The Body Problem Part 2-Advanced Classical and Relativistic Mechanics-Lecture Handout, Exercises of Classical and Relativistic Mechanics

This lecture handout is part of Advanced Classical and Relativistic Mechanics course. Prof. Manasi Singh provided this handout at Punjab Engineering College. It includes: Body, Problem, Particles, Potential, Conservation, Momentum, Symmetrical, Translation, Relative, Positions

Typology: Exercises

2011/2012

Uploaded on 07/19/2012

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bg1
E=1
2m|˙q|2+V(|q|).
Recall that the the kinetic energy of the i-th particle is given by
Ti=1
2m˙q2
i
and that the total kinetic energy for the system is T=T1+T2.Bearing this in mind, we calculate:
1
2m˙q2=1
2
m1m2
m1+m2
( ˙q1˙q2)2
=1
m1+m2
(m2T1+m1T2m1m2˙q1˙q2).(2)
Note that (1) implies that
˙qi=mj
mi
˙qj(3)
for qiequal to q1or q2. Hence
1
2m1m2˙q1˙q2=1
2m2
i˙q2
i=miTi
for i= 1,2.Thus (2) becomes
1
m1+m2
(m2T1+m1T2m1m2˙q1˙q2) = 1
m1+m2
(m2T1+m1T2+ (m1T1+m2T2)) = T1+T2.
Whence
1
2m˙q2=T1+T2=T,
and the energy of the system is given by
E=T+V(|q1q2|) = 1
2m˙q2+V(|q|).
3. Establish
J=mq ×˙q
where Jis the total angular momentum.
We work from the right hand side:
2
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E =

m| q˙|^2 + V (|q|).

Recall that the the kinetic energy of the i-th particle is given by

Ti =

m q˙^2 i

and that the total kinetic energy for the system is T = T 1 + T 2. Bearing this in mind, we calculate:

m q˙^2 =

m 1 m 2 m 1 + m 2

( ˙q 1 − q˙ 2 )^2

m 1 + m 2

(m 2 T 1 + m 1 T 2 − m 1 m 2 q˙ 1 q˙ 2 ). (2)

Note that (1) implies that

q˙i = −

mj mi

q ˙j (3)

for qi equal to q 1 or q 2. Hence

1 2

m 1 m 2 q˙ 1 q˙ 2 = −

m^2 i q˙ i^2 = −miTi

for i = 1, 2. Thus (2) becomes

m 1 + m 2

(m 2 T 1 + m 1 T 2 − m 1 m 2 q˙ 1 q˙ 2 ) =

m 1 + m 2

(m 2 T 1 + m 1 T 2 + (m 1 T 1 + m 2 T 2 )) = T 1 + T 2.

Whence

1 2

m q˙^2 = T 1 + T 2 = T,

and the energy of the system is given by

E = T + V (|q 1 − q 2 |) =

m q˙^2 + V (|q|).

  1. Establish

J = mq × q˙

where J is the total angular momentum. We work from the right hand side:

mq × q˙ =

m 1 m 2 m 1 + m 2

(q 1 − q 2 ) × ( ˙q 1 − q˙ 2 )

= m 1 m 2 m 1 + m 2

(q 1 × q˙ 1 − q 1 × q˙ 2 − q 2 × q˙ 1 + q 2 × q˙ 2 )

m 1 m 2 m 1 + m 2

q 1 × q˙ 1 +

m 1 m 2

q 1 × q˙ 1 +

m 2 m 1

q 2 × q˙ 2 + q 2 × q˙ 2

m 1 m 2 m 1 + m 2

m 1 + m 2 m 2

q 1 × q˙ 1 +

m 1 + m 2 m 1

q 2 × q˙ 2

= m 1 q 1 × q˙ 1 + m 2 q 2 × q˙ 2 = J 1 + J 2 = J;

where the third equality follows from (3).

Poisson brackets

Let R^2 n^ be the phase space of a particle in Rn, with coordinates qi, pi (1 ≤ i ≤ n). Let C∞(R^2 n) be the set of smooth real-valued functions on R^2 n, which becomes an commutative algebra using pointwise addition and multiplication of functions.

We define the Poisson bracket of functions F, G ∈ C∞(R^2 n) by:

{F, G} =

∑^ n

i=

∂F

∂pi

∂G

∂qi

∂G

∂pi

∂F

∂qi

  1. Show that Poisson brackets make the vector space C∞(R^2 n) into a Lie algebra. In other words, check the antisymmetry of the bracket:

{F, G} = −{G, F }

the bilinearity of the bracket:

{F, αG + βH} = α{F, G} + β{F, H}

{αF + βG, H} = α{F, H} + β{G, H}

and Jacobi identity: {F, {G, H}} = {{F, G}, H} + {G, {F, H}}

for all F, G, H ∈ C∞(R^2 n) and α, β ∈ R.

To make life a whole lot easier on ourselves we will impose the following conventions:

  1. Lower indices indicate differentiation with respect to pi, for instance:

Fi =

∂F

∂pi

  1. Upper indices indicate differentiation with respect to qi, that is:

Gi^ =

∂G

∂qi