Θ , ϕ )(2 ia ) − ia ) Φ( ia, Θ , ϕ )(2 ia ) (cid:27) == ˆ d Ω lim a → aiπ (cid:26) Φ( ia )2 ia + Φ ′ ( ia )4 − Φ( ia )4 ia (cid:27) == ˆ d Ω lim a → aiπ (cid:26) Φ( ia )4 ia + Φ ′ ( ia )4 (cid:27) = ˆ d Ω Φ(0)4 π = Φ(0) = ( δ ( ~p ) , Φ( ~p )) . The truth of the statement (b) follows particularly from the proven (с) and (d).It is also possible to obtain the relationship (b) using the representation of the δ -function from the statement (a).It is possible to prove the relations (a) – (d) in another way.Using Sokhotsky’s formula x + i x − i πδ ( x ) , athematical Physics: Problems and Solutions D ′ , it is easy to derive eq. (a): lim a → π aa + x = lim a → π (cid:18) x − i a − x + i a (cid:19) == 12 π i (cid:18) P 1 x + i πδ ( x ) − P 1 x + i πδ ( x ) (cid:19) = δ ( x ) . It is possible to transform eq. (b) into the following form: δ ( x ) = lim a → π ax ( a + x ) . (1)Transforming the right part of the last equation, we obtain by recalling eq. (a): lim a → π ax ( a + x ) = lim a → π aa + x (cid:18) − a a + x (cid:19) == 2 δ ( x ) − lim a → a π ( a + x ) . (2)For any function ϕ ( x ) ∈ D , the following relations are true: (cid:18) lim a → a π ( a + x ) , ϕ (cid:19) = lim a → ˆ R a ϕ ( x )d xπ ( a + x ) == lim a → π ˆ R ϕ ( at )d t (1 + t ) == lim a → π ˆ R [ ϕ ( at ) − ϕ (0)](1 + t ) d t + 2 ϕ (0) π ˆ R d t (1 + t ) . The second integral in the last expression can be computed with the help of thetransformation t = tg α , d t = ( t + 1) d α : + ∞ ˆ −∞ d t (1 + t ) = π ˆ − π cos α d α = π . The first integral can be estimated in the following way. Let us split the integralinto the two integrals: lim a → π ˆ R [ ϕ ( at ) − ϕ (0)](1 + t ) d t =lim a → ˆ [ − A ; A ] π [ ϕ ( at ) − ϕ (0)](1 + t ) d t + lim a → ˆ R \ [ − A ; A ] π [ ϕ ( at ) − ϕ (0)](1 + t ) d t. Modern Problems of Mathematical Physics. Special Issue № 3 The first integral on the right side of the last equation tends to zero at a → .For the upper limit of the second integral, it is possible to use the value of lim a → ˆ R \ [ − A ; A ] π M (1 + t ) d t = 4 Mπ ˆ R \ [ − A ; A ] d t (1 + t ) , independent of a . At A → ∞ , this integral also converges to zero. Thus (cid:18) lim a → a π ( a + x ) , ϕ (cid:19) = ϕ (0) . It follows that lim a → a π ( a + x ) = δ ( x ) . Recalling this result, eq. (2) converges to eq. (1), which proves the validity ofeq. (b).Eq. (c) should take place on the functions from D ( R ) , i. e. from that, the eq.should follow: ˆ R δ ( ~p ) ϕ ( ~p )d ~p = ˆ R δ ( p )2 πp ϕ ( ~p )d ~p, (3)where ϕ ( ~p ) ∈ D ( R ) . Conversing the right part of this equality, we obtain ˆ R δ ( p )2 πp ϕ ( ~p )d ~p = ˆ R δ ( p )2 πp ϕ ( ~p ) p d p dΩ == ˆ p > δ ( p ) ˆ | ~p | = p ϕ ( ~p ) dΩ4 π dp = ˆ R δ ( p ) ˜ ϕ ( p ) dp = ˜ ϕ (0) = ϕ (0) . Here, Ω is solid angle, ˜ ϕ ( p ) = ˆ | ~p | = p ϕ ( ~p ) dΩ4 π . In the last line, function ˜ ϕ was evenlyextended to negative p . The integral in the left part of eq. (3) also equals ϕ (0) whichproves the validity of that eq. (3) as well as of eq. (c).Equation (d) automatically follows from (b) and (c): δ ( ~p ) = δ ( p )2 πp = 12 π lim a → π a ( a + p ) = lim a → aπ ( a + p ) . Let us obtain the final approximate equation ( ⋆ ) for the wave function of theground state in Coulomb field using the formula δ ( ~p ) = δ ( p )2 πp , valid for the case of athematical Physics: Problems and Solutions ϕ c ( ~p ) = 8 παµ | ϕ c ( r = 0) | p + α µ ) , and, using the identity (d), we obtain lim α → ϕ c ( ~p ) = lim α → παµ | ϕ c (0) | ( ~p + α µ ) = 8 π | ϕ c ( r = 0) | δ ( ~p ) . Due to the fact that αµ ≪ , the function approximately equals its limit at αµ → .From the form of Coulomb wave function of the ground state it follows thatthe main contribution into the splitting of the energy levels is due to the momentumfrom the range that satisfies the condition ~p ∼ α µ . As a result, expansion of theintegrand by p/m would be equivalent to an expansion of the whole integral by α (under condition that the integral converges). 