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

$$V(x)=V_0(e^{-2ax}-2e^{-ax}),\qquad V_0,a>0,\qquad E>-V_0.$$

From the only other question I have found on this topic (here):

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.

An answer was not given. I have computed this integral in the form

$$\dfrac{\ln\left(\left|\sqrt{c\mathrm{e}^{2ax}+b\left(2\mathrm{e}^{ax}-1\right)}+\sqrt{c}\left(\mathrm{e}^{ax}-1\right)-\sqrt{c+b}\right|\right)-\ln\left(\left|\sqrt{c\mathrm{e}^{2ax}+b\left(2\mathrm{e}^{ax}-1\right)}+\sqrt{c}\left(1-\mathrm{e}^{ax}\right)-\sqrt{c+b}\right|\right)}{a\sqrt{c}}$$

but am not sure if my solution is correct (I have not plugged in the bounds yet), and Wolfram exceeds computation time. Googling the classical period yields no results, and so I am at a loss.

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closed as off-topic by G. Smith, Gert, John Rennie, GiorgioP, Jon Custer Apr 9 at 13:55

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  • $\begingroup$ The definite integral will be much simpler. I also suggest a change of variables to $y=e^{-ax}$. $\endgroup$ – G. Smith Apr 5 at 0:20
  • $\begingroup$ Hi. This isn't about physics, it's about solving an integral. Voted to close. $\endgroup$ – Gert Apr 5 at 0:57
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See my comments on @DinosaurEgg’s answer about why it is wrong in at least five ways.

The integral derived from energy conservation actually produces

$$T=\frac{\pi}{a}\sqrt{\frac{2m}{-E}}.$$

Note that there is periodic motion only for $-V_0<E<0$.

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The period of motion in a one-dimensional classical potential is given by:

$$T=\int\limits_{x_1}^{x_2}\frac{dx}{\sqrt{2m(E-V(x))}}$$

where $x_1, x_2$ are two adjacent solutions to the equation $E=V(x)$ (remember, the particle will only oscillate in a potential well between two endpoints).

Making the substitution $u=e^{-ax}$, Wolfram Alpha happily computes:

$$T=\frac{a}{\sqrt{V_0}}\int\limits_{u_1}^{u_2}\frac{du}{u\sqrt{\frac{E}{V_0}-u^2+2u}}=-\frac{a}{\sqrt{V_0}}\frac{\log u-\log(\sqrt{\Delta}\sqrt{\Delta-u^2+2u}+\Delta +u)}{\sqrt{\Delta}}\Big|_{u_1}^{u_2}$$

where $u_1=1-\sqrt{\Delta+1}~~,~~ u_2=1+\sqrt{\Delta+1}~~, ~~\Delta=\frac{E}{V_0}$.

The final result reads:

$$T=\frac{2a}{\sqrt{E}}\Bigg[ \tanh^{-1}\Big(\sqrt{1 + \frac{E}{V_0}}\Big) + \tanh^{-1}\Big(\frac{1}{\sqrt{1 + \frac{E}{V_0}}} \Big)\Bigg]$$

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  • $\begingroup$ This solution has numerous problems. The first equation has the wrong power of $m$, and is only half the period. The final answer has no $m$. The final answer doesn’t have the dimensions of time. The final answer is complex for, say, $E=-0.5V_0$, when it should be real. $\endgroup$ – G. Smith Apr 5 at 1:48
  • $\begingroup$ I found a very simple answer,$$T=\frac{\pi}{a}\sqrt{\frac{2m}{-E}}$$ for $-V_0<E<0$, by letting Mathematica do the definite integral. I verified this analytic result using numerical integration. I think @DinosaurEgg may have run into problems with branch cuts. $\endgroup$ – G. Smith Apr 5 at 2:03
  • $\begingroup$ You can also get this from @DinosaurEgg's logarithmic form of the indefinite integral. All four logs combine to give the log of -1, which is where the $\pi$ comes from. $\endgroup$ – G. Smith Apr 5 at 5:31

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