# Degrees of freedom of a diatomic molecule

#### Pushoam

1. The problem statement, all variables and given/known data

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3. The attempt at a solution
Considering the molecule as collection of two spherical atoms whose centers are joined such that they touch each other,

Then, as a rigid body the system has 6 degrees of freedom and around the axis joining the two centers, the two atoms could rotate in the opposite directions and if the two atoms are not considered symmetrical, then this motion also adds one more degrees of freedom. So, the total degrees of freedom of the system is 7.
Is this correct?

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#### DrClaude

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Considering the molecule as collection of two spherical atoms whose centers are joined such that they touch each other,

Then, as a rigid body the system has 6 degrees of freedom and around the axis joining the two centers, the two atoms could rotate in the opposite directions and if the two atoms are not considered symmetrical, then this motion also adds one more degrees of freedom. So, the total degrees of freedom of the system is 7.
Is this correct?
No. You have two atoms, each with 3 degrees of freedom, so the total system must also have 6 degrees of freedom.

But you don't really need that to answer the question. Just think about rotations. How many rotational axes does a rigid body have? Is it the same for a diatomic molecule?

#### Pushoam

No. You have two atoms, each with 3 degrees of freedom, so the total system must also have 6 degrees of freedom.

But you don't really need that to answer the question. Just think about rotations. How many rotational axes does a rigid body have? Is it the same for a diatomic molecule?
Are you taking the two atoms as two point particles centered at their center of mass connected by a massless rod?

If it is the case, then it is just a rigid body. So, it has degrees of freedom = 6.

#### DrClaude

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Are you taking the two atoms as two point particles centered at their center of mass connected by a massless rod?

If it is the case, then it is just a rigid body. So, it has degrees of freedom = 6.
As vibrations are allowed, it is not exactly a rigid body. Indeed, the diatomic molecule has six degrees of freedom. The question is how many of these are rotational degrees of freedom.

#### BvU

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Are you taking the two atoms as two point particles centered at their center of mass connected by a massless rod?

If it is the case, then it is just a rigid body. So, it has degrees of freedom = 6.
No. the exercise statement explicitly excludes a rotation around the axis joining the two point masses. Two point masses at fixed distance have one constraint (that distance) so one degreee of freedom less.

#### Charles Link

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The reason why diatomic molecules have molar heat capacity $C_v=(\frac{5}{2}) R$, (and $C_p=(\frac{7}{2}) R$), is because of the 5 degrees of freedom ,with internal energy $U= (\frac{5}{2}) kT$ per molecule, and $U=(\frac{5}{2}) RT$ per mole of molecules , that results from 5 degrees of freedom of the diatomic molecule, when the vibrational degree of freedom is excluded. Most diatomic molecules remain in the vibrational ground state at ordinary temperatures (e.g. room temperature), so the vibrational degree of freedom does not contribute to the heat capacity.

#### BvU

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Adding to this thread: I was surprised to find that this vibration degree of freedom then ups the $c_v$ to ${7\over 2}R$ instead of $3R$ as I expected. Anybody comment ?

#### Charles Link

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Adding to this thread: I was surprised to find that this vibration degree of freedom then ups the $c_v$ to ${7\over 2}R$ instead of $3R$ as I expected. Anybody comment ?
I believe the answer is that there are initially 6 degrees of freedom, but it gets reduced to 5 if the vibration degree of freedom is removed. The vibration degree of freedom is part of the 6 degrees of freedom. $\\$ In order to incorporate the distance between the particles as a degree of freedom, I believe it requires an orthogonal set of transformation equations on the coordinates. $\\$ e.g. In one dimension, if you have particle positions $x_1$ and $x_2$, you can write $u_1=x_2-x_1$, and $u_2=x_1+x_2$. You can then work with either $x's$ or $u's$. In both cases, you have two degrees of freedom. $u_1$ is the vibration coordinate/degree of freedom. If $u_1$ is fixed, you are left with one degree of freedom. $\\$ Alternatively, in working with rotations as degrees of freedom for the two atom case, if there is a fixed distance between atoms, there are only two degrees of freedom when considering a rotation as a degree of freedom, because a rotation about an axis of the line connecting the two atoms has zero effect. There are two independent and perpendicular axes about which a rotation of significance can take place, accounting for the two degrees of freedom of rotation.

