Why Does a Particle Not Remain at x(t)=0 in a Negative Quartic Potential?

[deuteron]

Homework Statement

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Relevant Equations

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This question is from Collection of Problems in Classical Mechanics by Kotkin & Serbo, here, the answer is given as the following:

However, the graph of $-Ax^4$ looks like:

so shouldn't the trajectory be just $x(t)=0$?

 

[TSny]

so shouldn't the trajectory be just $x(t)=0$?

How do you arrive at this conclusion? Wouldn't $x(t) = 0$ contradict the assumption that the initial value $x_0$ is greater than zero?

 

[deuteron]

b

How do you arrive at this conclusion? Wouldn't $x(t) = 0$ contradict the assumption that the initial value $x_0$ is greater than zero?

but there is no such assumption in the question, the second picture is from the solution, the first one is the question

My thought process was the following:
1. Total energy is $0$
2. The potential has a minimum at $U(x)=0$
3. Therefore the particle would require either a force (which is not present at $x=0$ since $U(0)=0$) or kinetic energy (which is also not present since $E=0$ and $U(0)=0$ thererfore $U(0)=E$, therefore $E-U(0)=T(0)=0$)
4. That's why particle would continue to stay there since that is a stable equilibrium if $-A>0$, and an unstable equilibrium if $-A<0$

 

[TSny]

b

but there is no such assumption in the question, the second picture is from the solution, the first one is the question

My thought process was the following:
1. Total energy is $0$
2. The potential has a minimum at $U(x)=0$
3. Therefore the particle would require either a force (which is not present at $x=0$ since $U(0)=0$) or kinetic energy (which is also not present since $E=0$ and $U(0)=0$ thererfore $U(0)=E$, therefore $E-U(0)=T(0)=0$)
4. That's why particle would continue to stay there since that is a stable equilibrium if $-A>0$, and an unstable equilibrium if $-A<0$

You are correct that if the particle is initially placed at x = 0 with zero initial velocity, then the particle would remain at x = 0. But I think the problem was asking for discussion of the general case where the particle could start at any initial position with an initial velocity such that the total energy is zero. The answer assumes that $A$ is positive. However, this is not specified in the problem statement.

If $A$ is negative, then the only solution for zero total energy is your answer $x(t) = 0$ for all $t$.

 

[deuteron]

You are correct that if the particle is initially placed at x = 0 with zero initial velocity, then the particle would remain at x = 0. But I think the problem was asking for discussion of the general case where the particle could start at any initial position with an initial velocity such that the total energy is zero. The answer assumes that $A$ is positive. However, this is not specified in the problem statement.

If $A$ is negative, then the only solution for zero total energy is your answer $x(t) = 0$ for all $t$.

Thanks!

 

[Steve4Physics]

To add to what @TSny has already said...

You are told that the potential energy $U(x) = -Ax^4$ and that the total energy is zero. So the kinetic energy is $Ax^4$. (Hint: this is the first step in deriving the equation for $x(t)$.)

We need to assume that $x(0) \ne 0$ otherwise the question is trivial (and would, give $x(t) = 0$).

For illustration purposes the model answer explains what happens when $x_0>0$ in 2 cases:
- when initial velocity > 0 (particle moves to $x=\infty$)
- when initial velocity < 0 (particle asymptotically moves to $x=0$)
A similar argument would apply for $x_0<0$.