08 | May | 2010 | Physics Town

$\left\langle \delta F[\varphi], f \right\rangle = \left.\frac{d}{d\epsilon}F[\varphi+\epsilon f]\right|_{\epsilon=0}$ .

Sometimes physicists write the definition in terms of a limit and the Dirac delta function, $\delta$ :

$\frac{\delta F[\varphi(x)]}{\delta \varphi(y)}=\lim_{\varepsilon\to 0}\frac{F[\varphi(x)+\varepsilon\delta(x-y)]-F[\varphi(x)]}{\varepsilon}$ .

Relationship between the mathematical and physical definitions

The mathematicians’ definition and the physicists’ definition of the functional derivative differ only in the physical interpretation. Since the mathematical definition is based on a relationship that holds for all test functions f, it should also hold when f is chosen to be a specific function. The only handwaving difficulty is that specific function was chosen to be a delta function — which is not a valid test function.

In the mathematical definition, the functional derivative describes how the entire functional, $F[\varphi(x)]$ , changes as a result of a small change in the function $\varphi(x)$ . Observe that the particular form of the change in $\varphi(x)$ is not specified. The physics definition, by contrast, employs a particular form of the perturbation — namely, the delta function — and the ‘meaning’ is that we are varying $\varphi(x)$ only about some neighborhood of y. Outside of this neighborhood, there is no variation in $\varphi(x)$ .