Why dynamical self-excitation is possible (Q1977821)

This paper is an answer (and rejection) of a criticism formulated by \textit{M. Bridger} and \textit{J. S. Alper} [see the foregoing entry] considering a previous contribution of the author. There seems to be quite a lot of misunderstanding culminating in the author's statement: if so, ``dynamics could not deal with interaction by contact between rigid solids, as such collisions originate discontinuities in force and velocity of the kind that Alper and Bridger say violate `any of the fundamental principles of Newtonian mechanics' ''. From the viewpoint of classical mechanics it seems that axioms and modeling aspects are pretty mixed up. Concerning the modeling, the ``supertask'' is defined by infinitely many unit point masses \(P_1, P_2, P_3, \cdots\) arrayed at rest along the \(x\)-axis of space at positions \( x_1=\frac{1}{2}, x_2=\frac{3}{4}, x_3=\frac{7}{8}, \cdots\). At time \(t=0\) a unit point mass \(P_o\), moving at unit velocity, passes \(x=0\) and approaches the remaining masses. Infinitely many perfectly elastic collisions happen to follow with the unit velocity passing successively from \(P_o\) to \(P_1\), from \(P_1\) to \(P_2\) etc. However, at \(t=1\), all collisions are completed, and all the masses are at rest. If, as in the present, the collisions are looked at as Dirac distributions (along with its discontinuties), this is only a consequence of mechanical modeling: the discontinuities occur from limiting the impact time to zero, which is never microscopic true but just a question of convenience. It can, however, not be the source of any physical contradiction. To be more precise: the mechanical model under consideration is prescribed. The question arises how this model behaves. Looking at ``trivial'' mathematics, one has the following situation: since \(v_o=1\), the \(n\)th impact takes place at \(t_n = 1 + 1/2 + 1/4 + \cdots+ (1/2)^{n-1} = 1 - (1/2)^n\), \(n=1,2,3,\dots\). Thus, the first collision takes place at \(t_1=1/2\) yielding \(v_o = 0\Rightarrow v_1=1\), the second at \(t_2=3/4\) yielding \(v_1 = 0\Rightarrow v_2=1\) and so on, finally the \(n\)th collistion is at \(t_n=1-(1/2)^n \) yielding \(v_{n-1} = 0\Rightarrow v_n=1\), no matter how large \(n\) is. The model under consideration does not yield any other result, and we are not asking wether the model makes sense: it is prescribed. It does not include the completion of collisions in the sense that for \(n \rightarrow \infty \Rightarrow t= 1\) no motion remains. This is only the case if a) an additional mechanism (force) enters consideration (how?) which stops the motion instantaneously for the last particle, or b) the last particle is removed from the system. Case b) is in accordance with the underlying classical (Newtonian, here additionally energy conserving) mechanics, while case a) may be anything else which could cause any desired or undesired behaviour just by intuition. All this is obviously independent from global (Hilbertian) or local considerations: embedding into Hilbert spaces means obviously to include every particle including the last one (\(n \rightarrow \infty\)) where then something goes in (i.e. particle, \(P_o\)) but nothing comes out. The same holds when prescribing \(v_n =0\), \(n\) being infinite or not. There arises, however, one suspicion: \(t=1\) is a limiting case. If, however, no particle leaves the control volume, it might be that the total process is not yet over. This corresponds to Bridger and Alper's ``singularity caused by the infinite density of particle collisions in the neighbourhood of \(t=1\)'' -- with emphasis on neighbourhood. Thus, taking the limit value for \(t\) along with anything which does not correspond to the same limit for \(v\), obviously corresponds to a wrong limitation process. The problem then seems to be ill-posed. In the reviewer's opinion, the model which is considered here needs more explanation, basicly the answer to: why (and how) should the motion of the entire system come to an end for \(t=1\)?