Expressing the Hamilitonian in c.o.m. coordinates

[Mechdude]

Homework Statement

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

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

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

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

Homework Equations

Newtons laws

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.

 

[fzero]

The original form of the Hamiltonian is written in terms of $$\vec{r}_A, \vec{r}_B$$. Solve for those in terms of $$\vec{R}, \vec{r}$$ and substitute. The algebra is not bad.

 

[vela]

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.

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.

 

[Mechdude]

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.

thanks but after "staring" at the problem for a while, this is what i did:
$$m_A \vec{r}_A + m_B \vec{r}_B =0$$
$$\vec{r} = \vec{r}_A + \vec{r}_B$$
from the above two :
$$\vec{r}_A = \frac{m_B}{m_A + m_B} \vec{r}$$
$$\vec{r}_B = -\frac{m_A}{m_A + m_B} \vec{r}$$

$$\dot{ \vec{r}}_A = \frac{m_B}{m_A + m_B}\dot{ \vec{r}}$$
$$\dot{ \vec{r}}_B = -\frac{m_B}{m_A + m_B}\dot{ \vec{r}}$$

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

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

$$H = \frac{m_A m_B}{2 (m_A + m_B)} \dot{r}^2 + V(r)$$

$$H = \frac{1}{2 } \mu \dot{r}^2 + V(r)$$

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 can't obtain( or think of a way of getting there -since simply adding the term to the equation seems illegitimate) , to quote you

The first term is the kinetic energy due to the molecule moving as a whole.

this i can't figure out how to factor into the expression

 

[vela]

Your equation

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

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.

 

[Mechdude]

so this
$$<br /> m_A \vec{r}_A + m_B \vec{r}_B =0<br />$$

ought to be replaced by some other expression that accounts for the motion of the com?

 

[vela]

Yes, by this one which was given to you

$$R = \frac{(m_A \vec{r_A} + m_B \vec{r_B})}{m_A + m_B}$$

 

[Mechdude]

thanks i'd not yet seen that, now the problem looks a lot more doable .i'l work on it