Conservation of Energy for the n-Body Problem
If Newton’s Second Law holds, then energy is conserved.
Solution: The energy is given by E=T+V, where Tis the kinetic energy and Vis the potential
energy. If there are nbodies in a system, then T=
n
X
i=1
1
2mi˙q2
i(t) and so ˙
T=
n
X
i=1
mi¨qi·˙qi=
n
X
i=1
Fi·˙qi,
by Newton’s Second Law. Since Fi=X
j6=i
Fij =X
j6=i
fij(|qi−qj|)qi−qj
|qi−qj|we have
˙
T=
n
X
i=1 X
j6=i
fij(|qi−qj|)qi−qj
|qi−qj|·˙qi=
n
X
i=1
X
j>i
fij(|qi−qj|)qi−qj
|qi−qj|·˙qi+X
j<i
fij (|qi−qj|)qi−qj
|qi−qj|·˙qi
=
n
X
i=1 X
j>i
qi−qj
|qi−qj|·(fij(|qi−qj|) ˙qi−fji (|qi−qj|) ˙qj) =
n
X
i=1 X
j>i
fij(|qi−qj|)qi−qj
|qi−qj|·( ˙qi−˙qj), since
fij =fji by Newton’s Third Law. Also V=
n
X
i=1
Vi=
n
X
i=1 X
j>i
Vij (|qi−qj|) so
˙
V=
n
X
i=1 X
j>i
˙
Vij(|qi−qj|)d
dt (|qi−qj|)) =
n
X
i=1 X
j>i
−fij(|qi−qj|)qi−qj
|qi−qj|·( ˙qi−˙qj), since V0
ij =−fij.
So ˙
E=˙
T+˙
V= 0 and energy is conserved.
Conservation of Energy for the n-Body Problem
If Newton’s Third Law holds, then angular momentum is conserved.
Solution: If there are nbodies in a system, the angular momentum is J(t) =
n
X
i=1
Ji(t) =
n
X
i=1
miqi×˙qi, so ˙
J(t) =
n
X
i=1
mi˙qi×˙qi+qi×m¨qi=
n
X
i=1
qi×Fi=
n
X
i=1
qi×
X
j6=i
fij(|qi−qj|)qi−qj
|qi−qj|
=
n
X
i=1 X
j6=i
fij (|qi−qj|)
|qi−qj|qi×(qi−qj) =
n
X
i=1 X
j6=i
fij(|qi−qj|)
|qi−qj|qj×qi= 0 since qi×qj=−qj×qi,fij =fji
and all the terms will cancel.
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