13. Heat conduction equation (heat source presents) The temperature on the ends of thin regular rod is maintained constant andequals zero. Lateral face of the rod is heat-insulated. The frame of reference is definedso that coordinate axis x is oriented along the rod, its ends have the coordinates x = 0 and x = l .Thermal diffusivity coefficient of the material of the rod equals a .Find spatial and time distribution T ( x, t ) ( x l , t > ) of temperaturealong the rod in two cases:(a) at time moment t = 0 the temperature of the rod is constant, T ( x, ≡ T at < x < l ;(b) in the center of the rod, point source of intensity Q is switched on at timemoment t = 0 , and T ( x, ≡ at ≤ x ≤ l .SOLUTIONCase (a). In this case, the decision function is defined as the solution of heatconduction equation T t = a T xx , (1)satisfying supplementary conditions T (0 , t ) = T ( l, t ) = 0 ( t > , (2) T ( x, 0) = T (0 < x < l ) . (3)0 Modern Problems of Mathematical Physics. Special Issue № 3 First, let us find the eigenvalues and eigenfunctions of equation (1) using separa-tion of variables (Fourier method).Representing the decision function in the form T ( x, t ) = X ( x ) U ( t ) and substituting the expression into (1), we obtain X ˙ U = a U X ′′ . Hence X ′′ X = ˙ Ua U = − λ ,X ′′ + λ X = 0 , (4) ˙ U + a λ U = 0 , (5)where λ = const. Solving eq. (4) and (5) with consideration of boundary condi-tions (2), we obtain: X n = r l sin (cid:16) πnxl (cid:17) ( x l , orthonormal system of functions) , (6) U n = C n exp (cid:18) − (cid:16) πnal (cid:17) t (cid:19) ( t > , (7)where n = 1 , , ... , and C n – real coefficients. It is possible to represent the generalsolution in the form: T ( x, t ) = ∞ X n =1 X n ( x ) U n ( t ) . Thus T ( x, t ) = ∞ X n =1 C n r l sin (cid:16) πnxl (cid:17) exp (cid:18) − (cid:16) πnal (cid:17) t (cid:19) . (8)According to (8), expansion of the distribution function of initial temperatureinto series by eigenfunctions (6) takes the form T ( x, 0) = ∞ X n =1 C n r l sin (cid:16) πnxl (cid:17) , where C n = ˆ l T ( ξ, r l sin (cid:18) πnξl (cid:19) dξ. athematical Physics: Problems and Solutions C n = r l T ˆ l sin (cid:18) πnξl (cid:19) dξ == r l T lπn [ − cos πn + 1] = √ lT πn (1 − ( − n ) == √ lπ (2 k + 1) T , n = 2 k + 1 , , n = 2 k + 2 ( k = 0 , , , ... ) . (9)Substituting the expression (9) into (8), we obtain the solution for the case (a): T ( x, t ) = 4 T π ∞ X k =0 k + 1 exp − (cid:18) π (2 k + 1) al (cid:19) t ! sin (cid:18) π (2 k + 1) xl (cid:19) . Case (b). Method I. In this case, the decision function is defined as the solution ofheat conductivity equation with the source, i. e. boundary problem with the densityof heat generation described by Dirac δ -function: T t = a T xx + Qc δ (cid:18) x − l (cid:19) , T ( x, 0) = 0 , T (0 , t ) = T ( l, t ) = 0 . (10)Here, c is heat capacity of unit length of the thin rod, Qδ (cid:18) x − l (cid:19) – intensity ofheat generation per unit length.We shall search for the solution in the form of the sum of a stationary one ( ω )with a non-stationary one ( v ): T ( x, t ) = ω ( x ) + v ( x, t ) . Stationary solution satisfies ω xx = − Qa c δ (cid:18) x − l (cid:19) , ω (0) = ω ( l ) = 0 . And non-stationary solution satisfies v t = a v xx , v (0 , t ) = v ( l, t ) = 0 , v ( x, 0) = − ω ( x ) . The common solution of stationary equation can be presented in the followingform (see [2]) ω = C (cid:12)(cid:12)(cid:12)(cid:12) x − l (cid:12)(cid:12)(cid:12)(cid:12) + C , Modern Problems of Mathematical Physics. Special Issue № 3 where indeed the first constant can be found from differential equation ω xx = − Qa c δ ( x − l ⇒ C = − Q a