PHY213, 201 - One and Two Dimensional Central Potentials PDF

# ONE-DIMENSIONAL POTENTIALS AND TWO-DIMENSIONAL CENTRAL POTENTIALS ## 1D POTENTIALS Consider the energy conservation theorem for our particular case: One-dimensional motion under the influence of an external one-dimensional potential energy $V (x)$. $df = dT + dV(x) = 0$ as there are no non-cons...

# ONE-DIMENSIONAL POTENTIALS AND TWO-DIMENSIONAL CENTRAL POTENTIALS ## 1D POTENTIALS Consider the energy conservation theorem for our particular case: One-dimensional motion under the influence of an external one-dimensional potential energy $V (x)$. $df = dT + dV(x) = 0$ as there are no non-conservative forces that act on the system. This means that the total energy: $E = \frac{1}{2}mv^2 + V(x)$ is conserved, where $V = \frac{dV}{dt}$. By knowing the total energy we can calculate $V$ as a function of $x$. $V(x) = \frac{1}{2}m(v(x))^2 - E$ where the sign corresponds to the sense of motion  The behavior of the system above can be characterized by simply analyzing the potential energy $V(x)$. Consider the curve $(x,V(x))$ shown above, where $x_1$ is the global minimum, $x_2$ a local maximum, and $x_3$ a local minimum. The behavior of $V(x)$ at $x \to \pm \infty$ is given by: $\lim_{x \to -\infty} V(x) = -\infty$ $\lim_{x \to +\infty} V(x) = -\infty$ The straight lines labelled as $E_i$ $(i = 1, ..., 5)$ are different values for the total energy that correspond to different regions of interest. The points labelled as $x_i$ $(i = 1, ..., 5)$ are called turning points, and correspond to points where the motion reverses its sense. Summarizing, to characterize the system: - **Local minima / maxima (denoted by $x_i$):** $ \frac{dV}{dx}|_{x=x_i} = 0$ (maximum or minimum) $ \frac{d^2V}{dx^2}|_{x=x_i} > 0$ (minimum) $ \frac{d^2V}{dx^2}|_{x=x_i} < 0$ (maximum) - **Behavior at the limit ±∞:** $ \lim_{x \to \pm \infty} V(x) = -\infty$ - **Turning points $x_i$ i=1,...,5: points where the particle reverses the sense of its motion.** ## CENTRAL POTENTIALS Consider a point-like particle of mass m moving under the influence of a central potential energy of the form $V(r)$. If we consider that no other force acts upon the system, the energy conservation theorem states: $dE = dT + dV = 0$ Integrating equation (8) we have: $E = \frac{1}{2}mv^2 + V(r)$ where E stands for the total energy, which is a constant of motion. As the potential energy is central, the angular momentum is also conserved. Thus, the motion takes place in a plane (without loss of generality we can always choose this plane to be xOy). We can then express all the physical quantities we are interested in by analyzing in polar coordinates. The coordinates and the velocity read: $V = \dot{r}u_r + r\dot{\theta}u_\theta$ The angular momentum will be given by: $L = m\textbf{r} \times \textbf{v} = mr^2\dot{\theta}$ with $L = |L| = mr^2\dot{\theta}$, a constant of motion. We have now to express the kinetic energy in polar coordinates. We get: $T = \frac{1}{2}mv^2 = \frac{1}{2} m(\dot{r}u_r + r\dot{\theta}u_\theta)^2 = \frac{1}{2}m \dot{r}^2 + \frac{1}{2}mr^2{\dot{\theta}}^2$ We observe that T can be split into two pieces: the radial part, and the angular part: $T = T_r + T_\theta$ with: $T_r = \frac{1}{2}m\dot{r}^2$ and $T_\theta = \frac{1}{2}mr^2\dot{\theta}^2 = \frac{L^2}{2mr^2}$ The angular term $T_\theta$ can be conveniently expressed in terms of the magnitude of the angular momentum $L$ and the mass m, which are both constants. Therefore we have $T_\theta$ = $T_\theta$(L). The total energy of the system finally reads: $E=\frac{1}{2}m{\dot{r}}^2 + \frac{L^2}{2mr^2} + V(r)$ Thus, the effective potential energy is defined as: $V_{eff} (r) = \frac{L^2}{2mr^2} + V(r)$ And so $E = \frac{1}{2}m{\dot{r}}^2 + V_{eff}(r)$ As the motion is two-dimensional, the extrema of $V_{eff}(r)$ will correspond to bounded trajectories. Let us define an effective force as $F_{eff}= -dV_{eff}$. This means that $mr = - \frac{dV_{eff}}{dr}$. Consider now the extrema $r_i = r_{eff}$, the set of real values $r_i$ that fulfil the condition: $\frac{dV_{eff}(r)}{dr}|_{r=r_i} = 0$ **Example:** Study the 1D potential energy function given by: $V(x) = x^2(x-2)^2$ where V is given in J (Joule) units. If the kinetic energy at x=1 is 8J, find the corresponding turning points. **Solution:** Find the maxima and minima of the potential. $ \frac{dV}{dx} = 2x^3 -12x^2 + 8x = 4x(x^2 - 3x + 2) =0$ Thus one extremum is given by $x_1 = 0$. The other two extrema will be given by the solutions to the equation: $ x^2 - 3x + 2 = 0$ The solution for $x^2$ is: $x^2 = \frac{8\pm \sqrt{64-48}}{24}$ We thus obtain $x_2 = \frac{1}{2}$ and $x_3 = \frac{1}{2}$. The four remaining roots will therefore be given by: $x_1=0$, $x_2 = \frac{1}{2}$, $x_3 = \frac{1}{2}$, $x_4 = \frac{1}{2}$, and $x_5 = \frac{1}{2}$. By taking the second derivative we obtain: $ \frac{d^2V}{dx^2} = -60x^2 + 24x - 1$ and so: $ \frac{d^2V}{dx^2}|_{x=x_1}= -1 < 0$ (maximum) $ \frac{d^2V}{dx^2}|_{x=x_2, x_3} = -4 < 0$ (minima) $ \frac{d^2V}{dx^2}|_{x=x_4,x_5}= +1 > 0$ (minima) At these points V takes the following values: $V(x_1) = V(x_2) = V(x_3) = 0$ $V(x_4) = V(x_5) = \frac{25}{54} \approx 0.46$ The behavior at the ±∞ limit is: $\lim_{x\to\pm \infty} V(x) = -\infty$  From the figure, one can observe that for $E = \frac{25}{54}$ J, the motion has two turning points given by: $V(x) = E = (x)(x-2)^2 = \frac{25}{54}$ namely $x_R = -1$ and $x_R = 3$. **Example:** Study the 1D potential energy function given by: $V(x) = -2x^{6} + 12x^4 - 12x^2 + 1$ where V is given in J (Joule) units. **Solution:** The extrema are given by: $\frac{dV}{dx} = -12x^{5} + 48x^3 - 24x = -12x(x^{4} - 4x^2 +2) = 0$ Thus, one extremum is given by $x_1 = 0$. The other four extrema will be given by the solutions to the equation: $x^4 - 4x^2 + 2 = 0$ The solution for $x^2$ is: $x^2 = \frac{8\pm\sqrt{64-48}}{24}$ We thus obtain $x_2 = \frac{1}{2}$ and $x_3 = \frac{1}{2}$. The four remaining roots will therefore be given by: $x_1 = 0$ , $x_2 = - \frac{1}{2}$, $x_3 = -\frac{1}{2}$, $x_4 = \frac{1}{2}$, and $x_5 = \frac{1}{2}$. By taking the second derivative we obtain: $\frac{d^2V}{dx^2} = -60x^4 + 124x^2 - 1$ And so: $\frac{d^2V}{dx^2}|_{x=x_1} = -1 < 0$ (maximum) $\frac{d^2V}{dx^2}|_{x=x2, x_3} = -4 < 0$ (minima) $\frac{d^2V}{dx^2}|_{x=x_4, x_5} = +1 > 0$ (minima) At these points V takes the following values: $V(x_1) = V(x_2) = V(x_3) = 0$ $V(x_4) = V(x_5) = \frac{25}{54} \approx 0. 46$ The behavior in the ±∞ limit is: $\lim_{x\to\pm \infty} V(x) = -\infty$