DOCSLIB.ORG

General Central Force

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
General Central Force General Central Force 1. Definition and Notations: Here vectors will be displayed in bold blue, unit vectors in bold plum, and scalars (and components) in regular black. Since force is a vector, F should be a vector, F. If the force depends on position only, we can express force as a function of position as F(r), which in three dimensions means F(x,y,z) or F(r,,) . A central force is one where the force is directed along the radial direction, r, and not in the other two directions, or . Its magnitude also depends only on the radial distance, r, and not on or . This can be expressed as F = F(r) r . Be aware, however, that the radial direction, r, does depend on and : F = F(r) r(,). r(x,y,z) = xx + yy + zz = r sin() cos() x + r sin() sin() y + r cos() z = rr where r = r(,) = sin()cos() x + sin() sin() y + cos() z . 2. Is a central force (always, sometimes, never) a conservative force? To answer this, recall Stokes Theorem: (F) n dS = closed loop F dl (where S is an area) lf so that if F = 0 everywhere, then closed loop F dl = 0, and so li F dl is independent of path, and so a potential energy can be defined: rf PE = - ri F dl . Let’s now calculate F for a central force (using spherical coordinates): F = [r /r + (1/r) / + (1/{r sin()}) / ] F(r) r = (F(r)/r) (r r) + F(r) (r r/r) + (1/r) (F(r)/) ( r) + (F(r)/r) ( (r/)) + (1/{r sin()}) (F(r)/) ( r) + (1/{r sin()}) F(r) ( r/) . There are six terms, so let’s look at these one by one. The first term is zero since (r r) = 0 (since any vector crossed itself is zero). The second term is zero since r/r = 0 (since r depends only on and ). The third term is zero since F(r)/ = 0 (since F(r) does not depend on ). The fourth term is zero since r/ = , and ( ) = 0. The fifth term is zero since F(r)/ = 0 (since F(r) does not depend on ). The sixth term is zero since r/ = sin() , and ( ) = 0. Therefore, F = 0 for F = F(r)r, so a central force is always a conservative force! We can then define a potential energy function: r PE(r) = - r-standard F(r) dr . 3. Torque and Angular Momentum for a Central Force Starting with the definition of torque as ≡ r F, for a central force, F = F(r)r, we have (recalling that r = rr): = r F(r)r = r F(r) r r = 0. Central forces can create no torque. Now we recall the relation between torque and angular momentum (L): = dL/dt . Since torque under a central force is zero, the angular momentum must remain constant. Let’s now get an expression for angular momentum in polar coordinates. We can use 2-D polar (actually 3-D cylindrical) coordinates since a central force has no component of either motion or force perpendicular to the (r v) plane, i.e., the z plane. By definition, L ≡ r p = r mv . In cylindrical form, v = dr/dt = d(rr)/dt = (dr/dt)r + r(dr/dt) = r'r + r' (where r' = dr/dt, and ' = d/dt) . We will use this ' notation to only denote time derivatives, not any other derivatives. Therefore, L = rr m(r'r + r') = mrr'(r r) + mr2'(r ) = 0 + mr2' z = constant. Note that if we know the initial conditions, ro and 'o, we can calculate the value of L. If we just consider the magnitude, since the direction will not change, L = mr2' = constant. Eq. 1 This equation provides a relation between the radius, r and the angular speed, ' (which we sometimes have called ). We can write this as: ' = L/mr2 . Eq. 1a 4. Newton’s Second Law for a Central Force F = ma is a vector equation, which means it is one equation for each dimension. For a central force, F = F(r)r, in polar coordinates we have Fr = F(r) , and F = 0 . The acceleration expressed 2 2 2 2 in polar form is: ar = r''– r' (where r'' = d r/dt is the usual linear acceleration in the radial direction, and r' 2 2 2 = r is the circular acceleration); and a = r'' + 2r'' (where r'' is