Classical Square Well: Hamiltonian Form & Elastic Collision

[jjustinn]

you have your equation for x(t) right here! For a given range of x values, you have:
\frac{d^2x(t)}{dt^2} = \mathrm{constant}
So for any motion within that range of x values, you simply have constant acceleration. And the motion of the particle will cause it to go into different regions, and have a different acceleration in each region. So yeah, it is a bit annoying to have to think about each region. But if you think about the different possible cases, for a given set of initial conditions, you can write down the entire function $x(t)$ for all $t>0$. (The different possible initial conditions will fall into one of a few 'categories', which will each have a different equation for $x(t)$). p.s. it will help to draw a few graphs of velocity against position. Also, it will help if you think about the kinetic energy of the particle.

So this seems like the same problem as before, except here the "constant" force is over a region instead of at a point...in both cases, you have an equation of the form $d^2x/dt^2 = -F(x(t))$...and in this case, the "regions" are over the dependent variable.

I'll try a few graphs later on today and see if anything clicks, but it seems like it's going to be a ridiculously-complicated equation highly dependent on the initial conditions (as opposed to the "point" case, where the solution has the trivial closed form $x(t) = |t|$)

Also, note that the velocity can only be given as a function of position up to a sign (in the case of a reflection it will pass previous points in the opposite direction) -- and even that is only because it's only one particle in 1D; if there was another particle or dimension, the total energy could be distributed among the extra degrees of freedom in an infinite number of ways...

 

[BruceW]

before, you had a force that was a Dirac-Delta in space... so yeah, I guess that force was not explicitly dependent on time either. It would be pretty unusual if you did have a force that was explicitly dependent on time. I think that would imply non-conservation of energy.

It's not that complicated to get an exact answer. But yes, it does depend on the initial conditions. And what do you mean the point case? you mean when the triangle is an isosceles triangle which is very tall and very thin? The equation would not be $x(t) = |t|$ (although, It is true for $t \rightarrow \infty$)

well, if you specify the initial conditions, then you will know the sign of the velocity at any given time. Yes, the sign may change with time in some cases, but this is not a problem, since the velocity does change with time.

in 2D It is not that much more complicated. I'm imagining the potential as a right circular cone. (where the height corresponds to potential and the x and y coordinate correspond to position). Yes, it is a bit more annoying to work out the function $(x(t),y(t))$ but it is possible as long as you specify the initial conditions. There is no possible ambiguity about which direction the particle will move in. Also, in 3D, it is still OK. Just more difficult to visualise the potential, since you'd need to imagine a 4D cone.

 

[jjustinn]

The "point case" was where the force was a delta function instead of a step function...the only point I was trying to make was that it seems the "simpler" finite triangle barrier (with step function force ) seems to have a much more complicated solution when compared to the step function barrier (with delta function force)...but before pre-judging any further, I'm going to take your advice and at least try to get a qualitative solution for the triangle case.