c , and C from boundary conditions ω (0) = ω ( l ) = 0 , exactly − Q a c l C = 0 ⇒ C = Qa c l . Finally ω ( x ) = − Q a c (cid:12)(cid:12)(cid:12)(cid:12) x − l (cid:12)(cid:12)(cid:12)(cid:12) + Qa c l . Non-stationary part of the problem is solved using separation of variables, as inthe (a) case, only with the boundary condition v ( x, 0) = − ω ( x ) . As a result, weobtain T ( x, t ) = ω ( x ) + v ( x, t ) .Note, that expressions for ω ( x ) and v ( x, t ) can be presented as Fourier seriesexpansions (see details in [2]): ω ( x ) = 2 Qlπ a c ∞ X k =0 ( − k (2 k + 1) sin (cid:18) π (2 k + 1) xl (cid:19) , (11) v ( x, t ) = − Qlπ a c ∞ X k =0 ( − k (2 k + 1) sin (cid:18) π (2 k + 1) xl (cid:19) exp − (cid:18) π (2 k + 1) al (cid:19) t ! . (12)The answer will be T ( x, t ) == 2 Qlπ a c ∞ X k =0 ( − k (2 k + 1) sin (cid:18) π (2 k + 1) xl (cid:19) " − exp − (cid:18) π (2 k + 1) al (cid:19) t ! . (13)Case (b). Method II. When solving (10), let us expand T ( x, t ) into Fourier series T ( x, t ) = ∞ X n =1 C n ( t ) sin (cid:16) πnxl (cid:17) , where C n ( t ) = 2 l l ˆ T ( x, t ) sin (cid:16) πnxl (cid:17) dx. athematical Physics: Problems and Solutions C n ( t ) , the equation (10) will take the form (use the expansion ofdelta-function) ˙ C n ( t ) = − (cid:16) πnal (cid:17) C n ( t ) + 2 Qlc sin (cid:16) πn (cid:17) . (14)The general solution of the eq. (14) is the sum of the general solution of theuniform equation and partial solution of the non-uniform equation, i. e. C n ( t ) = A n exp (cid:18) − (cid:16) πnal (cid:17) t (cid:19) + 2 Qlc sin (cid:16) πn (cid:17) (cid:16) πnal (cid:17) − . (15)Recalling the initial condition (10), we obtain C n (0) = 2 l l ˆ T ( x, 0) sin (cid:16) πnxl (cid:17) dx = 0 . (16)Considering (16), from (15) it follows that A n = − Qlc sin (cid:16) πn (cid:17) (cid:16) πnal (cid:17) − . (17)Substituting (17) into (15), we obtain C n ( t ) = 2 Qlc sin (cid:16) πn (cid:17) (cid:16) πnal (cid:17) − (cid:20) − exp (cid:18) − (cid:16) πnal (cid:17) t (cid:19)(cid:21) . (18)Taking into account (18), we can write the solution of eq. (10) in the form T ( x, t ) = 2 Ql ∞ X n =1 sin (cid:16) πn (cid:17) (cid:16) πnal (cid:17) − (cid:20) − exp (cid:18) − (cid:16) πnal (cid:17) t (cid:19)(cid:21) sin (cid:16) πnxl (cid:17) . (19)Because sin (cid:16) πn (cid:17) = ( ( − k , n = 2 k + 1 , , n = 2 k + 2 , where k = 0 , , , ... , the solution (19) coincides with (13).In the limit t → ∞ , the solution (19) would tend to the stationary solution ω ( x ) ,determined by (11). 14. Heat conduction equation with nonlinear add-on Burgers’ equation is a fundamental partial differential equa-tion from fluid mechanics and other areas of applied math-ematics. It bears the name of the Dutch physicist JohannesMartinus Burgers (1895 – 1981). For a given velocity of afluid u and its viscosity coefficient ν , the general form ofBurgers’ equation has the following form: v t + vv x = νv xx .4 Modern Problems of Mathematical Physics. Special Issue № 3 Show that it could be linearized by substitution v = − ν ∂∂x ln f, and reduced to the heat conductivity equation f t = νf xx .SOLUTIONSubstitute the expression v = − νf x /f (1)into the Burgers equation. The result of such substitution is that the all derivativesin the Burgers equation obtain the form: v t = − ν f xt f + 2 ν f x f t f , v x = − ν f xx f + 2 ν f x f ,v xx = − ν f xxx f + 6 ν f xx f x f − ν f x f . (2)Substituting the expressions (2) into the Burgers equation, we obtain − f xt f + f x f t f = ν (cid:16) − f xxx f + f xx f x f (cid:17) . The obtained equation can be transformed in the following way: ∂∂x (cid:16) f t f (cid:17) = ν ∂∂x (cid:16) f xx f (cid:17) . Then, for f we obtain almost the equation of heat conductivity (or diffusion): f ( x, t ) t = νf ( x, t ) xx + F( t ) f ( x, t ) , where F( t ) is an arbitrary time function. If F( t ) = 0 , we really obtain the heatconductivity (or diffusion) equation.Short reference. Suppose that in a certain region of space all particles are movingalong straight lines parallel to the X axis.Let us designate v = dx/dt – the projection of the medium velocity (being thefunction of the coordinate of the point x and time t ) on the X axis. The equationof free one-dimensional motion of incompressible fluid is written in the form: v t + vv x = 0 (3)and, as seen, is non-linear. It has a solution in the form of traveling waves the frontof which is becoming more steep with time and as a result the wave breaks. Thereare many examples of breaking waves from which perhaps the most visual would be athematical Physics: Problems and Solutions v t + vv x = νv xx . Here, ν is the viscosity factor. Within this model, it is possible to describe the wavesin which the competition takes place between the two opposite processes, steepingwave fronts due to non-linearity and quenching due to viscosity. As a consequenceof such competition, stationary motion can appear.The point of interest of the Burgers equation is the existence of exact solutionbuilt by Hopf [9] and Cole [10]. Transformation leading to linearization of the Burg-ers equation (recalled in the statement of this problem) is called in literature asCole–Hopf transformation. eferences [1] Thornton S.T., Marion J.B. Classical Dynamics of Particles and Sys-tems. 5ed. Thomson, 2004. P. 277; and in the 1st volume of "GeneralPhysics" by D.V. Sivukhin (Moscow, FIZMATLIT, 2005, p.148).[2] Mathematical Physics: Problems and Solutions of Distance All-Russian Stu-dents Training Contest Olympiad in Mathematical and Theoretical Physics(May 21st–24th, 2010) / [G.S. Beloglazov et al.]. Ser. Modern Problems inMathematical Physics. Special Issue No. 4. Samara : Samara University Press,2010. 84 p.[3] Weinberg S. Gravitation And Cosmology: Principles And Applications Of TheGeneral Theory Of Relativity. Wiley, 2008. 678 p.; Gravitation and Cosmology.1972; 2008.[4] Wald R.M. General Relativity. The University of Chicago Press, 1984.[5] Guth A.H. Inflationary Universe: A possible solution to the horizon and flatnessproblems // Phys. Rev. D. 1981. Vol. 23. №2.[6] Peskin M.E., Schroeder D.V. An Introduction to Quantum Field Theory. Read-ing. MA: Addison-Wesley, 1995. 842 p.[7] Nie S., Sher M. Vacuum stability bounds in the two Higgs doublet model.Phys. Lett. B. 1999. V. 449. P. 89–92.[8] Landau L.D., Lifshitz L.M. Quantum Mechanics. Non-Relativistic Theory. But-terworth-Heinemann; Third Edition: Volume 3. 1981. 689 p.[9] Hopf E. The partial differential equation u t + uu x = µu xx // Comm. Pure Appl.Math. 1950. V. 3. P. 201–230.[10] Cole J. D. On a quasi-linear parabolic equation occurring in aerodynamics //Q. Appl. Math. 1951. V. 9. P. 225–236. athematical Physics: Data on authors Series «Modern Problems in Mathematical Physics». Special Issue № 3 AnnexStatements of the Problems of the Second International Olympiadon Mathematical and Theoretical Physics«Mathematical Physics» September, 4 – 17, 2010 v t + vv x = νv xx canbe linearized using Coal–Hopf transformation v = − ν ∂∂x ln f. (1)Here, v ( x, t ) is the solution of Burgers equation, f ( x, t ) – solution of the heatconduction equation f t = νf xx . For the Burgers equation, the initial condition is given as: v ( x, (cid:12)(cid:12) t =0 = v ( x ) , ∞ ˆ −∞ v ( x ) dx < ∞ . (a) Using the transformation (1) for v ( x ) , find the corresponding function f ( x ) initial condition for the heat conduction equation. athematical Physics: Annex f ( x, t ) = 1 √ πνt ∞ ˆ −∞ f ( y ) exp " − ( x − y ) νt dy, x, y ∈ R , t ≥ . (2)Using transformation (1), obtain the solution v ( x, t ) of the Burgers equation.Hints1. To answer the