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9.1: Bond springs

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    24135
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    A ubiquitous spring is the bond between the electron and proton in hydrogen—the bond that is our model for all chemical bonds. In Section 9.1.1, we’ll build a spring model of hydrogen, giving us a physical model for the Young’s modulus (Section 9.1.2) and for the speed of sound (Section 9.1.3).

    9.1.1 Finding the spring

    In Section 8.3.2.2, we saw how hydrogen is a competition between electrostatics and quantum mechanics. When the electron–proton separation x is much smaller than the Bohr radius a0, quantum mechanics wins, and the net force on the electron is outward (positive). When the separation is much larger than the Bohr radius, electrostatics wins, and the net force is inward (negative). Between these two extremes, when the separation is the Bohr radius (when \(x = a_{0}\)), the force crosses the zero-force line (the x axis).

    This equilibrium point is an easy case that helps us describe the bond force simply. We magnify the force curve at the equilibrium point. The curve now looks straight, because any curve looks straight at a large-enough magnification. Equivalently, any curve can be approximated locally by its tangent line—which is an example of lumping shapes and graphs (Section 6.4) and is where spring models discard actual information and complexity.

    clipboard_ec9daa5fc954807bda68e1b66c845eeed.png

    Physically, the straight-line approximation means that, as long as the bond distance differs from a0 by only a small amount \(\Delta x\), the force is linearly proportional to the deviation \(\Delta x\). Furthermore, the force curve has a negative slope: A negative deviation produces a positive force, and vice versa. The force therefore opposes the deviation and is a restoring force. A linear restoring force is the force from an ideal spring, so the electron–proton bond is an ideal spring! It has an equilibrium length \(a_0\) at which \(F = 0\) and spring constant \(k\), where \(−k\) is the slope of the force curve.

    \[F = -k \Delta x.\]

    To make this small \(\Delta x\) stretch, the required energy \(\Delta E\) is

    \[\Delta E \sim \textrm{force} \times \Delta x\]

    Because the force varies from 0 to \(k \Delta x\), the typical or characteristic force is comparable to \(k \Delta x\). Then

    \[\Delta E \sim k \Delta x \cdot \Delta x = k (\Delta x)^{2}.\]

    The scaling exponent connecting \(\Delta E\) and \(\Delta x\) is 2. This quadratic dependence on the displacement is the energy signature of a spring. For small displacements around the equilibrium point (the minimum), the energy-versus-displacement curve is a parabola. (This analysis is a physical version of a Taylor-series approximation.)

    Because almost any energy curve has a minimum, almost every system contains a spring. For the bond spring, the energy relation \(\Delta E \sim k (\Delta x)^{2}\) gives us an estimate of the spring constant k. Pretend that the energy curve is exactly a parabola, even for large displacements, and increase \(\Delta x\) to the bond size a (for hydrogen, a is the Bohr radius a0). That stretch requires an energy \(\Delta E \sim k a^{2}\). It is a characteristic energy of the bond, so it must be comparable to the bond energy E0. Then \(E_{0} \sim ka^{2}\), and

    \[k \sim \frac{E_{0}}{a^{2}}.\]

    This relation is an estimate for the spring constant in terms of quantities that we know in other ways: E0 from the heat of vaporization and a from the density and atomic mass. The estimate is often off by a factor of 3 or 10, because of the inaccuracy in extending the parabolic, ideal-spring approximation to displacements comparable to the bond length. But it gives us an order of magnitude that will be useful in subsequent estimates.

    9.1.2 Young’s Modulus

    From the spring constant of one bond, we could find the spring constant of a block of material. But as we discussed in Section 5.5.4, a better measure is the Young’s modulus Y: It is an intensive quantity, so not dependent on the block’s dimensions. The Young’s modulus is measured by stretching a block of material with a force F at each end.

    clipboard_e58a8f733a12506f3d5617561b68ba4f3.png

    \[Y \equiv \frac{\textrm{stress}}{\textrm{strain}}.\]

    The stress is F/A, where A is the block’s cross-sectional area. Estimating the strain requires more steps. Fortunately, the estimate breaks into a tree. To grow it, imagine the block as a bundle of fibers, each a chain of springs (bonds) and masses (atoms). Because strain is the fractional length change, the strain in the block is the strain in each fiber and the strain in each spring of each fiber:

    clipboard_e7e4ff66bc37a1b3426b9fa117d7058f9.png

    \[\textrm{strain} = \frac{\textrm{spring extension}}{\textrm{bond length } a}.\]

    That’s the root of the tree. Here are the internal nodes:

    \[\textrm{spring extension} = \frac{\textrm{force/fiber}}{\textrm{spring constant } k};\]

    \[\frac{\textrm{force}}{\textrm{fiber}} = \frac{F}{N_{fibers}}.\]

    The number of fibers will be

    \[N_{fibers} = \frac{\textrm{cross-sectional area } A}{\textrm{per-fiber cross-sectional area } a^{2}}.\]

    Here is how the leaf values propagate to the root.

