# Infinite time for an object to get K=0 conservative Force

• titansarus
In summary, the conversation discusses a question where only conservative forces are present and a particle is moving from one point to another in a position diagram. The goal is to prove that it takes infinite time for the particle to reach its final destination. The equations used include the conservation of energy, kinetic energy, and force. Various attempts are made to find a mathematical proof for this, including using the Taylor series and the equation for velocity. However, a conclusive proof is not found.
titansarus

## Homework Statement

Think that we have only conservative force in this question and no nonconservative force like friction exists. Also## U ##means potential energy (eg Gravitational Potential Energy),##K## means Kinetic Energy (##1/2 mv^2##) and ##E = K + U = constant##

We have a arbitrary Potential energy (U) in terms of position (x) diagram like the picture. The object is at first, in a position like ##x_0## and wants to get to point ##x_1##. At ##x_1## we have ##dU / dx = 0## and Also the initial Mechanical Energy (E = K + U) of the object at first is equal to the U at point ##x_1##. We want to prove that it takes infinite time -theoretically- for this object to get to the point ##x_1##. (The movement is 1 Dimensional)

## Homework Equations

##E = K + U = Constant##
##K = 1/2 m v^2##
##F = m a##
##F = -dU/dx##
Taylor Series at ##x = x_1## for function ##f(x)##: ##f(x) = f(x_1) + f'(x) (x - x_1) + f''(x) (x - x_1)^2 / 2! + ...##
All forces are conservative in this question.

3. The Attempt at a Solution

At first we know that the K at ##x_1## equals to zero. Intuitively, I can understand the at the end of the path, the acceleration goes to zero and also the velocity goes to zero (because ## K -> 0 ==> V -> 0##). But I don't know how to prove this. I know that F = -dU/dx.

For the start, I get this from taylor series but I don't know what to do with it:

##U(x) = U(x1) + dU/dx (x - x1) + O(x^2)## -> ##U(x) = U(x1) + (-m a) (x - x1)

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What would happen to a particle that started at ##x_1## with ##v = 0##?

titansarus and haruspex
PeroK said:
What would happen to a particle that started at ##x_1## with ##v = 0##?
It will not move at all. because ##dU/dx = 0 -> F =0 -> a= 0 , v =0##. But I want to derive something that mathematically and not intuitively shows for particles started at ##x=x_0## with ##v>0##, the time that they will get to ##x = x_1## is going to infinite. (something like ## t = 1 / (x-x_1)##). Probably using Taylor Series.

titansarus said:
It will not move at all. because ##dU/dx = 0 -> F =0 -> a= 0 , v =0##. But I want to derive something that mathematically and not intuitively shows for particles started at ##x=x_0## with ##v>0##, the time that they will get to ##x = x_1## is going to infinite. Probably using Taylor Series.

To make things easier, you could let ##E = 0## and note that, for a local maximum, ##U'(x) = 0## and ##U''(x) \le 0##.

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titansarus
PeroK said:
To make things easier, you could let ##E = 0## and note that, for a local maximum, ##U'(x) = 0## and ##U''(x) \ge 0##.

I think ##U''(x) \leq 0## at local maximum. However, I don't know how to relate this equations to time. We don't know anything about how U is related to t so we can't easily integrate something like ##(U(x) - U(x_1) )dt^2 = - m (x - x_1)d^2 x##

titansarus said:
I think ##U''(x) \leq 0## at local maximum. However, I don't know how to relate this equations to time. We don't know anything about how U is related to t so we can't easily integrate something like ##(U(x) - U(x_1) )dt^2 = - m d^2 x (x - x_1)##

Yes, of course. I've corrected that.

What about using ##v^2 = \frac2m(E -U(x))##?

titansarus
PeroK said:
Yes, of course. I've corrected that.

What about using ##v^2 = \frac2m(E -U(x))##?
Maybe using this We can get:

##E = U(x_1) => U(x) = E + 0 + U''(x_1)(x-x_1)^2 / 2 + ...##

So ##v^2 = U''(x_1) (x-x_1)^2 / m)##. Now what to do with that? Maybe I can say that t at the last part approximately equals to ##t = (x-x_1) / v## so I will get something like ##t= m / \sqrt {(-U''(x_1))}## Which is not what we want.

Even if I say something like ##1/2 a t^2 + v t = 0## (##v## of the last moments), we get t =0 or ##t = -2U''(x_1) (x-x_1)^2 / ( -m^2 U'(x))## and I don't get something reasonable from this.

So what can I do from Here?

titansarus said:
Maybe using this We can get:

##E = U(x_1) => U(x) = E + 0 + U''(x_1)(x-x_1)^2 / 2 + ...##

So ##v^2 = U''(x_1) (x-x_1)^2 / m)##. Now what to do with that? Maybe I can say that t at the last part approximately equals to ##t = (x-x_1) / v## so I will get something like ##t= m / \sqrt {(-U''(x_1))}## Which is not what we want.

Even if I say something like ##1/2 a t^2 + v t = 0## (##v## of the last moments), we get t =0 or ##t = -2U''(x_1) (x-x_1)^2 / ( -m^2 U'(x))## and I don't get something reasonable from this.

So what can I do from Here?

First, a mathematical observation.

It's possible, although I haven't checked it out, that you could construct a pathological function ##U(x)## that breaks this rule. So, you may need ##U(x)## to be "well-behaved": i.e. a physically possible potential. One common assumption to make in these cases is that the derivatives of ##U## are uniformly bounded. I.e.

