Download Poisson Brackets and Two-Body System: Deriving the Equations of Motion and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity!
Solution to 1:
From (2) we can immediately see that q¨ = q¨ 1 − q¨ 2. We can re-write q¨ 1 and ¨q 2 using their respective equations of motion. This leaves us with:
q ¨ 1 = f (|q 1 − q 2 |) q 1 − q 2 m 1 |q 1 − q 2 |
and q¨ 2 = f (|q 2 − q 1 |) q 2 − q 1 m 1 |q 2 − q 1 |
Subtracting these two equations we get an expression for ¨q as follows:
q ¨ = f (|q 1 − q 2 |)
q 1 − q 2 m 1 |q 1 − q 2 |
− f (|q 2 − q 1 |)
q 2 − q 1 m 2 |q 2 − q 1 |
Now since |q 1 − q 2 | = |q 2 − q 1 | the above expression can be written as:
q ¨ = f (|q|) q 1 − q 2 m 1 |q|
− f (|q|) q 2 − q 1 m 2 |q|
f (|q|) |q|
q 1 − q 2 m 1
q 2 − q 1 m 2
Combining qi terms of equal i we get:
q ¨ =
f (|q|) |q|
[(
m 1 + m 2 m 1 m 2
q 1 −
m 1 + m 2 m 1 m 2
q 2
]
The coefficients in front of q 1 and q 2 are just the inverse of the reduced mass.
1 m
m 1 + m 2 m 1 m 2
Thus we have shown that:
q ¨ =
m
f (|q|) |q|
(q 1 − q 2 )
⇒ mq¨ = f (|q|) q |q|
- Recall that the total energy E of the 2-particle system is the sum of the kinetic energies of the particles plus the potential energy. Express E in terms of q and the reduced mass. Show that
E =
m | q˙|^2 + V (|q|)
Note that this looks exactly like the energy of a single particle!
Solution to 2:
The total energy E for the two particle system is given by:
E =
m 1 q˙ 1 2 +
(3) m 2 q˙ 2 2 + V (|q 1 − q 2 |)
From (1) and (2) we express q˙ 1 and q˙ 2 in terms of q˙, the mass m 1 and the mass m 2 by noting the following:
m 1 q˙ 1 + m 2 q˙ 2 = 0 ⇒ q˙ 1 = − m 2 m 1
q ˙ 2 and q˙ 2 = − m 1 m 2
q ˙ 1
q ˙ 1 = −
m 2 m 1
q ˙ 2 and q˙ 1 − q˙ 2 = ˙q ⇒ q˙ 2 +
m 2 m 1
q ˙ 2 = − q˙ ⇒ q˙ 2 =
−m 1 m 1 + m 2
(4a) q ˙
q ˙ 2 = −
m 1 m 2 q ˙ 1 and q˙ 1 − q˙ 2 = ˙q ⇒ q˙ 1 +
m 1 m 2 q ˙ 1 = q˙ ⇒ q˙ 1 =
m 2 m 1 + m 2 (4b) q ˙
Substituting the right most equalities of (4a) and (4b) for their respective values in (3) and noting that |q 1 − q 2 | = |q| we get:
E =
m 1 m^22 (m 1 + m 2 )^2
q˙ 2 +
m 2 m^21 (m 1 + m 2 )^2
(5) q˙ 2 + V (|q|).
Finally, summing the two kinetic terms explicitly gives the desired result.
E =
m | q˙|^2 + V (|q|)
Where as before m is the reduced mass of the system.
- Let J be the total angular momentum of the 2-particle system. Show that
J = mq × q˙
Note that this looks exactly like the angular momentum of a single particle!
Solution to 3:
Solving for the positions q 1 and q 2 , gives identical equations to (4a) and (4b) only without time derivatives. We summarize the results below:
q 1 =
m 2 m 1 + m 2
q q˙ 1 =
m 2 m 1 + m 2
(6a) q ˙
q 2 =
−m 1 m 1 + m 2
q q˙ 2 =
−m 1 m 1 + m 2
(6b) q ˙
The total angular momentum, J, is the sum of the angular momentum of each particle. We give the full expression below.
(7) J = J 1 + J 2 = m 1 q 1 × q˙ 1 + m 2 q 2 × q˙ 2
Again, inserting (6a) and (6b) into (7) in favor of q we get:
J = m 1 q 1 × q˙ 1 + m 2 q 2 × q˙ 2
= m 1
m 2 m 1 + m 2
q ×
m 2 m 1 + m 2
q ˙ + m 2
−m 1 m 1 + m 2
×
−m 1 m 1 + m 2
q ˙
m 1 m^22 (m 1 + m 2 )^2
q × q˙ +
m 2 m^21 (m 1 + m 2 )^2
× q˙
m 1 m 2 (m 1 + m 2 )
m 1 + m 2 (m 1 + m 2 )
q × q˙
J = mq × q˙
This proves the claim in question 3.
At this point we’re back to a problem you’ve already solved: a single particle in a central force. The only difference is that now q stands for the relative position and m stands for the reduced mass!
