সাম্যাবস্থার আশেপাশে ছোট স্পন্দন (ক্যালকুলাস আবশ্যক)

The general problem of a particles moving in a conservative (no friction) potential $V(x)$ is quite difficult. There are a few common situations in which we can understand the motion of a particle in a complicated potential. In previous sections, you learned about static potentials. You can solve this by finding points where the force is zero, i.e. solve for the position $x_0$ such that $F = -\frac{dV}{dx} = 0$.

The next simplest situations is called small oscillations. If we place a particle very close to the minimum, but not quite at it, it generally feels a force towards the minimum. This force is approximately linear with position if we are very close, i.e. $F \approx – k x$. When this is true, we already know what happens: we have a simple harmonic oscillator! The particle will oscillate around $r_0$ at some frequency $\omega$.

$$k = -\left. \frac{dF}{dx} \right|_{x=x_0} \ = \left. \frac{d^2V}{dx^2} \right|_{x=x_0}$$ $$\omega = \sqrt\frac{k}{m} = \sqrt{\frac{1}{m}\left. \frac{d^2V}{dx^2} \right|_{x=x_0}}$$

Let’s work through a simple example. The energy of two atoms colliding is often similar to a Lennard-Jones potential. This has the following complicated form as a function of radius $r$.

$$V = \epsilon \left[ \left( \frac{a}{r} \right)^{12} - \left( \frac{a}{r} \right)^6 \right]$$

What can we say about such a complex problem? Well, first, we can determine the minimum, which occurs when the force is zero. $F = -\frac{dV}{dr} = 0$.

$$F = - \frac{\epsilon}{a} \left[ 12 \left( \frac{a}{r} \right)^{13} - 6 \left( \frac{a}{r} \right)^7 \right]$$

When r is very close to r_0, we can approximate the force as nearly linear. This is the same as a Taylor series in calculus.

$$F = - \epsilon \frac{9*2^{2/3}}{a^2} (r – r_0)$$ $$V \approx V_0 + \frac{1}{2} k x^2 \qquad k = \epsilon \frac{9*2^{2/3}}{a^2}$$

This looks nearly the same as a harmonic oscillator! The extra term $V_0$ is not of concern.