Where are the maxima and minima? Optimization in one variable.
Two argon atoms. At what distance is the energy minimized?
$$V(r) = 4\varepsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6} \right]$$
You could plot it and eyeball the minimum. But chemistry demands precision.
The derivative tells you exactly where. At a minimum, the function stops decreasing and starts increasing. At that instant, the rate of change is zero: V'(r) = 0.
| Problem | What You're Optimizing |
|---|---|
| Equilibrium bond length | Minimize potential energy |
| Transition state | Saddle point (max in one direction) |
| Equilibrium constant | Minimize Gibbs free energy |
| Most probable speed | Maximize Maxwell-Boltzmann distribution |
| Crystal structure | Minimize lattice energy |
| Protein folding | Minimize free energy |
Finding extrema is how nature finds equilibrium.
Critical point: A value x = c where f'(c) = 0 (horizontal tangent) or f'(c) does not exist (corner, cusp).
Critical points are candidates for extrema, but not every critical point is an extremum.
Local maximum: f(c) ≥ f(x) for all x near c.
Global maximum: f(c) ≥ f(x) for all x in the entire domain.
Key insight: Every global extremum is a local extremum, but local extrema are not necessarily global. A critical point may not be an extremum at all (inflection point).
Let c be a critical point where f is continuous.
Why it works: f' > 0 means increasing, f' < 0 means decreasing. If f goes from increasing to decreasing, it peaked (maximum). If from decreasing to increasing, it bottomed out (minimum).
Let f'(c) = 0 and f''(c) exist.
Why it works: f'' > 0 means the curve bends upward (like a cup) — minimum. f'' < 0 means it bends downward (like a cap) — maximum.
Setting V'(r) = 0:
$$\frac{12\sigma^{12}}{r^{13}} = \frac{6\sigma^{6}}{r^{7}} \implies r^6 = 2\sigma^6 \implies r_{eq} = 2^{1/6}\sigma \approx 1.122\sigma$$
At the minimum: V(req) = −ε
The distribution: f(v) ∝ v²e−av² where a = m/(2kBT)
Setting d/dv[v²e−av²] = 0:
$$2ve^{-av^2}(1 - av^2) = 0 \implies v_{mp} = \sqrt{\frac{1}{a}} = \sqrt{\frac{2RT}{M}}$$
If f is continuous on [a, b], then f attains both a global maximum and minimum on [a, b].
| Test | Condition | Result |
|---|---|---|
| First Derivative | f' changes + to − | Local maximum |
| First Derivative | f' changes − to + | Local minimum |
| First Derivative | f' no sign change | Not an extremum |
| Second Derivative | f''(c) > 0 | Local minimum |
| Second Derivative | f''(c) < 0 | Local maximum |
| Second Derivative | f''(c) = 0 | Inconclusive |