Transformation has $\delta \mathcal{L}=0$ but $\delta S \neq 0$ and $\delta S \neq \epsilon \int_{\partial\Omega} d^3 x f $

Question

I saw a answer for Why are these two definitions for symmetries in the Lagrangian equivalent?

I have an example has $\delta \mathcal{L}=0$ but $\delta S \neq 0$ and $$\delta S \neq \epsilon \int_{\partial\Omega} d^3 x f .\tag{1} $$

Action $$S=\int_\Omega d^4x~\mathcal{L},\tag{2}$$ $$\mathcal{L}=\frac{1}{2}(\partial_t \phi(t,\mathbf{x}))^2 -\frac{1}{2}(\partial_\mathbf{x} \phi(t,\mathbf{x}))^2.\tag{3}$$

Transformation: $$x\rightarrow x'=\lambda x,\tag{4}$$ $$\phi(x)\rightarrow\phi'(x')=\lambda\phi(x).\tag{5}$$

This transformation has $$\delta\mathcal{L}=\mathcal{L}(\partial_{x'}\phi'(x'),\phi'(x'))-\mathcal{L}(\partial_{x}\phi(x),\phi(x))=0, \tag{6}$$ $$\delta S= \frac{1}{2}\int_{\Omega'} d^4x' ((\partial_{t'} \phi(t',\mathbf{x'}))^2 -(\partial_{\mathbf{x}'} \phi(t',\mathbf{x}'))^2) - \frac{1}{2}\int_\Omega d^4x ((\partial_t \phi(t,\mathbf{x}))^2 -(\partial_\mathbf{x} \phi(t,\mathbf{x}))^2)=(\lambda^4-1)S\tag{7}$$

Note: $$\int_{\Omega'}d^4x'=\int_{\Omega}|\frac{\partial x'^\mu}{\partial x^\nu}|d^4x=\int_{\Omega}\lambda^4 d^4 x.\tag{8}$$

My question:

According to this answer, this transformation is L$1$ but is not S$1$ or S$2$. So does this transformation has conserved current? I feel like it's not a symmetry.

PS: According to Qmechanic's answer: there're two different definition of $\delta \mathcal{L}$

The first one:

$$\delta_1 \mathcal{L}=\mathcal{L}(\phi'(x'),\partial'_\mu\phi'(x'),x')-\mathcal{L}(\phi(x ),\partial _\mu\phi (x ),x )= [\mathcal{L}]_\phi \bar\delta \phi+\partial_\mu(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\bar\delta \phi)+ (\partial_\mu \mathcal{L})\delta x^\mu$$

The second one:

$$\delta_2 \mathcal{L}=J \mathcal{L}(\phi'(x'),\partial'_\mu\phi'(x'),x')-\mathcal{L}(\phi(x ),\partial _\mu\phi (x ),x )$$ with Jacobian $J$ $$J=\det(\frac{\partial x'^\mu}{\partial x^\nu})$$

$$\delta_2 \mathcal{L}=[\mathcal{L}]_\phi \bar\delta \phi+\partial_\mu(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\bar\delta \phi)+ \partial_\mu (\mathcal{L}\delta x^\mu) = \delta_1 \mathcal{L}+ \mathcal{L} (\partial_\mu \delta x^\mu)$$

When we consider the symmetry transformation, we need to compute $\delta_2 \mathcal{L}$ instead of $\delta_1 \mathcal{L}$. After using the definition of $\delta_2 \mathcal{L}$, the symmetry defined by $\delta_2 \mathcal{L}$ is always consistent to the symmetry defined by $\delta S$.

Answer 1

I) The infinitesimal version of OP's transformation is $$\begin{align}\text{Total:}&\qquad \phi^{\prime}(x^{\prime})-\phi(x)~=~\delta \phi~=~\varepsilon~\phi, \tag{A}\cr \text{Horizontal:}&\qquad x^{\prime\mu} -x^{\mu} ~=~\delta x^{\mu}~=~\varepsilon~ x^{\mu}, \tag{B}\cr \text{Vertical:}&\qquad\phi^{\prime}(x)-\phi(x)~=~ \delta_0 \phi~=~\varepsilon~ (1-x^{\mu}d_{\mu})\phi, \tag{C} \end{align}$$ where $\varepsilon$ is an infinitesimal parameter.

II) The action is $$S[\phi]~=~\int_{\Omega} \mathbb{L}; \tag{D}$$ the Lagrangian 4-form is $$ \mathbb{L}~=~{\cal L}~ d^4x; \tag{E}$$ and the Lagrangian density is $$ {\cal L}~=~\frac{1}{2}d_{\mu}\phi ~d^{\mu}\phi. \tag{F} $$

III) The Lagrangian density ${\cal L}$ naively transforms as $$ \delta {\cal L}~=~\delta_0 {\cal L}+ \delta x^{\mu} d_{\mu}{\cal L}~=~0 . \qquad\longleftarrow \text{Naive calculation.} \tag{G}$$ However, in case of non-zero horizontal transformations, formula (G) is not a decisive quantity for Noether's theorem, and one should rather look at the action $$ \delta S~=~4\varepsilon ~S ,\tag{H}$$ or at least the Lagrangian 4-form $$ \delta \mathbb{L}~=~4\varepsilon ~\mathbb{L} .\tag{I}$$ Eq. (H) agrees with OP's eq. (7). The issue is that the Lagrangian density ${\cal L}$ is not a scalar but a density, so there is a non-trivial contribution ${\cal L}d_{\mu}\delta x^{\mu}$ coming from the Jacobian of the integration measure $d^4x$. Therefore the decisive quantity for Noether's theorem is not formula (G) but rather $$ "\delta {\cal L}"~=~\delta {\cal L}+{\cal L}d_{\mu}\delta x^{\mu} ~=~\delta_0 {\cal L}+ d_{\mu}(\delta x^{\mu}{\cal L})~=~4\varepsilon ~{\cal L} .\tag{J}$$

IV) In conclusion, the infinitesimal transformation (A)-(C) is not a quasisymmetry for the action (D) nor the Lagrangian density (F), and Noether's theorem does not apply.

Answer 2

Consider an infinitesimal version of this transformation, given by $\lambda=1+\epsilon$. With this choice, the field and coordinate variations are

$$\delta\phi = \epsilon\phi,\hspace{0.5cm}\delta x=\epsilon x.$$

Under these variations, the action transforms as

$$\delta S=\int_{\Omega'}\mathrm{d}^4x\,\mathcal{L}-\int_{\Omega}\mathrm{d}^4x\,\mathcal{L}.$$

Note that I allowed the same variable $x$ to parametrize both the regions $\Omega$ and $\Omega'$, since these are dummy variables. The only thing that changes is the domain. Thus, the change in the action is

$$\delta S=\int_{\Omega'\setminus\Omega}\mathrm{d}^4x\,\mathcal{L}.$$

If $\Omega$ is a convex set in $\mathbb{R}^4$, then $\Omega'\setminus\Omega$ is a set that is essentially $\partial\Omega$ with thickness $\epsilon$ (this idea can be generalized to non-convex sets, but it becomes a bit more complicated and I really don't want to think about it that much). Thus, in the infinitesimal limit, the integral reduces to

$$\delta S=\epsilon\int_{\partial\Omega}\mathrm{d}^4x\,\mathcal{L},$$

et voilà! Your transformation is now type S2!

I hope this helps!