question (b) use the result obtained in (a) for the presentproblem.2. It is convenient to express the answer to (b) using the function ψ ( x, t ; y ) = y ˆ v ( x ′ ) dx ′ + 12 t ( x − y ) . (c) For Burgers equation, the initial condition is given: v ( x, 0) = v ( x ) = 11 + ( x − + 11 + ( x + 5) . Use for this case the solution scheme developed above in paragraphs (a)and (b), find system’s response v ( x, t ) . Follow and analyse evolution of theobtained solution in time. Utilize the ’Mathematica’ package.2) Harmony of a flute.In woodwind and brass musical instruments, the source of the sound isthe oscillating column of air. In a pipe, the standing waves emerge. Suchvibrations occur at certain eigen frequencies.Oscillations of pressure in a pipe of length L are described by the wave equation ∂ p∂x = ρ β ∂ p∂t , where p is the overpressure (relative to the atmospheric), ρ – density of air inthe pipe, β – modulus of volume elasticity, x – coordinate along the pipe axis(see fig. 1), t is time.A specific solution of that equation is the function p ( x, t ) = ( A cos kx + B sin kx ) cos ωt. (a) Find the values of A, B, k, ω when both ends of the pipe are open, andalso the condition p ( x = L/ , t = 0) = p > is met.0 Series «Modern Problems in Mathematical Physics». Special Issue № 3 Fig. 1 (b) On the basis of the solution obtained, analyse the time evolution of the gaspressure p ( x, t ) in the pipe. For this purpose, you are encouraged to usethe graphics features of the ’Mathematica’ software package.For your information:1) air density under normal conditions is ρ = 1 , kg/ m , and the modulusof volume elasticity is β = 1 , × Pa;2) the length of a flute may vary widely, so for illustration, it is possible tochoose L =0.5 m.(с) Illustrate the obtained solution using sound synthesis features of the ’Math-ematica’ software package. Stipulate an opportunity to hear the fundamentaltone and some overtones of the pipe of variable length. How would the tone ofthe pipe depend on the following parameters: ρ , β, L ?3) From the history of LHC: LEP.At the end of the XXth century, the colliding beams experiments on electron-positron accelerator have been held in CERN. Such collider is known as theLEP-collider (Large Electron-Positron). The detectors (see figure) recordingcollisions of particles with anti-particles were placed at the intersections ofthe colliding beams.Mounting the ALEPH detector Fig. 2 ALEPH, end view Fig. 3 To study the collision pattern, it is required not only to find out which par-ticles are born but also to measure their characteristics with high precision,reconstruct the particles’ trajectories, find out their momenta and energies. athematical Physics: Annex Fig. 4. Curvature of the particlestracks in a magnetic field Such measurements are held with the aidof various types of detectors that coaxiallysurround the place of the collision of theparticles. In the area of magnetic field, cur-vature of a trajectory (see fig. 4) enables tofind out the momenta of the products of areaction.(a)On the figure, the event of the birthof a neutral K – meson (kaon) isshown. The length of its trajectoryis 0.1542208 m, the momentum equals . · − kg m/s (or 2.240160GeV/ c ), the speed of meson is . c ,where c is the speed of light in vacuum. Us-ing these data, find out intrinsic lifetime of K – meson, its total and kinetic energies(in GeV). Fig. 5 (b)In a magnetic field with induction B = 1 . T, K – meson decays into π + and π − – mesons with momenta . · − kg m/s and . · − kgm/s, respectively. Analyse maximum possible value of radii of the circles oflateral motion (with respect to ~B ) of π ± -mesons. Also, find the angle of theirdivergence. The elementary charge e = 1 . · − Clmb.4) Virial of gravitational collapse. In classical mechanics of systems executingfinite