201907181500 Homework 2 (Q3)

Two mass points, $m_1$ and $m_2$ , move under the influence of a mutual central force, where the central force potential is given by $U(\mathbf{r}_1,\mathbf{r}_2)=U(|\mathbf{r}_1-\mathbf{r}_2|)$ . Assume the center of mass is at the rest, please find the equivalent one-body problem and show that the corresponding Lagrangian can be written as

$\mathcal{L}=\displaystyle{\frac{1}{2}}\mu\dot{r}^2-U_{\textrm{eff}}$

where $r=|\mathbf{r}_1-\mathbf{r}_2|$ is the relative distance between the two mass points and $\mu=\displaystyle{\frac{m_1m_2}{m_1+m_2}}$ is the reduced mass.

Solution.

(Reference: https://www.physics.rutgers.edu/~shapiro/507/book4(DOT)pdf)

Let $\mathbf{R}\stackrel{\textrm{def}}{=}\displaystyle{\frac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2}}$ and $\mathbf{r}\stackrel{\textrm{def}}{=}\mathbf{r}_2-\mathbf{r}_1$ . Expressing $\mathbf{r}_1$ and $\mathbf{r}_2$ in terms of $\mathbf{R}$ and $\mathbf{r}$ , we write

$\mathbf{r}_1=\mathbf{R}-\displaystyle{\frac{m_2}{M}}\mathbf{r}$ ,

$\mathbf{r}_2=\mathbf{R}+\displaystyle{\frac{m_1}{M}}\mathbf{r}$

where $M=m_1+m_2$ .

The kinetic energy $T$ is computed as follows:

$\begin{aligned} T & = \displaystyle{\frac{1}{2}m_1\dot{r}_1^2}+\displaystyle{\frac{1}{2}m_2\dot{r}_2^2} \\ & = \displaystyle{\frac{1}{2}m_1\Bigg[ \displaystyle{\frac{\mathrm{d} }{\mathrm{d}t}} \bigg( \mathbf{R}-\displaystyle{\frac{m_2}{M}}\mathbf{r} \bigg)\Bigg]^2}+\displaystyle{\frac{1}{2}m_2\Bigg[ \displaystyle{\frac{\mathrm{d} }{\mathrm{d}t}} \bigg( \mathbf{R}+\displaystyle{\frac{m_1}{M}}\mathbf{r}\bigg)\Bigg]^2} \\ & = \displaystyle{\frac{1}{2}}(m_1+m_2)\dot{R}^2+\displaystyle{\frac{1}{2}\frac{m_1m_2}{M}}\dot{r}^2 \\ & = \displaystyle{\frac{1}{2}}M\dot{R}^2+\displaystyle{\frac{1}{2}\mu\dot{r}^2} \\ \end{aligned}$

where $\mu$ is the reduced mass $\displaystyle{\frac{m_1m_2}{m_1+m_2}}$ .

From its formula above, $T$ can be seen as the sum of the kinetic energy of the motion of the centre of mass, i.e., $\displaystyle{\frac{1}{2}}M\dot{R}^2$ , and the kinetic energy of motion about the centre of mass, i.e., $\displaystyle{\frac{1}{2}\mu\dot{r}^2}$ .

And from the fact that the Lagrangian $\mathcal{L}=\displaystyle{\frac{1}{2}}M\dot{R}^2+\displaystyle{\frac{1}{2}\mu\dot{r}^2}-U(r)$ is cyclic on $R$ , the centre of mass is either at rest or in uniform motion.

Thus the equation of motion for $r$ will not contain terms involving $\mathbf{R}$ or $\dot{\mathbf{R}}$ . We may hence ignore the first term, and what remains in the Lagrangian is

$\mathcal{L}=\displaystyle{\frac{1}{2}\mu\dot{r}^2}-U(r)$ .

Now that we introduce a spherical coordinate system given by its equation of transformation from the Cartesian as:

$\begin{aligned} x & =r\sin\theta\cos\phi \\ y & =r\sin\theta\sin\phi \\ z & =r\cos\theta \\ \end{aligned}$

we can write the kinetic energy as

$\begin{aligned} T & =\displaystyle{\frac{1}{2}}\mu (\dot{x}^2+\dot{y}^2+\dot{z}^2) \\ & = \displaystyle{\frac{1}{2}}\mu[(\dot{r}\sin\theta\cos\phi+\dot{\theta}r\cos\theta\cos\phi -\dot{\phi}r\sin\theta\sin\phi)^2\\ & \qquad +(\dot{r}\sin\theta\sin\phi+\dot{\theta}r\cos\theta\sin\phi+\dot{\phi}r\sin\theta\cos\phi)^2\\ & \qquad +(\dot{r}\cos\theta -\dot{\theta}r\sin\theta)^2]\\ & = \displaystyle{\frac{1}{2}}\mu [\dot{r}^2+r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2]\\ \end{aligned}$

(Note that the kinetic energy is cyclic on the coordinate $\phi$ and the conjugate momentum $P_\phi=\displaystyle{\frac{\partial\mathcal{L}}{\partial\dot{\phi}}}=\mu r^2\sin^2\theta\dot{\phi}=\textrm{constant}$ . Observe that $r\sin\theta$ is the distance between the particle and the $z$ -axis, and it can be easily seen that $P_\phi$ is the $z$ -component of the angular momentum $\mathbf{L}$ .)

To simplify things, we choose the direction of angular momentum $\mathbf{L}$ as the $z$ -direction. It follows that $\theta=\pi/2$ , $\dot{\theta}=0$ , and $L=\mu r^2\dot{\phi}$ .

I wish to obtain the Euler-Lagrange equation for $r$ . Hence I compute the following:

$\begin{aligned} \displaystyle{\frac{\partial \mathcal{L}}{\partial r}} & = \mu r\dot{\phi}^2- \partial_r U\\ \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{r}}} & = \mu\dot{r}\\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial \mathcal{L}}{\partial \dot{r}}}\bigg) & = \mu\ddot{r}\\ \end{aligned}$

Then one can express the one-body problem as:

$\mu\ddot{r}-\mu r\dot{\phi}^2+\displaystyle{\frac{\mathrm{d}U}{\mathrm{d}r}}=0$ ,

or,

$\mu\ddot{r}-\displaystyle{\frac{L^2}{\mu r^3}}+\displaystyle{\frac{\mathrm{d}U}{\mathrm{d}r}}=0$ ,

or,

$\mu\ddot{r}+\displaystyle{\frac{\mathrm{d}}{\mathrm{d}r}U_{\textrm{eff}}(r)}=0$ ,

where $U_{\textrm{eff}}(r)=U(r)+\displaystyle{\frac{L^2}{2\mu r^2}}$ is the effective potential.

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