# Classical particle in a box [closed]

I'm trying to work out some of the details for this system. A particle with mass $\mu$, initial velocity $v_0$ at $x_0$ and moving freely between two walls located at $\pm L/2$, with which it bounces elastically. Heuristically, I've got the equations of motion:

\begin{align}x(t) &= \frac{L}{\pi} \arcsin \{\sin\left[\frac{\pi}{L}\left(v_0 t + x_0\right)\right]\} \\ \dot x(t) &= v_0 \mathrm{sgn}\{\cos\left[\frac{\pi}{L}\left(v_0 t + x_0\right)\right]\}\end{align}

But I'm wondering if there's a way to get there from the Hamiltonian:

$$\frac{\mu \dot x^2}{2}+\alpha \mathrm{\theta}\left(x^2-\frac{L^2}{4}\right)=\frac{\mu v_0^2}{2}$$

Where $\theta(x)$ is the Heaviside function and $\alpha > \frac{\mu v_0^2}{2}$. Or from Newton's second law:

$$\mu\ddot x=-2\alpha x\delta\left(x^2-\frac{L^2}{4}\right)$$

My attempts so far:

1) solving for the velocity from the energy equation (assuming the motion starts at $t=0$):

\begin{align} \mathrm{sgn}(v_0)\sqrt{\frac{2}{\mu}}t&=\int_{x_0}^{x(t)}\frac{dx}{\sqrt{\frac{\mu v_0^2}{2}-\alpha\theta\left(x^2-\frac{L^2}{2}\right)}}\\ v_0t&=\int_{x_0}^{x(t)}\frac{dx}{\sqrt{1-\frac{2\alpha}{\mu v_0^2}\theta\left(x^2-\frac{L^2}{2}\right)}} \\ v_0t+x_0&=\mathrm{sgn}(v_0)\frac{L}{2}+\mathrm{sgn}(v_0)L\sum_{n=1}^{\lfloor\frac{v_0t+x_0}{L}\rfloor} (-1)^n +x(t)-(-1)^{\lfloor\frac{v_0t+x_0}{L}\rfloor}\mathrm{sgn}(v_0)\frac{L}{2}\\ \therefore x(t)&=v_0t+x_0-\mathrm{sgn}(v_0)L\left(\frac{1-(-1)^{\lfloor\frac{v_0t+x_0}{L}\rfloor}}{2}+\sum_{n=1}^{\lfloor\frac{v_0t+x_0}{L}\rfloor} (-1)^n\right)\end{align}

This can't be true since clearly the rhs is allowed to be greater than $L/2$

2) working with Laplace transform

\begin{align}\mathcal{L}[\mu\ddot x &=-2\alpha x\delta\left(x^2-\frac{L^2}{4}\right)]\\ s^2\tilde{x}-sv_0-x_0 &=-2\frac{\alpha} {\mu} \int_0^\infty x\delta\left(x^2-\frac{L^2}{4}\right)\mathrm{e}^{st}dt\\ &=-\frac{2\alpha}{\mu L}\int_0^\infty \left(\sum_n\frac{\delta\left[t-x^{-1}_n(L/2)\right]}{|\dot x[x^{-1}_n(L/2)]|}+\sum_m\frac{\delta\left[t-x^{-1}_m(-L/2)\right]}{|\dot x[x^{-1}_m(-L/2)]|}\right)x\mathrm{e}^{st}dt\\ &= \frac{\alpha} {\mu} \left(\sum_n\frac{\exp\left[sx^{-1}_m(-L/2)\right]}{|\dot x[x^{-1}_m(-L/2)]|}-\sum_m\frac{\exp\left[sx^{-1}_n(L/2)\right]}{|\dot x[x^{-1}_n(L/2)]|}\right)\end{align} $$\therefore x(t)=v_0t+x_0+ \frac{\alpha} {\mu} \left(\sum_n\frac{\left[t-x^{-1}_m(-L/2)\right]\theta\left[t-x^{-1}_m(-L/2)\right]}{|\dot x[x^{-1}_m(-L/2)]|}\\-\sum_m\frac{\left[t-x^{-1}_n(L/2)\right]\theta\left[t-x^{-1}_n(L/2)\right]}{|\dot x[x^{-1}_n(L/2)]|}\right)$$ Here $x_n^{-1}[x(t)]=t$ $\forall n$.

I don't have any idea on how to work out the inverse functions.

3) working with the Fourier transform:

In a similar way as the previous case, the final result is:

$$x(t)= \frac{\alpha} {\mu} \left(\sum_n\frac{|t-x^{-1}_m(-L/2)|}{|\dot x[x^{-1}_m(-L/2)]|}\\-\sum_m\frac{|t-x^{-1}_n(L/2)|}{|\dot x[x^{-1}_n(L/2)]|}\right)$$

Any input is really appreciated.

• If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Feb 18 '16 at 11:46

Ok, sorry for not just commenting. Anyway, I think the answer can be obtained from integrating your energy formula, since $$t-t_0 = \sqrt{\frac{\mu}{2}}\int_{x_0}^x \frac{\pm \mathrm{d}x}{\sqrt{E-\alpha\theta(x^2-L^2/4)}}$$ the time during the particle goes from $x_0$ to $x$. But in the box, we have $x \in [-L/2,L/2]$, so $V(x)=0$, so it is a trivial integral then, and you can invert it to have $x(t)$. At the wall, only the sign of dx changes in the above integral formula. So the particle moves backward then. So you can give the trajectory between any successional turning points using the above formula.
• Your expressions and mine are equivalent, granted $\alpha = V_0$ – GeoArt Feb 17 '16 at 23:30
• Ok, so maybe I misunderstand you, but $x(t)=\frac{L}{\pi}\mathrm{arcsin}(\mathrm{sin}(\frac{\pi}{L}(v_0 t + x_0))$ is simply $x_0 + v_0 t$ and that is what comes from the integral, when you use $t_0 = 0, x(t=t_0)=x_0$ and $E=\frac{\mu}{2}v_0^2$. – fermion Feb 18 '16 at 17:51