motion, the following relationship takes place: h K i = − D X i ~F i · ~r i E . (3)Here, h K i is the mean (for sufficiently long time interval) kinetic energy ofthe system of point particles defined by radius vectors ~r i and exposed to theaction of the forces ~F i .2 Series «Modern Problems in Mathematical Physics». Special Issue № 3 A planet revolves around the Sun. Interac-tion between the planet and the Sun obeysthe law of universal gravitation. The massof the Sun is much larger than the massof the planet so the heliocentric referenceframe can be considered inertial.(a)Obtain the relationship between the meankinetic h K i and mean potential h U i ener-gies of the planet directly from the VirialTheorem (3). Fig. 6 to the problem 4) (a)(b)A planet revolves around the Sun alongthe circular orbit of the radius R . Showthat the kinetic K and potential U ener-gies of the planet on its circular orbit arerelated in the following way K = − U. (4)(c)A planet revolves around the Sun alongthe circular orbit of the radius R . If themass of the Sun would instantly diminishby 2 times, what will be the trajectory ofthe planet? What relationship would begiven by the Virial Theorem in this case? Fig. 7 to problems 4) (b), (c)5) Waves on Moebius strip. A Moebius strip is a rectangular block ≤ x ≤ a , ≤ y ≤ b , where points with coordinates (0 , y ) and ( a, b − y ) are gluedtogether (see fig.).For sufficiently large ratio a/b , the Moebius strip can be implemented nearlywithout stretching as a surface with an edge in three-dimensional space.Let the oscillations of the surface of Moebius strip be described by the waveequation for the function u ( x, y, t ) u tt − △ u = 0 . The edge of the Moebius strip is free, and hence Neuman’s boundary conditionis set (see fig. 8) u y ( x, , t ) = u y ( x, b, t ) = 0 . athematical Physics: Annex Fig. 8 1) State boundary conditions on the gluing line (points with the coordinates (0 , y ) and ( a, y ) , y ∈ [0 , b ] ) corresponding to the longitudinal vibrations ( u is asmall displacement along the surface). Find eigen harmonic oscillations as thesolutions of the wave equation with corresponding boundary conditions.2) State boundary conditions on the gluing line (points with the coordinates (0 , y ) and ( a, y ) , y ∈ [0 , b ] ) corresponding to the transverse vibrations ( u isa small displacement perpendicular to the surface). Find eigen harmonic os-cillations as the solutions of the wave equation with corresponding boundaryconditions.6) Collapse of a bubble. Smooth 2-dimensional surface without self-intersectionsin 3-dimensional space is topologically equivalent to a sphere. At the initialtime moment, the surface bounds the volume V . Points of the surface aremoving with normally oriented variable velocities. At each time moment, theprojection of the velocity on the internal normal equals the Gauss curvature(product of the two main curvatures) of the surface. Let the surface remainsmooth during the process of the motion, self intersections do not occur. Atcertain time moment, the surface collapses into a point. What time will it takethe surface to collapse into a point?7) Random problem. Let ξ and ξ be positive random variables on probabilityspace { Ω , F , P } , and such that for all real p ∈ [ a, b ] , < a < b , E ξ p = E ξ p < ∞ . E is denoted as the operator of mathematical expectation value: E ξ p := ˆ Ω ξ p d P = + ∞ ˆ −∞ x p dF ξ ( x ) . Prove that their distribution functions coincide: F ξ ( x ) = P { ξ x } = P { ξ x } = F ξ ( x ) for all x ∈ R . Series «Modern Problems in Mathematical Physics». Special Issue № 3 8) Maximal domain for a matrix. Find maximal domain in which the Cauchyproblem U ( x, t ) | t =0 = T ( x ) , ∂U∂t | t =0 = N ( x ) , (5)for the system of equations U tt − AU xx = 0 , (6)with the matrix A = (cid:18) (cid:19) , U = (cid:18) u u (cid:19) has a unique solution for any x ∈ (0 , . 