Eli Botkin said:
Hurkyl:
How would one rearrange the terms of the infinite string 0 +0 +0 +0 +0 +0 +... to change the value of the sum?
___________
Eli
We're not talking about an infinite sum of zeroes here. We're talking about the sum of all of the internal forces. You only get a string of zeroes when you rearrange and regroup this summation so that they pairwise cancel.
But, in the course of writing this post, I realized that it's a moot point -- in Tomaz's problem, the "force" on the center of mass
cannot even be computed as a sum of forces on the individual particles.
Let's go back to first principles. First, some definitions:
x
i = the x-coordinate of the
i-th particle. (1 \leq i)
m
i = the mass of the
i-th particle.
x
c = the center of mass
m = the total mass
F
i = the net force on the
i-th particle
F
c = the net "force" on the center of mass
By definition, the center of mass is given by
<br />
x_c = \sum_{1 \leq i} \frac{m_i x_i}{m}.<br />
(Furthermore, the definition of center of mass requires that this sum converges absolutely)
The "force" on the center of mass is
<br />
F_c = m \frac{d^2 x_c}{dt^2}<br />
If we plug in the definition of the center of mass, we get
<br />
F_c = m \frac{d^2}{dt^2} \sum_{1 \leq i} \frac{m_i x_i}{m}<br />
= \frac{d^2}{dt^2} \sum_{1 \leq i} m_i x_i.<br />
Under
good circumstances, we can pull the derivative into the summation, giving us F_c = \sum_{1 \leq i} F_i. But this is not a good circumstance -- the theorem that would normally allow us to swap the order of summation and differentiation requires that
<br />
\sum_{1 \leq i} m_i \frac{d x_i}{dt}<br />
and
<br />
\sum_{1 \leq i} m_i \frac{d^2 x_i}{dt^2}<br />
= \sum_{1 \leq i} F_i<br />
both to be uniformly convergent (in
t). But we don't have that here -- in fact, the latter sum
doesn't even converge when
t = 0.
This is all fine, since I'm pretty sure the center of mass doesn't exist for this collection of particles when
t is nonzero, so it would be surprising if these sums were well-behaved.
