To expand on what I said in post 273, here's one way to formalize Newton's theory:
We define a system as a set of point masses at positions x_\alpha(t) \in R^3, where t \in R and with masses m_\alpha \in (0,\infty), indexed by a set A (ie, \alpha takes on values in A, which could be, eg, the natural numbers). Then we assign to each pair (\alpha,\beta) \in A^2 an interaction force f_{\alpha \beta}(t) \in R^3. Then the two main axioms of Newtonian mechanics are:
1. m_\alpha \frac{d^2 x_\alpha}{d t^2} = \sum_{\beta \in A} f_{\alpha \beta}(t)
2. f_{\alpha \beta}(t) = - f_{\beta \alpha}(t), \mbox{ all } t \in R
So far we have made no other restrictions on the parameters of the model, which to summarize, are:
m_\alpha, x_\alpha(t), f_{\alpha \beta} (t), A.
For (1) to be well defined, the sum on the LHS should converge absolutely (since we haven't specified an order to take the sum). So we require:
3. \sum_{\beta \in A} ||f_{\alpha \beta}(t)|| < \infty, \mbox{ all } \alpha \in A, t \in R
At this point we potentially have an infinite number of coupled differential equations, and I'm not familiar with the theory of such equations. The model might already be inconsistent. But we can forget about actually solving these equations for now and just treat d/dt as an arbitrary linear operator (ie, we'll drop t dependence and just look at this as a fixed configuration of masses and forces).
Next let's consider a situation where the only force is gravity, ie, we define:
f_{\alpha \beta} = \frac{G m_\alpha m_\beta}{||x_\alpha - x_\beta||^3} (x_\beta- x_\alpha)
Which satisfies (2) (assuming all point masses are at distinct positions). For the configuration in post 1, (3) is also satisfied, as is easily checked. So we see that that system is perfectly valid under this model, which we'll call (1,2,3).
The question is whether conservation of momentum holds in this model. Specifically, if we define (of course, now we need to limit to systems where these are defined, but they are for the system in question):
M_{tot} = \sum_{\alpha \in A} m_\alpha
x_{CM} = \frac{1}{M_{tot}}\sum_{\alpha \in A} m_\alpha x_\alpha
p_{tot} = M_{tot} \frac{d x_{cm}}{dt} = \sum_{\alpha} m_\alpha \frac{d x_\alpha}{dt}
(where we have freely interchanged the derivative with the infinite sum, which again may not be valid, but turns out not to be the problem) then we want:
\frac{d p_{tot}}{dt} =0
The way this is usually proved is to write:
\frac{d p_{tot}}{dt} = \frac{d}{dt} \sum_{\alpha} m_\alpha \frac{d x_\alpha}{dt}= \sum_{\alpha} m_\alpha \frac{d^2 x_\alpha}{dt^2}=\sum_{\alpha} \sum_{\beta} f_{\alpha \beta}
If at this point we impose the following axiom (strictly stronger than (3)):
4. \sum_{\alpha} \sum_{\beta} ||f_{\alpha \beta}|| < \infty
Then the sum above converges absolutely, so we can rearrange it so that f_{\alpha \beta} sits next to f_{\beta\alpha}, these will cancel by (2), and we have proven conservation of momentum in the model (1,2,4).
However, the system in post 1 does not satisfy (4), so does not fit into a model where conservation of momentum holds.
The piramid you constructed was not of uniform density, no ? Its density diverged at its vertex ?
