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I am currently studying the textbook Infrared and Raman Spectroscopy: Principles and Spectral Interpretation, second edition, by Peter J. Larkin. Section 10. Calculating the Vibrational Spectra of Molecules, says the following:

In the case of vibrational spectroscopy, the polyatomic molecule is considered to oscillate with a small amplitude about the equilibrium position and the PE expression is expanded in a Taylor series and takes the form: $$V = V_0 + \sum_{i = 1}^{3N} \left( \dfrac{\partial{V}}{\partial{q_i}} \right)_e d q_i + \dfrac{1}{2} \sum_{i = 1}^{3N}\sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e d q_i d q_j + \dots$$ The above expression is expressed in internal coordinates, $q_i$ and $q_j$ which are directly connected to the internal bond lengths and angles. The above expression is simplified since:

  1. The first term $V_0 = 0$ since the vibrational energy is chosen as vibrating atoms about the equilibrium position.
  2. At the minimum energy configuration the first derivative is zero of definition.
  3. Since the harmonic approximation is used all terms in the Taylor expansion greater than 2 can be neglected. This leaves only the second term in the PE expression for $V$. Using Newton's second law the above is expressed as $$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$

Even after taking the points above into account, I don't see how we get from the first equation to the second. I'm assuming that $V$ is the potential energy (PE), right? And Newton's second law in differential equation form is $F = m \dfrac{dV}{dt}$. So how does this and points 1., 2., and 3. then lead to $$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$?

I would greatly appreciate it if people would please take the time to explain this.

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1 Answer 1

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It's probably simplest to start with the familiar example of the 1D harmonic oscillator. Here

$$F=-\frac{dV}{dx}=-\frac{d\left(\frac{1}{2}kx^2\right)}{dx}=-kx\tag{1}$$

which is often called Hooke's law, $F=-kx$, where x is the displacement from the equilibrium position, and the potential is assumed quadratic (parabolic) in the position coordinate. The derivation above requires the substitution $V(x) = \frac12 kx^2$, which can be obtained by trimming a Taylor expansion of the potential about the energy minimum (assumed = 0) at the quadratic term, and the force constant is defined as

$$k=\left(\frac{d^2V}{dx^2} \right)_e\tag{2}$$

Also,

$$F=ma=m\frac{d^2x}{dt^2}\tag{3}$$

If we transform to mass-weighted coordinates (or equivalently set m=1), we therefore can write that

$$\frac{d^2x}{dt^2} = -\frac{dV}{dx}=-\left(\frac{d^2V}{dx^2} \right)_e x\tag{4}$$

The derivation in your book does essentially the same thing. Note that since the potential is assumed quadratic and the other listed conditions are imposed, the potential can be written as

$$V = \frac12 \sum_{i = 1}^{3N}\sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e q_i q_j \tag{5}$$

The double summation can be collapsed into a single one when you take the derivative with respect to a spatial coordinate, say $q_k$:

$$\begin{align} \dfrac{d^2 q_k}{d t^2} &=-\left( \dfrac{\partial{V}}{\partial{q_k}} \right) \\&=-\left(\dfrac{\partial{}}{\partial{q_k}} \left[ \frac12 q_k^2 \left( \dfrac{\partial^2{V}}{\partial{q_k^2}} \right)_e + \frac12 q_k\sum_{j \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k}\partial{q_j}} \right)_e q_j + \frac12 q_k \sum_{i \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_k}} \right)_e q_i + \frac12 \sum_{i \neq k }^{3N}\sum_{j \neq k}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e q_i q_j \right] \right) \\ &= -\left( \left[ \left( \dfrac{\partial^2{V}}{\partial{q_k^2}} \right)_e q_k + \frac12 \sum_{j \neq k}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k}\partial{q_j}} \right)_e q_j + \frac12 \sum_{i \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_k}} \right)_e q_i \right] \right) \\ &= - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k} \partial{q_j}} \right)_e q_j \end{align}$$

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  • $\begingroup$ Thanks for the answer. What are "mass-weighted coordinates"? And where did the $\dfrac{d^2 x}{dt^2}$ come from? $\endgroup$ Jul 14, 2020 at 15:49
  • $\begingroup$ From equation 3 by setting the mass=1 (more correctly by transforming to mass-weighted coordinates). The latter combine mass with the displacement coordinate (normal mode). The substitution removes explicit mass terms from the equations and makes things look a little tidier. See for instance also gaussian.com/wp-content/uploads/dl/vib.pdf $\endgroup$
    – Buck Thorn
    Jul 14, 2020 at 15:54
  • $\begingroup$ Hmm, I see. What happened to the sum for $i$ in $$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$? It only has the sum for $j$. $\endgroup$ Jul 14, 2020 at 16:09
  • $\begingroup$ I agree it's a tad unclear, I think it would be better to take the derivative wrt say $q_k$, but in the end if you do that you can swap labels $k\rightarrow i$ and you obtain the same result. $\endgroup$
    – Buck Thorn
    Jul 14, 2020 at 16:19
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    $\begingroup$ Ahh, yes, it is now all clear to me. Thank you very much for taking the time to explain this to me. $\endgroup$ Jul 15, 2020 at 10:27

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