In the previous article, we obtained the equipotential surfaces and electric field lines of an electric dipole, which should give us a general understanding of the force field of an electric dipole. Of course, we still hope to determine how a charged particle moves under such a force. For simplicity, in the following discussion, we assume that the mass and charge of the charged particle are both 1. As for the sign of the charge, it can be controlled by changing the sign of the k value in U=-\frac{k \cos\theta}{r^2}. The tool we use is still the Euler-Lagrange equations from theoretical mechanics.
Many readers may find formulas a headache, let alone the profound field of theoretical mechanics. But please believe me, if you spend a little effort to understand the basic ideas and steps of using variational methods to study mechanics (or other physical systems), it will be of great benefit to your physical research. In my eyes, after learning even a little bit of theoretical mechanics, all I see is the simplicity and harmony of physics. Interested friends can look at my articles such as "Natural Extremum."
First, write the expression for kinetic energy: T=\frac{1}{2} (\dot{r}^2+r^2 \dot{\theta}^2)
And potential energy: U=-\frac{k \cos\theta}{r^2}
The Lagrangian is: L=T-U=\frac{1}{2}(\dot{r}^2+r^2 \dot{\theta}^2)+\frac{k \cos\theta}{r^2}
Find the differential equation with respect to r: \begin{aligned}\frac{\partial L}{\partial \dot{r}}&=\dot{r} \\ \frac{\partial L}{\partial r}&=r\dot{\theta}^2-\frac{2k\cos\theta}{r^3}\end{aligned}
We get: \ddot{r}=r\dot{\theta}^2-\frac{2k\cos\theta}{r^3} \tag{10}
We could continue to find the partial derivative with respect to \theta to obtain the differential equation for \theta, but we do not do so because energy conservation must exist in a conservative force field: T+U=h=\frac{1}{2}(\dot{r}^2+r^2 \dot{\theta}^2)-\frac{k \cos\theta}{r^2} \tag{11}
Combining (10) and (11) allows us to find all solutions. The problem is essentially a system of second-order differential equations for two functions, so it should contain 4 integration constants. Since equation (11) already contains one integration constant, three remain.
After manipulating the equations for a while, we find a very coincidental thing: h=\frac{1}{2}\dot{r}^2+\frac{1}{2}\left(r^2 \dot{\theta}^2-\frac{2k \cos\theta}{r^2}\right)=\frac{1}{2}(\dot{r}^2+r\ddot{r}) \tag{12} Namely: h=\frac{1}{2}(\dot{r}^2+r\ddot{r}) \tag{12}
This is so coincidental, and the result is so concise! Following this lead, we get: \begin{aligned}r\dot{r}&=2ht+C_1 \\ \frac{1}{2}r^2&=ht^2+C_1 t+C_2\end{aligned} \tag{13}
At this point, we have obtained three integration constants. Let’s analyze this solution carefully. From equation (13), we know that C_2 > 0 (it cannot start at the origin). If h > 0, the charged particle will eventually fly to infinity, which necessarily requires \Delta=C_1^2-4hC_2 \leq 0; if h < 0, the particle will eventually return to the origin and collide with the electric dipole. This also conforms to our knowledge: a system with positive energy can escape the bound, while a system with negative energy cannot. When h is 0, further discussion is needed.
One last integration constant remains. We have already found the evolution of r with respect to time t, and energy conservation gives the evolution of velocity. What is the last one? There are many answers, but what we are interested in is—what is the shape of the orbit? Here is the simple process:
\begin{aligned}\dot{r}^2+r^2 \dot{\theta}^2&=2h+\frac{2k\cos\theta}{r^2} \\ 1+r^2\left(\frac{d\theta}{dr}\right)^2&=\frac{2h+\frac{2k\cos\theta}{r^2}}{\dot{r}^2}\end{aligned} \tag{14}
Next, we find a way to eliminate \dot{r}^2 in equation (14). According to (13), we have: \begin{aligned}r^2 \dot{r}^2&=4h^2 t^2+C_1^2+4hC_1 t \\ 2hr^2&=4h^2 t^2 +4hC_1 t+4hC_2 \\ r^2 \dot{r}^2-C_1^2&=2hr^2-4hC_2 \\ \dot{r}^2&=2h+\frac{C_1^2-4hC_2}{r^2}\end{aligned}
Substituting this in: \begin{aligned}1+r^2\left(\frac{d\theta}{dr}\right)^2&=\frac{2hr^2+2k\cos\theta}{2hr^2+C_1^2-4hC_2} \\ r^2\left(\frac{d\theta}{dr}\right)^2&=\frac{2k\cos\theta-C_1^2+4hC_2}{2hr^2+C_1^2-4hC_2} \\ \left(\frac{d\theta}{dr}\right)^2&=\frac{2k\cos\theta-C_1^2+4hC_2}{(2hr^2+C_1^2-4hC_2)r^2} \\ \frac{dr}{r\sqrt{2hr^2+C_1^2-4hC_2}}&=\frac{d\theta}{\sqrt{2k\cos\theta-C_1^2+4hC_2}}\end{aligned} \tag{15}
The last equation already contains the shape of the orbit; integrating both sides will yield it. Unfortunately, this is not a simple explicit solution; it can only be expressed using elliptic integrals. An easy way is to look at some simple special cases to quickly see the shape of the curve. Let’s set k=0.5, -C_1^2+4hC_2=1, h=0.5, then (15) becomes: \frac{dr}{r\sqrt{r^2-1}}=\frac{d\theta}{\sqrt{\cos\theta+1}}
We obtain: \arccos\left(\frac{1}{r}\right)=\frac{1}{\sqrt{2}}\ln\left(\tan\left(\frac{\pi}{4}+\frac{\theta}{4}\right)\right)+Const
That is: r=\frac{\sqrt{2}}{\cos\left(\ln\left(\tan\left(\frac{\pi}{4}+\frac{\theta}{4}\right)\right)\right)}
The shapes of these curves also give me a headache... Since my research on the relevant knowledge is not deep enough, I will not say more... Consider this a cliffhanger, and I hope interested readers will continue to complete it.
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