# Example: Degrees of freedom

## Understanding the situation

The critical idea for understanding entropy is that energy is continually moving around, being shared and exchanged through the collisions in a thermal material. Thermal equilibrium results from equipartition -- each place that can hold energy holding, on the average (though fluctuating), the same amount of energy. The result is like a chemical equilibrium. Things are continually happening, but if there's the same amount of energy flowing in each direction the result looks static. The key then, is where are the places we can put energy? These are called degrees of freedom.

It's a little tricky to think about this because energy is hard to think about. That's because energy is a rather abstract concept. We don't have concrete direct experience with the concept of energy; it's really only something that we calculate. There are a lot of different forms -- ways of calculating an energy -- and the total energy is just the sum of these terms. (An very nice description of this is given by Richard Feynman in The Feynman Lectures, Vol. 1, Section 4-1. This is highly recommended.)

When we have a cart rolling along a track, it can have the kinetic energy of its motion along the track, $½mv^2$, and its potential energy, $mgh$. Such a system would have two kinds of energy that need to be calculated, and therefore, two degrees of freedom. Its total energy is given by

$$E = ½mv^2 + mgh$$

For more systems of many particles this gets more complicated. Let's do an example to show how this counting takes place and what it can tell you.

## Presenting a sample problem

Consider a gas. We'll consider a variety of models for that gas and see how the number of degrees of freedom change, and what that tells us about the specific heat of the gas. Consider a dilute gas consisting of $N$ molecules moving freely in 3D in a small container. Since the molecules are very far apart, they rarely interact so we will ignore the potential energy of their interactions. Since the molecules are moving very fast, they have little time to fall between collisions so we will ignore gravitational potential energy.

A. Assume that we can treat the gas molecules as if they are point particles with no dimensions or structure. How many degrees of freedom are there? What would be the total energy of this gas if it were in thermal equilibrium?

B. Now consider a gas that consists of diatomic molecules that can be treated as if they are ball-and-stick dimers. How many degrees of freedom are there? What would be the total energy of this gas if it were in thermal equilibrium?

C. Finally, suppose that we have a gas consisting of diatomic molecules are not rigid but can vibrate — but only along the line joining the two atoms. How many degrees of freedom are there? What would be the total energy of this gas if it were in thermal equilibrium?

D. For the above three cases, what would happen to the temperature of the three different gases if we added a thermal energy $Q$ to it?

## Solving this problem

A. If the gas only consists of moving point particles, each particle can have kinetic energy. But since the particles can have independent motions in each of three directions, we have three degrees of freedom — three numbers are needed to specify the actual motion of each particle: their x, y, and z velocities. The total energy for one particle is $KE = ½mv_x^2 + ½mv_y^2 + ½mv_z^2$. That makes 3 degrees of freedom per particle. If there are $N$ particles, there will be $3N$ degrees of freedom. In equilibrium, the average energy in each degree of freedom is $\frac{1}{2}k_BT$, so the total energy in the gas will be

$$E = \frac{3}{2} Nk_BT$$

B. A ball-and-stick dimer can not only move independently in three different directions, as shown in the figure in part A, it can rotate about its various axes as shown at the right. This means that we now have 6 independent degrees of freedom for each molecule. If there are $N$ particles, there will be $6N$ degrees of freedom. In equilibrium, the average energy in each degree of freedom is $\frac{1}{2}k_BT$, so the total energy in the gas will be

$$E = \frac{6}{2}Nk_BT = 3Nk_BT$$

C. In our third model, we allow it to vibrate but only up and down along the line joining the spring, as shown in the figure at the right. Since a spring has both kinetic and potential energy, this adds two degrees of freedom to what we had above. If there are $N$ particles, there will be $8N$ degrees of freedom. In equilibrium, the average energy in each degree of freedom is $\frac{1}{2}k_BT$, so the total energy in the gas will be

$$E = \frac{8}{2}Nk_BT = 4Nk_BT$$

D. This example shows clearly that each material translates energy into temperature in its own way. In the first case, total energy and the temperature of the gas are related by $E = \frac{3}{2}k_BT$.  If we added heat $Q$ to the total energy we would now have a new temperature, $T_1$ given by

$$E + Q = \frac{3}{2}Nk_BT_1$$

Solving for $T_1$ (and putting in the expression for $E$ in terms of the original temperature $T$) we get

$$\frac{3}{2}Nk_BT_1 = E + Q$$

so

$$T_1 = \frac{E + Q}{\frac{3}{2}Nk_B} = \frac{\frac{3}{2}Nk_BT+ Q}{\frac{3}{2}Nk_B} = T + \frac{2Q}{3Nk_B}$$

giving

$$ΔT = T_1 - T = \frac{2Q}{3Nk_B}$$

Since we have defined the heat capacity by $Q = CΔT$, or equivalently, $ΔT = Q/C$, this result tells us that for the gas in case A

$$C_A = \frac{3}{2}Nk_B$$

Similarly, for case B we get $ΔT = \frac{2Q}{6Nk_B}$ and so

$$C_B = \frac{6}{2}Nk_B = 3Nk_B$$

For case C we get $ΔT = \frac{2Q}{8Nk_B}$ so

$$C_C = \frac{8}{2}Nk_B = 4Nk_B$$

Different kinds of materials change their temperatures by different amounts — by less, the more places (degrees of freedom) there are to put the energy.

Interestingly, this is not the whole story. Although it is true that the heat capacity of a dilute inert gas is indeed measured to be $C = \frac{3}{2}Nk_B$, it seems a bit strange. Can't a spherical atom also rotate about three axes? What's worse is that molecules that are well approximated as rigid dimers don't have the expected heat capacity of $C = \frac{6}{2}Nk_B$ but only $\frac{5}{2}Nk_B$. It seems like one degree of freedom is missing! And if I have a dimer, why shouldn't it have the vibration degrees of freedom as well?

The answer turns out to be that we are dealing with atoms and they are not classical particles; they obey the laws of quantum mechanics. These tell us two things: first, that in quantum mechanics, rotating a symmetric object like a sphere is not a physical change. It cannot take any energy. So the rotation of the atoms in case A don't count and the rotation about the y axis in case B doesn't count! Furthermore, as we will see in the reading Quantum Oscillator: Discrete States, vibrations have discrete energy levels. If the temperature is too low, there isn't enough energy to excite the vibration to the first level. This predicts that a dimer gas will have a heat capacity $\frac{5}{2}Nk_B$ at low temperatures and $\frac{7}{2}Nk_B$ for high enough temperatures! All of these properties are in fact observed.

This shows the interesting result that some of the weird predictions of quantum mechanics can be confirmed through classical macroscopic measurements!

Joe Redish 1/11/16

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