Carbon dioxide

Classical Mechanics Level 5

A C O X 2 \ce{CO_2} molecule consists of two oxygen atoms of mass m 1 = m 3 ≈ 16 u m_1 = m_3 \approx 16 \, \text{u} and a carbon atom of mass m 2 ≈ 12 u m_2 \approx 12 \, \text{u} . In order to model the electrostatic forces between the atoms, we imagine elastic springs with the force constant k k between the oxygen and the carbon. In equilibrium, the atoms each have a distance of d d from each other. The forces acting on the individual atoms result in F 1 = m 1 x ¨ 1 = − k ( x 1 − x 2 + d ) F 2 = m 2 x ¨ 2 = − k ( x 2 − x 1 − d ) − k ( x 2 − x 3 + d ) F 3 = m 3 x ¨ 3 = − k ( x 3 − x 2 − d ) \begin{aligned} F_1 = m_1 \ddot x_1 &= -k(x_1 - x_2 + d) \\ F_2 = m_2 \ddot x_2 &= -k(x_2 - x_1 - d) - k(x_2 - x_3 + d)\\ F_3 = m_3 \ddot x_3 &= -k(x_3 - x_2 - d) \end{aligned} along the x x -axis. We assume harmonic oscillations of the atoms: x 1 = ξ 1 e i ω t x 2 = ξ 2 e i ω t + d x 3 = ξ 3 e i ω t + 2 d \begin{aligned} x_1 &= \xi_1 e^{i\omega t} \\ x_2 &= \xi_2 e^{i\omega t} + d\\ x_3 &= \xi_3 e^{i\omega t} + 2d \end{aligned} with amplitudes ξ n \xi_n and frequency ω \omega . Show that there are only two possible solutions ω symmetric \omega_\text{symmetric} and ω asymmetric \omega_\text{asymmetric} for the frequency, that correspond to symmetric and asymmetric stretching vibrations, respectively.

What is the ratio α = ω asymmetric ω symmetric \displaystyle \alpha = \frac{\omega_\text{asymmetric}}{\omega_\text{symmetric}} of the frequencies?


Bonus question: What is the role of these vibrations for the greenhouse effect of carbon dioxide?


The answer is 1.915.

