Force of Particle in Step Potential

[bitty]

Homework Statement

We have a particle in a step potential. Consider it as a plane wave traveling left to right and hitting a step potential V at x=0. Assuming it behaves like a fluid, what is the force it exerts on the wall in terms of F'[0], where F'[x] is the derivative of the wave function?

Homework Equations

The Attempt at a Solution

The wall has to exert a force equal to the force exerted on it so as to not move.
Since we assume fluid behavior, we know the force on the wall is equal to the change in momentum over the change in time. We are given that the wave hits the wall and bounces back, so the change in momentum for a particle is just 2*h_bar*k, where we assume it's final velocity has same magnitude as the initial velocity.

I don't know where to go from here. The question asks that we express F in terms of the derivative of the wave function at x=0, but I don't see it's connection with force.

 

[mark726]

I would approach this problem by first considering the basic principles of mechanics and fluid dynamics. In this situation, we can treat the particle as a fluid, as it is moving in a wave-like manner. The force exerted by the particle on the wall can be calculated using Newton's second law: F = ma, where m is the mass of the particle and a is its acceleration.

In this case, we know that the particle is experiencing a change in momentum when it hits the wall and bounces back. We can use the equation p = mv, where p is momentum, m is mass, and v is velocity, to calculate the change in momentum. Since the particle is bouncing back, its final velocity will have the same magnitude as its initial velocity, but in the opposite direction. This means that the change in momentum is 2mv.

Now, we can relate this change in momentum to the change in time using the definition of force: F = Δp/Δt. Since the particle is hitting the wall and bouncing back in a very short amount of time, we can approximate this change in time as Δt = 0. Therefore, we can say that the force exerted by the particle on the wall is F = 2mv/0, which is undefined.

However, we can use the derivative of the wave function to help us make sense of this situation. The derivative of the wave function, F'[x], represents the change in the wave's amplitude at a specific point x. In this case, we are interested in the derivative at x=0, where the particle hits the wall. This derivative can give us information about the rate of change of the wave at that point, and we can use this information to estimate the force exerted by the particle on the wall.

To do this, we can use the equation F = ma = m(dv/dt), where a = dv/dt is the acceleration of the particle. We can then substitute in the change in momentum, Δp = 2mv, and the change in time, Δt = 0, to get F = 2mv/0 = 2m(dv/dt). Since we know that the particle is behaving like a fluid, we can also use the equation F = ρAv^2, where ρ is the density of the fluid, A is the cross-sectional area of the particle, and v is its velocity.