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#### Pushoam

If I see the diatomic molecule as a collection of two point atoms such that the distance between the atoms are fixed ( which is a constraint), then each atom has 3 degrees of freedom and so the degrees of freedom of the system is 3*2 -1 = 5. In this case, since the atoms themselves are point masses, I don’t understand what is meant by their vibration.

If I see the diatomic molecule as a collection of two point atoms joined by a massless line such that the two atoms along with the line could vibrate, then the center of mass has 3 coordinates, and the system could rotate about the axes perpendicular to the line giving two rotational coordinates and the given vibrational motion gives one vibrational coordinate. So, this way it has degrees of freedom = 6.

#### Charles Link

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The diatomic atom consists of two positive nuclei and a collection of electrons distributed in such a manner, that at least in a classical description, the forces balance and the nuclei are a fixed distance apart in the ground state. If the inter-nuclei distance is changed, there is, in the classical model, a restoring force proportional to the displacement from equilibrium. (See also post 8 for how this inter-nuclei distance can be defined as coordinate $u_1$). The vibration modes of the diatomic molecule consists of exciting the modes of oscillation of what is essentially very much like two masses (the nuclei) connected by a spring whose equilibrium length is the inter-nuclei distance. This simple harmonic oscillator system is often in the ground state at temperatures near room temperature, and the molecules don't go into the excited vibration states very often. Thereby this degree of freedom contributes little to the heat capacity, and the approximation can often be made that the molecule remains in the ground state for the vibration modes at a fixed inter-nuclei distance. $\\$ The rotation modes do, however, get excited from thermal excitation, because their excited states are individually of lower energy than the excited vibration states.

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#### DrClaude

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Adding to this thread: I was surprised to find that this vibration degree of freedom then ups the $c_v$ to ${7\over 2}R$ instead of $3R$ as I expected. Anybody comment ?
Vibrations are approximately harmonic oscillators, so they contribute two quadratic degrees of freedom (potential and kinetic energies).

#### DrClaude

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I think that @Pushoam needs to clarify the context of the question. I was reading it from the point of view of basic molecular physics, while @Charles Link approached it from the point of view of thermodynamics.

#### BvU

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Vibrations are approximately harmonic oscillators, so they contribute two quadratic degrees of freedom (potential and kinetic energies).
Is what I read in the link, which I don't contest, but don't understand either: I need some explanation. I see only 1 DOF there because a vibration state has a specific energy that continuously interchanges potential energy with kinetic energy, the sum of which is constant. Is it that an excitation is an even multiple of ${1\over 2} \hbar\omega$ and not just any multiple that corresponds to this increase of 2 in DOF ?

#### Charles Link

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Is what I read in the link, which I don't contest, but don't understand either: I need some explanation. I see only 1 DOF there because a vibration state has a specific energy that continuously interchanges potential energy with kinetic energy, the sum of which is constant. Is it that an excitation is an even multiple of ${1\over 2} \hbar\omega$ and not just any multiple that corresponds to this increase of 2 in DOF ?
This is interesting. I believe from a statistical physics viewpoint, the harmonic oscillator has a boson like character, but in the high temperature regime the Bose factor $\bar{n}_s=\frac{1}{e^{E/kT}-1}$ becomes $e^{-E/kT}$. (Editing: Not $e^{- E/kT}$ , but instead $\bar{n_s} \approx \frac{kT}{E}$, where $E=\hbar \omega_o$, so that $\bar{E}=\bar{n}_s \hbar \omega_o=kT$), so the result is the mean energy of a harmonic oscillator in the higher temperature regime is $\bar{E}=kT$, while the translational momentum states have mean energy $(\frac{1}{2})kT$ for each of 3 dimensions x,y, and z . For the translational states, the density of states per energy interval is proportional to $E^{1/2}$. [Edit:] For the harmonic oscillator, there is essentially one single mode that can contain multiple phonons, resulting in multiple vibration states that are possible.

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#### DrClaude

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Is what I read in the link, which I don't contest, but don't understand either: I need some explanation. I see only 1 DOF there because a vibration state has a specific energy that continuously interchanges potential energy with kinetic energy, the sum of which is constant. Is it that an excitation is an even multiple of ${1\over 2} \hbar\omega$ and not just any multiple that corresponds to this increase of 2 in DOF ?
I don't know the best way to explain it. The fact that the energy is written as the sum of terms in $p^2$ and $x^2$ is taken to represent two quadratic degrees of freedom, even though the two terms are related.

