Chemistry - Does a diatomic gas have one or two vibrational degrees of freedom?
Solution 1:
I think this confusion arises from quite different definitions of degrees of freedom. Here are three different definitions that could reasonably arise:
- Positional degrees of freedom. This leads to the common $3N$ degrees of freedom based on movement in three directions. Refer to Jan's answer. (Question: Bonds constrain atoms, so why don't they reduce the number of degrees of freedom? Answer: Bonds are not rigid constraints; the atoms don't necessarily remain a fixed distance apart.)
- Phase space degrees of freedom. Here we have $6N$ degrees of freedom: $3N$ from the positions of all particles, $3N$ from the velocities of all particles. Positions and velocities are independent, so to specify a system of $N$ particles exactly requires $6$ degrees of freedom per particle. (We also presume no internal structure; as example, for rigid bodies, there are $12$ degrees of freedom: the $6$ mentioned above, $3$ more for specifiying the orientation of the body, $3$ more for describing the speed of rotation of the body.)
- Quadratic degrees of freedom. This may seem like a strange distinction to make, but is a key requirement of the equipartition theorem, which is one of the areas where the concept of degrees of freedom is first introduced. These degrees of freedom are the subset of the phase space degrees of freedom whose elements all contribute quadratically to the system energy.
We will clarify these degrees of freedom with the example of a diatomic gas. In this case, there are $6$ positional degrees of freedom, $12$ phase space degrees of freedom, and $7$ quadratic degrees of freedom. Jan's answer describes the $6$ positional degrees of freedom well; specifying the $6$ analogous velocity degrees of freedom gives the $12$ phase space degrees of freedom. Let us enumerate the $7$ quadratic degrees of freedom: $$\dot{x}, \dot{y}, \dot{z}, \dot{\theta}, \dot{\phi}, q, \dot{q} \quad \Longleftrightarrow \quad \frac{1}{2}m\dot{x}^2, \frac{1}{2}m\dot{y}^2, \frac{1}{2}m\dot{z}^2, \frac{1}{2}I\dot{\theta}^2, \frac{1}{2}I\dot{\phi}^2, \frac{1}{2}kq^2, \frac{1}{2}\mu\dot{q}^2$$ Positional degrees of freedom do not play into this, because there is no quadratic energy contribution associated with them. Instead, we have the six velocity degrees of freedom from the two atoms, $\{\dot{x}_1, \dot{y}_1, \dot{z}_1, \dot{x}_2, \dot{y}_2, \dot{z}_2\}$, which we replace by the conventional center-of-mass coordinates $x, y, z$ and the two angles $\theta, \phi$ that describe the orientation of the molecule, $\{\dot{x}, \dot{y}, \dot{z}, \dot{\theta}, \dot{\phi}\}$. One degree of freedom has "gone missing", because by symmetry there is no (quadratic) energy associated with rotation about the bond axis. The bond also produces two more quadratic degrees of freedom, represented by the vibrational coordinate along the bond, $q$, corresponding to the potential and kinetic energies of vibration.
Hence in one sense there exists only one vibrational degree of freedom; in another sense, two.
Remark. One might argue that the two vibrational degrees of freedom $q, \dot{q}$ are not independent, as they should be related by the equation of motion for a harmonic oscillator; this is important for the statement of the equipartition theorem. This is true for an isolated system, but presumably collisions decorrelate $q$ and $\dot{q}$---I don't have a better answer for this.
At any rate, experimental evidence (heat capacities) does corroborate seven quadratic degrees of freedom for diatomic molecules at high enough temperatures, when vibrational modes are active.
Solution 2:
You can think of the degrees of freedom with respect to rotation, translation and vibration in the following way:
Each atom can, in theory, move in all three orthogonal directions of your standard coordinate system ($x, y, z$). Depending on which coordinate system you use (e.g. polar coordinates would be a valid alternative) you may have different ‘directions’ but the core principle remains: each atom has three ‘ways to go’. Therefore, the entire molecule has $3N = 6$ degrees of freedom.
Next we can try to analyse how this molecule can exercise its freedom. For example, there are three different ways in which both atoms move in the same direction at the same speed. These are the degrees of translation, and there are always three ($x, y, z$ or whichever coordinates you chose).
