implies that indeed the total energy of the system is conserved. Hence substituting equations (2)
and (3) into (1) gives:
dE(t)
dt =
n
X
i= 1
˙qi(t)
X
j6=i
fij(|qi(t)−qj(t)|)qi(t)−qj(t)
(|qi(t)−qj(t)|)+X
j6=i
V0
ij
qi(t)−qj(t)
|qi(t)−qj(t)|
.(5)
Now substituting (4) into (5), we find the desired result:
dE(t)
dt =
n
X
i= 1
˙qi(t)X
j6=ifij(|qi(t)−qj(t)|)−fij (|qi(t)−qj(t)|)qi(t)−qj(t)
(|qi(t)−qj(t)|)= 0
We could have just as well arrived at the same valid conclusion by omitting (4) and noting that
∂Vij
∂qi=−∂Vij
∂qjbecause the quantity qi(t)−qj(t) is anti-symmetric on interchange of iand j. This
is why fij =fji but Fij =−Fji . This is of course what makes (4) true when the assumption of
the existence of a conservative force interaction connected to a potential with the precise argument
dependence of |qi(t)−qj(t)|is made.
Conservation of Angular Momentum in the n-Body Problem:
Show that d
dt J(t) = 0 using Newton’s 2nd law and
Fij(t) = fij(|qi(t)−qj(t)|)qi(t)−qj(t)
|qi(t)−qj(t)|
where fij =fji.
Solution:
We start with the following:
J(t) =
n
X
i=1
Ji(t) =
n
X
i=1
miqi(t)×˙qi(t)
Differentiating the above equation gives:
dJ(t)
dt =d
dt "n
X
i=1
miqi(t)×˙qi(t)#
=
n
X
i=1
mi
d
dt [qi(t)×˙qi(t)]
=
n
X
i=1
[mi˙qi(t)×˙qi(t) + qi(t)×mi¨qi(t)]
=
n
X
i=1
[qi(t)×Fi(t)] (since: a×a≡0)
=
n
X
i=1
qi(t)×X
j6=i
Fij (t)
2
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