I have a continuous transformation on the field $\phi$ of the form

$$\phi(x)\rightarrow \phi'(x)=\phi(x)+\alpha\Delta\phi(x),\tag{1}$$

where $\alpha$ is a constant infinitesimal parameter and $\Delta\phi$ is a deformation of the field. Note that in this notation (Peskin and Schroeder one) $\delta\phi=\alpha\Delta\phi$

To have a symmetry my action should be invariant up to a surface term, so my lagrangian has to be invariant up to a 4-divergence:

\begin{equation} \mathcal{L}\rightarrow\mathcal{L}+\alpha\partial_\mu J^\mu.\tag{2} \end{equation}

Now we proceed varying the Lagrangian:

\begin{equation} \delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu(\delta\phi)=\alpha\frac{\partial\mathcal{L}}{\partial\phi}\Delta\phi+\alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\Big)-\alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Big)\Delta\phi.\tag{3} \end{equation}

Now, the first and the third term cancel out because of Euler-Lagrange equations.

If I want to satisfy my symmetry, the variation I just calculated has to be equal to the 4-divergence:

\begin{equation} \alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\Big)=\alpha\partial_\mu J^\mu \rightarrow \alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi-J^\mu\Big)=0 .\tag{4} \end{equation}

Thus, the quantity

$$j^\mu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi-J^\mu\tag{5}$$ is conserved.

And this is pretty clear to me. But what if $\alpha=\alpha(x)$?

$\textbf{My attempt}$:

Since $\alpha$ is a function of $x$, the quantity $\partial_\mu(\delta\phi)=\partial_\mu(\alpha\Delta\phi)$ becomes $\Delta\phi\partial_\mu \alpha+\alpha\partial_\mu \Delta\phi$

\begin{equation} \delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu(\delta\phi)=\frac{\partial\mathcal{L}}{\partial\phi}\alpha\Delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\partial_\mu \alpha+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\alpha\partial_\mu \Delta\phi.\tag{6} \end{equation}

The first and the third term give rise to a term $\alpha\Delta\mathcal{L}$, exactly like the one we get with constant $\alpha$.

And from here my ideas start to get fuzzy.

So, I got my action varying like:

$$ \delta S=\int d^4x\Big(\alpha\Delta\mathcal{L}+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\partial_\mu \alpha\Big)=\int d^4x\Big(\alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\Big)+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi(\partial_\mu \alpha)\Big)$$ $$=\int d^4x \partial_\mu \Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\alpha\Big).\tag{7} $$

Now, if I have a symmetry, my action is invariant up to a boundary term, hence an integral of $\alpha\partial_\mu J^\mu$. So I get the same result as conserved current $j^\mu$.

$\textbf{My question}$:

In Tong notes (http://www.damtp.cam.ac.uk/user/tong/qft/one.pdf) at page 19 there is a total different approach. He says that the Lagrangian varies like \begin{equation} \delta\mathcal{L}=(\partial_\mu \alpha)h^\mu\tag{8} \end{equation} where $h^\mu$ is the conserved current. Using my notation, I guess that this true only for the cases where $\delta\mathcal{L}=0$ for $\alpha$=const, hence for $J^\mu=0$.

But My professor, and also some other notes, says that since

\begin{equation} \delta S=\int d^4x(\partial_\mu \alpha) h^\mu\tag{9} \end{equation}

if I want to find the conserved current all I have to do is vary the Lagrangian I'm given assuming $\alpha=\alpha(x)$ and then simply search for the quantity "next to" $\partial_\mu \alpha$. Isn't this true only for the cases where

$$\delta\mathcal{L}=0\quad\text{for}\quad \alpha=\text{const}? \tag{10}$$

He didn't say anything about it, suggesting that this is the most general case.


1 Answer 1


In the context of Noether's first theorem, the following comments apply:

  1. OP's eq. (10) is not necessarily assumed. It is in general only assumed that the action functional $S$ is invariant up to possible boundary terms.

  2. OP's eq. (9) holds up to possible boundary conditions. This is e.g. explained in this Phys.SE post.

  3. OP's eq. (8) is only true in special cases. The general form is OP's eq. (9) (up to possible boundary terms).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.