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Alternate approach to proving the virial theorem?

  1. Apr 10, 2008 #1
    Hi guys,
    In quantum mechanics, the virial theorem for a system in its ground state is proved by a very nice scaling technique (Nielsen and Martin, PRB, 1985). I was trying to do something similar in classical mechanics and arrived at the virial theorem but i am not sure about why it should work.


    Typically, the virial theorem (in 1-D) is proven as follows. Consider a set of interacting point masses Pi (i=1,2,3…N). The motion of Pi in an inertial frame is governed by

    [tex]
    \[
    f_i = \frac{d}{{dt}}\left( {m_i v_i } \right)
    \]


    \[\Rightarrow\
    \sum\limits_i {x_i f_i } = \sum\limits_i {x_i \frac{d}{{dt}}\left( {m_i v_i } \right)}
    \]


    now since,



    \[
    \sum\limits_i {x_i \frac{d}{{dt}}\left( {m_i v_i } \right)} = \frac{d}{{dt}}\left( {\sum\limits_i {x_i \left( {m_i v_i } \right)} } \right) - \sum\limits_i {v_i \left( {m_i v_i } \right)}
    \]

    we can write

    \[
    \sum\limits_i {m_i v_i^2 } + \sum\limits_i {x_i f_i } = \frac{d}{{dt}}\left( {\sum\limits_i {m_i x_i v_i } } \right)
    \]

    Further if we assume that the position-momentum product on the right side of the above equation remains bound in time, then by taking a sufficiently long time average we can say

    \[
    \left\langle {\sum\limits_i {m_i v_i^2 } } \right\rangle + \left\langle {\sum\limits_i {x_i f_i } } \right\rangle = 0
    \]

    [/tex]

    where <> denotes the time-average.
    This is the virial theorem.

    Now, Please take a look at my approach. Please note that though virial theorem holds even if the system is not in equilibrium, i will consider a system of particles in its minimum energy configuration.

    -------------------------------------------------------------------------------------------------------------
    The Hamiltonian for the n-particle system is given by

    [tex]
    H=
    \[
    \sum\limits_i^N {\frac{{p_i^2 }}{{2m_i }}} + V\left( {x_1 ,x_2 ,...,x_N } \right)
    \]

    [/tex]

    Now i will do a canonical transformation such that
    [tex]
    \[
    \begin{array}{l}
    x'_i \to \left( {1 + \upsilon } \right)x_i \\
    p'_i \to p_i /\left( {1 + \upsilon } \right) \\
    \end{array}
    \]
    [/tex]

    I know that i am just scaling the coordinates, but let us say i put these new coordinates in the expression for the Hamiltonian. Then, i will further say that since i started with a minimum energy configuration, the derivative of the energy w.r.t the parameter [tex]\upsilon[/tex] should be zero. Hence i should have,

    [tex]
    \[
    \frac{{\partial \left( {\sum\limits_i^N {\frac{{p_i^2 }}{{2m_i \left( {1 + \upsilon } \right)^2 }}} + V\left( {\left( {1 + \upsilon } \right)x_1 ,\left( {1 + \upsilon } \right)x_2 ,...,\left( {1 + \upsilon } \right)x_N } \right)} \right)}}{{\partial \upsilon }} = 0
    \]
    [/tex]

    On simplification this gives the virial theorem back.

    [tex]
    \[
    \left\langle {\sum\limits_i {m_i v_i^2 } } \right\rangle + \left\langle {\sum\limits_i {x_i f_i } } \right\rangle = 0
    \]
    [/tex]

    Why is this happening? What is the significance of the canonical transformation? And how come, i dont have to take any time-averaging? Is this calculation wrong?


    I am a mechanical engineer..so please excuse me if there are any errors!
     
  2. jcsd
  3. Apr 11, 2008 #2
    hi schuldiner, just want to share some thoughts with you. i am not very familiar with the virial theorm, but i know it takes a meaning of averaging the spatial coordinates for a system. so it seems natural that when you scales the xi axes, they are averaged out eventially and any scaling of them does not matters. time is not averaged due to its definition (averaging in space is the only thing virial theorem concern, if i remember correctly). for the velocity space, since those derivatives are gone, it also does not participate in the final results. But for non-minimum E configuration, i think stuffs will be different. sorry i have not try to go thru those math steps by myself. that is interesting, though :smile:

    thanks for the Q. just curious, are there any reason why you like to transform in that way? :smile:
     
    Last edited: Apr 11, 2008
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