Angular Momentum

© Fernando Caracena, 5 November 2012

In addition to the conservation of energy and momentum for closed systems, there is another quantity, called angular momentum, that is important, not only in astrophysics, but also in atomic and molecular physics. In classical physics, we can follow orbital and rotational motion, which on Earth produce the variations of night and day and the seasons of the year. We can see these motions and their effects. The behavior of tops and gyroscopes is amazing. [See also, a street act in Barcelona] On the scale of atoms and molecules, we see the effects of such motion, which is consistent with the effects of rotation, but the rotation cannot seen directly because of the restrictions of quantum mechanics.

The effects of rotation in tops and gyroscopes is an important topic in classical physics, which however, we will not discuss here. We are interested in the dynamics of orbiting and rotation bodies because they are important in quantum physics, which is to be discussed later.


Newton's laws of motion (the first and second) involve the momentum vector,

p = mv.                                   (1)

For a rotating system of particles, we can also define an angular momentum vector,

L = r x p                                (2)

and a torque vector,

N  = r x F.                                (3)

Newton's laws of motion,

F = dp/dt,                                (4)

have a counterpart between angular momentum and torque,

N = dL/dt .                               (5)

A very small, test mass moving in a central force field, does not experience any torque from that field,

F = -ur f(r),                                (6)

where ur is a unit vector pointing away from the center of the force field, parallel to the the radius.

Note that

rur   =0,                                  (7)

because both vectors point in the same direction. Therefore, (3) results in

N  = 0,                                        (8a)

and as a result of the vanishing of torque, the angular momentum vector does not change with time

dL/dt = 0.                                   (8b)

In other words, angular momentum is a constant vector,

L = a constant vector.                  (8c)

The force that holds planets in orbit is the universal attraction between all masses in the universe, i.e., gravity, which has the following form in (6):

f(r) = G M m /r2                                    (9a)


f(r) = G M m r -2,                          (9b)

where M is the mass of the attracting body such as the Sun , m is a very small, test mass and negative exponents are reciprocals of positive ones; e. g.,

r -2  =1 /r2.


Note about the idea of a reduced mass

If a small test mass orbits a heavy body, you can neglect the orbital effect that the test mass has on the body that it orbits; however, when the orbiting body's mass is not negligible, it causes the larger body to also orbit. Astronomers looking at a distant star can tell that it has a dark orbiting companion when they see it wobbling. This is the result of Newton's Third Law of motion: for every force acting on an object, that object reacts back with an equal and opposite force.

Consider an isolated system consisting of two orbiting bodies interacting through gravity and orbiting around a common center a short distance apart compared to the distances to the surrounding stars. Approximately, no other force acts on these bodies from the outside because the force of gravity falls off so fast with distance that the effects of remote stars are minuscule. If an astronomical body (of mass,m1) acts on another (m2), pulling it into an orbit, the other reacts back, to pull the first body into an orbit. Actually, both orbit about a common center that is called the center of mass of the two body system.  Each mass orbits a different distance (r'1 and r'2) from the center of mass of the binary system, which expressible in terms of the distance between centers of the two masses (r). In terms of position vectors, the relationship is expressed as follows:

r'1= -m2/(m1 + m2) r                                          (10a)


r'2=m1/(m1 + m2r   ,                                       (10b)

which are necessary conditions for the center of mass to not change its state of momentum for the two isolated bodies. To see this; differentiate (10a and b) with respect to time; multiply by the corresponding masses; and sum these momenta to find the total momenta as follows:

m1dr'1/dt= -m1m2/(m1 + m2) dr/dt                                          (11a)


m2dr'2/dt=m1m2/(m1 + m2)  dr/dt   ,                                        (11b)

m1dr'1/dt + m2dr'2/dt= 0  .                                                       (11c)

According to (11c), the assumption of position vectors of the forms of the (10a and b) is equivalent to choosing a coordinate system in which the center of mass is at rest.

By explicitly plugging into the sum of the kinetic energies of the two bodies, we can find that it is equivalent to a one body problem orbiting a center of force:

½ m1(dr'1/dt)2+ ½ m2(dr'2/dt)2 = ½ μ (dr/dt)2 ,                      (12a)


μ = m1 m2 /( m1 + m2  )

is called the reduced mass of the system.

The total energy of the system is

ET = ½ μ (dr/dt)2 - G m1 m2 |r|-2 .                                          (12b)

Suppose that m1 is very much greater than m2 . In that case the effects of the orbiting body on the parent body are almost nothing and the reduced mass approaches the value of the orbiting mass,

μ ≈  m2 .                                                                                 (12c)


Before proceeding further, we should rewrite the kinetic energy in terms of radial motion and angular motion. To begin, rewrite (12b) as follows:

ET = ½ p2 /m2 - G m1 m2 r-2 .                                          (12d)

Using (2) and a vector identity, we can evaluate the cross product, L x r , as follows:

L x r = p r2 - p•r r                                                           (13a)


L x r /r2 = p - ur pr                                                                                                     (13b)


pr = urp                                                                                                                  (13c)

and ur is the outward pointing unit, radial vector

u= r /r .                                                                                                                   (13d)

Now, rewrite (13b) by solving for the momentum vector,

pur p+    L x r /r2 .                                                     (13e)

The first term on the RHS of (13e) describes the in and out motion of an orbit along a radial line. The second term describes the angular motion about the center of rotation. (12d)

Squaring the both sides of the momentum equation given by (13e), and plugging the results into (12d) results in an expression that contains a variable radial momentum and a constant angular momentum,

ET = ½ p2 /m2 +½ L2/(m2 r-2) - G m1 m2 r-1 .                     (14)

The above equation (14) treats the radial motion as if it were in a potential energy field (on the RHS) composed of an attractive, gravitational part (last term) and a repulsive, centrifugal barrier (next to last term). The centrifugal term represents energy that goes elsewhere, really as kinetic energy of rotation. In other words, the angular velocity varies as an orbiting object moves in and out as it orbits about a center of force.




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