- Two-Body Problem Analysis

1. Center of Mass $\mathbf{R}$ #

Total Mass $M$ $$M = m_1 + m_2$$ Center of Mass $\mathbf{R}$ $$M \mathbf{R} = m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2$$ Therefore, $$\mathbf{R} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{m_1 + m_2}\tag{1}$$

2. Relative Distance $\mathbf{r}$ #

The relative distance between two bodies $$\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2\tag{2}$$

3. Analysis of Two-Body System Motion #

The motion can be analyzed as a combination of the center of mass motion and the relative motion between the two bodies. From equations (1) and (2), we can express $\mathbf{r}_1$ and $\mathbf{r}_2$ in terms of $\mathbf{R}$ and $\mathbf{r}$: $$\mathbf{r}_1 = \mathbf{R} + \frac{m_2}{m_1 + m_2}\mathbf{r}\tag{3}$$ $$\mathbf{r}_2 = \mathbf{R} - \frac{m_1}{m_1 + m_2}\mathbf{r}\tag{4}$$ Therefore, if we know $\mathbf{R}$ and $\mathbf{r}$, we can determine $\mathbf{r}_1$ and $\mathbf{r}_2$.

4. Center of Mass Inertial Reference Frame #

(1) Motion Analysis in the Center of Mass Reference Frame #

When using the center of mass as the inertial reference frame, $\mathbf{R}=0$, and equations (3) and (4) simplify to: $$\mathbf{r}_1 = \frac{m_2}{m_1 + m_2}\mathbf{r}\tag{5}$$ $$\mathbf{r}_2 = -\frac{m_1}{m_1 + m_2}\mathbf{r}\tag{6}$$

(2) Kinetic Energy #

In the center of mass reference frame, the kinetic energy of the center of mass is 0. By calculating and summing the kinetic energies derived from equations (5) and (6), $$K.E. = \frac{1}{2}\frac{m_1m_2}{m_1+m_2}|\mathbf{v}_1 - \mathbf{v}_2|^2$$ Defining the reduced mass $\mu = \frac{m_1m_2}{m_1+m_2}$, $$K.E. = \frac{1}{2} \mu |\mathbf{v}|^2, \mathbf{v}= \mathbf{v}_1 - \mathbf{v}_2\tag{7}$$

5. Definition of Relative Momentum $\mathbf{p}$ #

From equations (5) and (6), $$m_1 \mathbf{r}_1 = \frac{m_1 m_2}{m_1 + m_2}\mathbf{r}$$ $$m_2 \mathbf{r}_2 = -\frac{m_1m_2}{m_1 + m_2}\mathbf{r}$$ Differentiating with respect to time, $$\mathbf{p}_1 = \frac{m_1 m_2}{m_1 + m_2}\mathbf{v}\tag{8-1}$$ $$\mathbf{p}_2 = -\frac{m_1m_2}{m_1 + m_2}\mathbf{v}\tag{8-2}$$ From this, we can define the relative momentum $\mathbf{p}$ as: $$\mathbf{p}=\frac{m_1 m_2}{m_1 + m_2}\mathbf{v}=\mu\mathbf{v\tag{8-3}} $$ Therefore, equation (7) becomes: $$K.E. = \frac{|\mathbf{p}|^2}{2\mu}\tag{9}$$

6. Relative Motion Analysis ($\mu$, $\mathbf{r}$, $\mathbf{p}$) #

Since the radial motion $\mathbf{p}_r$ is perpendicular to the rotational motion $\mathbf{p}_l$, $$|\mathbf{p}|^2 = |\mathbf{p}_r|^2 + |\mathbf{p}_l|^2$$ Therefore, equation (9) can finally be written as: $$K.E. = \frac{|\mathbf{p}_r|^2}{2\mu} + \frac{|\mathbf{p}_l|^2}{2\mu}\tag{10}$$

7. Hamiltonian in the Center of Mass Reference Frame $H$ #

$$H = \frac{|\mathbf{p}_r|^2}{2\mu} + \frac{|\mathbf{p}_l|^2}{2\mu} + V(|\mathbf{r}|)\tag{11}$$

8. Hamiltonian of the Hydrogen Atom #

In the case of the hydrogen atom, since $m_e \ll m_p$, $\mu \approx m_e = m$, and $V(|\mathbf{r}|) = -\frac{e^2}{|\mathbf{r}|}$ (in cgs units). $$H = \frac{|\mathbf{p}_r|^2}{2m} + \frac{|\mathbf{p}_l|^2}{2m} -\frac{e^2}{|\mathbf{r}|}\tag{12}$$ It’s worth noting that the value of the reduced mass $\mu$ is approximately equal to the electron’s mass $m_e$, so this can be viewed as describing the motion of the electron with the hydrogen nucleus at the origin.