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Hamiltonian with two masses

  1. Jul 6, 2009 #1
    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?

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


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


    \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.
  4. Jul 6, 2009 #3


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

    x = x_1 - x_2 \\
    X = \frac{{m_1 x_1 + m_2 x_2 }}{{m_1 + m_2 }} \\

    however the problem is stated in terms of [tex] x_1[/tex] and [tex]x_2[/tex]. You're simply switching to these new variable [tex] x [/tex] and [tex] X[/tex]. 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 [tex]\frac{\partial }{{\partial x_1 }}[/tex] and [tex]\frac{\partial }{{\partial x_2 }}[/tex], you're looking for [tex]\frac{\partial }{{\partial x}}[/tex] and[tex]\frac{\partial }{{\partial X}}[/tex]. Simple chain rule shows for example [tex]
    \frac{\partial }{{\partial X}} = \frac{\partial }{{\partial x_1 }}\frac{{dx_1 }}{{dX}} + \frac{\partial }{{\partial x_1 }}\frac{{dx_2 }}{{dX}}
    [/tex] which is about what's going on except you're going the other way and looking for the what [tex]\frac{\partial }{{\partial x_1}}[/tex] is and [tex]\frac{\partial }{{\partial x_2}}[/tex] is
    Last edited: Jul 6, 2009
  5. Jul 6, 2009 #4
  6. Jul 6, 2009 #5


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