r(d /dt ), the acceleration in the circular direction, and 2r'' is that strange Coriolis acceleration). Therefore, in component form, Newton’s Law becomes: r component: F(r) = m(r'' – r'2) , or F(r) + mr'2 = mr'' Eqs. 2 2 2 [where the term mr’ = m r can be considered the “centrifugal” force] component: F = 0, so 0 = m(r'' + 2r'') . Notice that both equations have both r and so they are simultaneous differential equations. However, if we use the angular momentum equation, ' = L/mr2 in the radial equation, we get: F(r) + mr(L/mr2)2 = mr'', or F(r) + L2/mr3 = mr'' Eq. 2a which is an expression that is equivalent to a one-dimensional problem since there is no mention of . If we can solve this for r(t), we can then use the component equation to solve for '(t) and then q(t). We could also use ' = L/mr2 with r(t) and both L and m constant to solve for '(t) and then q(t). 5. Conservation of Energy Since a central force is a conservative force, we can define a potential energy. If there are no other forces, then energy is conserved and we can write this in an equation: KE + PE = E = constant 2 2 2 2 2 v = vrr + v , and v = vr + v = (r') + (r') so we get: ½ m v2 + PE(r) = ½ m r'2 + ½ m r2 '2 + PE(r) = E . If we use the angular momentum expression, ' = L/mr2, this becomes: ½ m r'2 + ½ m r2 (L/mr2)2 + PE(r) = E = ½ m r'2 + ½L2/mr2 + PE(r) . If we treat the term ½L2/mr2 as an additional potential energy which is equivalent to treating the centripetal acceleration as a centrifugal force (since it only depends on r), we can define an effective potential energy: 2 2 PEeff(r) = PE(r) + L /2mr . Eq. 3 We now have an equation that is essentially one-dimensional: 2 ½ m r' + PEeff(r) = E = constant . Eq. 4 As always, this is a first order differential equation whose solution is r(t). Then we can use ' = L/mr2 to solve for q'(t) and then q(t). 6. Straightforward solutions a) Use the Conservation of Energy (1-D) equation with the equivalent potential energy, and use the technique we used earlier in the course to first get r(t) and then to get (t): 2 2 2 ½ mr' + PEeff(r) = E where PEeff(r) = PE(r) + L /2mr ½ dr/dt = r' = [2(E-PEeff(r))/m ] t r ½ 0 dt = ro {1 / [2(E-PEeff(r))/m ] } dr Assuming we can actually do the integration over r, we get an expression for t(r). Then we must find the inverse to get r(t). This is sometimes very difficult. If we can get r(t), we can then use the angular momentum expression: L = mr2' to get an expression for ': '(t) = L/mr(t)2 . We can then use ' = d/dt to get: t t 2 0 d = 0 ' dt = 0 [L/mr(t) ] dt . If we can do the integration, we then have an expression for (t) and the problem is solved. b) Use Newton’s Second Law to try to get the trajectory r() [instead of r(t) and (t)]. While this will lose some information (time information), it has the advantage of giving us the orbits directly. We start with the equivalent 1-D relation (see part 4 above): F(r) + L2/mr3 = mr'' . Eq. 2a Since we want r(), we still have to deal with dr/dt, but we note that (recalling that ' = L/mr2) : r' = dr/dt = (dr/d)*(d/dt) = (dr/d)*' = (dr/d)*(L/mr2) . r'' = d(r')/dt = (dr'/d)*(d/dt) = d[(dr/d)*(L/mr2)]/d * ' = (d2r/d2)*(L/mr2)*' + (dr/d)*(L/m)*(d[1/r2]/d)*' note: d[1/r2]/d = (d[1/r2]/dr)*(dr/d) = (-2/r3)*(dr/d) r'' = (d2r/d2)*(L/mr2)*' + (dr/d)*(L/m)*(-2/r3)*(dr/d)*' (again using ' = L/mr2 ) r'' = (d2r/d2)*(L2/m2r4) - (2L2/m2r5)*(dr/d)2 . Using this, we can now write Newton’s 2nd Law (r component equation, Eq. 2a) as: F(r) + L2/mr3 = (d2r/d2)*(L2/mr4) - (2L2/mr5)*(dr/d)2 . Eq. 5 This doesn’t look all that easy to solve for r(q). 7. Tricky way to solve for r() Sometimes things look nicer in one form than in another. We will try a trick that involves substitution: let’s define u = 1/r (or r = 1/u). L = mr2' = m'/u2, or ' = Lu2/m r' = dr/dt = (dr/du)*(du/dt) = (dr/du)*(du/d)*(d/dt) = (d(1/u)/du)*(du/d)*' = (-1/u2)*(du/d)*' = (-1/u2)*(du/d)*(Lu2/m) = (-L/m)*(du/d). r'' = dr’/dt = (dr'/d)*(d/dt) = d[(-L/m)*(du/d)]/d * ' = (-L/m)*(d2u/d2)*(Lu2/m) = (-L2u2/m2)*d2u/d2 . Newton’s Second Law (r component equation) F(r) + L2/mr3 = mr’'' Eq. 2a thus becomes: F(1/u) + L2u3/m = m*(-L2u2/m2)*d2u/d2 , or (-m/L2u2)*F(1/u) + (-m/L2u2)*(L2u3/m) = d2u/d2 , or (-m/L2u2)*F(1/u) + -u = d2u/d2 .
