1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    [tex]M=m_1+m_2[/tex][tex]mu=\frac{m_1m_2}{m_1+m_2}[/tex][tex]\mathbf{P}=\mathbf{p}_1+\mathbf{p}_2[/tex][tex]\mathbf{p}_{\text{rel}}=\frac{m_1\mathbf{p}_2-m_2\mathbf{p}_1}{m_1+m_2}[/tex]

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Separating a hamiltonian into C.O.M and relative hamiltonians
  1. Hamiltonian Mechanics (Replies: 0)

Loading...