# Change of Coordinates (Two-Body Problem)

1. Feb 17, 2007

### cepheid

Staff Emeritus
Context: reducing a two-body problem to a one body problem (each body is a quantum particle, although this is not relevant to the parts of the problem I am stuck on

1. The problem statement, all variables and given/known data

Griffiths QM, Problem 5.1

Typically the interaction potential depends only on the vector $\mathbf{r} = \mathbf{r_1} - \mathbf{r_2}$ between the two particles. In that case, the Schrodinger equation separates if we change variables from $\mathbf{r_1}, \mathbf{r_2}$ to $\mathbf{r}$ and $\mathbf{R} \equiv (m_1\mathbf{r_1} + m_2\mathbf{r_2})/m_1+m_2$ (the centre of mass).

(a) Show that $\mathbf{r_1} = \mathbf{R} + (\mu/m_1)\mathbf{r}$, $\mathbf{r_2} = \mathbf{R} - (\mu/m_2)\mathbf{r}$, and $\nabla_1 = (\mu/m_2)\nabla_R + \nabla_r$, $\nabla_2 = (\mu/m_1)\nabla_R - \nabla_r$, where

$$\mu \equiv \frac{m_1m_2}{m_1 + m_2}$$

is the reduced mass of the system.

2. Relevant equations

3. The attempt at a solution

(a)

(i) $$\mathbf{r_1} = \mathbf{r} + \mathbf{r_2} = \mathbf{r} + \left(\frac{m_1 + m_2}{m_2}\mathbf{R} - \frac{m_1}{m_2}\mathbf{r_1} \right)$$

$$\mathbf{r_1}\left(1 + \frac{m_1}{m_2}\right) = \mathbf{r_1}\left(\frac{m_1 + m_2}{m_2}\right) = \left(\frac{m_1 + m_2}{m_2}\right)\mathbf{R} + \mathbf{r}$$

$$\longRightarrow \mathbf{r_1} = \mathbf{R} + \frac{m_2}{m_1 + m_2}\mathbf{r} = \mathbf{R} + \frac{\mu}{m_1}\mathbf{r}$$

(ii) $$\mathbf{r_2} = \mathbf{r_1} - \mathbf{r} = \mathbf{R} + \frac{\mu}{m_1}\mathbf{r} - \mathbf{r}$$

$$= \mathbf{R} + \left(\frac{\mu}{m_1} - 1 \right) \mathbf{r} = \mathbf{R} + \left(\frac{m_2}{m_1 + m_2} - \frac{m_1 + m_2}{m_1 + m_2} \right)\mathbf{r}$$

$$= \mathbf{R} - \left(\frac{m_1}{m_1 + m_2}\right)\mathbf{r} = \mathbf{R} - \frac{\mu}{m_2}\mathbf{r}$$

(iii) Now here's where I got stuck. I thought to start in cartesian coordinates:

$$\nabla_1 \equiv \mathbf{\hat{x}} \frac{\partial}{\partial x_1} + \mathbf{\hat{y}} \frac{\partial}{\partial y_1} + \mathbf{\hat{z}} \frac{\partial}{\partial z_1}$$

and then based on the relationships between $x_1$, $x_R$, and $x_r$ obtained in the first part, I could find the answer. But if I do that, I just get:

$$\nabla_1 = \nabla_R + \frac{\mu}{m_1}\nabla_r$$

which is clearly not the right answer. Any tips for this section?

Last edited: Feb 17, 2007
2. Feb 17, 2007

### quasar987

It works fine over here. For instance,

$$\frac{\partial}{\partial x_1}=\frac{\partial R_x}{\partial x_1}\frac{\partial}{\partial R_x}+\frac{\partial r_x}{\partial x_1}\frac{\partial}{\partial r_x}=\frac{\mu}{m_2}\frac{\partial}{\partial R_x}+\frac{\partial}{\partial r_x}$$

3. Feb 23, 2007

### cepheid

Staff Emeritus
You're right. It turns out I was just being dumb. Thanks.