field theory - Variation of Lagrangian density $mathcal{L}$ w.r.t $x^{mu}$

If a function $f(x(t),y(t))$ has no explicit dependence on the variable $t$, then $\frac{\partial f}{\partial t}=0$.

In quantum field theory, the Lagrangian density $\mathcal{L}(\phi,\partial_\mu\phi)$ has no explicit dependence of $x^{\mu}$, and therefore, as I understand $\frac{\partial \mathcal{L}}{\partial x^{\mu}}\equiv \partial_\mu\mathcal{L}$ being a partial derivative should also vanish.

However, in deriving Noether's current, in almost all books (for example, in W. Greiner, Bjorken and Drell, or Lewis Ryder) didn't set this term to zero. Why is that?

On the other hand, if this term is really set equal to zero, we do not reach at the correct expression for the Noether's current. But I do not understand why should $\partial_\mu\mathcal{L}\neq 0$?

Answer

There are two kinds of derivatives we should distinguish: $$\frac{\mathrm d\mathcal L}{\mathrm dx}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x+h),\phi'(x+h),x+h)-\mathcal L(\phi(x),\phi'(x),x)\big]\tag{1}$$ and $$\frac{\partial\mathcal L}{\partial x}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x),\phi'(x),x+h)-\mathcal L(\phi(x),\phi'(x),x)\big]\tag{2}$$

In general, it is $(2)$ which is zero, while it is $(1)$ which is used in Noether's theorem (to derive, for example, the energy-momentum tensor).

For example, the Klein-Gordon lagrangian reads $$\mathcal L\sim (\partial\phi)^2-m^2\phi^2$$ which doesn't explicitly depend on $x$. The $(2)$ derivative is clearly zero, that is, $\partial L=0$. On the other hand, upon a translation, $$\mathcal L\ \big|_{x\to x+a}=\mathcal L\ \big|_x+a\ \mathrm d\mathcal L\ \big|_x+\cdots$$ where $\mathrm d\mathcal L\neq 0$ because $\mathcal L$ depends implicitly on position through the fields. It is $\mathrm d\mathcal L$ what you have in the definition of, say, $T^{\mu\nu}$.