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.

 

[Nickie]

As a scientist, it is important to approach problems in a systematic and logical manner. In this case, we are trying to show that the two-body Hamiltonian can be separated into center of mass and relative Hamiltonians. To do this, we can approach the problem in two ways: using momentum operators in abstract and using momentum operators in the position representation.

a) Using momentum operators in abstract:

We start by writing the Hamiltonian in terms of momentum operators:

H_{\text{sys}}=\frac{\mathbf{p}_1^2}{2m_1}+\frac{\mathbf{p}_2^2}{2m_2}+V( \mathbf{r}_1,\mathbf{r}_2)

Next, we can rewrite the momentum operators in terms of the center of mass and relative momentum operators, as given in the homework equations:

\mathbf{p}_1=\mathbf{P}-\frac{m_2}{M}\mathbf{p}_{\text{rel}}\mathbf{p}_2=\mathbf{P}+\frac{m_1}{M}\mathbf{p}_{\text{rel}}

Substituting these expressions into the Hamiltonian, we get:

H_{\text{sys}}=\frac{(\mathbf{P}-\frac{m_2}{M}\mathbf{p}_{\text{rel}})^2}{2m_1}+\frac{(\mathbf{P}+\frac{m_1}{M}\mathbf{p}_{\text{rel}})^2}{2m_2}+V( \mathbf{r}_1,\mathbf{r}_2)

Expanding and simplifying, we get:

H_{\text{sys}}=\frac{\mathbf{P}^2}{2M}+\frac{\mathbf{p}_{\text{rel}}^2}{2\mu}+V( \mathbf{r}_1,\mathbf{r}_2)-\frac{\mathbf{P}\cdot\mathbf{p}_{\text{rel}}}{M}

Now, we can see that the last term in the expression is dependent on both the center of mass and relative coordinates, and hence it cannot be separated into the two Hamiltonians we are looking for. Therefore, we can redefine the center of mass Hamiltonian as:

H_{\