On Newton's $2^{\mathrm{nd}}$ law and singularities

Question

A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force. $${\bf{F}} = \frac{ {\rm d}\, \mathbf{p}}{{\rm d} \, t}, \qquad {\bf p }= m {\bf v}.$$

Consider a collision process involving two rigid bodies. Their momentum changes instantaneously when they bump into each other, and still the net force has to be some finite number. So why isn't it necessary to introduce some Dirac delta function so "fix" this singularity?

Of course but... the collision time is negligible compared to that of the whole phenomenon (at least in a non-relativistic scenario). So it might as well be considered instantaneous — ric.san, Nov 03 '21 at 21:41
And yet the folks with giant finite element models happily simulate what happens in car crashes, missile launches, and asteroid impacts all the time using Newton's laws. The real question is just what does 'rigid' mean in this context where real material parameters are used. — Jon Custer, Nov 03 '21 at 21:59
Well there aren't rigid bodies in physics so the situation you describe isn't a real situation. However you can solve newton's second law with a delta impact if you want to, it's just not a real situation. — Triatticus, Nov 03 '21 at 22:03
There's no instant processes in classical physics. Apart from an "instant action" on entangled particles in QM which is non-classical way. Though, it's questionable if one can call it "action on one entangled particle", because pair of entangled particles may be described as a common wave-function distributed spatially. So term "instant" may not be applicable here as similarly like you don't say that on photon reflection from a mirror- action on one side of photon instantly propagates to another photon side :-D — Agnius Vasiliauskas, Nov 03 '21 at 22:24
As per $\delta p = (nF) (\delta t)/n $, if you want for momentum change to be finite and const, then in case you approach $n \to \infty$ (time span to zero), force must approach infinity too (nothing finite here as you tried to imply). However things get more complicated in this case, because $ 0 \cdot \infty$ is indeterminate form, thus undefined. So if you set impact time to zero, either momentum change will be zero too, or undefined if you'll take infinite force. — Agnius Vasiliauskas, Nov 03 '21 at 22:31
The two assertions of changes instantaneously and has to be some finite number don't go together. Eiter it's instantaneous (idealized), then the force will diverge, or the force is somehow bounded, then it's not instantaneous. — noah, Nov 03 '21 at 22:32
Related: In a collision shouldn't objects of different mass have same acceleration? — Sandejo, Nov 04 '21 at 03:55

score 2 · Accepted Answer · answered Nov 03 '21 at 22:42

An instantaneous collision is an approximation to reality. Objects really deform as they smash together. This takes time. The objects travel a short distance as this happens.

If the collision is elastic, they push each other away as they spring back.

See Ball hits curve of same curvature

On Newton's $2^{\mathrm{nd}}$ law and singularities

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