    1. The force per fiber becomes \(Fa^{2}/A.\)
    2. The per-spring extension becomes \(Fa^{2}/kA\).
    3. The strain becomes \(Fa/kA\).

    Finally, the Young's modulus becomes \(k/a\):

    \[Y \equiv \frac{\textrm{stress}}{\textrm{strain}} = \frac{F/A}{Fa/kA} = \frac{k}{a}.\]

    Because \(k \sim E_{0}/a^{2}\) (Section 9.1.1), \(Y \sim E_{0}/a^{3}\), which confirms with a spring model our dimensional-analysis prediction in Section 5.5.4.

    clipboard_e772559118865755dc621360cfaf6a32c.png

    9.1.3 Sound speed in solids and liquids

    The spring model of solids and even liquids also gives a physical model for the speed of sound. Start with a fiber of atoms and bonds:

    clipboard_e723d1f34a9424fc8781e1911e64a4784.png

    The sound speed is the speed at which a vibration signal travels along the fiber. Let’s study the simplest lumped signal: In one instant, the first mass moves to the right by a distance \(x\), the signal amplitude.

    clipboard_e5d90fcdf8182b47c37db4a380ad7142f.png

    The compressed bond pushes the second atom to the right. When the second atom has moved to the right by the signal amplitude \(x\), the signal has traveled one bond length. The sound speed is the propagation distance \(a_{bond}\) divided by this one-bond propagation time.

    clipboard_ec8d719cbc4e03741ac0ef64171c9c666.png

    What is the propagation time? That is, roughly how much time does the second mass require to move to the right by the signal amplitude \(x\)?

    The second mass moves because of the spring force from the first spring. At t = 0, the spring force is kx. As the mass moves, the spring loses compression, and the force on the second mass falls. So an exact calculation of the propagation time requires solving a differential equation. But lumping will turn the calculation into algebra: Just replace the changing spring force with its typical or characteristic value, which is comparable to \(kx\).

    The force produces a typical acceleration \(a \sim kx/m\). After a time t, the mass will have a velocity comparable to \(at\) and therefore will have moved a distance comparable to \(at^{2}\). This force acts for long enough to move the mass by a distance x, so \(at^{2} \sim x\). Using \(a \sim kx/m\) gives \(kxt^{2}/m \sim x\). The amplitude x cancels, as it always does in ideal-spring motion. (Thus, the sound speed does not depend on loudness.) The propagation time is then

    \[t \sim \sqrt{\frac{m}{k}}.\]

    This characteristic time is just the reciprocal of the natural frequency \(\omega_{0} = k/m\). In this time, the signal travels a distance \(a_{bond}\), so

    \[c_{s} \sim \frac{\textrm{distance traveled}}{\textrm{propagation time}} \sim \frac{a_{bond}}{\sqrt{m/k}} = \sqrt{\frac{ka_{bond}^{2}}{m}}.\]

    To make this expression more meaningful, let’s convert the numerator and denominator, which are in terms of microscopic (atomic) quantities, into macroscopic quantities. To do so, divide by \(a_{bond}^{3}/a_{bond}^{3}\) bond in the square root:

    \[c_{s} \sim \sqrt{\frac{ka_{bond}^{2}/a_{bond}^{3}}{m/a^{3}_{bond}}} = \sqrt{\frac{k/a_{bond}}{m/a^{3}_{bond}}}.\]

    The numerator \(k/a_{bond}\) is, as we saw in Section 9.1.2, the Young’s modulus Y. The denominator is the mass per molecular volume, so it is the substance’s density \(\rho\). Thus, \(c_{s} \sim Y/\rho\). Our physical, spring model therefore confirms our estimate for cs in Section 5.5.4 based on dimensions analysis.


    This page titled 9.1: Bond springs is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Sanjoy Mahajan (MIT OpenCourseWare) .

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