1
$\begingroup$

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?

Attempt:

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

$${\phi\left(x\right)\rightarrow\left(1-\omega+\ldots\right)\phi=\phi-\omega\phi}\quad\Rightarrow\delta\phi=-\omega\phi\tag{4}$$

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

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,

$$\partial_{\mu}j^{\mu}=-\partial_{\mu}\left(\phi\partial^{\mu}\phi\right)=-\partial_{\mu}\phi\partial^{\mu}\phi-\phi\underset{=0\left(EL\right)}{\underbrace{\partial_{\mu}\partial^{\mu}\phi}=}-\partial_{\mu}\phi\partial^{\mu}\phi?\tag{7}$$

Can you please see what am I doing wrong?

$\endgroup$
  • $\begingroup$ Eq. (6) is incomplete in its current form (v9). $\endgroup$ – Qmechanic Jun 13 '18 at 2:50
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '18 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 '18 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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