Load more

Recommended publications
  • Central Forces Notes
    Central Forces Corinne A. Manogue with Tevian Dray, Kenneth S. Krane, Jason Janesky Department of Physics Oregon State University Corvallis, OR 97331, USA [email protected] c 2007 Corinne Manogue March 1999, last revised February 2009 Abstract The two body problem is treated classically. The reduced mass is used to reduce the two body problem to an equivalent one body prob- lem. Conservation of angular momentum is derived and exploited to simplify the problem. Spherical coordinates are chosen to respect this symmetry. The equations of motion are obtained in two different ways: using Newton’s second law, and using energy conservation. Kepler’s laws are derived. The concept of an effective potential is introduced. The equations of motion are solved for the orbits in the case that the force obeys an inverse square law. The equations of motion are also solved, up to quadrature (i.e. in terms of definite integrals) and numerical integration is used to explore the solutions. 1 2 INTRODUCTION In the Central Forces paradigm, we will examine a mathematically tractable and physically useful problem - that of two bodies interacting with each other through a force that has two characteristics: (a) it depends only on the separation between the two bodies, and (b) it points along the line con- necting the two bodies. Such a force is called a central force. Perhaps the 1 most common examples of this type of force are those that follow the r2 behavior, specifically the Newtonian gravitational force between two point masses or spherically symmetric bodies and the Coulomb force between two point or spherically symmetric electric charges.
    [Show full text]
  • 4. Central Forces
    4. Central Forces In this section we will study the three-dimensional motion of a particle in a central force potential. Such a system obeys the equation of motion mx¨ = V (r)(4.1) r where the potential depends only on r = x .Sincebothgravitationalandelectrostatic | | forces are of this form, solutions to this equation contain some of the most important results in classical physics. Our first line of attack in solving (4.1)istouseangularmomentum.Recallthatthis is defined as L = mx x˙ ⇥ We already saw in Section 2.2.2 that angular momentum is conserved in a central potential. The proof is straightforward: dL = mx x¨ = x V =0 dt ⇥ − ⇥r where the final equality follows because V is parallel to x. r The conservation of angular momentum has an important consequence: all motion takes place in a plane. This follows because L is a fixed, unchanging vector which, by construction, obeys L x =0 · So the position of the particle always lies in a plane perpendicular to L.Bythesame argument, L x˙ =0sothevelocityoftheparticlealsoliesinthesameplane.Inthis · way the three-dimensional dynamics is reduced to dynamics on a plane. 4.1 Polar Coordinates in the Plane We’ve learned that the motion lies in a plane. It will turn out to be much easier if we work with polar coordinates on the plane rather than Cartesian coordinates. For this reason, we take a brief detour to explain some relevant aspects of polar coordinates. To start, we rotate our coordinate system so that the angular momentum points in the z-direction and all motion takes place in the (x, y)plane.Wethendefinetheusual polar coordinates x = r cos ✓, y= r sin ✓ –48– Our goal is to express both the velocity and acceleration y ^ ^ θ r in polar coordinates.
