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This might be more of an applied math question. Why is the energy of a system typically able to be described using quadratic expressions. Is there an underlying mechanic that drives this?

Qmechanic
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a_hipster_peter_pan
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    Related: http://physics.stackexchange.com/q/159021/2451 , http://physics.stackexchange.com/q/78500/2451 and links therein. – Qmechanic Apr 25 '15 at 18:56

1 Answers1

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For most systems, if you are operating near equilibrium you are at a point where the net force is zero. That means that for small displacements, there will be a small force proportional to the displacement which restores the system to its equilibrium position (Taylor expansion - for small displacements, only first order effects matter).

$$F(x + dx) = F(x) + dx\cdot F'(x) + O(dx^2)$$

And if force is linear with displacement, then energy (the integral of force times displacement) goes with displacement squared.

It follows that for systems near equilibrium, the potential well has a quadratic shape (and the system behaves as a simple harmonic oscillator).

Note - I did state "most systems". The above is not always true, as was pointed out in the comment by Fernando Randisi. In some systems there may not be a linear component of the force, and the first coefficient might be cubic. Such things occur in certain nonlinear crystals. It is not the norm for most mechanical systems.

Floris
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    IMHO this answer is very good, but I would also mention that not in all system the potential near equilibrium can be approximated by a quadratic well: for example, if in your system F'(x) = 0, then your system will be described by something that goes as a higher power of x, for example $dx^3$ or even $dx^4$. This happens, for example, in some crystals that are used in non-linear optics. An example of a force that does not have a quadratic expansion next to equilibrium is something that goes as $F = -k, dx^3$. – Ferdinando Randisi Apr 25 '15 at 18:32