# Functional derivative for the action $S$

From Lancaster and Blundell's Quantum Field Theory for the Gifted Amateur, p. 15:

Example 1.3

The Lagrangian $$L$$ can be written as a function of both position and velocity. Quite generally, one can think of it as depending on a generalized position coordinate $$x(t)$$ and its derivative $$\dot{x}(t)$$, called the velocity. Then the variation of $$S$$ with $$x(t)$$ is $$\delta S/ \delta x(t)$$ and can be written as \begin{align} \frac{\delta S}{\delta x(t)} &= \int \mathrm{d}u \left[\frac{\delta L}{\delta x(u)}\frac{\delta x(u)}{\delta x(t)} + \frac{\delta L}{\delta \dot{x}(u)}\frac{\delta \dot{x}(u)}{\delta x(t)}\right] \\ &= \int \mathrm{d}u \left[\frac{\delta L}{\delta x(u)}\delta(u-t) + \frac{\delta L}{\delta \dot{x}(u)}\frac{\mathrm{d}}{\mathrm{d}t}\delta(u-t)\right] \\ &= \frac{\delta L}{\delta x(t)} + \left[\delta(u-t) \frac{\delta L}{\delta \dot{x}(u)}\right]^{t_f}_{t_i} - \int \mathrm{d}u \,\delta(u-t) \frac{\mathrm{d}}{\mathrm{d}t}\frac{\delta L}{\delta \dot{x}(u)} \\ &= \frac{\delta L}{\delta x(t)} - \frac{\mathrm{d}}{\mathrm{d}t}\frac{\delta L}{\delta \dot{x}(t)}, \tag{1.27} \end{align}

I have trouble understanding why $$\frac{\delta x(u)}{\delta x(t)}=\delta(u-t), \frac{\delta \dot{x}(u)}{\delta x(t)}=\frac{d}{dt}\delta(u-t).$$

I learned how to take derivatives of functionals, but $$x(u)$$ is not a functional. I also read that $$\delta x(u)$$ is defined as $$\delta x(u)=\epsilon\delta(u-t).$$

I'm guessing I have to take the fraction of $$\delta x(u)/ \delta x(t)$$, but what is $$\delta x(t)$$? Also, the differentiation in $$\delta \dot{x}(u)$$ seems to have swapped with the $$\delta$$ sign, i.e. $$\delta \dot{x}(u)=\frac{d}{dt}\delta x(u)$$. What is happening here?

• Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. I edited your question to adhere to these standards. Feb 2, 2022 at 6:39
• @NíckolasAlves Sorry. I will take note next time. Thank you for the edit. Feb 2, 2022 at 8:13

While you are correct that $$x(u)$$ is not in principle a functional, it can be seen as one if we employ the Dirac delta to write $$x(u) = \int \delta(u-v) x(v) \mathrm{d}v. \tag{1}$$
Now, if $$F_x[f]$$ is given by $$F_t[f] = \int K(t,t') f(t') \mathrm{d}t',$$ we know that the functional derivative of $$F_t$$ with respect to $$f$$ is $$\frac{\delta F_t}{\delta f(u)} = K(t,u).$$ Applying this expression to Eq. (1) leads us to $$\frac{\delta x(u)}{\delta x(t)} = \delta(u-t).$$
As for the swapping of $$\delta$$ with $$\frac{\mathrm{d}}{\mathrm{d}t}$$, that is because the change in the function (associated with $$\delta$$) is assumed to occur in fixed instants of time, so that the two operations commute.
For further detail, I particularly recommend Lemos' Analytical Mechanics. Sec. 10.3 deals with functional derivatives and addresses the first part of your question, about how to differentiate $$x(u)$$ with respect to $$x(t)$$ (my approach in this answer is essentially taken from there). Secs. 2.1–2.2 deal with variational calculus and, in particular, the matter of why $$\delta$$ and $$\frac{\mathrm{d}}{\mathrm{d}t}$$ commute (see Eq. (2.37)).