##\exists \ M > 0 \ ## such that ##\forall x, n \ |U^{(n)}(x)| < M##

Alternatively, and less rigorously, you could note that, for small enough ##x-x_1##

##E - U(x) \approx E - U(x_1) - (x-x_1)U'(x_1) - \frac12 (x-x_1)^2U''(x_1)##

And in this case, therefore, you have:

##E - U(x) \approx - \frac12 (x-x_1)^2U''(x_1) = k^2(x-x_1)^2 ##

For some ##k## and ##x## close to ##x_1##. Let's ignore the case where ##U''(x_1) = 0##. You could do that as an additional piece of work if you want.

Now assume that the particle gets close enough to ##x_1## in a finite time for the above to hold - some point ##x_2##, say. Again, you may require ##U## to be well-behaved for this assumption to hold (edit: this assumption is safe enough, although you might want to prove it!). Then we have:

##\Delta t = \int_{x_0}^{x_1}\frac{dx}{v} = \int_{x_0}^{x_2}\frac{dx}{v} + \int_{x_2}^{x_1}\frac{dx}{v}##

This is the time to get from ##x_0## to ##x_1##. Can you finish things from there?

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titansarus
PeroK said:
First, a mathematical observation.

It's possible, although I haven't checked it out, that you could construct a pathological function ##U(x)## that breaks this rule. So, you may need ##U(x)## to be "well-behaved": i.e. a physically possible potential. One common assumption to make in these cases is that the derivatives of ##U## are uniformly bounded. I.e.

##\exists \ M > 0 \ ## such that ##\forall x, n \ |U^{(n)}(x)| < M##

Alternatively, and less rigorously, you could note that, for small enough ##x-x_1##

##E - U(x) \approx E - U(x_1) - (x-x_1)U'(x_1) - \frac12 (x-x_1)^2U''(x_1)##

And in this case, therefore, you have:

##E - U(x) \approx - \frac12 (x-x_1)^2U''(x_1) = k^2(x-x_1)^2 ##

For some ##k## and ##x## close to ##x_1##. Let's ignore the case where ##U''(x_1) = 0##. You could do that as an additional piece of work if you want.

Now assume that the particle gets close enough to ##x_1## in a finite time for the above to hold - some point ##x_2##, say. Again, you may require ##U## to be well-behaved for this assumption to hold (edit: this assumption is safe enough, although you might want to prove it!). Then we have:

##\Delta t = \int_{x_0}^{x_1}\frac{dx}{v} = \int_{x_0}^{x_2}\frac{dx}{v} + \int_{x_2}^{x_1}\frac{dx}{v}##

This is the time to get from ##x_0## to ##x_1##. Can you finish things from there?
I think the integral is something in form of ##C ln (x_1-x)## so when x goes to x_1, log goes to -infinity and because of this, the time is infinity. Is this reasoning correct?

Also I must note that because the problem has a diagram like the one above, we can assume that the U(x) is well-behaved.

titansarus said:
I think the integral is something in form of ##C ln (x_1-x)## so when x goes to x_1, log goes to -infinity and because of this, the time is infinity. Is this reasoning correct?

Yes, it hinges on the fact that ##\int_0^1 \frac{dx}{x^s}## diverges for ##s \ge 1##.

titansarus
Is it possible to construct a shape of the potential energy "hill" such that the time is finite rather than infinite?

For convenience, choose the origin of the x-y coordinate system at the top of the hill. In this coordinate system, suppose the hill is described by

##U(x) = -|x|^{1.8}##. The first derivative of U exists at x = 0 with U'(x) = 0. But higher order derivatives do not exist at the origin. I don't know if this would be considered "pathological" according to @PeroK 's comments in post #8.

The graph of U(x) vs. x looks like an OK hill.

If the particle starts at ##x_1## with just enough initial speed to reach ##x = 0## with zero speed, then I get that the time to travel from ##x_1## to ##x = 0## is

t = ##10\sqrt{\frac{m}{2}} x_1^{0.1}##

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titansarus and PeroK

## 1. What is "Infinite time for an object to get K=0 conservative Force"?

"Infinite time for an object to get K=0 conservative Force" refers to a scenario in which an object is subjected to a conservative force, such as gravity, and eventually comes to a complete stop with no kinetic energy (K=0) after an infinite amount of time has passed.

## 2. How is this scenario possible?

This scenario is possible because conservative forces, as opposed to non-conservative forces, do not dissipate energy. This means that the total mechanical energy of the object, which is the sum of its potential and kinetic energy, remains constant. As the object moves and its kinetic energy decreases, its potential energy increases until it reaches a point where its kinetic energy is zero and its potential energy is at a maximum.

## 3. Is this scenario realistic?

In theory, this scenario is possible for an object subject to a conservative force. However, in reality, there are always other factors at play that will prevent the object from coming to a complete stop after an infinite amount of time. These factors can include external forces, such as friction, or the object encountering an obstruction.

## 4. What is the significance of "Infinite time for an object to get K=0 conservative Force"?

This scenario is significant because it demonstrates the conservation of energy, a fundamental principle in physics. It also highlights the difference between conservative and non-conservative forces and their effects on an object's energy.

## 5. How does this scenario relate to real-life situations?

While this scenario may not occur exactly as described in real-life, it can be observed in various situations. For example, a pendulum will eventually come to a complete stop due to the effects of air resistance, but if this resistance is minimized, the pendulum can continue to swing for a longer period of time until it eventually reaches a state of K=0. Similarly, a satellite orbiting Earth will eventually lose its kinetic energy and come to a stop, but this could take an infinite amount of time if there are no other forces acting upon it.

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