We might also assume that one molecule travels in $x$ and $-z$ direction while the other one travels in $-x$ and $z$ direction; this would be a rotation around the $y$ axis. The same with reverse coordinates can be constructed around the $x$ axis (assuming the $z$ axis coincides with the bond direction). These are the only two rotational modes, any other attempted rotational mode is actually non-movement (around the $z$ axis) or a combination of the former two. Thus, there are two rotational degrees of freedom ($R_x$ and $R_y$).
Finally, you can think of a degree of freedom in which one atom moves in $z$ direction while the other moves in $-z$ direction; this is the vibrational mode, characterised by the bond length being streched or shortened. There is only one possible vibrational mode $V_z$.
These degrees of freedom add up to $6$, so we have them all. Any other movement you can construct can be broken up into a linear combination of the previous six modes.
Now let’s assume for a second the statement of two vibrational modes were true. Let’s attempt to find the second vibrational mode by assuming it to be a flip-flop vibration (one atom moves in $x$ direction, the other remains still). Careful analysis will reveal that this hypothetical $V_x$ mode is actually a linear combination of $R_y + V_z + T_x$ — rotation around the $y$ axis (moving the $x$-displaced atom one way and the other one another way while moving them both closer to the centre in $z$ direction), $z$-directional vibration (countering the movement in $z$ direction giving an apparant movement in $+x$ for the $x$-displaced atom and $-x$ for the other) and $x$-directional translation (putting the non-displaced atom back on its original position).
If the reader wants to take up the exercise, they can attempt to find any other, potentially more complex ‘second vibrations’. They will find that all can be written as a linear combination of the six previously mentioned modes. Without formal proof, we can therefore reasonably assume that the statement ‘diatomic molecules have two degrees of vibrational freedom’ is false.
On the other hand, the reader may attempt the exercise of trying to describe any of the six modes presented before the horizontal line as a linear combination of any of the other five. I hope I am not spoiling the fun if I tell them they will fail, since these six modes are indeed orthogonal to each other.
Solution 3:
There are two ways to think about the degrees of freedom for a vibration. One of them is to think only about the number of vibrational modes. The number of vibrational modes for a linear molecule is given by the formula $$3N-5$$
Where N is the number of atoms. For a diatomic molecule, $N=2$. Hence, the number of vibrational modes is considered to be 1. This is why the degrees of freedom for a molecule would be considered to be 1.
Another way to think about the degrees of freedom Or they may be talking about the energy degrees of freedom associated with each of the vibrational modes. Vibrating atoms will have two forms of energy: potential and kinetic. The kinetic energy comes from the fact that vibrating atoms are moving closer and farther from each other. The potential term comes from the fact that the atoms are "bound" to each other; they will experience a force that depends on the positions of the atoms. If we treat the bond as a harmonic oscillator, the potential energy will be equal to
$$k(x-x_0)^2$$
Where $x_0$ is the equilibrium length of the bond, $k$ is force constant, and $x$ is the position. Hence, each vibrational mode will contribute two degrees of freedom. Therefore a diatomic molecule would have 2 energy degrees of freedom since it has one vibrational mode. A linear triatomic molecule would have 4 normal modes.
$$3N-5=3(3)-5=4$$
And it would have 8 energy degrees of freedom associated with it
However, the first interpretation is more common.
TLDR: Both of them are correct. It depends on what type of degree of freedom you are looking at (as answered by a-cyclohexane-molecule).
Solution 4:
Degrees of freedom are related to the quadratic terms in final energy expressions, but also are related to the number of variables needed to describe the movement of a system. To specify the movement you need 3N variables and to specify vibration (movement) you need 3N-5=1 (only one possible movement). But talking in equipartition language, it has 2 degrees of freedom. Why? Because the energy of a spring has the following expression:
$$\frac{1}{2}mv^2 + kx^2$$ so we have 2 degrees of freedom. It answers your question.
I will add something related to quantum chemistry.
The previous statements are derived from classical statistical mechanics (which result on equipartition principle). However, if the system is at low temperatures (below $T_{amb}$ for some diatomic molecules as $\ce{N2}$) vibrational modes aren't excited. In that condition the molecule only store energy in rotation and translation, but not in vibration. You can check it here. And classical mechanics fails to describe molecules.
The previous phenomena is explained by quantum chemistry. When can we safely neglect quantum effects? When $kT>>\Delta E_n$ being $\Delta E_n= E_n -E_{n-1}$ and being $E_i$ the energy of the $n^{th}$ state. A proof of this can be done by measuring heat capacity of gases.