Consider a simple scalar field and its Lagrangian $L=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi$. Then say you have the following transformation

$$x^{\mu}\rightarrow e^{\omega}x^{\mu},\tag{1}$$

$$\phi\left(x\right)\rightarrow e^{-\omega}\phi\left(e^{\omega}x\right).\tag{2}$$

What is the associated conserved current?


I compute the Euler-Lagrange (EL) equations,

$$\partial_{\mu}\left(\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}\right)=\frac{\partial L}{\partial\phi}\quad\Leftrightarrow\quad\partial_{\mu}\partial^{\mu}\phi=0,\tag{3}$$

write the transformations infinitesimal as


$${\delta L=0}\tag{5}$$

and compute the conserved current, factoring out $\omega$, from

$$j^{\mu}=\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}\delta\phi-\delta L=-\phi\partial^{\mu}\phi\tag{6}$$

but something must be wrong because I can't show it is conserved,


Can you please see what am I doing wrong?

  • $\begingroup$ Eq. (6) is incomplete in its current form (v9). $\endgroup$
    – Qmechanic
    Jun 13, 2018 at 2:50

2 Answers 2


1) OP's action for a free scalar particle in $n$ spacetime dimensions is $$ S~=~\int \! \mathbb{L}(x), \qquad \mathbb{L}(x)~=~d^nx ~{\cal L}(x), \qquad {\cal L}(x) ~:=~\frac{1}{2} \frac{\partial \phi(x)}{\partial x^{\mu}} \eta^{\mu\nu}\frac{\partial \phi(x)}{\partial x^{\nu}}.\tag{A}$$ It is not hard to check that the Lagrangian $n$-form $\mathbb{L}(x)$ is invariant under the scaling $$x^{\mu}\quad\longrightarrow\quad x^{\prime\mu}~=~\lambda x^{\mu}, \qquad \phi(x)\quad\longrightarrow\quad \phi^{\prime}(x^{\prime})~=~\lambda^{1-n/2} \phi(x) \tag{B}$$ with a positive parameter $\lambda > 0$. Of course, this is nothing but the well-known fact that in absolute units $\hbar=1=c$, the mass dimensions are $$ [x]~=~ -1 , \qquad [S]~=~0, \qquad [\phi]~=~n/2-1. \tag{C}$$ It seems OP assumes that the spacetime dimension is $n=4$.

2) Next let us consider the corresponding infinitesimal transformation. Assume that $\lambda=1+\varepsilon$, where $\varepsilon$ is infinitesimal. The so-called horizontal infinitesimal variation is

$$\delta x^{\mu} ~:=~x^{\mu \prime}-x^{\mu} ~=~ \varepsilon \cdot x^{\mu}. \tag{D}$$

The infinitesimal variation of the dynamical variable $\phi$ is

$$ \delta \phi(x)~:= ~\phi^{\prime}(x^{\prime})-\phi(x)~=~\varepsilon \cdot (1-n/2) \phi(x), \tag{E}$$

so the vertical infinitesimal variation is

$$ \delta_0 \phi(x)~:= ~\phi^{\prime}(x)-\phi(x)~=~\varepsilon \cdot (1-n/2 -x^{\nu}d_{\nu} )\phi(x).\tag{F}$$

In other words, the transformation (B) has horizontal generator $x^{\mu}$ and vertical generator $(1-n/2 -x^{\nu}d_{\nu} )\phi$.

3) The bare Noether current $j^{\mu}$ is defined as (the partial derivative of the Lagrangian density wrt. field derivatives) times the vertical generator plus the Lagrangian density times the horizontal generator:

$$ j^{\mu}~:=~ \frac{\partial {\cal L}}{\partial d_{\mu}\phi}\cdot(1-n/2 -x^{\nu}d_{\nu} )\phi+ {\cal L}\cdot x^{\mu} ~=~ d^{\mu}\phi\cdot(1-n/2 -x^{\nu}d_{\nu} )\phi + {\cal L}\cdot x^{\mu}.\tag{G} $$

It is straightforward to verify the on-shell continuum equation

$$ d_{\mu}j^{\mu}~\approx~0. \tag{H}$$

4) See also this related Phys.SE post.


Why did you write $\frac{\partial L}{\partial (\partial_\mu\phi)}=-\phi\partial^\mu\phi$? It should be $=\partial^\mu\phi$.

update: everthing's correct, $\partial_\mu\phi\partial^\mu\phi$ vanishes because it is equal to $\partial_\mu\partial^\mu\phi$ up to a surface term. Whenever equation of motion is satisfied ($\partial_\mu\partial^\mu\phi=0$), $\partial_\mu\phi\partial^\mu\phi$ vanishes too.

  • $\begingroup$ That's what I've done, you can see it above in the EL equation. But then, when computing the current I have to multiply that by $\delta\phi$ which I evaluated as $\omega\phi$. $\endgroup$
    – johani
    Jun 11, 2018 at 19:14
  • $\begingroup$ surface term means total derivative in a Lagrangian. When you integrate it (to obtain action) you get integration limits. It is in any textbook that deals with least action $\endgroup$
    – Kosm
    Jun 11, 2018 at 20:39

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