ENSENANZA˜ REVISTA MEXICANA DE FISICA´ E 54 (2) 212–215 DICIEMBRE 2008 An alternative solution to the general tautochrone problem R. Gomez´ and V. Marquina Facultad de Ciencias, Departamento de F´ısica, Universidad Nacional Autonoma´ de Mexico,´ Av. Universidad 3000. Mexico´ D.F., 04510, Mexico,´ e-mail: [email protected], [email protected] S. Gomez-A´ ´ıza Escuela Nacional Preparatoria 6, Antonio Caso, Universidad Nacional Autonoma´ de Mexico,´ Corina 3, Mexico,´ D.F., 04100 Mexico,´ e-mail: [email protected] Recibido el 3 de julio de 2008; aceptado el 2 de septiembre de 2008 In 1658, Blaise Pascal put forward a challenge for solving the area under a segment of a cycloid and also its center of gravity. In 1659, motivated by Pascal challenge, Huygens showed experimentally that the cycloid is the solution to the tautochrone problem, namely that of finding a curve such that the time taken by a particle sliding down to its lowest point, under uniform gravity, is independent of its starting point. Ever since, this problem has appeared in many books and papers that show different solutions. In particular, the fractional derivative formalism has been used to solve the problem for an arbitrary potential and also to put forward the inverse problem: what potential is needed in order for a particular trajectory to be a tautochrone? Unfortunately, the fractional derivative formalism is not a regular subject in the mathematics curricula for physics at most of the Universities we know. In this work we develop an approach that uses the well-known Laplace transform formalism together with the convolution theorem to arrive at similar results. Keywords: Tautochrone; Laplace transform; convolution theorem. En 1658, Blaise Pascal lanzo´ el reto de determinar el area´ debajo de la curva de un segmento de cicloide, as´ı como su centro de gravedad. En 1659, motivado por el reto de Pascal, Huygens demuestra experimentalmente que la cicloide es la solucion´ al problema de la tautocrona,´ es decir, al problema de encontrar una curva tal que, si una part´ıcula engarzada en ella se mueve por la accion´ del campo gravitacional uniforme, su tiempo de descenso es independiente de la posicion´ inicial. Desde entonces, este problema ha sido tratado en muchos libros y art´ıculos con diferentes soluciones. En particular, el formalismo de derivadas fraccionales ha sido utilizado para resolver el problema en el caso de un potencial arbitrario as´ı como el problema inverso: ¿que´ potencial se requiere para que una trayectoria, en particular, sea una tautocrona?´ Desafortunadamente, el formalismo de derivadas fraccionales no forma parte de la curr´ıcula de la carrera de F´ısica de muchas de las Universidades que conocemos. En este trabajo desarrollamos un calculo´ que utiliza el bien conocido formalismo de la transformada de Laplace, que junto con el teorema de convolucion,´ nos lleva a resultados similares. Descriptores: Tautocrona;´ transformada de Laplace; teorema de convolucion.´ PACS: 45.20.-d; 02.30.Uu 1. Introduction The problem has been, and still is, of interest, espe- cially because of its similarity with the brachistochrone prob- As stated in the abstract, Huygens showed experimentally lem [3-5]. Great mathematicians such as Joseph Louis La- that a cycloid is the tautochrone curve for a particle sliding grange and Leonhard Euler looked for an analytical solution without friction in the uniform gravitational field. He pub- to the problem and Niels H. Abel used the Laplace formalism lished this result in his book Horologium oscillatorium [1] to solve it [6]. Ever since, many papers have been published and some years later, when J. Bernoulli found that the cy- dealing with tautochrone curves under special physical con- cloid is also a brachistochrone, he wrote [2]: ditions, such as: the tautochrone with friction [7], the rela- tivistic tautochrone [8], the tautochrone in rotating frames of reference [9] and the tautochrone under an arbitrary poten- Before I end I must voice once more the admi- tial [10]. In this last paper, Flores and Osler introduced the ration I feel for the unexpected identity of Huy- fractional derivatives formalism to solve the problem and to gens’ tautochrone and my brachistochrone. I generalize to an arbitrary potential energy function, and also consider it especially remarkable that this coin- to solve the inverse problem. However, the fractional deriva- cidence can take place