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Oct 17, 2017

Substituting x n = ξ n e i ω t + ( n − 1 ) d x_n = \xi_n e^{i \omega t} + (n-1)d in the equations of motions results F 1 = − m 1 ω 2 ξ 1 e i ω t = − k ( ξ 1 − ξ 2 ) e i ω t F 2 = − m 2 ω 2 ξ 2 e i ω t = − k ( 2 ξ 2 − ξ 1 − ξ 3 ) e i ω t F 3 = − m 1 ω 2 ξ 3 e i ω t = − k ( ξ 3 − ξ 2 ) e i ω t \begin{aligned} F_1 = - m_1 \omega^2 \xi_1 e^{i \omega t} &= -k(\xi_1 - \xi_2)e^{i \omega t} \\ F_2 = - m_2 \omega^2 \xi_2 e^{i \omega t} &= -k(2 \xi_2 - \xi_1 - \xi_3)e^{i \omega t}\\ F_3 = - m_1 \omega^2 \xi_3 e^{i \omega t} &= -k(\xi_3 - \xi_2) e^{i \omega t} \end{aligned} This is set of linear equations for the displacement vector ( ξ 1 , ξ 2 , ξ 3 ) (\xi_1, \xi_2,\xi_3) . After dividing by e i ω t e^{i \omega t} and putting all terms to one side of the equation we get a homogeneous matrix equation ( k m 1 − ω 2 − k m 1 0 − k m 2 2 k m 2 − ω 2 − k m 2 0 − k m 1 k m 1 − ω 2 ) ( ξ 1 ξ 2 ξ 3 ) = 0 \left( \begin{array}{ccc} \frac{k}{m_1} - \omega^2 & -\frac{k}{m_1} & 0 \\ -\frac{k}{m_2} & \frac{2k}{m_2} - \omega^2 & -\frac{k}{m_2} \\ 0 & -\frac{k}{m_1} & \frac{k}{m_1} - \omega^2 \end{array}\right) \left( \begin{array}{c} \xi_1 \\ \xi_2 \\ \xi_3 \end{array} \right) = 0 This equation has a nontrivial solution ξ n ≠ 0 \xi_n \not= 0 , only if the determinant of the matrix equals zero: 0 = ∣ k m 1 − ω 2 − k m 1 0 − k m 2 2 k m 2 − ω 2 − k m 2 0 − k m 1 k m 1 − ω 2 ∣ = ( k m 1 − ω 2 ) 2 ( 2 k m 2 − ω 2 ) − 2 k 2 m 1 m 2 ( k m 1 − ω 2 ) = ( k m 1 − ω 2 ) [ ( k m 1 − ω 2 ) ( 2 k m 2 − ω 2 ) − 2 k 2 m 1 m 2 ] = ( k m 1 − ω 2 ) ( ω 2 − ( k m 1 + 2 k m 2 ) ) ω 2 \begin{aligned} 0 &= \left| \begin{array}{ccc} \frac{k}{m_1} - \omega^2 & -\frac{k}{m_1} & 0 \\ -\frac{k}{m_2} & \frac{2k}{m_2} - \omega^2 & -\frac{k}{m_2} \\ 0 & -\frac{k}{m_1} & \frac{k}{m_1} - \omega^2 \end{array}\right| \\ &= \left(\frac{k}{m_1} - \omega^2\right)^2 \left(\frac{2k}{m_2} - \omega^2\right) - 2 \frac{k^2}{m_1 m_2} \left(\frac{k}{m_1} - \omega^2\right) \\ &= \left(\frac{k}{m_1} - \omega^2\right) \left[ \left(\frac{k}{m_1} - \omega^2\right) \left(\frac{2k}{m_2} - \omega^2\right) -2 \frac{k^2}{m_1 m_2} \right]\\ &= \left(\frac{k}{m_1} - \omega^2\right) \left( \omega^2 - \left(\frac{k}{m_1} + \frac{2k}{m_2}\right) \right) \omega^2 \end{aligned} The solution ω = 0 \omega = 0 corresponds to a rigid movement of the whole molecule. Therefore, there are only to relevant zeros, that corresponds to natural frequencies of the molecule ω 1 = k m 1 , ω 2 = k m 1 + 2 k m 2 \omega_1 = \sqrt{\frac{k}{m_1}}, \quad \omega_2 = \sqrt{\frac{k}{m_1} + \frac{2k}{m_2}} Substituting ω = ω 1 \omega = \omega_1 in the matrix equation, we get ( 0 − k m 1 0 − k m 2 2 k m 2 − k m 1 − k m 2 0 − k m 1 0 ) ( ξ 1 ξ 2 ξ 3 ) = 0 ⇒ ( ξ 1 ξ 2 ξ 3 ) = ( ξ 1 0 − ξ 1 ) \left( \begin{array}{ccc} 0 & -\frac{k}{m_1} & 0 \\ -\frac{k}{m_2} & \frac{2k}{m_2} - \frac{k}{m_1} & -\frac{k}{m_2} \\ 0 & -\frac{k}{m_1} & 0 \end{array}\right) \left( \begin{array}{c} \xi_1 \\ \xi_2 \\ \xi_3 \end{array} \right) = 0 \quad \Rightarrow \quad \left( \begin{array}{c} \xi_1 \\ \xi_2 \\ \xi_3 \end{array} \right) = \left( \begin{array}{c} \xi_1 \\ 0 \\ -\xi_1 \end{array} \right) Therefore, we have a symmetric vibration with $\omega 1 = \omega \text{symmetric}$. For the case ω = ω 2 = ω asymmetric \omega = \omega_2 = \omega_\text{asymmetric} it follows ( 2 k m 2 − k m 1 0 − k m 2 k m 1 − k m 2 0 − k m 1 2 k m 2 ) ( ξ 1 ξ 2 ξ 3 ) = 0 ⇒ ( ξ 1 ξ 2 ξ 3 ) = ( ξ 1 − 2 μ ξ 1 ξ 1 ) \left( \begin{array}{ccc} \frac{2 k}{m_2} & -\frac{k}{m_1} & 0 \\ -\frac{k}{m_2} & \frac{k}{m_1} & -\frac{k}{m_2} \\ 0 & -\frac{k}{m_1} & \frac{2 k}{m_2} \end{array}\right) \left( \begin{array}{c} \xi_1 \\ \xi_2 \\ \xi_3 \end{array} \right) = 0 \quad \Rightarrow \quad \left( \begin{array}{c} \xi_1 \\ \xi_2 \\ \xi_3 \end{array} \right) = \left( \begin{array}{c} \xi_1 \\ -2\mu \xi_1 \\ \xi_1 \end{array} \right) with μ = m 1 / m 2 \mu = m_1/m_2 , so that the results corresponds to the asymmetric vibration. Finally, the frequency ratio results α = ω 2 ω 1 = 1 + 2 m 1 m 2 ≈ 1 + 8 3 ≈ 1.915 \alpha = \frac{\omega_2}{\omega_1} = \sqrt{1 + \frac{2 m_1}{m_2}} \approx \sqrt{1 + \frac{8}{3}} \approx 1.915

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