- #1
chuckschuldiner
- 15
- 0
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
<br /> \[<br /> f_i = \frac{d}{{dt}}\left( {m_i v_i } \right)<br /> \]<br /> <br /> <br /> \[\Rightarrow\<br /> \sum\limits_i {x_i f_i } = \sum\limits_i {x_i \frac{d}{{dt}}\left( {m_i v_i } \right)} <br /> \]<br /> <br /> <br /> now since,<br /> <br /> <br /> <br /> \[<br /> \sum\limits_i {x_i \frac{d}{{dt}}\left( {m_i v_i } \right)} = \frac{d}{{dt}}\left( {\sum\limits_i {x_i \left( {m_i v_i } \right)} } \right) - \sum\limits_i {v_i \left( {m_i v_i } \right)} <br /> \]<br /> <br /> we can write<br /> <br /> \[<br /> \sum\limits_i {m_i v_i^2 } + \sum\limits_i {x_i f_i } = \frac{d}{{dt}}\left( {\sum\limits_i {m_i x_i v_i } } \right)<br /> \]<br /> <br /> 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 <br /> <br /> \[<br /> \left\langle {\sum\limits_i {m_i v_i^2 } } \right\rangle + \left\langle {\sum\limits_i {x_i f_i } } \right\rangle = 0<br /> \]<br /> <br />
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.
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The Hamiltonian for the n-particle system is given by
<br /> H=<br /> \[<br /> \sum\limits_i^N {\frac{{p_i^2 }}{{2m_i }}} + V\left( {x_1 ,x_2 ,...,x_N } \right)<br /> \]<br /> <br />
Now i will do a canonical transformation such that
<br /> \[<br /> \begin{array}{l}<br /> x'_i \to \left( {1 + \upsilon } \right)x_i \\ <br /> p'_i \to p_i /\left( {1 + \upsilon } \right) \\ <br /> \end{array}<br /> \]<br />
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 \upsilon should be zero. Hence i should have,
<br /> \[<br /> \frac{{\partial \left( {\sum\limits_i^N {\frac{{p_i^2 }}{{2m_i \left( {1 + \upsilon } \right)^2 }}} + V\left( {\left( {1 + \upsilon } \right)x_1 ,\left( {1 + \upsilon } \right)x_2 ,...,\left( {1 + \upsilon } \right)x_N } \right)} \right)}}{{\partial \upsilon }} = 0<br /> \]<br />
On simplification this gives the virial theorem back.
<br /> \[<br /> \left\langle {\sum\limits_i {m_i v_i^2 } } \right\rangle + \left\langle {\sum\limits_i {x_i f_i } } \right\rangle = 0<br /> \]<br />
Why is this happening? What is the significance of the canonical transformation? And how come, i don't have to take any time-averaging? Is this calculation wrong?
I am a mechanical engineer..so please excuse me if there are any errors!
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
<br /> \[<br /> f_i = \frac{d}{{dt}}\left( {m_i v_i } \right)<br /> \]<br /> <br /> <br /> \[\Rightarrow\<br /> \sum\limits_i {x_i f_i } = \sum\limits_i {x_i \frac{d}{{dt}}\left( {m_i v_i } \right)} <br /> \]<br /> <br /> <br /> now since,<br /> <br /> <br /> <br /> \[<br /> \sum\limits_i {x_i \frac{d}{{dt}}\left( {m_i v_i } \right)} = \frac{d}{{dt}}\left( {\sum\limits_i {x_i \left( {m_i v_i } \right)} } \right) - \sum\limits_i {v_i \left( {m_i v_i } \right)} <br /> \]<br /> <br /> we can write<br /> <br /> \[<br /> \sum\limits_i {m_i v_i^2 } + \sum\limits_i {x_i f_i } = \frac{d}{{dt}}\left( {\sum\limits_i {m_i x_i v_i } } \right)<br /> \]<br /> <br /> 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 <br /> <br /> \[<br /> \left\langle {\sum\limits_i {m_i v_i^2 } } \right\rangle + \left\langle {\sum\limits_i {x_i f_i } } \right\rangle = 0<br /> \]<br /> <br />
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
<br /> H=<br /> \[<br /> \sum\limits_i^N {\frac{{p_i^2 }}{{2m_i }}} + V\left( {x_1 ,x_2 ,...,x_N } \right)<br /> \]<br /> <br />
Now i will do a canonical transformation such that
<br /> \[<br /> \begin{array}{l}<br /> x'_i \to \left( {1 + \upsilon } \right)x_i \\ <br /> p'_i \to p_i /\left( {1 + \upsilon } \right) \\ <br /> \end{array}<br /> \]<br />
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 \upsilon should be zero. Hence i should have,
<br /> \[<br /> \frac{{\partial \left( {\sum\limits_i^N {\frac{{p_i^2 }}{{2m_i \left( {1 + \upsilon } \right)^2 }}} + V\left( {\left( {1 + \upsilon } \right)x_1 ,\left( {1 + \upsilon } \right)x_2 ,...,\left( {1 + \upsilon } \right)x_N } \right)} \right)}}{{\partial \upsilon }} = 0<br /> \]<br />
On simplification this gives the virial theorem back.
<br /> \[<br /> \left\langle {\sum\limits_i {m_i v_i^2 } } \right\rangle + \left\langle {\sum\limits_i {x_i f_i } } \right\rangle = 0<br /> \]<br />
Why is this happening? What is the significance of the canonical transformation? And how come, i don't have to take any time-averaging? Is this calculation wrong?
I am a mechanical engineer..so please excuse me if there are any errors!