only under the hypothe- tive formalism is not a regular subject in the under graduate sis of Galileo, so that we even obtain from this syllabus of physics or mathematics studies. In this paper we a proof of its correctness. Nature always tends develop the same type of generalization, but using the more to act in the simplest way, and so it here lets accessible and well-known formalism of Laplace transform. one curve serve two different functions, while under any other hypothesis we should need two curves. AN ALTERNATIVE SOLUTION TO THE GENERAL TAUTOCHRONE PROBLEM 213 2. The Laplace transform formalism T), where the ending point has been taken, without lost of generality as y = 0, so that equation can be written as: Before we move toward the tautochrone problem, we shall ZT Zy=0 µ ¶ recall a pair of definitions and the convolution theorem [11]: p p dσ 2gdt = 2gT = ¡ (y ¡ y)¡1=2 dy 0 dy 1.- The Laplace transform f(s) of a function F (t) is de- 0 y=y0 fined as: Zy0 µ ¶ 1 ¡1=2 dσ Z = (y0 ¡ y) dy (7) ¡st dy f(s) = L fF (t)g = e F (t)dt (1) 0 0 For a tautochrone, the time of descent T is to be constant and the convolution between two functions as: independent of y0, so that the upper limit of Eq. (7) is ar- bitrary and this fact makes it possible to take the integrand Zt as the convolution of dσ=dy. The convolution theorem then F1(z)F2 (t ¡ z) dz = F1 ¤ F2 (2) states that p dσ 0 2gT = y¡1=2 ¤ ; (8) dy 2.- The convolution theorem: and calculating the corresponding Laplace transforms one If f1(s) and f2(s) are the Laplace transforms of F1(t) obtains ½ ¾ n o ½ ¾ r and F2(t), respectively, then p 1 dσ dσ ¼ 2gT = L L y¡1=2 = L (9) 8 9 s dy dy s <Zt = f1 (s) f2 (s) = L F1 (z) F2 (t ¡ z) dz because thep Laplace transform of a constant A is A/s and that : ; of y¡1=2 is ¼=s. Applying the inverse transform yields: 0 p dσ 2gT ¡1=2 = L (F1 ¤ F2) (3) = y (10) dy ¼ Squaring, separating variables, and integrating we arrive 3. The Laplace formalism in the Tautochrone at the parametric equations of a cycloid passing through the problem origin [11]. This is the usual method for solving the tau- tochrone in the homogeneous gravitational field. The same The path-time employed in a given trajectory defined by an method can be used to solve the general case, in which the arc element dσ potential energy function is arbitrary. We give the solution in q p p the next section. dσ= dx2 + dy2= 1 + (dx=dy)2dy= 1 + x02dy (4) with a velocity v 4. General case p v = (2=m)[U (y0) ¡ U (y)]; (5) In the general case for an arbitrary potential, we cannot use the convolution theorem because the left side of Eq. (6) does where m is the mass of the particle and U(y) is the potential not contain a term y0 ¡ y [as in Eq. (7)]. However, making energy function, can be written as: z = U(y) we can write Zt Z dσ dσ dz dσ dU 0 dσ dy= dy= dz and dz= dy = U dy (11) dt = ¡ dy dz dy dz dy v so then Eq. (6) takes the form t0 σ r T r z Zy Z Z µ ¶ dσ p 2 2 ¡1=2 dσ = ¡ (2=m)[U (y ) ¡ U (y)]dy; (6) dt= T =¡ (z0 ¡ z) dz (12) dy 0 m m dz 0 z0 y0 Under the isochronal condition, the left side of Eq. (12) the minus sign is because dσ=dt < 0. As stated before, the is a constant, so the right side does not depend on the integra- standard way to solve the problem using the Laplace formal- tion limits and we can apply the convolution theorem; then ism was put forward by Abel [6], and it can be seen, for in- (r ) r ½ ¾ 2 2 T n o dσ stance, in Arfken’s book [11]. However, we include it here L T = = L z¡1=2 L for the sake of completeness: m m s dz In the constant and uniform gravitational potential case, r ½ ¾ ¼ dσ U(y) =mgy, the last integral of Eq. (6) is from the top of = L (13) s dz the trajectory (y = y0, t= t0 = 0) to its bottom; (y = 0, t = Rev. Mex. F´ıs. E 54 (2) (2008) 212–215 214 R. GOMEZ,´ V. MARQUINA, AND S. GOMEZ-A´ IZA´ because thep Laplace transform of a constant A is A/s and that Proceeding as in the Cartesian coordinates case, we call of z¡1=2 is ¼=s. Solving for L fdσ=dzg we obtain: z = U(r) so that Ãr ! ½ ¾ r r Zz0 dσ 2 T ¼ 2 dσ (z) L = : (14) T = (z ¡ z)¡1=2 dz (23) dz m ¼ s m 0 dz z Applying the inverse transform, and substituting z = Repeating the previous steps one obtains: U(y), yields r r dσ 2 T dσ 2 T = z¡1=2: (24) = U ¡1=2 (15) dz m ¼ dU m ¼ So then so that r p r ZU Z 2 T 1=2 2 T σ = 2 U = B U (25) U 1=2dU = dσ; (16) m ¼ m ¼ and because 0 σ r U = 0 dσ dσ dr 1 dσ 2 T where is taken at the end point of the trajectory.