Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

Physics for IIT JAM, UGC - NET, CSIR NET

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Physics : Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

The document Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
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 Reduction to the Equivalent One-body Problem – the Reduced Mass 

We consider a system consisting of two point masses, m1 and m2 , when the only forces are those due to an interaction potential U . We will assume that U is a function of the distance between the two particles  Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRevSuch a system has six degrees of freedom. We could choose, for example, the three components of each of the two vectors, r1 and r(see Figure 6-1). However, since the potential energy is solely a function of the distance between the two particles, i.e., U = U ( r ) , it is to our advantage if we also express the kinetic energy as function of r (that is, of Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev ). Let’s first define the center of mass R of the system as

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev         (6.1)

We also consider the distance between each particle and the center of mass

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev       (6.2)

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

with r = r2 - r1 .

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

Figure 6-1 – The different vectors involved in the two-body problem.

We now calculate the kinetic energy of the system

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

and inserting equations (6.2) in equation (6.3) we get

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.4)

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

We introduce a new quantity µ , the reduced mass, defined as

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev               (6.5)

which can alternatively be written as

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev         (6.6)

We can use equation (6.5) to write the kinetic energy as

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.7)
We have, therefore, succeeded in expressing the kinetic energy as a function of Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev . We are now in a position to write down the Lagrangian for the central force problem

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev           (6.8)

where the potential energy U ( r ) is yet undefined, except for the fact that it is solely a function of the distance between the two particles. It is, however, seen from equation (6.8) that the three components of the center of mass vector are cyclic. More precisely,

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev       (6.9)

withCentral Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRevThe center of mass is, therefore, either at rest or moving uniformly since the equations of motion for X , Y , and Z can be combined into the following vector relation

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.10)

Since the center of mass vector (and its derivative) does not appear anywhere else in the Lagrangian, we can drop the first term of the right hand side of equation (6.8) in all subsequent analysis and only consider the remaining three degrees of freedom. The new Lagrangian is therefore

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.11)

What is left of the Lagrangian is exactly what would be expected if we were dealing with a problem of a single particle of mass µ subjected to a fixed central force. Thus, the central force motion of two particles about their common center of mass is reducible to an equivalent one-body problem.


The First Integrals of Motion 

Since we are dealing with a problem where the force involved is conservative, where the potential is a function of the distance r of the reduced mass to the force center alone, the system has spherical symmetry. From  Noether’s theorem,  we know that for such systems the angular momentum is conserved . That is,

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev      (6.12)

From the definition of the angular momentum itself we know that L is always parallel to vectors normal to the plane containing r and p . Furthermore, since in this case L is fixed, it follows that the motion is at all time confined to the aforementioned plane. We are, therefore, fully justified to use polar coordinates as the two remaining generalized coordinates for this problem (i.e., we can set the third generalized coordinate, say, z to be a constant since the motion is restricted to a plane). Since Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev we have

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.13)

In the last equation we have used the following relation (with the usual transformation between the ( r,θ ) polar and the ( x, y ) Cartesian coordinates)

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.14)

from which it can be verified that

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.15)

We can now rewrite the Lagrangian as a function of r and θ

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.16)

We notice from equation (6.16) that θ is a cyclic variable. The corresponding generalized momentum is therefore conserved, that is

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.17)

The momentum Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev is a first integral of motion and is seen to equal the magnitude of the angular momentum vector. It is customarily written as

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.18)

As the particle (i.e., the reduced mass) moves along its trajectory through an infinitesimal angular displacement dθ within an amount of time dt , the area dA swept out by its radius vector r is given by

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.19)

Alternatively, we can define the areal velocity as

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.20)

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

Thus, the areal velocity is constant in time. This result, discovered by Kepler for planetary motion, is called Kepler’s Second Law. It is important to realize that the conservation of the areal velocity is a general property of central force motion and is not restricted to the inverse-square law force involved in planetary motion.

Another first integral of motion (the only one remaining) concerns the conservation of energy. The conservation is insured because we are considering conservative systems.
Writing E for the energy we have

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.21)

or using equation (6.18)

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.22)

 

The Equations of Motion 

We will use two different ways for the derivation of the equations of motion. The first one consists of inverting equation (6.22) and express Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev as a function of E, l, and U ( r ) such that

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.23)

or alternatively

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.24)

Equation (6.24) can be solved, once the potential energy U ( r ) is defined, to yield the solution t = t ( r ) , or after inversion r = r (t ) . We are, however, also interested in determining the shape of the path (or orbit) taken by the particle. That is, we would like to evaluate r = r (θ) or θ = θ ( r ) . To do so we use the following relation

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.25)

Inserting equations (6.18) and (6.23) into equation (6.25), we get

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev (6.26)

It is important to note that the integral given by equation (6.26) can be solved analytically only for certain forms of potential energy. Most importantly, if the potential energy is of the form  Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev for n = 2, - 1, and - 2 the solution is expressible in terms of circular functions.

The second method considered here for solving the equations of motion uses the Lagrange equations

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.27)

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

The second of these equations was already used to get equation (6.18) for the conservation of angular momentum. Applying the first of equations (6.27) to the Lagrangian (equation (6.16)) gives

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.28)

We now modify this equation by making the following change of variable

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.29)

We calculate the first two derivatives of u relative to θ

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev (6.30)

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

where we have used the fact that Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev , and for the second derivative

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

From this equation we have

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.32)

and from equation (6.18)

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.33)

Inserting equations (6.32) and (6.33) in equation (6.28) yields

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.34)

which can be rewritten as

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.35)

Equation (6.35) can be used to find the force law that corresponds to a known orbit r = r (θ) .

 

Examples

1. Let’s consider the case where the orbit is circular, i.e., r = cste . Then from equation (6.35) we find that

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.36)

Equation (6.36) implies that

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.37)

which is the functional form of the gravitational force.

2. Let’s assume that we have an orbit given by

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.38)

From the second derivative of equation (6.38) relative to θ we have

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev

and using equation (6.35), we get

Central Force Motions (Part - 1)- Classical Mechanics, UGC - NET Physics Physics Notes | EduRev(6.40)

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