# Expressing the Hamilitonian in c.o.m. coordinates

• Mechdude
In summary, Homework Equations state that the center of mass of the molecule is at 0 and the kinetic energy due to the molecule moving as a whole and relative to the atoms is due to the second term in the target answer.
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 $\frac{m_A m_B}{ m_A + m_B }$

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.

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.

Mechdude said:
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.

vela said:
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
vela said:
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

Last edited:

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

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.

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

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

Yes, by this one which was given to you

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

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

## What is the significance of expressing the Hamiltonian in center-of-mass (c.o.m.) coordinates?

Expressing the Hamiltonian in c.o.m. coordinates allows for simplification of the equations of motion and a clearer understanding of the underlying physical principles.

## How is the Hamiltonian expressed in c.o.m. coordinates?

The Hamiltonian in c.o.m. coordinates is obtained by substituting the c.o.m. coordinates and momenta into the original Hamiltonian, and then expressing the remaining coordinates and momenta in terms of the c.o.m. quantities.

## What are the benefits of using c.o.m. coordinates in Hamiltonian mechanics?

Using c.o.m. coordinates in Hamiltonian mechanics can lead to more efficient and elegant calculations, as well as providing insight into the symmetries and conservation laws of the system.

## Is it always possible to express the Hamiltonian in c.o.m. coordinates?

No, the Hamiltonian can only be expressed in c.o.m. coordinates for systems with a well-defined center of mass.

## Can the Hamiltonian be expressed in c.o.m. coordinates for systems with multiple particles?

Yes, the Hamiltonian can be expressed in c.o.m. coordinates for systems with multiple particles, as long as the particles are interacting through conservative forces.

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