    [Show full text]
  • Central Forces
    Central Forces Corinne A. Manogue with Tevian Dray, Kenneth S. Krane, Jason Janesky Department of Physics Oregon State University Corvallis, OR 97331, USA [email protected] c 2007 Corinne Manogue ! March 1999, last revised February 2009 Abstract The two body problem is treated classically. The reduced mass is used to reduce the two body problem to an equivalent one body prob- lem. Conservation of angular momentum is derived and exploited to simplify the problem. Spherical coordinates are chosen to respect this symmetry. The equations of motion are obtained in two different ways: using Newton’s second law, and using energy conservation. Kepler’s laws are derived. The concept of an effective potential is introduced. The equations of motion are solved for the orbits in the case that the force obeys an inverse square law. The equations of motion are also solved, up to quadrature (i.e. in terms of definite integrals) and numerical integration is used to explore the solutions. 1 2 INTRODUCTION In the Central Forces paradigm, we will examine a mathematically tractable and physically useful problem - that of two bodies interacting with each other through a force that has two characteristics: (a) it depends only on the separation between the two bodies, and (b) it points along the line con- necting the two bodies. Such a force is called a central force. Perhaps the 1 most common examples of this type of force are those that follow the r2 behavior, specifically the Newtonian gravitational force between two point masses or spherically symmetric bodies and the Coulomb force between two point or spherically symmetric electric charges.
    [Show full text]
  • Motion in a Central Force Field
    Motion in a Central Force Field We now study the properties of a particle of (constant) mass moving in a particular type of force field, a central force field. Central forces are very important in physics and engineering. For example, the gravitational force of attraction between two point masses is a central force. The Coulomb force of attraction and repulsion between charged particles is a central force. Because of their importance they deserve special consideration. We begin by giving a precise definition of central force, or central force field. Central Forces: The Definition. Suppose that a force acting on a particle of mass has the properties that: • the force is always directed from toward, or away, from a fixed point O, • the magnitude of the force only depends on the distance from O. Forces having these properties are called central forces. The particle is said to move in a central force field. The point O is referred to as the center of force. Mathematically, is a central force if and only if: (1) where is a unit vector in the direction of . If , the force is said to be attractive towards O. If , the force is said to be repulsive from O. We give a geometrical illustration in Fig. 1. Properties of a Particle Moving under the Influence of a Central Force. If a particle moves in a central force field then the following properties hold: 1. The path of the particle must be a plane curve, i.e., it must lie in a plane. 2. The angular momentum of the particle is conserved, i.e., it is constant in time.
