A How was the symmetry used here? ("QFT and the SM" by Schwartz)

[Hill]

TL;DR Summary

In this derivation of energy-momentum conservation, on what step the symmetry of space-time translation was used?

Here are steps.
Consider shift of field $\phi$ by a constant 4-vector $\xi$:
(1) $\phi(x) \rightarrow \phi(x+\xi)=\phi(x)+\xi^{\nu} \partial_{\nu} \phi(x) + ...$
The infinitesimal transformation,
(2) $\frac {\delta \phi} {\delta \xi^{\nu}} = \partial_{\nu} \phi$
and
(3) $\frac {\delta \mathcal L} {\delta \xi^{\nu}} = \partial_{\nu} \mathcal L$
Using the E-L equations, the variation of Lagrangian is
(4) $\frac {\delta \mathcal L[\phi, \partial_{\mu} \phi]} {\delta \xi^{\nu}}=\partial_{\mu} (\frac {\partial \mathcal L}{\partial (\partial_{\mu} \phi)} \frac {\delta \phi} {\delta \xi^{\nu}})$
Using (2) and (3),
(5) $\partial_{\nu} \mathcal L =\partial_{\mu} (\frac {\partial \mathcal L}{\partial (\partial_{\mu} \phi)} \partial_{\nu} \phi)$
or equivalently
(6) $\partial_{\mu} (\frac {\partial \mathcal L}{\partial (\partial_{\mu} \phi)} \partial_{\nu} \phi - g_{\mu \nu} \mathcal L) = 0$
The conclusion is,
"The four symmetries have produced four Noether currents, one for each $\nu$:
$\mathcal T_{\mu \nu} = \frac {\partial \mathcal L}{\partial (\partial_{\mu} \phi)} \partial_{\nu} \phi - g_{\mu \nu} \mathcal L$
all of which are conserved: $\partial_{\mu} \mathcal T_{\mu \nu}=0$."

My question: where in this derivation the assumption was used that the transformation is a symmetry?

 

[Demystifier]

The assumption of symmetry is hidden in the use of EL equation. If you did not use EL equation, that would mean that $\phi(x)$ is treated as a non-dynamical fixed background, the explicit $x$-dependence of which would violate the symmetry. Indeed, the EL equation can be written as
$$\frac{\delta A}{\delta \phi(x)}=0$$
where $A$ is the action, which can be interpreted as a statement that you only consider such $\phi(x)$ for which the action does "not depend" on $\phi(x)$. Since the dependence on $x$ arises only from dependence on $\phi(x)$, this means that EL equation expresses also independence on $x$, which is nothing but the symmetry.

 

[Hill]

The assumption of symmetry is hidden in the use of EL equation. If you did not use EL equation, that would mean that $\phi(x)$ is treated as a non-dynamical fixed background, the explicit $x$-dependence of which would violate the symmetry. Indeed, the EL equation can be written as
$$\frac{\delta A}{\delta \phi(x)}=0$$
where $A$ is the action, which can be interpreted as a statement that you only consider such $\phi(x)$ for which the action does "not depend" on $\phi(x)$. Since the dependence on $x$ arises only from dependence on $\phi(x)$, this means that EL equation expresses also independence on $x$, which is nothing but the symmetry.

Thank you.
I think also that the symmetry validates the equation (3), because this equation makes the variation of Lagrangian a total derivative, and this makes the variation of action vanish: $$\delta A = \int d^4 x \delta \mathcal L = \delta \xi^{\nu} \int d^4x \partial_{\nu} \mathcal L = 0$$ IOW, without the symmetry, we can't go from (4) to (5).

 

[jbergman]

TL;DR Summary: In this derivation of energy-momentum conservation, on what step the symmetry of space-time translation was used?

Here are steps.
Consider shift of field $\phi$ by a constant 4-vector $\xi$:
(1) $\phi(x) \rightarrow \phi(x+\xi)=\phi(x)+\xi^{\nu} \partial_{\nu} \phi(x) + ...$
The infinitesimal transformation,
(2) $\frac {\delta \phi} {\delta \xi^{\nu}} = \partial_{\nu} \phi$
and
(3) $\frac {\delta \mathcal L} {\delta \xi^{\nu}} = \partial_{\nu} \mathcal L$
Using the E-L equations, the variation of Lagrangian is
(4) $\frac {\delta \mathcal L[\phi, \partial_{\mu} \phi]} {\delta \xi^{\nu}}=\partial_{\mu} (\frac {\partial \mathcal L}{\partial (\partial_{\mu} \phi)} \frac {\delta \phi} {\delta \xi^{\nu}})$
Using (2) and (3),
(5) $\partial_{\nu} \mathcal L =\partial_{\mu} (\frac {\partial \mathcal L}{\partial (\partial_{\mu} \phi)} \partial_{\nu} \phi)$
or equivalently
(6) $\partial_{\mu} (\frac {\partial \mathcal L}{\partial (\partial_{\mu} \phi)} \partial_{\nu} \phi - g_{\mu \nu} \mathcal L) = 0$
The conclusion is,
"The four symmetries have produced four Noether currents, one for each $\nu$:
$\mathcal T_{\mu \nu} = \frac {\partial \mathcal L}{\partial (\partial_{\mu} \phi)} \partial_{\nu} \phi - g_{\mu \nu} \mathcal L$
all of which are conserved: $\partial_{\mu} \mathcal T_{\mu \nu}=0$."

My question: where in this derivation the assumption was used that the transformation is a symmetry?

I believe that the symmetry is used in step 3. Under the translation symmetry the Lagrangian change is a total derivative. I think this is essentially what @Demystifier is saying.

Incidentally, i wrote a long blog post on this derivation but omitted this question. Thanks for bringing it up.

 

[Hill]

I believe that the symmetry is used in step 3. Under the translation symmetry the Lagrangian change is a total derivative. I think this is essentially what @Demystifier is saying.

Incidentally, i wrote a long blog post on this derivation but omitted this question. Thanks for bringing it up.

Thank you. That was my understanding also, in the post #3.

 

[haushofer]

I believe that the symmetry is used in step 3. Under the translation symmetry the Lagrangian change is a total derivative. I think this is essentially what @Demystifier is saying.

Incidentally, i wrote a long blog post on this derivation but omitted this question. Thanks for bringing it up.

Step 3 is just saying the Lagrangian is a scalar (density) under translations, right?

I also find this stuff treacherous, and many textbooks completely miss these subtleties and make it look easy. My confusion for a long time was that these symmetries are derived "on shell", but then EL=0 for any variation of the field. And I know from experience that you can even confuse post docs working on string theory with this stuff.