Peskin & Schroeder's expression of the Noether current If a (quasi-)symmetry is defined as a transformation that changes the action by a surface term i.e. $$S\to S'=S+\int d^4x \partial_\mu K^\mu(\phi_a),\tag{1}$$ or equivalently, the Lagrangian changes by a 4-divergence, $$\mathscr{L}\to\mathscr{L}'=\mathscr{L}+\partial_\mu K^\mu,\tag{2}$$ then considering transformations on the fields only, the expression of the Noether current turns out to be $$j^\mu(x)=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a -K^\mu.\tag{3}$$ P & S give the example of an internal transformation where $K^\mu=0$, and a spacetime transformation (namely, spacetime translation) under which $K^\mu\neq 0$.
Lewis Ryder's expression of the Noether current Here, the symmetry of the action is defined as a transformation that leaves the action invariant i.e. $$S\to S'=S.\tag{4}$$ Considering transformations $$x^\mu\to x^{\mu'}=x^\mu+\delta x^\mu,\\ \phi(x)\to\phi'(x)=\phi(x)+\delta\phi(x),$$ they derive the following expression for the Noether current $$j^\mu(x)=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a -\Theta^{\mu\nu}\delta x_\nu\tag{5}$$ where $\Theta^{\mu\nu}$ is the stress-energy tensor given by $$\Theta^{\mu\nu}=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\partial^\nu\phi_a -\eta^{\mu\nu}\mathscr{L}.\tag{6}$$
Question $1$ Between $(3)$ and $(5)$ which expression of the Noether's current is more general?
Question $2$ By generalizing Ryder's definition of symmetry $(4)$ (to a quasi-symmetry, i.e., $(1)$), we will obtain $$j^\mu(x)=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a -\Theta^{\mu\nu}\delta x_\nu-K^\mu.\tag{7}$$ Should $(7)$ be regarded as the most general expression of the Noether current?