    [Show full text]
  • 08. Central Force Motion I Gerhard Müller University of Rhode Island, [email protected] Creative Commons License
    View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by DigitalCommons@URI University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 2015 08. Central Force Motion I Gerhard Müller University of Rhode Island, [email protected] Creative Commons License This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License. Follow this and additional works at: http://digitalcommons.uri.edu/classical_dynamics Abstract Part eight of course materials for Classical Dynamics (Physics 520), taught by Gerhard Müller at the University of Rhode Island. Documents will be updated periodically as more entries become presentable. Recommended Citation Müller, Gerhard, "08. Central Force Motion I" (2015). Classical Dynamics. Paper 14. http://digitalcommons.uri.edu/classical_dynamics/14 This Course Material is brought to you for free and open access by the Physics Course Materials at DigitalCommons@URI. It has been accepted for inclusion in Classical Dynamics by an authorized administrator of DigitalCommons@URI. For more information, please contact [email protected]. Contents of this Document [mtc8] 8. Central Force Motion I • Central force motion: two-body problem [mln66] • Central force motion: one-body problem [mln67] • Central force problem: formal solution [mln18] • Orbits of power-law potentials [msl21] • Unstable circular orbit [mex51] • Orbit of the inverse-square potential at large angular momentum [mex46] • Orbit of the inverse-square potential
    [Show full text]
  • Chapter 7 Gravitation and Central-Force Motion
    Chapter 7 Gravitation and Central-force motion In this chapter we describe motion caused by central forces, especially the orbits of planets, moons, and artificial satellites due to central gravitational forces. Historically, this is the most important testing ground of Newtonian mechanics. In fact, it is not clear how the science of mechanics would have developed if the Earth had been covered with permanent clouds, obscuring the Moon and planets from view. And Newton's laws of motion with central gravitational forces are still very much in use today, such as in designing spacecraft trajectories to other planets. Our treatment here of motion in central gravitational forces is followed in the next chapter with a look at motion due to electromagnetic forces, which can also be central in special cases, but are commonly much more varied, partly because they involve both electric and magnetic forces. 7.1 Central forces A central force on a particle is directed toward or away from a fixed point in three dimensions and is spherically symmetric about that point. In spherical coordinates (r; θ; ') the corresponding potential energy is also spherically symmetric, with U = U(r) alone. For example, the Sun, of mass m1 (the source), exerts an attractive central force m m F = −G 1 2 r^ (7.1) r2 263 7.1. CENTRAL FORCES FIGURE 7.1 : Newtonian gravity pulling a probe mass m2 towards a source mass m1. on a planet of mass m2 (the probe), where r is the distance between their centers and r^ is a unit vector pointing away from the Sun (see Figure 7.1).
    [Show full text]
  • Central Force Motion Challenge Problems
    Central Force Motion Challenge Problems Problem 1: Elliptic Orbit A satellite of mass ms is in an elliptical orbit around a planet of mass mp which is located at one focus of the ellipse. The satellite has a velocity va at the distance ra when it is furthest from the planet. a) What is the magnitude of the velocity vp of the satellite when it is closest to the planet? Express your answer in terms of ms , m p , G , va , and ra as needed. b) What is the distance of nearest approach rp ? c) If the satellite were in a circular orbit of radius r0 = rp , is it’s velocity v0 greater than, equal to, or less than the velocity vp of the original elliptic orbit? Justify your answer. Problem 1 Solution: The angular momentum about the origin is constant therefore L ! Lp = La . (28.1.1) Because ms << m p , the reduced mass µ ! ms and so the angular momentum condition becomes L = ms r p v p = ms r a v a (28.1.2) Thus rp = ra v a/ v p . (28.1.3) Note: We can solve for v p in terms of the constants G , m p , ra and va as follows. Choose zero for the gravitational potential energy for the case where the satellite and planet are separated by an infinite distance, U (r = !)= 0. The gravitational force is conservative, so the energy at closest approach is equal to the energy at the furthest distance from the planet, hence 1 2 Gms m p 1 2 Gms m p ms va ! = ms v p ! .
    [Show full text]
  • Physics 3550, Fall 2012 Angular Momentum. Relevant Sections in Text: §3.4, 3.5
    Angular Momentum. Physics 3550, Fall 2012 Angular Momentum. Relevant Sections in Text: x3.4, 3.5 Angular Momentum You have certainly encountered angular momentum in a previous class. The impor- tance of angular momentum lies principally in the fact that it is conserved in appropriate circumstances, as we shall explore. It is worth noting that the angular momentum of elec- trons in atoms is one of the most important physical features of atoms and largely makes them what they are. The angular momentum of a single particle is defined by* ~l = ~r × ~p; where usually the momentum is just ~p = m~v (more on this later). Notice that the angular momentum involves the position vector, which is defined relative to a choice of origin. Thus the angular momentum is defined relative to an origin. It is easy to forget this. Don't. As I mentioned, we focus on angular momentum because of its conservation under appropriate circumstances. What are these circumstances? Well, let the particle move according to Newton's second law, we have d~l = ~v × (m~v) + ~r × F:~ dt The first term involves the cross product of two parallel vectors and so it vanishes. We then get d~l = ~γ; dt where ~γ = ~r × F~ is the torque. Like the position vector and the angular momentum, the torque is always defined relative to a choice of origin. We tacitly assume the torque is defined relative to the same origin as the angular momentum. We see then that the angular momentum of a particle is conserved if and only if the torque vanishes.
