# Hamiltonian with two masses

1. Jul 6, 2009

### marmot

I am not posting this in the homework section because it is not really a homework problem. Its from the schaum outline and I am stumped in this:

http://img379.imageshack.us/img379/688/67356569.jpg [Broken]
I have NO idea about 3.2.4 and 3.2.5. Its black magic! How the hell does that substitution work?

Thanks

Last edited by a moderator: May 4, 2017
2. Jul 6, 2009

### Cyosis

In 3.2.4 they simply take the derivatives of x and X with respect to x1 and x2.

Example:

$$\frac{\partial x}{\partial x_1}=\frac{\partial x_1-x_2}{\partial x_1}=\frac{\partial x_1}{\partial x_1}-\frac{\partial x_2}{\partial x_1}=1$$

Because x2 is a constant with respect to x1. The others go the same way. In 3.2.5 they use the chain rule.

3. Jul 6, 2009

### Pengwuino

It's a simple partial derivative. It has nothing to do with the problem. You're looking at

$$\begin{array}{l} x = x_1 - x_2 \\ X = \frac{{m_1 x_1 + m_2 x_2 }}{{m_1 + m_2 }} \\ \end{array}$$

however the problem is stated in terms of $$x_1$$ and $$x_2$$. You're simply switching to these new variable $$x$$ and $$X$$. When you want to switch variable in a problem that involves derivatives, you'll need to determine how the derivatives act. What you're looking for is instead of $$\frac{\partial }{{\partial x_1 }}$$ and $$\frac{\partial }{{\partial x_2 }}$$, you're looking for $$\frac{\partial }{{\partial x}}$$ and$$\frac{\partial }{{\partial X}}$$. Simple chain rule shows for example $$\frac{\partial }{{\partial X}} = \frac{\partial }{{\partial x_1 }}\frac{{dx_1 }}{{dX}} + \frac{\partial }{{\partial x_1 }}\frac{{dx_2 }}{{dX}}$$ which is about what's going on except you're going the other way and looking for the what $$\frac{\partial }{{\partial x_1}}$$ is and $$\frac{\partial }{{\partial x_2}}$$ is

Last edited: Jul 6, 2009
4. Jul 6, 2009

5. Jul 6, 2009

### Cyosis

Multiply equation 14 by -1 on both sides, then plug the result into equation 13 to obtain equation 16.