From the point of view of statistical physics, one can start with an energy of the form $E = c q^2$, where $q$ is a generic degree of freedom, and show that this leads to $\bar{E} = k T / 2$. Then taking $E_n = \hbar \omega (n + 1/2)$, we can calculate the partition function
$$Z = \sum_{n=0}^{\infty} e^{-\beta E_n} = \frac{e^{\frac{\beta \hbar \omega }{2}}}{e^{\beta \hbar \omega }-1}$$
then one gets the energy
$$\bar{E} = - \frac{\partial \ln Z}{\partial \beta} = \frac{\hbar \omega \left(e^{\beta \hbar \omega }+1\right)}{2 \left(e^{\beta \hbar \omega }-1\right)}$$
which, when Taylor expanded in the $\beta \rightarrow 0$ limit (high $T$), gives $\bar{E} = k T$. Comparing with the generic result above, one concludes that 2 such degrees of freedom must be present.

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#### vela

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Is what I read in the link, which I don't contest, but don't understand either: I need some explanation. I see only 1 DOF there because a vibration state has a specific energy that continuously interchanges potential energy with kinetic energy, the sum of which is constant. Is it that an excitation is an even multiple of ${1\over 2} \hbar\omega$ and not just any multiple that corresponds to this increase of 2 in DOF ?
The molecules have some distribution of oscillator energies, and there are two contributions—potential and kinetic energy—to the total energy of an oscillator. You don't want to start with the assumption the energy is fixed, which seemingly leaves you with only one degree of freedom. If you did the same with translational kinetic energy, you'd conclude there are only two degrees of freedom there because once you know two components of velocity, you have no choice for the third for a given amount of energy.

#### Charles Link

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The molecules have some distribution of oscillator energies, and there are two contributions—potential and kinetic energy—to the total energy of an oscillator. You don't want to start with the assumption the energy is fixed, which seemingly leaves you with only one degree of freedom. If you did the same with translational kinetic energy, you'd conclude there are only two degrees of freedom there because once you know two components of velocity, you have no choice for the third for a given amount of energy.
The harmonic oscillator seems to be something that doesn't fit this "standard mode of thinking", because if the energy of the oscillator is fixed, instead of having 2 degrees of freedom for the oscillator, it seems, at least from a thermodynamic (heat capacity) viewpoint, there are then zero degrees of freedom to be attributed to the harmonic oscillator. $\\$ Perhaps it is worthwhile to compare this case to that of a crystalline solid. The heat capacity of the solid (by Dulong and Petit) is approximately $3 R$ because of the 3 quadratic kinetic and the three quadratic potential energy terms, each contributing $(\frac{1}{2})RT$ to the energy. If you fix the internal energy of the crystalline solid, you can't really do anything to change its temperature, and the energies are divided equally among the six contributions. Calling it a "degree of freedom" and expecting it to behave in a precisely mathematical sense, (where applying one equation of constraint reduces the degrees of freedom by 1), is perhaps expecting too much from the designation "degree of freedom". $\\$ And for the OP @Pushoam : It is a very interesting question that you have, but it seems like there are at least a couple of answers to this question that have some degree of validity. [Edit:] The question still seems to persist: Do we count the vibration states as having one degree of freedom or two?

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#### Pushoam

I was studying degrees of freedom and I got this question as an exercise under this topic. So, @DrClaude it is asked neither from molecular physics point of view nor from the thermodynamics point of view.

And for the OP @Pushoam : It is a very interesting question that you have, but it seems like there are at least a couple of answers to this question that have some degree of validity. [Edit:] The question still seems to persist: Do we count the vibration states as having one degree of freedom or two?
This is how I see the answer:
If I see the diatomic molecule as a collection of two point atoms joined by a massless line such that the two atoms along with the line could vibrate, then the center of mass has 3 coordinates, and the system could rotate about the axes perpendicular to the line giving two rotational coordinates and the given vibrational motion gives one vibrational coordinate. So, this way it has degrees of freedom = 6.
I, too, have read in books that harmonic oscillator has two degrees of freedom from energy point of view as it is said above.
According to my understanding from the book, degrees of freedom = minimum no. of generalized co - ordinates needed to specify the system ..(1)

To specify the vibrational motion of diatomic molecule, we need only one generalized coordinate i.e. $u_1$ as given in post # 8. So, accordingly, degrees of freedom for vibrational mode should be one.

