What happens first

1. Nov 17, 2005

vaishakh

When the mechanical equilibrium of an object is disturbed, its velocity, acceleration, rate of change of acceleration and so on everything starts changing continuously. For a change of something the rate of change has to change and the same procedure gets repeated for rate of change. Thus what starts first is doubtful. In short, Newton’s laws cannot explain when does an object at rest start moving when force is applied on it. It only states that the object does not start moving instantaneously.

2. Nov 17, 2005

Kurdt

Staff Emeritus
I think what you're getting at here is that higher orders of position derivatives wrt time would continue infinitely when something is disturbed? Assuming I have the jist of your problem, there are situations in which these higher order derivatives are zero. Take for example, a car travelling at a constant velocity v. We know v is simply the first time derivative of the cars position. If we integrate with respect to time then wes see that s=v*t+c. We can get rid of the arbitrary constant by choosing the coordinates correctly. Now in this example if you want to find the rate of change of velocity you would differentiate twice wrt to time and you will clearly see that the acceleration will be equal to 0.

That was just a simple example but it is very rare that we ever need to use higher order derivatives. The only people i know of that use higher than 3rd derivatives are rolllercoaster engineers. Even then simple laws can be derived by assuming that whatever order derivative is constant. I assume you have seen the equations of motion for constant acceleration.

s=ut+1/2 a*t^2

which you can derive from assuming that d2s/dt2= a constant.

3. Nov 17, 2005

kleinwolf

I had the same problem at high school...in my view this problem was : if an object starts moving from rest...does it mean infinite acceleration....but i was told this is wrong...let's take Newton's law : let say if $$t_-<0\quad t_+ >0$$ ...then the hypothesis is $$\ddot{x}(t_-)=0\quad \ddot{x}(t_+)=\delta\neq0\Rightarrow x(t_-)=at+b\quadx(t_+)=\delta/2t^2+dt+e$$...from the continuity hypothesis of movement : $$x(0^-)=x(0^+) \Rightarrow e=b$$ this gives that the speed is continuous at t=0...however the accelaration is not, which means the jerk (triple derivative) is Dirac-delta shaped (derivative of a Heaviside step function in this case). However, if the system is taking as being subject to another force (non step-like), then obviously the jerk is not singular...but I don't if there exist a general theory about this...It distrubed me a lot of time indeed... But I think this has to do with the microscopic description of the force acting on the body : deformation and temperature and other thermodynamical concepts have to be taken into account in order to find the right depedence of the force towards time for the "in movement making" out of the equilibrium position.

Last edited: Nov 17, 2005
4. Nov 17, 2005

vanesch

Staff Emeritus
The error is to think that the only possible time dependencies are analytic functions. Penrose has a nice argument about that in his book "The Road to Reality" (a super read in my opinion!).

Indeed, it is impossible for an analytic function to be 0 over a finite time lapse and then to start doing something from t>t0 onward.

5. Nov 18, 2005

kleinwolf

So does this book explain how the Force has to change in a schock similar process...it should be something of the type : $$F(t)=\left\{\begin{array}{ll} 0 & t<-t_0 \\ e^{-\frac{1}{|t^2-t_0^2|^2} & -t_0<t<0\\ e^{-\frac{1}{t_0^4} & t>0 \end{array}\right.$$
It is easy to verify that $$F(t)\in C^\infty(\mathbb{R})$$ (analytic function different than holomorph)...and $$F(t<-t_0)=0\quad F(t>t_0)>0$$...
Can you please send me a copy of the page indicating the rate of change of the force as given in your reference...? We call those Schwarz Function or "compact support dense in L^2 function subset"...if you change slightly with F(t>t0)=0....For remembrance this subset is used in theorem proofs for Fourier transform similar operators properties on separable Hilbert spaces like L^2...aso

Last edited: Nov 18, 2005
6. Nov 20, 2005

vanesch

Staff Emeritus
The eternal nomenclature. With analytic I meant: can be written in a powerseries around each point in an open domain, with finite convergence radius. Your example function is a C-inf function which is not analytic (I follow the definition that C-inf is "holomorphic" and analytic is series convergence, but I know some authors inverse these terms ; in the complex plane, both concepts are equivalent of course).

The point of analytic functions over R is that a very small segment of them determine the entire function. As such, the small part left of -t0 that is identical to 0 would determine the entire function over R, which would be 0 everywhere. Clearly this is not the case. I think the issue clears out if you continue your function in the complex plane. Then you will see that your smooth function is not at all so smooth.

The entire chapter 9 of "the road to reality" is devoted to this issue (although he does not treat exactly the function you propose). He proposes as "reasonable class of real functions to be used" the difference between a holomorphic function in a part of the upper half plane (down to the real axis) and another holomorphic function in a part of the lower half plane (modulo a holomorphic function over both which will not contribute to the difference of course). Such a "hyperfunction" (an equivalence relation on pairs of holomorphic functions) can then be manipulated by derivation...

(I'm typing this very quickly... may contain some inaccuracies).

cheers,
Patrick.