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Homework Help: Separating a hamiltonian into C.O.M and relative hamiltonians

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the two-body hamiltonian[tex]H_{\text{sys}}=\frac{\mathbf{p}_1^2}{2m_1}+\frac{\mathbf{p}_2^2}{2m_2}+V( \mathbf{r}_1,\mathbf{r}_2)[/tex]can be separated into centre of mass and relative hamiltonians[tex]H_{\text{sys}}=\frac{\mathbf{P}^2}{2M}+\frac{\mathbf{p}_{\text{rel}}}{2\mu}+V(r)[/tex]Do this in two ways:

    a)with momentum operators in abstract
    b)momentum operators in the position representation

    2. Relevant equations
    I'm assuming this one, the text does not actually say

    3. The attempt at a solution

    I have tried to do this by plugging the definitions into the equations. tried working backwards too. not really sure where to start here!
  2. jcsd
  3. Nov 27, 2013 #2
    You can write [itex]r_2-r_1=r[/itex] and [itex]R=\frac{m_1r_1+m_2r_2}{m1 + m2}[/itex].

    Using these, you can solve for [itex]r_1[/itex] and [itex]r_2[/itex] in terms of [itex]R[/itex] and [itex]r[/itex]. Then, obviously, you can find momentum or the two objects in terms of [itex]R[/itex] and [itex]r[/itex].

    It is important to remember that [itex]R[/itex] is the center of mass. If you end up with a term that says [itex]\frac{1}{2}(m_1+m_2)\dot{R}^2[/itex] then you can say "Not interested in translational energy of the entire system, thus, this term can be neglected.. Etc etc. Use your imagination :)

    Lastly, don't forget that these r's are vectors. I'm not sure how much of a difference it will make in your math, but it's still an important point.
  4. Nov 27, 2013 #3
    sorry. I am already lost. how can I obviously find the momentum in terms of r and R?
  5. Nov 27, 2013 #4
    My apologies, I assumed that if you are studying Hamiltonian mechanics you would understand what I was saying.

    Let me reiterate:
    From my example above [itex]r[/itex] is the vector that points from mass 1 to mass 2. Also [itex]R[/itex] is the vector that points to the center of mass from the origin.

    Using those two definitions, you can solve for [itex]r_1[/itex] in terms of [itex]R[/itex] and [itex]r[/itex]. Meaning that you can get [itex]r_1=...[/itex] where the only thing on the right would be [itex]R[/itex]'s, [itex]r[/itex]'s, and masses. No [itex]r_1[/itex]'s or [itex]r_2[/itex]'s.

    Total momentum is mass times velocity. Therefore, [itex]p_1=m_1v_1=m_1\dot{r}_1[/itex] where [itex]\dot{r}_1[/itex] is the derivative of [itex]r_1[/itex] with respect to time, AKA, the velocity of mass 1.

    Anyways, the problem is asking you to rewrite the Hamiltonian in terms of [itex]R[/itex]'s and [itex]r[/itex]'s. Since you know what [itex]p_1[/itex] is and you know [itex]r_1[/itex] in terms of [itex]R[/itex]'s and [itex]r[/itex]'s, you can find momentum in terms of [itex]R[/itex]'s and [itex]r[/itex]'s. The same can be said for [itex]p_2[/itex]. Thus, the problem is pretty much solved.
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