Separating a hamiltonian into C.O.M and relative hamiltonians

[richyw]

Homework Statement

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

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

Homework Equations

I'm assuming this one, the text does not actually say
$$M=m_1+m_2$$$$mu=\frac{m_1m_2}{m_1+m_2}$$$$\mathbf{P}=\mathbf{p}_1+\mathbf{p}_2$$$$\mathbf{p}_{\text{rel}}=\frac{m_1\mathbf{p}_2-m_2\mathbf{p}_1}{m_1+m_2}$$

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!

 

[Hertz]

You can write $r_2-r_1=r$ and $R=\frac{m_1r_1+m_2r_2}{m1 + m2}$.

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

It is important to remember that $R$ is the center of mass. If you end up with a term that says $\frac{1}{2}(m_1+m_2)\dot{R}^2$ 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.

 

[richyw]

sorry. I am already lost. how can I obviously find the momentum in terms of r and R?

 

[Hertz]

sorry. I am already lost. how can I obviously find the momentum in terms of r and R?

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 $r$ is the vector that points from mass 1 to mass 2. Also $R$ is the vector that points to the center of mass from the origin.

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

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

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