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# Alternate approach to proving the virial theorem?

1. Apr 10, 2008

### chuckschuldiner

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

$$$f_i = \frac{d}{{dt}}\left( {m_i v_i } \right)$ $\Rightarrow\ \sum\limits_i {x_i f_i } = \sum\limits_i {x_i \frac{d}{{dt}}\left( {m_i v_i } \right)}$ now since, $\sum\limits_i {x_i \frac{d}{{dt}}\left( {m_i v_i } \right)} = \frac{d}{{dt}}\left( {\sum\limits_i {x_i \left( {m_i v_i } \right)} } \right) - \sum\limits_i {v_i \left( {m_i v_i } \right)}$ we can write $\sum\limits_i {m_i v_i^2 } + \sum\limits_i {x_i f_i } = \frac{d}{{dt}}\left( {\sum\limits_i {m_i x_i v_i } } \right)$ Further if we assume that the position-momentum product on the right side of the above equation remains bound in time, then by taking a sufficiently long time average we can say $\left\langle {\sum\limits_i {m_i v_i^2 } } \right\rangle + \left\langle {\sum\limits_i {x_i f_i } } \right\rangle = 0$$$

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

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

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

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,

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

On simplification this gives the virial theorem back.

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

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. Apr 11, 2008

### luben

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

thanks for the Q. just curious, are there any reason why you like to transform in that way?

Last edited: Apr 11, 2008


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# A body rotating on a plane with friction, being pulled by another body

Two bodies are attached with a string. One body is rotating on a plane centering a point. A hole is present at the center of rotation and the string is passed through the hole and attached with a hanging body. The plane has friction. I need to find out the graph of the velocity of the hanging body. I have described the problem (with the masses, coefficient of friction, radius of rotation) in the picture. [$$m_1=1.5$$ kg (rotating body), $$m_2=2$$ kg (hanging body), $$r=1$$ m (initial radius of rotation), $$\mu=0.2$$ (coefficient of kinetic friction)]

For the reason mentioned at the bottom, I wrote the solution on paper and took pictures. I have uploaded the images in this link. The link goes to a folder in my google drive.

Below, I briefly present what I have done on paper:

$$m_2g \cdot dr = m_2a_r \cdot dr + m_1a_r\cdot dr + m_1a_x\cdot dx + \mu m_1 g \sqrt{(dx)^2 + (dr)^2}$$

Dividing by $$dt$$, then by $$v_r$$, then putting $$\tan(\theta) = \frac{v_x}{v_r}$$, it is possible to obtain:

$$v_r' = \frac{m_2}{m_1+m_2}g - \frac{m_1}{m_1+m_2} \mu g \frac{v_r}{\sqrt{v_x^2 + v_r^2}}$$

$$v_x' = -\mu g \frac{v_x}{\sqrt{v_x^2 + v_r^2}}$$

My questions are:

1. In my solution, I got a system of differential equations. Is the system correct? Does it actually describe what is going on in the problem?
2. I solved the system numerically. Is it possible solve the system in some other way? Is it possible to get an explicit expression for $$v_r$$?
3. Is there an easier way to solve this problem?
4. In page 9 (the pages are labeled in top right), I obtained another equation for $$a_r$$. But the equation does not hold when $$t=0$$. What have I done wrong in that page?

*I unfortunately had to write the solution on paper and upload the pictures. Writing down the solution would have made the question too long. I hope this isn't a huge inconvenience. I also included the tag differential-equations because I am asking if there is a way to solve the system (not numerically).

• Duplicate : physics.stackexchange.com/q/659844/305718
– ACB
Aug 19 at 9:50
• @ACB Wait, I added a solution, that question doesn't have it. Aug 19 at 9:52
• It is not a good thing to present solutions as images through links. Briefly type it here.
– ACB
Aug 19 at 9:57

I do not think your equations of motion are correct. The equations of motion can be expressed in polar coordinates on the flat surface \begin{align} &x = r \, \cos(\theta)\\ &y = r \, \sin(\theta) \end{align}

and I think they look something like this:

\begin{align} (m_1+m_2)\,&\frac{d^2 r}{dt^2} \,=\, m_1\,r\,\left(\frac{d\theta}{dt}\right)^2 \, -\, m_2g\, - \, \mu m_1 g\,\, \frac{\frac{dr}{dt}}{\sqrt{\left(\frac{dr}{dt}\right)^2 \, +\, r^2\left(\frac{d\theta}{dt}\right)^2}} \\ m_1\,r^2 \, &\frac{d^2 \theta}{dt^2} \, = \, - \,2 \, m_1\,r \, \frac{d r}{dt} \left(\frac{d \theta}{dt}\right)^2\, - \, \mu m_1 g\,\, \frac{r\,\frac{d\theta}{dt}}{\sqrt{\left(\frac{dr}{dt}\right)^2 \, +\, r^2\left(\frac{d\theta}{dt}\right)^2}} \\ \end{align}

I have serious doubts that these equations are explicitly solvable, because I do not think that they are integrable in the first place (they do not have conservation laws because the friction breaks them, except for the fact that these equations are not changed by rotation around the center of the plane, i.e. when $$\theta \mapsto \theta + c$$). Hence I would not expect explicit expression for the solutions.

You could also write the system as a system of first order differential equations as follows:

\begin{align} &\frac{dr}{dt} \, = \,v \\ &\\ &\frac{d \theta}{dt} \, = \,\omega \\ &\\ &\frac{d v}{dt} \,=\, \frac{m_1}{m_1+m_2}\,r\,\omega^2 \, -\, \frac{m_2g}{m_1+m_2}\, - \, \frac{\mu m_1 g}{m_1+m_2}\,\, \frac{v}{\sqrt{v^2 \, +\, r^2\omega^2}} \\ &\\ &\frac{d \omega}{dt} \, = \, - \, 2\, \frac{v \, \omega^2}{r} \, - \, \mu g\,\, \frac{\omega}{r \, \sqrt{v^2 \, +\, r^2\omega^2}} \\ \end{align}

and use this formulation for numeric simulations.

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