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Homework Help: Expressing the Hamilitonian in c.o.m. coordinates

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    first this is indeed a class assignment , but for some reason i can not remember how to do it since its a review fro a previous semesters course.
    here it is
    Consider a diatomic molecule with a toms labelled A and B and witha classical Hamilitonian given by

    [tex] H =\frac{1}{2} M_A (\dot{x}^2_A + \dot{y}^2_A + \dot{z}^2_A ) + \frac{1}{2} M_B (\dot{x}^{2}_{B} + \dot{y}^{2}_{B} + \dot{z}^{2}_{B}) + V(\vec{r})
    [/tex]
    where [itex] r = \left[ (x_A - x_B)^2 + (y_A-y_B)^2 + (z_A-z_B)^2 \right]^\frac{1}{2} [/itex] is the distance between the atoms and [itex] \vec{r_A} = (x_A,y_B,z_B) [/itex] and [itex] \vec{r_B} = (x_B,,y_B,z_B) [/itex] are vectors that locate each atom.

    a ) Show using the variable [itex] R= (X,Y,Z) [/itex] and [itex] \vec{r} = (x,y,z) [/itex] defined by [itex] R = \frac{(m_A \vec{r_A} + m_B \vec{r_B})}{m_A + m_B} [/itex] and [itex] \vec{r} = \vec{r_A} - \vec{r_B} [/itex] that

    [tex] H = \frac{1}{2}M (\dot{X}^2 + \dot{Y}^2 + \dot{Z}^2 ) + \frac{1}{2} \mu (\dot{x}^2 +\dot{y}^2 + \dot{z}^2 ) + V(\vec{r})
    [/tex]
    where [itex] M =m_A + m_B [/itex] and [itex] \frac{m_A m_B}{ m_A + m_B } [/itex]


    2. Relevant equations
    newtons laws


    3. The attempt at a solution

    i really need a clue to get started
    but i think my problem is getting the total energy in c.o.m. coordinates, i can not figure out where the second term in the c.o.m. hamilitonian comes from.
     
  2. jcsd
  3. Sep 29, 2010 #2

    fzero

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    The original form of the Hamiltonian is written in terms of [tex]\vec{r}_A, \vec{r}_B[/tex]. Solve for those in terms of [tex]\vec{R}, \vec{r}[/tex] and substitute. The algebra is not bad.
     
  4. Sep 30, 2010 #3

    vela

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    The first term is the kinetic energy due to the molecule moving as a whole. The second term is the kinetic energy due to the relative motion of the atoms within the molecule.
     
  5. Sep 30, 2010 #4
    thanks but after "staring" at the problem for a while, this is what i did:
    [tex] m_A \vec{r}_A + m_B \vec{r}_B =0 [/tex]
    [tex] \vec{r} = \vec{r}_A + \vec{r}_B [/tex]
    from the above two :
    [tex] \vec{r}_A = \frac{m_B}{m_A + m_B} \vec{r} [/tex]
    [tex] \vec{r}_B = -\frac{m_A}{m_A + m_B} \vec{r} [/tex]

    [tex]\dot{ \vec{r}}_A = \frac{m_B}{m_A + m_B}\dot{ \vec{r}} [/tex]
    [tex]\dot{ \vec{r}}_B = -\frac{m_B}{m_A + m_B}\dot{ \vec{r}} [/tex]

    squaring and substituting into the classical Hamiltonian (first equation in my post above)
    [tex]H = \frac{m_A m_{B}^2}{2 (m_A + m_B)^2} \dot{r}^2 + \frac{m_B m_{A}^2}{2 (m_A + m_B)^2} \dot{r}^2 + V(r) [/tex]

    [tex]H = \frac{m_A m_B}{2 (m_A + m_B)^2} \dot{r}^2 [m_B + m_A] + V(r) [/tex]

    [tex] H = \frac{m_A m_B}{2 (m_A + m_B)} \dot{r}^2 + V(r) [/tex]

    [tex] H = \frac{1}{2 } \mu \dot{r}^2 + V(r) [/tex]

    this looks like the second term of the target answer , at first i thought this is what i could not get , now it seems its the first term in the target answer that i cant obtain( or think of a way of getting there -since simply adding the term to the equation seems illegitimate) , to quote you
    this i cant figure out how to factor into the expression
     
    Last edited: Sep 30, 2010
  6. Sep 30, 2010 #5

    vela

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    Your equation

    [tex]
    m_A \vec{r}_A + m_B \vec{r}_B =0
    [/tex]

    is saying the center of mass of the molecule is fixed at R=0. That's why you don't get the first term. If the molecule isn't moving, the only kinetic energy there is is due to the relative motion of the atoms.
     
  7. Sep 30, 2010 #6
    so this
    [tex]

    m_A \vec{r}_A + m_B \vec{r}_B =0

    [/tex]

    ought to be replaced by some other expression that accounts for the motion of the com?
     
  8. Sep 30, 2010 #7

    vela

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    Yes, by this one which was given to you

    [tex]R = \frac{(m_A \vec{r_A} + m_B \vec{r_B})}{m_A + m_B}[/tex]
     
  9. Sep 30, 2010 #8
    thanks i'd not yet seen that, now the problem looks alot more doable .i'l work on it
     
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