    [Show full text]
  • ROYAL ARMY MEDICAL CORPS. ,---- (Torps 1Rews
    J R Army Med Corps: first published as 10.1136/jramc-26-03-20 on 1 March 1916. Downloaded from JOURNAL OF THIll ROYAL ARMY MEDICAL CORPS. ,---- (torps 1Rews. MARCH, 1916. Protected by copyright. EXTRACT FROM THE "LONDON GAZETTE," FEBRUARY 22, 1916. War Office, February 24, 1916. The President of the French Republic has bestowed the decoration of the Legion of Honour, with the approval of His Majesty the King, on the undermentioned Officers, in recognition of their distinguished service during the campaign. CROIX Dill COMMANDEUR. Surgeon-General William Grant Macpherson, C.B., C.M.G., M.B., K.H.P. Surgeon-General Tom Percy Woodhouse, C.B. Colonel Sir William Boog Leishman, C.B., F.R.S., M.B., F.R.C.P., K.H.P., Army Medical Service. CROIX D'OFFlCIER. Colonel Charles Henry Burtchaell, C.M.G., M.B., Army Medical Service. http://militaryhealth.bmj.com/ CROIX DE CHEVA.LIER. Major WaIter Rothney Battye, D.S.O., M.B., F.R.C.S., Indian Medical Service. Major Robert Barclay Black, M.B., Reserve of Officers, Royal Army Medical Corps. Major James Sydney Pascoe, Royal Army Medical Corps. Major Eugene Ryan, D.S.O., Royal Army Medical Corps. Major John Weir West, M.B., Royal Army Medical Oorps. Oaptain Francis Oasement, M.B., Royal Army Medical Oorps. Captain Douglas Murray McWhae, Australian Army Medical Corps. Captain (temporary Major) John Morley, M.B., F.R.O.S., Royal Army Medical Corps (Territorial Force). Lieutenant (temporary Oaptain) Joseph Wilfrid Craven, M.B., Royal Army Medical Oorps (Territorial Force). The President of the French Republic has bestowed the decoration" Croix de on September 28, 2021 by guest.
    [Show full text]
  • THE HYDROGEN ATOM (1) Central Force Problem (2) Rigid Rotor (3) H Atom
    THE HYDROGEN ATOM (1) Central Force Problem (2) Rigid Rotor (3) H Atom CENTRAL FORCE PROBLEM: The potential energy function has no angular (θ, φ) dependence V = V(r) The force is a vector & is defined as the gradient of the potential F = - ∇ V(x,y,z) ∇ is a vector quantity, the gradient operator: ∇ = i ∂/∂x + j ∂/∂y + k ∂/∂z F = - (i ∂V/∂x + j ∂V/∂y + k ∂V/∂z) Or, by transforming coordinates from Cartesian (x,y,z) to Spherical Polar (r, θ, φ), one can give the gradient operator in spherical polar coordinates. Since V depends only on r, not θ& ϕ, the resulting force depends only on r: F = - ( dV(r)/dr ) ( r/r) for V = V(r). To express the Hamiltonian for a Central Force problem, use spherical polar coordinates where ∇2 = ∂2/∂r2 + (2/r) ∂/∂r + (1/ r2) ∂2/∂θ2 + + (1/r2) cot θ ∂/∂θ + 1/(r2 sin2 θ) ∂2/∂φ2 = ∂2/∂r2 + (2/r) ∂/∂r - 1/( r2h2) L2 where L2 = - h2 (∂2/∂θ2 + cot θ ∂/∂θ + (1/ sin2 θ) (∂2/∂φ2). Then H = T + V = - (h2/2m) ∇2 + V(r) = - (h2/2m) (∂2/∂r2 + (2/r) ∂/∂r ) + L2/(2m r2) + V(r) Will the angular momentum be conserved (constant)? Can we specify both angular momentum & energy (i.e. does H commute 2 with L and Lz? Let’s see: [H, L2] = [T + V, L2] = [T , L2] + [V, L2] = [- (h2/2m) (∂2/∂r2 + (2/r) ∂/∂r ) + L2/(2m r2) , L2] + [V, L2] = [- (h2/2m) (∂2/∂r2 + (2/r) ∂/∂r ) , L2] + [L2/(2m r2) , L2] + [V, L2] In the first & third commutators above, the left-hand side depends only on r; the right, only on θ & φ.
