1
$\begingroup$

I clearly am interpreting the Virial Theorem incorrectly, but I don't know how. In dipole gases, the molecules can exhibit five kinetic modes, while they can only experience 2 potential modes. Doesn't the Virial Theorem say the two must have an equal distribution of energy?

$\endgroup$
7
  • $\begingroup$ I don't understand your concern. Could you please elaborate? $\endgroup$
    – Bubble
    Sep 10 '13 at 15:38
  • $\begingroup$ Don't dipole gases have an unequal distribution of energy? 5/7 kinetic? $\endgroup$
    – user24082
    Sep 10 '13 at 16:13
  • $\begingroup$ Since when? Can you provide a reference? Are you confusing modes and expectation values? $\endgroup$
    – Bubble
    Sep 10 '13 at 17:19
  • 2
    $\begingroup$ No, there is less energy per each mode so that the relation between total kinetic and potential energy is still given by the virial theorem. For energy per mode see equipartition theorem. $\endgroup$
    – Bubble
    Sep 10 '13 at 20:54
  • 1
    $\begingroup$ Are you certain? I was taught that there are the $3*(1/2)KT$ Joules of of KE, then a set of $2*(1/2)KT$ kinetic Joules from rotations and another set for potential within the bond. Upon rereading the Virial Theorem, it says that N particles held by a potential, so might it be that the potential only "sees" the rotational kinetic? I'm confused. $\endgroup$
    – user24082
    Sep 10 '13 at 21:29
1
$\begingroup$

The short answer is that the motion of a diatomic gas is not periodic, so the virial theorem does not apply. Let's see exactly what is going on.

In fact, let's consider the even simpler case of a monatomic ideal gas. Here there is also an apparent contradiction, because there is kinetic energy but no potential energy, while you would say the virial theorem implies energy should be evenly distributed amoung kinetic and potential. $\newcommand{\v}{\mathbf{v}}\newcommand{\s}{\boldsymbol{\sigma}}\newcommand{\r}{\mathbf{r}}\newcommand{\f}{\mathbf{f}}$ To discover the source of the contradiction, let's look at what the virial theorem says. The virial theorem comes from the observation that for a periodic system, things look the same after a very long time. This means that there can't be a net flow of momentum in the long run. The flow of momentum, which I will call momentum current is the sum of two pieces. The first piece is just caused by moving objects carrying momentum along with them. This piece is given by $\rho \v \otimes \v$, where $\rho$ is the mass density and $\v$ is the velocity field. The other way momentum can flow is to be transferred by forces. The second piece of the momentum current is (minus) the stress, $\s$. The net flow of momentum can be measured by the spatial integral of the momentum current, $\int_\textrm{space} \ \ \rho \v \otimes \v - \s d \r$. Now since we know that there can be no net flow of momentum in the long term, the statement of the virial theorem becomes $\langle \int_\textrm{space} \ \ \rho \v \otimes \v - \s d \r \rangle = 0$, where $\langle \cdots \rangle$ denotes a time average.

We can rewrite the virial theorem as $\langle \int_\textrm{space} \ \ \rho \v \otimes \v d \r \rangle = \langle \int_\textrm{space} \ \ \s d \r \rangle$. Next notice that $\int_\textrm{space} \ \ \sigma_{ij} \ d \r = \int_\textrm{space} \ \ \sigma_{ik} \partial_k r_j \ d \r = \int_\textrm{space} \ \ \partial_k \sigma_{ik} r_j \ d \r - \int_\textrm{space} \ \r_j \partial_k \sigma_{ik} \ d \r$. On the right hand side, the first term is a surface term, so it can be ignored. The second term can be transformed using $\partial_k \sigma_{ik} = f_i$, where $f_i$ is the body force per unit volume. Thus we find $\langle \int_\textrm{space} \ \ \rho \v \otimes \v d \r \rangle = -\langle \int_\textrm{space} \ \ \r \otimes \f d \r \rangle$.

If we have a system of point particles, the virial theorem becomes $\langle \sum_\alpha m_\alpha \v_\alpha \otimes \v_\alpha \rangle = -\langle \sum_\alpha \r_\alpha \otimes \f_\alpha \rangle$, where $\alpha$ indexes over all particles. Taking the trace, we get $\langle \sum_\alpha m_\alpha \v_\alpha \cdot \v_\alpha \rangle = -\langle \sum_\alpha \r_\alpha \cdot \f_\alpha \rangle$, and finally identifying $\sum_\alpha m_\alpha \v_\alpha \cdot \v_\alpha$ with twice the kinetic energy, $2T$, we get $2T = -\langle \sum_\alpha \r_\alpha \cdot \f_\alpha \rangle$. Now the right hand side of this equation can be related to the potential energy for power law forces, and for a harmonic force, you get that potential and kinetic energies are equal.

So in summary, we saw that for periodic systems there is no net momentum current, so the "kinetic" piece $\rho \v \otimes \v$ must cancel the stress piece $-\s$. Doing in integration by parts, we were able to relate the $-\s$ piece to $\r \f$, which we then related to potential energy, and thus we basically got that kinetic energy was equal to potential energy up to some factor.

Now the reason that our monatomic gases don't satisfy the virial theorem is that the atoms just go in straight lines. Therefore the system is not periodic and there is no stress piece to balance the kinetic piece. Therefore we shouldn't expect there to be a potential energy.

Now what about the more complicated case of a diatomic gas? Consider first a single molecule. It's center of mass will just move in a straight line and so its motion is not periodic. But if we look in the center of mass frame, then the motion is periodic, and in the case of a harmonic potential, we find the kinetic energy must be equal to the potential. Now boosting back to the lab frame we get our three unbalanced translational modes back. So we have three translational plus two rotational kinetic modes, giving us $\frac{5}{2} kT$ of kinetic energy. On the other hand, the potential modes only balance out the rotational kinetic modes, so we still just have $\frac{2}{2} kT$ of potential energy. Since energies just sum, this same analysis works a whole gas of molecules.

$\endgroup$
0
$\begingroup$

The Virial Theorem does not require an equal distribution of energy amongst the kinetic and potential modes. It doesn't even require an equal distribution of energy amongst the kinetic or potential modes! Kinetic energy tends to surround wherever potential energy used to be, kind of.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy