We know that newtonian mechanics is described by a set of second order differential equations. This leads also to the principle of determinism that every trajectory is determined by initial conditions on position and velocity. How good is the experimental evidence on that that there is no third derivative term? (In the same spirit like ppl try to look for an electric field inside a conductor to test Coloumbs law).

Edit: to the downvoters: please comment what is the problem with this question? Using experiments to determine null-results is part and parcel of physics.

Edit: my question is not a duplicate of Why are there only derivatives to the first order in the Lagrangian? but related. The other question asks about theoretical grounds. I am asking about experiments conducted and the limits they impose on higher order derivative terms.

Edit: To clarify the question: we generally accept Newtons second law of motion $$ m \ddot {x} = F(x, \dot{x}, t) $$ this implies that specification of position and velocity uniquely defines a solution (so called principle of determination). How precisely do we know this to be true in the sense of what if we assume instead $$ const\dddot {x} + m \ddot{x} = F $$ then what are the experimental limits on the constant 'const'. How close to zero is it determined from actual experiments.

Note: I still diagree this is a duplicate. For example one answer which goes in the direction from the linked question: Why $F=ma$ and not $F=m\dot{a}$? is 'Because the second derivative is fully determined by the external force, it is an experimental fact.' This goes in the right direction, but my question really is: which experiment gives the highes accuracy on this, where was it published and to what precision does it prove this fact. My question is in the same spirit like questions 'what is the current upper bound on the photon mass' except for a different topic in physics.

Thought experiment to test this: build a rail-gun, place it in a vacuum-chamber and shoot a parabola using the earth gravitational field. Use constant acceleration while still on the rail. Now place the object at a different start position on the rail but also change the constant acceleration so that at the release point (exit nozzle of the gun) the object has the same velocity. Will the object alway hit the same spot? (of course, if the equation of motion is a second order differential equation; it wont if the trajectory depends on the initial acceleration). Of course we know that it will not depend on initial acceleration, but to what precision do we know that?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – rob Mar 23 '17 at 13:59

Apply time vary-ing force in situation with second derivative and you get one

Example- Let a force F=kt(k is constant) is applied. Differentiate it to get third derivative equal to constant.

This third derivative equation only tells rate of change of acceleration of particle is constant analogous to your example which tells that rate of change of velocity is constant.

On side note, in schools Mechanics MOSTLY deals with gravitational force, damping force and spring force which can be represented using second derivative equation.

  • $\begingroup$ Can you write this down? To me it looks like it would be still F=ma not F=ma + const dot{a} for example $\endgroup$ – lalala Mar 21 '17 at 20:13
  • $\begingroup$ Write me a second derivative situation and equation, I can help you improve the situation to make third derivative one $\endgroup$ – monster Mar 21 '17 at 20:32
  • $\begingroup$ a second derivative eq for a falling body would be ma=g with g constant. $\endgroup$ – lalala Mar 21 '17 at 21:01
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    $\begingroup$ I see what you are saying, but I think you misunderstood the question. What you are doing is that Newtows second law with a Force F=kt implies a differential equation like m\dot{a} =k. But this differential equation is not equivalent to the original one. It actually allows for trajectories which are not fulfilled by Newtons second law (since it is a third order diffeq, you are allowed and required to specifiy position, velocity and initial acc as integration constants.). Anyway, my question is about experimental bounds on the third order term. $\endgroup$ – lalala Mar 22 '17 at 12:19

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