# Turning points of particle

A particle of mass $m$ and energy $E<0$ moves in a one-dimensional Morse potential:

$$V(x)=V_0(e^{-2ax}-2e^{-ax}),\qquad V_0,a>0,\qquad E>-V_0.$$ Determine the turning points of the movement and the period of the oscillation of the particle.

I have started learning for my exam and this is one of the exercises in my textbook. Never dealt with this type of question, so these are my thoughts so far:

To get the turning points I was thinking of solving the equation $E=V_0(e^{-2ax}-2e^{-ax})$. I was doing the arithmetics with the absolute value of $E$. But still I couldn't seem to find the values for $x$. At the end I used Wolfram Alpha to find the values but it gave me results with complex values. Is there a simple way to solve this type of equations for $x$?

Anyway, about the period, I assume it's the time it takes for the particle to get from $x_1$ to $x_2$. But how am I supposed to approach this? How would I get a time value just out of the equation for the potential?

I hope someone can help me out here.

## 2 Answers

Energy conservation dictates $$E = \frac{1}{2}m\dot{x}^2 + V(x) = \text{const}$$ With some arithmetic it follows $$\dot{x} = \frac{dx}{dt} = \sqrt{2m^{-1}(E-V(x))}$$ This ODE can be solved via separation of variables, yielding $$\int_{t_1}^{t_2}dt = \int_{x_1}^{x_2} \frac{dx}{\sqrt{2m^{-1}(E-V(x))}}$$ The integral on the left hand side can be evaluated immediately, where $t_1$ and $t_2$ are understood as the times when the particle is at $x_1$ or $x_2$ respectively. So it is simply half the period.

Observe that the integral on the right diverges when $x$ approaches the turning point $E=V(x)$.

This method of solving Newtons equations in a 1d potential should be treated in any textbook on mechanics.

For a general potential it is in general hard or impossible to find the turning points in closed form. Here however, you can substitute $y=\exp(-ax)$ and solve the corresponding quadratic equation. I'll leave the explicit calculation to you.

• Thanks for the fast response. But I still can't seem to find $x_1$ and $x_2$ from the equation. I just can't seem to solve for x. Am I missing something? – Rafa Fafa Jun 4 '15 at 10:18
• @RafaFafa Had to think for a bit, but see my edit on how to do the calculation. – Nephente Jun 4 '15 at 10:54
• I did the calculations and got $y_{1,2}=y\pm \sqrt{e^{-2ax}+\frac{E}{V_0}}$. Can I just plug this in as boundaries instead of $x_1$ and $x_2$? Because I can't seem to solve this for x. – Rafa Fafa Jun 4 '15 at 12:54
• @RafaFafa You went wrong somewhere... Substituting $y=exp(-ax)$ yields $E/V_0 = y^2 - 2y$, which is simply a quadrativ equation in $y$ that has two solutions $y_{1,2}$ for negative $E$. – Nephente Jun 4 '15 at 15:07
• Yes, which should yield $y_{1,2}=y\pm \sqrt{y^2+\frac{E}{V_0}}$, right? – Rafa Fafa Jun 4 '15 at 15:17

nephente's answer solves for the period. My answer is just made to make you see how to go on from the point you were, solving:

$$e^{-2ax}-2e^{-ax}=\frac{E}{V_0}$$

We make the change $y=e^{-ax}$, which yields:

$$y^2-2y-\frac{E}{V_0} = 0$$

Second grade equation. Solve for $y$, and have in mind that $E/V_0$ is a negative number. It may give you complex solutions otherwise.

• Oh, so when I get the two values for y I can just put it in $y=e^{-ax}$ and solve for x. I get it now. Thanks. – Rafa Fafa Jun 4 '15 at 11:03