When motion begins, do objects go through an infinite number of derivatives of position?

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This might be a very vague and unclear question, but let me explain. When an object at rest moves, or moves from point A to point B , we know the object must have had some velocity (1st derivative of position) during that trip. It's also true that the object had to have accelerated to gain that velocity (2nd derivative)... if we extend this argument, is it logical to say that the object must have experienced jerk (3rd deriv.) to gain that acceleration? And so on and so forth... 3rd, 4th, 5th position derivatives? Or is my logic flawed somewhere?
 

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  • #2
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If an object changes position (x) then it must have had a non-zero velocity (x'), it need not have a non-zero acceleration (x''). Now, if you further constrain the velocity (x') to begin at zero then you will need to have a non-zero acceleration (x''), but you need not have a non-zero jerk (x''').

Basically, the constraints that you place on the motion are what determine how high a derivative you know must have occured (at a minimum). There is nothing about moving from A to B itself that requires anything other than a velocity (x').
 
  • #3
jbriggs444
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According to the mean value theorem if a function f defined on an interval (a,b) has a defined derivitive f' at every point in (a,b) then there is a point x within (a,b) where ##f^{\prime}(x) = \frac{f(b)-f(a)}{b-a}##

Apply this iteratively and it follows that if your position function has a defined derivitive at all levels and if it starts from a state of rest where all such derivitives are zero then it must have a non-zero derivitive at every level sometime before it reaches a new position. However, that's a very big "if". The position of an object is not a precisely defined physical measurement that can be made. Even if it were, there is no guarantee that its derivitives to all levels would be defined.
 
  • #4
Stephen Tashi
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is it logical to say that the object must have experienced jerk (3rd deriv.) to gain that acceleration? And so on and so forth... 3rd, 4th, 5th position derivatives? Or is my logic flawed somewhere?
If a function f(t) "has a derivative", the derivative f'(t) might be the constant function f'(t) = 0. Are you trying to argue that all the higher derivatives must be non-zero?
 
  • #5
jbriggs444
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The OP indicates that the object starts "at rest". So f'(0)=0. If f(t) is different from f(0) for some t, that eliminates the possibility that f is a constant function and consequently, the possibility that f' is a constant function.
 
  • #6
Stephen Tashi
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The OP indicates that the object starts "at rest". So f'(0)=0. If f(t) is different from f(0) for some t, that eliminates the possibility that f is a constant function and consequently, the possibility that f' is a constant function.
The OP asks about higher derivatives. In my post, the function f need not denote a position function.
 
  • #7
Stephen Tashi
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Or is my logic flawed somewhere?
Your reasoning can't be evaluated unless you say exactly what you are assuming about the object's condition when it is at point A. You could say "if the object is at rest at point A..." However, the meaning of that statement is not entirely clear.

If you see a statement in a physics textbook like "An object is initially at rest at time t= 0 ..." the only thing you can safely conclude is that it's velocity is zero at time zero. You can't conclude anything about the value of it's acceleration or the higher derivatives of its position function at t = 0.

By contrast, a common speech interpretation of "The object is at rest at time t = 0..." might be that the object had been in the same position for a finite interval of time, which ended at t = 0.

There is the additional technicality of whether you are assuming all derivatives of the position function exist. For example if the position function is:
[itex] f(t) = 5 - t^2 [/itex] when [itex] t \ge 0 [/itex] and [itex] f(t) = 5 [/itex] for [itex] t < 0 [/itex] then the acceleration [itex] f"(t) [/itex] does not exist at [itex] t = 0 [/itex] since the graph of the velocity [itex] f'(t) [/itex] doesn't have a unique slope at [itex] t= 0 [/itex].

You can take the viewpoint that physical changes must be infinitely differentiable and conclude that the above example is impossible, but you need to make it clear if that is one of the assumptions.
 
  • #8
mathman
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The original question confuses mathematics and physics. The object is in motion (physics). The motion is represented by some (approximate) formula. Whether or not various derivatives for this formula exist is a question of mathematics, not physics.
 
  • #9
Andrew Mason
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d2s/dt2 = F/m so if force and mass are constant, d/t(d2s/dt2)=0.

It appears that the OP is questioning whether F can go from 0 to some finite constant value instantaneously. Of course in the real physical world, nothing is really instantaneous and the actual forces in an interaction can be complicated. The question is how small a time element you want to consider and how close to constant you want to make the force during that time interval.

AM
 
  • #10
A.T.
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It appears that the OP is questioning whether F can go from 0 to some finite constant value instantaneously
I think jbrigs444 got best what the question is. Assume the object is at rest for a while before it starts moving, so all derivatives of position are zero. Then, to start moving, all derivatives must have been non zero at some point, not just acceleration. It's kind of a differential version of Zenos paradox.
 
  • #11
Andrew Mason
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I think jbrigs444 got best what the question is. Assume the object is at rest for a while before it starts moving, so all derivatives of position are zero. Then, to start moving, all derivatives must have been non zero at some point, not just acceleration. It's kind of a differential version of Zenos paradox.
If [itex]f(t)=0 \text{ for }t\le{0} \text{ and }f(t)=k\gt{0} \text{ for }t\gt{0}[/itex] then third and higher order derivatives of position are 0 for all [itex]t
\ne{0}[/itex] and undefined for t=0. So isn't the issue whether f(t) can be a real force?

AM
 
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  • #12
A.T.
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[itex]f(t)=k\gt{0} \text{ for }t\gt{0}[/itex]
Why should it be constant for t > 0? The only thing needed for the object to move from a to b is that velocity is non zero for some time.

undefined for t=0.
Which would be unphysical. I think the OP asks about a case without such discontinuities.
 
  • #13
Andrew Mason
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Why should it be constant for t > 0? The only thing needed for the object to move from a to b is that velocity is non zero for some time.
The OP was suggesting that all higher order derivatives of position must be non-zero for some time when a force is applied. I was just giving an example of a force for which the higher order derivatives of position with respect to time are not non-zero at any time.

Which would be unphysical. I think the OP asks about a case without such discontinuities.
While it may be unphysical, the time period required for the force to go from 0 to k can be made arbitrarily small.

AM
 

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