So, we instantly conclude that two bodies orbiting each other due to the force of gravity will both have an orbit that’s either an ellipse, or a parabola, or a hyperbola... when viewed in the center-of-mass frame.
Next, we investigate the bilinearity, writing {F, αG + βH} explicitly gives:
{F, α G + β H} =
∑^ n
i=
∂F
∂pi
∂(α G + β H) ∂qi
∂(α G + β H) ∂pi
∂F
∂qi
∑^ n
i=
∂F
∂pi
[
α
∂G
∂qi
∂H
∂qi
]
[
α
∂G
∂pi
∂H
∂pi
]
∂F
∂qi
∑^ n
i=
α
∂F
∂pi
∂G
∂qi
∂F
∂pi
∂H
∂qi
− α
∂G
∂pi
∂F
∂qi
− β
∂H
∂pi
∂F
∂qi
∑^ n
i=
α
∂F
∂pi
∂G
∂qi
− α
∂G
∂pi
∂F
∂qi
∂F
∂pi
∂H
∂qi
− β
∂H
∂pi
∂F
∂qi
= α
∑^ n
i=
∂F
∂pi
∂G
∂qi
∂G
∂pi
∂F
∂qi
∑^ n
i=
∂F
∂pi
∂H
∂qi
∂H
∂pi
∂F
∂qi = α {F, H} + β {G, H}}
For comlpeteness we must now check, {αF + βG, H}, as well.
{αF + βG, H} =
∑^ n
i=
∂(α F + β G) ∂pi
∂H
∂qi
∂H
∂pi
∂(α F + β G) ∂qi
∑^ n
i=
[
α
∂F
∂pi
∂G
∂pi
]
∂H
∂qi
∂H
∂pi
[
α
∂F
∂qi
∂G
∂qi
]
∑^ n
i=
α
∂F
∂pi
∂H
∂qi
∂G
∂pi
∂H
∂qi
− α
∂H
∂pi
∂F
∂qi
− β
∂H
∂pi
∂G
∂qi
∑^ n
i=
α
∂F
∂pi
∂H
∂qi
− α
∂H
∂qi
∂F
∂qi
∂G
∂pi
∂H
∂qi
− β
∂H
∂qi
∂G
∂qi
= α
∑^ n
i=
∂F
∂pi
∂H
∂qi
∂H
∂qi
∂F
∂qi
∑^ n
i=
∂G
∂pi
∂H
∂qi
∂H
∂qi
∂G
∂qi = α {F, H} + β {G, H}}
Thus the bracket fulfills the proptery of bilinearity. Lastly, we show that the Ja- cobi Identity holds.
{F, {G, H}} =
∑^ n
i=
∂F
∂pi
∂{G, H}
∂qi
∂{G, H}
∂pi
∂F
∂qi
∑^ n
i=
∂F
∂pi
∂qi
∑^ n
j=
∂G
∂pj
∂H
∂qj
∂H
∂pj
∂G
∂qj
∂pi
∑^ n
j=
∂G
∂pj
∂H
∂qj
∂H
∂pj
∂G
∂qj
∂F
∂qi
∑^ n
i=
∑^ n
j=
[
∂F
∂pi
∂qi
∂G
∂pj
∂H
∂qj
∂H
∂pj
∂G
∂qj
∂pi
∂G
∂pj
∂H
∂qj
∂H
∂pj
∂G
∂qj
∂F
∂qi
]
∑^ n
i=
∑^ n
j=
∂F
∂pi
∂^2 G
∂pj ∂qi
∂H
∂qj
∂G
∂pj
∂^2 H
∂qj ∂qi
∂^2 H
∂pj ∂qi
∂G
∂qj
∂H
∂pj
∂^2 G
∂qj ∂qi
∂^2 G
∂pj ∂pi
∂H
∂qj
∂G
∂pj
∂^2 H
∂qj ∂pi
∂^2 H
∂pj ∂pi
∂G
∂qj
∂H
∂pj
∂^2 G
∂qj ∂pi
∂F
∂qi
∑^ n
i=
∑^ n
j=
∂F
∂pi
∂^2 G
∂pj ∂qi
∂H
∂qj
∂F
∂pi
∂G
∂pj
∂^2 H
∂qj ∂qi
∂F
∂pi
∂^2 H
∂pj ∂qi
∂G
∂qj
∂F
∂pi
∂H
∂pj
∂^2 G
∂qj ∂qi
∂^2 G
∂pj ∂pi
∂H
∂qj
∂F
∂qi
∂G
∂pj
∂^2 H
∂qj ∂pi
∂F
∂qi
∂^2 H
∂pj ∂pi
∂G
∂qj
∂F
∂qi
∂H
∂pj
∂^2 G
∂qj ∂pi
∂F
∂qi
=
∑^ n
i=
∑^ n
j=
∂F
