Particle Takes Infinite Time to Reach Top of Potential Hill

In summary, the conversation discusses a particle moving in a potential with a well at R = 1 and a hump at R = 2. The question is posed on how long it would take for the particle to escape from the potential barrier if it has an energy of 1/32. The equations used to solve the problem are F=-grad(V) and the Euler-Lagrange equation, which leads to a differential equation that needs to be solved. Ultimately, the goal is to find an equation for the velocity of the particle and integrate it to find the time to escape. The standard trick suggested is to either multiply both sides by R' or write R' = V, to get a result similar to 1/2 mv
  • #1
Proofrific
12
0

Homework Statement



A particle is moving in a potential
[tex]V(R) = \frac{1}{2} \left( \frac{1}{R} - \frac{1}{R^2} \right)^2.[/tex]
If you plot this, is has a well at R = 1 with height V(1) = 0 and a hump at R = 2 with height V(2) = 1/32. Question: If a particle has energy 1/32, show that it takes log infinite time to escape from the potential barrier.

Homework Equations



See above

The Attempt at a Solution



Using either F=-grad(V) or the Euler-Lagrange equation, I get
[tex]m\ddot{R} = \frac{-1}{R^3} + \frac{3}{R^4} - \frac{2}{R^5}.[/tex]
How do I solve this differential equation?

Ultimately, I want to find an equation for the velocity of the particle so that I can integrate it to find the time to escape. That is, assuming that we're at R = 1 at t = 0, I want to find the time to reach R = 2 by solving:
[tex]2 = \int_0^t \dot{R}(t) dt[/tex]
 
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  • #2
Hi Proofrific! :smile:

Standard trick :wink:

either multiply both sides by R' (so eg the LHS is m(R'2)'/2)

or write R' = V, then R'' = dV/dt = dV/dR dR/dt = v dV/dR :smile:

(same result either way)

(and btw, this is where 1/2 mv2 comes from in KE)
 

1. What is a potential hill in physics?

A potential hill, also known as a potential energy barrier, is a concept in physics that refers to a region in space where the potential energy of a particle is higher than its surrounding areas. This creates a barrier that the particle must overcome in order to reach the other side.

2. Why does it take infinite time for a particle to reach the top of a potential hill?

According to classical mechanics, the time it takes for a particle to move from one point to another is dependent on its initial velocity. In the case of a potential hill, the particle's initial velocity is zero at the bottom of the hill. As it moves towards the top, its kinetic energy increases and its potential energy decreases. However, at the top of the hill, the particle's kinetic energy becomes zero again, meaning it has no velocity. Therefore, it will never be able to reach the top of the hill and will take infinite time to do so.

3. Can a particle ever overcome a potential hill?

In classical mechanics, a particle with a non-zero initial velocity can overcome a potential hill and reach the other side. However, if the initial velocity is zero, the particle will not be able to overcome the hill and will take infinite time to do so. In quantum mechanics, there is a small probability that a particle can tunnel through a potential hill, but this is a rare occurrence.

4. Is the concept of a potential hill applicable in all branches of physics?

Yes, the concept of a potential hill is applicable in various branches of physics, including classical mechanics, quantum mechanics, and thermodynamics. It is used to explain the behavior of particles in different systems, such as in potential energy diagrams and in the study of chemical reactions.

5. What are some real-life examples of a potential hill?

One example of a potential hill is a ball rolling up and down a hill due to the force of gravity. Another example is an electron moving through an electric potential barrier. In chemistry, potential hills can be seen in the activation energy required for a chemical reaction to occur. In biology, potential hills play a role in the movement of molecules across cell membranes.

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