Could anyone please explain using (1), why degrees of freedom for vibration should be 2?

The book gives the answer as 6.

#### VISHAL RAJ

as the atoms are joined in a line so they are restricted to in that direction so one of its freedom is destroyed.

#### DrClaude

Mentor
I, too, have read in books that harmonic oscillator has two degrees of freedom from energy point of view as it is said above.
According to my understanding from the book, degrees of freedom = minimum no. of generalized co - ordinates needed to specify the system ..(1)

To specify the vibrational motion of diatomic molecule, we need only one generalized coordinate i.e. $u_1$ as given in post # 8. So, accordingly, degrees of freedom for vibrational mode should be one.

Could anyone please explain using (1), why degrees of freedom for vibration should be 2?
There are two conflicting uses of the phrase "degrees of freedom." From the point of view of mechanics (classical and quantum), a degree of freedom is
A concise dictionary of physics said:
The number of independent parameters required to specify the configuration of a system.
According to the answer from the book, this is what is intended here, and how I read the question initially. So there are 6 degrees of freedom for a diatomic molecule (3 for the position of the centre of mass, 2 for rotations, and 1 for vibration).

In the context of thermodynamics, the equipartition theorem states that the average energy of a thermodynamic system is given by
$$\bar{E} = \frac{f}{2} k T$$
with $f$ the number of quadratic degrees of freedom (see my post #15 above). The word "quadratic" is often omitted here, hence the confusion. There are 5 or 7 quadratic degrees of freedom for a diatomic molecule: at room temperature, 3 for translation and 2 for rotations; at higher temperatures, when there is non-negligible vibrational excitation, one needs to add 2 degrees of freedom for the vibration of the bond (counting for 2 due to its harmonic-oscillator-like behaviour).

#### Oxvillian

In the context of thermodynamics, the equipartition theorem states that the average energy of a thermodynamic system is given by
$$\bar{E} = \frac{f}{2} k T$$
with $f$ the number of quadratic degrees of freedom (see my post #15 above). The word "quadratic" is often omitted here, hence the confusion. There are 5 or 7 quadratic degrees of freedom for a diatomic molecule: at room temperature, 3 for translation and 2 for rotations; at higher temperatures, when there is non-negligible vibrational excitation, one needs to add 2 degrees of freedom for the vibration of the bond (counting for 2 due to its harmonic-oscillator-like behaviour).
I wouldn't call those degrees of freedom though. The correct statement of the equipartition theorem would be something along the lines that you get a $kT/2$ contribution to the heat capacity for every term that appears quadratically in the Hamiltonian, rather than for every degree of freedom.

#### DrClaude

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I wouldn't call those degrees of freedom though.
Sorry to be blunt, but what does it matter what you want to call them? "Degrees of freedom" is the accepted expression. You will find in textbooks on thermodynamics as well as in dictionaries.

#### Oxvillian

Ha well it's all just words of course

But in an effort to convince you, I put it to you that the number of degrees of freedom of a system should be a kinematical characterization, independent of dynamics, and therefore independent of how many quadratic terms there happen to be in the Hamiltonian. The "$kT/2$ per degree of freedom" idea is an oversimplification that's useful for students studying thermal physics before they've seen Hamiltonian mechanics, the virial theorem, etc.

#### Pushoam

There are two conflicting uses of the phrase "degrees of freedom." From the point of view of mechanics (classical and quantum), a degree of freedom is

According to the answer from the book, this is what is intended here, and how I read the question initially. So there are 6 degrees of freedom for a diatomic molecule (3 for the position of the centre of mass, 2 for rotations, and 1 for vibration).

In the context of thermodynamics, the equipartition theorem states that the average energy of a thermodynamic system is given by
$$\bar{E} = \frac{f}{2} k T$$
with $f$ the number of quadratic degrees of freedom (see my post #15 above). The word "quadratic" is often omitted here, hence the confusion. There are 5 or 7 quadratic degrees of freedom for a diatomic molecule: at room temperature, 3 for translation and 2 for rotations; at higher temperatures, when there is non-negligible vibrational excitation, one needs to add 2 degrees of freedom for the vibration of the bond (counting for 2 due to its harmonic-oscillator-like behaviour).
Thanks for clarifying it.
Thanks to all for replying.
There are two kinds of degrees of freedom.

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