    [Show full text]
  • The Central Force Problem in N Dimensions∗
    GENERAL ARTICLE The Central Force Problem in n Dimensions∗ V Balakrishnan, Suresh Govindarajan and S Lakshmibala The motion of a particle moving under the influence of a central force is a fundamental paradigm in dynamics. The problem of planetary motion, specifically the derivation of The authors are with the Department of Physics, IIT Kepler’s laws motivated Newton’s monumental work, Prin- Madras, Chennai. Their cipia Mathematica,effectively signalling the start of modern current research interests physics. Today, the central force problem stands as a ba- are: sic lesson in dynamics. In this article, we discuss the clas- sical central force problem in a general number of spatial dimensions n, as an instructive illustration of important as- pects such as integrability, super-integrability and dynamical symmetry. The investigation is also in line with the realisa- V Balakrishnan (Stochastic processes and tion that it is useful to treat the number of dimensions as a dynamical systems) variable parameter in physical problems. The dependence of various quantities on the spatial dimensionality leads to a proper perspective of the problems concerned. We consider, first, the orbital angular momentum (AM) in n dimensions, and discuss in some detail the role it plays in the integrability Suresh Govindarajan of the central force problem. We then consider an important (String theory, black holes, statistical physics) super-integrable case, the Kepler problem, in n dimensions. The existence of an additional vector constant of the motion (COM) over and above the AM makes this problem maxi- mally super-integrable. We discuss the significance of these COMs as generators of the dynamical symmetry group of the SLakshmibala Hamiltonian.
    [Show full text]
  • Arxiv:Physics/0410149V1 [Physics.Ed-Ph] 19 Oct 2004 Ftepsil Rjcoisi Eta Oc Oin Most Motion
    Orbits in a central force field: Bounded orbits Subhankar Ray∗ Dept of Physics, Jadavpur University, Calcutta 700 032, India J. Shamanna Physics Department, Visva Bharati University, Santiniketan 731235, India (Dated: August 1, 2003) The nature of boundedness of orbits of a particle moving in a central force field is investigated. General conditions for circular orbits and their stability are discussed. In a bounded central field orbit, a particle moves clockwise or anticlockwise, depending on its angular momentum, and at the same time oscillates between a minimum and a maximum radial distance, defining an inner and an outer annulus. There are generic orbits suggested in popular texts displaying the general features of a central orbit. In this work it is demonstrated that some of these orbits, seemingly possible at the first glance, are not compatible with a central force field. For power law forces, the general nature of boundedness and geometric shape of orbits are investigated. I. INTRODUCTION B. Newtonian Synthesis The central force motion is one of the oldest and widely Almost 100 years later Newton realized that the plan- studied problems in classical mechanics. Several familiar ets go about in their nearly circular orbits around the sun force-laws in nature, e.g., Newton’s law of gravitation, under the influence of the same force that causes an ap- Coulomb’s law, van-der Waals force, Yukawa interaction, ple to fall to the ground, i.e., gravitation. Newton’s law and Hooke’s law are all examples of central forces. The of gravitation gave a theoretical basis to Kepler’s laws.
    [Show full text]