Editing Legendre transformation (section)

===Variable capacitor===
As another example from [[physics]], consider a parallel conductive plate [[capacitor]], in which the plates can move relative to one another. Such a capacitor would allow transfer of the electric energy which is stored in the capacitor into external mechanical work, done by the [[force]] acting on the plates. One may think of the electric charge as analogous to the "charge" of a [[gas]] in a [[cylinder (engine)|cylinder]], with the resulting mechanical [[force]] exerted on a [[piston]].

Compute the force on the plates as a function of {{math|'''x'''}}, the distance which separates them. To find the force, compute the potential energy, and then apply the definition of force as the gradient of the potential energy function.

The [[Electric potential energy|electrostatic potential energy]] stored in a capacitor of the [[capacitance]] {{math|''C''('''x''')}} and a positive [[electric charge]] {{math|+''Q''}} or negative charge {{math|-''Q''}} on each conductive plate is (with using the definition of the capacitance as <math display="inline">C = \frac{Q}{V}</math>),

<math display="block"> U (Q, \mathbf{x}) = \frac{1}{2} QV(Q,\mathbf{x}) = \frac{1}{2} \frac{Q^2}{C(\mathbf{x})},~</math>

where the dependence on the area of the plates, the dielectric constant of the insulation material between the plates, and the separation {{math|'''x'''}} are abstracted away as the [[capacitance]] {{math|''C''('''x''')}}. (For a parallel plate capacitor, this is proportional to the area of the plates and inversely proportional to the separation.)

The force {{math|'''F'''}} between the plates due to the electric field created by the charge separation is then
<math display="block"> \mathbf{F}(\mathbf{x}) = -\frac{dU}{d\mathbf{x}} ~. </math>

If the capacitor is not connected to any electric circuit, then the ''[[electric charge|electric charges]]'' on the plates remain constant and the voltage varies when the plates move with respect to each other, and the force is the negative [[gradient]] of the [[electrostatics|electrostatic]] potential energy as
<math display="block"> \mathbf{F}(\mathbf{x}) = \frac{1}{2} \frac{dC(\mathbf{x})}{d\mathbf{x}} \frac{Q^2}{{C(\mathbf{x})}^2}
= \frac{1}{2} \frac{dC(\mathbf{x})}{d\mathbf{x}}V(\mathbf{x})^2 </math>

where <math display="inline"> V(Q,\mathbf{x}) = V(\mathbf{x})  </math> as the charge is fixed in this configuration.

However, instead, suppose that the ''[[volt]]age'' between the plates {{math|''V''}} is maintained constant as the plate moves by connection to a [[battery (electricity)|battery]], which is a reservoir for electric charges at a constant potential difference. Then the amount of ''charges'' <math display="inline"> Q </math> ''is a variable'' instead of the voltage; <math display="inline"> Q </math> and <math display="inline"> V </math> are the Legendre conjugate to each other. To find the force, first compute the non-standard Legendre transform <math display="inline">U^*</math> with respect to <math display="inline"> Q </math> (also with using <math display="inline">C = \frac{Q}{V}</math>),

<math display="block">U^* = U - \left.\frac{\partial U}{\partial Q} \right|_\mathbf{x} \cdot Q =U - \frac{1}{2C(\mathbf{x})} \left. \frac{\partial Q^2}{\partial Q} \right|_\mathbf{x} \cdot Q = U - QV = \frac{1}{2} QV - QV = -\frac{1}{2} QV= - \frac{1}{2} V^2 C(\mathbf{x}).</math>

This transformation is possible because <math display="inline"> U </math> is now a linear function of <math display="inline"> Q </math> so is convex on it. The force now becomes the negative gradient of this Legendre transform, resulting in the same force obtained from the original function <math display="inline"> U </math>,
<math display="block"> \mathbf{F}(\mathbf{x}) = -\frac{dU^*}{d\mathbf{x}} = \frac{1}{2} \frac{dC(\mathbf{x})}{d\mathbf{x}}V^2 .</math>

The two conjugate energies <math display="inline"> U </math> and <math display="inline"> U^* </math> happen to stand opposite to each other (their signs are opposite), only because of the [[linear]]ity of the [[capacitance]]—except now {{math|''Q''}} is no longer a constant. They reflect the two different pathways of storing energy into the capacitor, resulting in, for instance, the same "pull" between a capacitor's plates.