∂pi
∂^2 G
∂pj ∂qi
∂H
∂qj
∂^2 F
∂pi∂pj
∂G
∂qi
∂H
∂qj
∂^2 F
∂pi∂pj
∂G
∂qi
∂H
∂qj
∂F
∂pi
∂G
∂pj
∂^2 H
∂qj ∂qi
∂F
∂pi
∂^2 H
∂pj ∂qi
∂G
∂qj
∂F
∂pi
∂H
∂pj
∂^2 G
∂qj ∂qi
∂^2 F
∂pj ∂qi
∂G
∂pi
∂H
∂qj
∂^2 F
∂pj ∂qi
∂G
∂pi
∂H
∂qj
∂^2 G
∂pj ∂pi
∂H
∂qj
∂F
∂qi
∂G
∂pj
∂^2 H
∂qj ∂pi
∂F
∂qi
∂^2 F
∂pi∂qj
∂G
∂qi
∂H
∂pj
∂^2 F
∂pi∂qj
∂G
∂qi
∂H
∂pj
∂^2 H
∂pj ∂pi
∂G
∂qj
∂F
∂qi
∂H
∂pj
∂^2 G
∂qj ∂pi
∂F
∂qi
∂^2 F
∂qi∂qj
∂G
∂pi
∂H
∂pj
∂^2 F
∂qi∂qj
∂G
∂pi
∂H
∂pj
∑^ n
i=
∑^ n
j=
∂F
∂pi
∂^2 G
∂pj ∂qi
∂H
∂qj
∂^2 F
∂pi∂pj
∂G
∂qi
∂H
∂qj
∂^2 G
∂pj ∂pi
∂H
∂qj
∂F
∂qi
∂^2 F
∂pj ∂qi
∂G
∂pi
∂H
∂qj
∂^2 F
∂pi∂qj
∂G
∂qi
∂H
∂pj
∂F
∂pi
∂H
∂pj
∂^2 G
∂qj ∂qi
∂H
∂pj
∂^2 G
∂qj ∂pi
∂F
∂qi
∂^2 F
∂qi∂qj
∂G
∂pi
∂H
∂pj
∂^2 F
∂pi∂qj
∂G
∂qi
∂H
∂pj
∂^2 H
∂pj ∂pi
∂G
∂qj
∂F
∂qi
∂F
∂pi
∂^2 H
∂pj ∂qi
∂G
∂qj
∂^2 F
∂pi∂pj
∂G
∂qi
∂H
∂qj
∂^2 F
∂pj ∂qi
∂G
∂pi
∂H
∂qj
∂F
∂pi
∂G
∂pj
∂^2 H
∂qj ∂qi
∂G
∂pj
∂^2 H
∂qj ∂pi
∂F
∂qi
∂^2 F
∂qi∂qj
∂G
∂pi
∂H
∂pj
∑^ n
i=
∑^ n
j=
∂F
∂pi
∂^2 G
∂pj ∂qi
∂^2 F
∂pi∂pj
∂G
∂qi
∂H
∂qj
∂^2 G
∂pj ∂pi
∂F
∂qi
∂^2 F
∂pj ∂qi
∂G
∂pi
∂H
∂qj
∂H
∂pj
∂^2 F
∂pi∂qj
∂G
∂qi
∂F
∂pi
∂^2 G
∂qj ∂qi
∂H
∂pj
∂^2 G
∂qj ∂pi
∂F
∂qi
∂^2 F
∂qi∂qj
∂G
∂pi
∂G
∂pj
∂^2 F
∂pi∂qj
∂H
∂qi
∂F
∂pi
∂^2 H
∂qj ∂qi
∂G
∂pj
∂^2 H
∂qj ∂pi
∂F
∂qi
∂^2 F
∂qj ∂qi
∂H
∂pi
∂^2 F
∂pj ∂qi
∂H
∂pi
∂^2 H
∂pj ∂pi
∂F
∂qi
∂G
∂qj
∂F
∂pi
∂^2 H
∂pj ∂qi
∂^2 F
∂pj ∂pi
∂H
∂qi
∂G
∂qj
=
∑^ n
i=
∑^ n
j=
∂pj
∂F
∂pi
∂G
∂qi
∂G
∂pi
∂F
∂qi
∂H
∂qj
∂H
∂pj
∂qj
∂F
∂pi
∂G
∂qi
∂G
∂pi
∂F
∂qi
∂G
∂pj
∂qj
∂F
∂pi
∂H
∂qi
∂H
∂pi
∂F
∂qi
∂pj
∂F
∂pi
∂H
∂qi
∂H
∂pi
∂F
∂qi
∂G
∂pj
=
∑^ n
j=
∂{F, G}
∂pj
∂H
∂qj
∂H
∂pj
∂{F, G}
∂qj
∂G
∂pj
∂{F, H}
∂qj
∂{F, H}
∂pj
∂G
∂pj
= {{F, G}, H} + {G, {F, H}}.
This completes the proof of the algebraic properties.
- Show that Poisson brackets and ordinary multiplication of functions make the vector space C∞(R^2 n) into a Poisson algebra. This is a Lie algebra that is also a commutative algebra, with the bracket {F, G} and the product F G related by the Leibniz identity:
{F, GH} = {F, G}H + G{F, H}.
