Download Derivation of Angular Momentum Conservation in n-Body Problem: Newton's Second Law and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity!
Solution: First note from the chain rule and our definition of Vij , we have
d dt Vij (||x(t)||) = d d||x(t)|| Vij (||x(t)||) d||x(t)|| dt = −fij (||x(t)||) x(t) ||x(t)|| x˙(t).
Therefore, from the above and Newton’s 3rd^ law, we have
d dt Vi(t) =
∑^ n j=
d dt Vij (||qi(t) − qj (t)||)
∑^ n j=
−fij (||qi(t) − qj (t)||) qi(t)^ −^ qj^ (t) ||qi(t) − qj (t)|| ( ˙qi(t) − q˙j (t))
∑^ n
j=
−Fij (||qi(t) − qj (t)||)( ˙qi(t) − q˙j (t))
∑^ n j=
−Fij (||qi(t) − qj (t)||) ˙qi(t) +^1 2
∑^ n j=
Fij (||qi(t) − qj (t)||) ˙qj (t)
∑^ n
j=
−Fij (||qi(t) − qj (t)||) ˙qi(t) +
∑^ n
j=
−Fji(||qi(t) − qj (t)||)( ˙qj (t)
and so
d dt V (t) = 1 2
∑^ n i=
∑^ n j=
−Fij (||qi(t) − qj (t)||) ˙qi(t) +^1 2
∑^ n i=
∑^ n j=
−Fji(||qi(t) − qj (t)||) ˙qj (t)
∑^ n
i=
∑^ n
j=
−Fij (||qi(t) − qj (t)||) ˙qi(t) +
∑^ n
j=
∑^ n
i=
−Fji(||qi(t) − qj (t)||) ˙qj (t)
∑^ n i=
∑^ n j=
−Fij (||qi(t) − qj (t)||)
(^) q˙i(t) +^1 2
∑^ n j=
( (^) n ∑ i=
−Fji(||qi(t) − qj (t)||)
q ˙j (t)
∑^ n i=
−Fi(t) ˙qi(t) +
∑^ n j=
−Fj (t) ˙qj (t)
∑^ n
i=
−Fi(t) ˙qi(t).
Therefore, by Newton’s 2nd^ law
d dt V (t) =
∑^ n i=
−mi q˙i(t)¨qi(t)
d dt T (t),
and thus d dt
E = 0.
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Conservation of Angular Momentum in the n-body Problem
Let pi(t) = mi q˙(t) be the momentum of the ith^ particle. We’ve shown that Newton’s 2nd^ law gives us conservation of momentum. The angular momentum of the ith^ particle is
Ji(t) = qi(t) × pi(t).
The total angular momentum is:
J(t) =
∑^ n i=
Ji(t).
Show that Newton’s 2nd^ gives us conservation of angular momentum: d dt J(t) = 0.
Solution: We have d dt J(t) =
∑^ n
i=
d dt qi(t) × pi(t)
∑^ n i=
q ˙i(t) × pi(t) + qi(t) × p˙i(t)
∑^ n i=
qi(t) × Fi(t)
∑^ n i=
qi(t) ×
∑^ n j=
Fij (||qi(t) − qj (t)||)
∑^ n i=
∑^ n j=
qi(t) × Fij (||qi(t) − qj (t)||)
∑^ n
i=
∑^ n
j=
qi(t) × fij (||qi(t) − qj (t)||) ||qi(t) − qj (t)|| qi(t) − qi(t) × fij (||qi(t) − qj (t)||) ||qi(t) − qj (t)|| qj (t)
∑^ n i=
∑^ n j=
−qi(t) × fij^ (||qi(t)^ −^ qj^ (t)||) ||qi(t) − qj (t)|| qj (t)
= 0
where the last equality follows from the identity:
−qi(t) × fij (||qi(t) − qj (t)||) ||qi(t) − qj (t)|| qj (t) = qj (t) × fji(||qi(t) − qj (t)||) ||qi(t) − qj (t)|| qi(t).
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