# Conserved currents in Mandl & Shaw (section 2.4)

I am currently reading "Quantum Field Theory" by Mandl & Shaw (2nd edition). In section 2.4, they give an equation for obtaining a conserved quantity (which leave the Lagrangian density invariant) :

$$\frac{\delta f^\alpha}{\delta x^\alpha} = 0$$ (2.36 in Mandl & Shaw)

In other textbooks that I have consulted, I usually find the following expression for a conserved quantity :

$$\frac{\partial f^\alpha}{\partial x^\alpha} = 0$$

I would like to know why is Mandl & Shaw using what seems to be a functional derivative instead of a simple partial derivative.

Furthermore, does anyone know how to work with the functional derivative used by Mandl & Shaw and how to obtain equation (2.36a) with it? Can I treat it as a simple derivative? It seems wrong to do so...

samalkhaiat
I think it is a mistake. You don’t take functional derivative with respect to coordinates. If that section of the book is about Noether’s theorem, then the following might help you understand the connection functional and partial derivatives;
Let $s(x)$ be a space-like hyper surface passes through a specific point x. Let us define a 4-vector differential area at the point x by
$$ds^{\mu}=(dx^{1}dx^{2}dx^{3},dx^{0}dx^{1}dx^{2},dx^{0}dx^{1}dx^{3},dx^{0}dx^{2}dx^{3})$$
Now, let $Q$ be a functional of a surface $s$. The functional derivative at a point x is defined by
$$\frac{\delta Q}{\delta s(x)}= \lim_{v(x) \rightarrow 0} \frac{Q[\bar{s}]-Q}{v(x)}$$
where $v(x)$ is the volume enclosed between the two surfaces $\bar{s}$ and $s$.
Now let $Q$ be given by
$$Q= \int_{s} ds_{\mu} f^{\mu}(x),$$
where $f_{\mu}(x)$ is some smooth vector field, then according to Gauss’ theorem we find
$$\frac{\delta Q}{\delta s}= \lim_{v \rightarrow 0} \frac{\int_{\bar{s}}ds_{\mu}f^{\mu}(x) - \int_{s} ds_{\mu}f^{\mu}(x)}{v(x)}=\partial_{\mu}f^{\mu}$$
Using this, Schwinger was able to prove the following (very, very useful) identity
$$\int ds_{\mu}\partial_{\nu}F(x) = \int ds_{\nu}\partial_{\mu}F(x),$$
where
$$|\vec{x}|^{2}F(x)\rightarrow 0,\ \ \mbox{as}\ \ |\vec{x}| \rightarrow \infty$$

sam