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Homework Help: Change of Coordinates (Two-Body Problem)

  1. Feb 17, 2007 #1

    cepheid

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    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 [itex] \mathbf{r} = \mathbf{r_1} - \mathbf{r_2} [/itex] between the two particles. In that case, the Schrodinger equation separates if we change variables from [itex] \mathbf{r_1}, \mathbf{r_2} [/itex] to [itex] \mathbf{r} [/itex] and [itex] \mathbf{R} \equiv (m_1\mathbf{r_1} + m_2\mathbf{r_2})/m_1+m_2 [/itex] (the centre of mass).

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

    [tex] \mu \equiv \frac{m_1m_2}{m_1 + m_2} [/tex]

    is the reduced mass of the system.

    2. Relevant equations

    Already given

    3. The attempt at a solution

    (a)

    (i) [tex] \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)[/tex]

    [tex] \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} [/tex]

    [tex] \longRightarrow \mathbf{r_1} = \mathbf{R} + \frac{m_2}{m_1 + m_2}\mathbf{r} = \mathbf{R} + \frac{\mu}{m_1}\mathbf{r} [/tex]

    (ii) [tex] \mathbf{r_2} = \mathbf{r_1} - \mathbf{r} = \mathbf{R} + \frac{\mu}{m_1}\mathbf{r} - \mathbf{r} [/tex]

    [tex] = \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} [/tex]

    [tex] = \mathbf{R} - \left(\frac{m_1}{m_1 + m_2}\right)\mathbf{r} = \mathbf{R} - \frac{\mu}{m_2}\mathbf{r} [/tex]

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

    [tex] \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} [/tex]

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

    [tex] \nabla_1 = \nabla_R + \frac{\mu}{m_1}\nabla_r [/tex]

    which is clearly not the right answer. Any tips for this section?
     
    Last edited: Feb 17, 2007
  2. jcsd
  3. Feb 17, 2007 #2

    quasar987

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    It works fine over here. For instance,

    [tex]\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}[/tex]
     
  4. Feb 23, 2007 #3

    cepheid

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