9) Abel’s Analogue. In 1823 Abel has been working on the generalization of theTautochrone Problem (to find a curve along which a heavy particle movingwithout friction would reach its lowest position for the same time independingon its initial position). Abel has reached to the equation x ˆ f ( t ) dt √ x − t = ϕ ( x ) , (7)where f ( x ) is the decision function, ϕ ( x ) – given function.Solution of the equation has the form f ( x ) = 1 π ddx x ˆ ϕ ( t ) dt √ x − t . In the present problem, it is offered to find a solution of the trigonometricanalogue of the equation (7) ϕ ( x ) = x ˆ f ( t ) dt p sin( x − t ) , < t < x < π , (8)where ϕ ( x ) = 1 √ cos x . 10) Problem of energy decomposition. Let u ( x, t ) ∈ C ( R × [0 , ∞ )) be a solutionof the Cauchy initial value problem for one-dimensional wave equation u tt − a u xx = 0 in R × (0 , ∞ ) , with initial conditions u ( x, 0) = g ( x ) , u t ( x, 0) = h ( x ) , athematical Physics: Annex g ( x ) , h ( x ) are finite functions.Kinetic energy K ( t ) = 12 + ∞ ˆ −∞ u t ( x, t ) dx. Potential energy P ( t ) = 12 + ∞ ˆ −∞ u x ( x, t ) dx. Prove thata) K ( t ) + P ( t ) a is constant for any t. b) K ( t ) = P ( t ) a for rather large t. 11) Dirac Problem. When deriving so called «Dirac equation» in relativistic quan-tum mechanics, Dirac has been driven by an idea of «square-rooting» froma second order differential operator.Find out in terms of square operator of the first order:а) a wave one-dimensional operator;б) Laplace operator in R . 12) Certain process for a wave equation.Some process is simulated by a function u ( x, t ) , that satisfies the initial condi-tions u ( x, 0) = (cid:20) sin πx, ≤ x ≤ , , x < and x > , ∂u∂t ( x, 0) = 0 . It is known that even part of this function u r ( x, t ) satisfies the wave equation u rtt − a u rxx = 0 in half plane t > . Odd part of this function u n satisfies the wave equation u ntt − b u nxx = 0 . Find the distance between x -coordinates at which u ( x, T ) has minimal valuesat sufficiently large T .13) Maximal domain and a square. For the equation u xx + √ yu xy = 0 (9)find maximal domain area on the x - y plane, where u ( x, x ) = ϕ ( x ) , ∂u∂x − ∂u∂y = ψ ( x ) , < x < . Show that this domain can be divided into 3 parts by straight linear cuts fromwhich it is possible to make a square block. What will be the area of suchsquare block?6 Series «Modern Problems in Mathematical Physics». Special Issue № 3 14) Evaluation of the solution of the ultrametric diffusion type of equation withfractional derivative. When solving equations of the ultrametric diffusion type(such equations are related to describing conformation dynamics of compoundsystems such as biomacromolecules), the solutions are often represented in theform of exponent series. One of such series is presented below: S ( t ) = ∞ X i =0 a − i E β ( − b − i t β ) . Here, E β ( z ) = ∞ X n =0 z n Γ( βn + 1) is Mittag-Leffler function, < β , and t istime, S ( t ) – probability of finding the system in definite state groups, a > b > S ( t ) at t → ∞ and find its asymp-totic evaluation by t -depending elementary functions. ATHEMATICAL PHYSICS PROBLEMS AND SOLUTIONSThe Students Training Contest Olympiadin Mathematical and Theoretical Physics( On May 21st – 24th, 2010 )Special Issue № 3of the Series of Proceedings «Modern Problems of Mathematical Physics»Authors:G.S. Beloglazov, A.L. Bobrick, S.V. Chervon, B.V. Danilyuk,M.V. Dolgopolov, M.G. Ivanov, O.G. Panina, E.Yu. Petrova, I.N. Rodionova,E.N. Rykova, I.S. Tsirova, M.Y. Shalaginov, I.V. Volovich, A.P. ZubarevTitle Editing T.A. MurzinovaComputer Design M.V. DolgopolovArt drawings Jy.A. NovikovaCover Art Design L.N. Zamamykina Signed for printing: 31.11.2010. Format 70 ××