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

[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)

 

[PeroK]

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

 

[titansarus]

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.

 

[PeroK]

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 $U''(x) \le 0$.

 

[titansarus]

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 $(U(x) - U(x_1) )dt^2 = - m (x - x_1)d^2 x$

 

[PeroK]

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]

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?

 

[PeroK]

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?

 

[titansarus]

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.

 

[PeroK]

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 $s \ge 1$.

 

[TSny]

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