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# Equipartition Theorem

The equipartition theorem, also known as the law of equipartition, equipartition of energy or simply equipartition, states that every degree of freedom that appears only quadratically in the total energy has an average energy of ½kBT in thermal equilibrium and contributes ½kB to the system's heat capacity. Here, kB is the Boltzmann constant, and T is the temperature in Kelvin.

### Introduction

First, the theorem is derived in its general form; then its application for monatomic diatomic and polyatomic gases is discussed. How this affects the heat capacity of gasses follows and, finally, the specific example for two different tri-atomic gasses is given. The internal energy of a monatomic ideal gas, containing, N particles, is 3/2 NkBT. This means that each particle possesses, on average, 3/2 kBT units of energy. Monatomic particles have only three translational degrees of freedom, corresponding to their motion in three dimensions. They possess no internal rotational or vibrational degrees of freedom. Thus, the mean energy per degree of freedom in a monatomic ideal gas is ½ kBT. In fact, this is a special case of a rather general result. In an attempt to prove this:

Suppose that the energy of a system is determined by some n, generalized coordinates by qk; and corresponding n generalized momenta by pk, so that

$E=E(q_1, q_2, ... q_n, p_1, p_2, ...p_n) \tag{1}$

Suppose, further, that (1) the total energy splits additively into the form,

$E=\epsilon_i(p_i)+ E'(q_1, q_2, ... q_n, p_1, p_2, ...p_n) \tag{2}$

where $$\epsilon_i$$ involves only one variable$$\_i$$ , and the remaining part  does not depend on $$p_i$$, and (2) the $$\epsilon_i$$ function is quadratic in $$p_i$$, so that

$\epsilon_i(p_i)=bp_i^2$

where $$b$$ is a constant.

The most common situation in which the above assumptions are valid is where $$p_i$$ is a momentum. This is because the kinetic energy is usually a quadratic function of each momentum component, whereas the potential energy does not involve the momenta at all. However, if a coordinate $$q_i$$ were to satisfy assumptions (1) and (2) then the theorem we are about to establish would hold just as well.

The mean value of $$\epsilon_i$$ is in thermal equilibrium if conditions (1) and (2) are satisfied. If the system is in equilibrium at absolute temperature: $$T=(k\beta)^{-1}$$, then it is distributed according to the Boltzmann distribution. In the classical approximation, the mean value of $$\epsilon_i$$ is expressed in terms of integrals over all phase-space:

$\bar{\epsilon_i}= \dfrac{ \int_{\infty}^{\infty} e^{-\beta E(q_1,... p_n} \epsilon_i dq_1 ... dp_n }} {\int_{\infty}^{\infty} e^{-\beta E(q_1,... p_n} dq_1 ... dp_n } dp_i}$

Condition (1) gives

$\text{equation}$

where use has been made of the multiplicative property of the exponential function, and where the last integrals in both the numerator and denominator extend over all variables and except . These integrals are equal and, thus, cancel. Hence,

$\bar{\epsilon_i}= \dfrac{ \int_{\infty}^{\infty} e^{-\beta \epsilon_i}\epsilon_i dp_i }{\int_{\infty}^{\infty} e^{-\beta \epsilon_i} dp_i}$

This expression can be simplified further since

$\int_{\infty}^{\infty} e^{-\beta \epsilon_i}\epsilon_i dp_i = \dfrac{\partial}{\partial \beta} \int_{\infty}^{\infty} e^{-\beta \epsilon_i}dp_i$

So

$\bar{\epsilon_i}=-\dfrac{\partial}{\partial \beta} \ln{ \left[ \int_{\infty}^{\infty} e^{-\beta \epsilon_i}dp_i \right]}$

According to condition (2), and if $$y=\sqrt{\beta} p_i$$,

$\text{equation}$

Note that the last term (integral on the right-hand side) does not depend on $$\beta$$ at all. It follows

$\bar{\epsilon_i}=- \dfrac{\partial}{\partial \beta} \left[ -\dfrac{1}{2} \ln{\beta} \right] = \dfrac{1}{2\beta}=\dfrac{1}{2}kT$

This is the famous equipartition theorem of classical physics. It states that the mean value of every independent quadratic term in the energy is equal to ½kBT. If all terms in the energy are quadratic, then the mean energy is spread equally over all degrees of freedom, from which comes the name “equipartition”.

### Degrees of Freedom

The law of equipartition of energy states that each quadratic term in the classical expression for the energy contributes ½kBT to the average energy. For instance, the motion of an atom has three degrees of freedom (number of ways of absorbing energy), corresponding to the x, y and z components of its momentum. Since these momenta appear quadratically in the kinetic energy, every atom has an average kinetic energy of 3/2kBT in thermal equilibrium.

The number of degrees of freedom of a polyatomic gas molecule is 3N where N is the number of atoms in the molecule. This is equal to number of coordinates for the system; e.g. for two atoms you would have x, y, z for each atom.

Simpler coordinate systems are available for connected atoms. Three coordinates are attributed to the position of the center of the mass of the molecule (translational). Two or three coordinates are necessary to describe the rotation of the molecule, depending on whether or not the molecule is linear (rotational). The rest of the motions are classified as vibrational degrees of freedom. These coordinates are related to the stretching and bending of the bonds and are 3N-5 for linear and 3N-6 for nonlinear molecules.

### Contribution to the heat capacity

Heat capacity at constant volume $$C_v$$,  is defined as

$C_v = \left(\dfrac{\partial U}{\partial T} \right)_v$

Equipartition theorem requires that each degree of freedom that appears only quadratically in the total energy has an average energy of ½kBT in thermal equilibrium and, thus, contributes ½kB to the system's heat capacity. Thus the three translational degrees of freedom each contribute ½R to (3/2 R). The contribution of rotational kinetic energy will be R for the linear, and 3/2R for the nonlinear molecules. For the vibration, an oscillator has quadratic kinetic and potential terms, making the contribution of each vibrational mode R. However, kBT has to be much greater than the spacing between the quantum energy levels. If this is not satisfied, the heat capacity will be reduced and which drop to zero at low temperatures. The corresponding degree of freedom is said to be frozen out; this is the situation for the vibrational degrees of freedom at room temperature and that is why the usual assumption is that they will not contribute.

### CO2 vs NO2

For comparing the molar heat capacities of nitrogen dioxide and carbon dioxide at constant volume (at room temperature), let us use the law of equipartition and assume the vibrations to be frozen out at room temperature. The predicted molar for the linear $$CO_2$$ (with three translational and two rotational degrees of freedom) is 5/2R (20.8 JK-1mol-1). The estimated molar for $$NO_2$$ (a bent molecule, with three translational and three rotational degrees of freedom) is 3R(25.0 JK-1mol-1) in which R is the universal gas constant (8.314 JK-1mol-1). The values are close to the experimental values (30.1 JK-1mol-1 for $$CO_2$$; 29.5 JK-1mol-1 for NO2) but, especially for $$CO_2$$, the deviation is significant. This suggests that, although not all vibrational degrees of freedom are available, they cannot be totally ignored. The bigger deviation in the prediction of molar heat capacities is probably due to the existence of the lower frequency-bending vibration in carbon dioxide.

### References

1.    Thermodynamics and Statistical Mechanics, Greiner W., Neise, Stocker, Springer, 2001
2.    Physical Chemistry, 5th ed., Atkins P., W. H. Freeman and Company, 1994.

### Contributors

• Zane Sterkewolf (UC Davis)

Viewing 2 of 2 comments: view all
Posted 11:54, 16 Dec 2013
Dr. Larsen, appears this page has broken equation inputs. The same issue while using Safari (8.0.2), Firefox (34.0.5). No question marks as I witnessed on Uncertainty Principle webpage until you corrected it.
Posted 14:13, 11 Feb 2015
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16:11, 11 Feb 2015

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This material is based upon work supported by the National Science Foundation under Grant Numbers 1246120, 1525057, and 1413739.