In Weinberg's QFT book, Vol-I (page 311), he derives the conserved current for translational invariance. He considers the transformations of the form $$\psi^l(x)\to \psi^l(x+\epsilon (x)).\tag{7.3.30}$$ Then he writes down the variation of the action as $$\delta I=\int d^4 x \left(\frac{\partial \cal{L}}{\partial \psi^l}\epsilon^\mu\partial_\mu \psi^l+\frac{\partial \cal{L}}{\partial (\partial_\nu \psi^l)} \partial_\nu\left[\epsilon^\mu \partial_\mu \psi^l \right]\right)\tag{7.3.31}$$ which, as I understand, comes form the replacement (7.3.30). Now what I am confused about is that why he did not consider the change in the integration measure $\int d^4 x$. I thought we should also replace $\int d^4 x \to \int J d^4 x $, where $J$ is the Jacobian for the transformation $x^\mu \to x^\mu +\epsilon^\mu (x)$. If someone can clarify why the Jacobian is not considered it will be very helpful.

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    $\begingroup$ If epsilon depends on $x$, then it is not a translation in a global sense. When epsilon is a constant four-vector, then the Jacobian is unity. $\endgroup$ Mar 12, 2021 at 6:43
  • $\begingroup$ You are not making a change of variable of integration here, but just transforming the field. The idea is to see what happens when your field varies at any given point as a result of an arbitrary translation, that in general can depend on the point itself. $\endgroup$
    – Phoenix87
    Mar 12, 2021 at 8:09
  • $\begingroup$ Although it might be obvious, I have checked for a scalar field theory that if we choose to change the integration measure we also have to change the space-time derivatives. If we do that the action becomes trivially invariant. To get the conserved current we have to transform only the fields. $\endgroup$ Mar 19, 2021 at 6:14

2 Answers 2


Conceptually, changing the measure $\int d^4 x$ would defeat the point. The action is a functional of an input field, $$ S = S[\psi] $$ and a symmetry transformation $\delta \psi$ is a transformation which satisfies $$ S[\psi + \delta \psi] = S[\psi] + \mathrm{boundary \; term}. $$ This means that if $\psi$ extremizes the action, then $\psi + \delta \psi$ must also extremize the action, meaning that our symmetry transformation maps one solution to the equations of motion, $\psi$, to another solution to the equations of motion, $\psi + \delta \psi$.

Now, note that the symmetry transformation always acts on the field, and not spacetime! You just plug $\psi$ into the action and get out a number. There is no other input to the action. In other words, if $\delta \psi = \epsilon^\mu \partial_\mu \psi$, for $\epsilon^\mu$ a constant infinitesimal translation, then this symmetry transformation translates the field in spacetime, say $0.1$ cm to the left. The points of space don't vary, it's just the field that is moving.

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    $\begingroup$ Yes. I understand it now. I thought space time it self is changing. This will make impossible to compare two fileds which are supposed to defined on the same space-time. $\endgroup$ Mar 12, 2021 at 8:49

OP asks a good question. In eq. (7.3.30) it may at first seems like Weinberg considers

  • (i) an infinitesimal spacetime (=horizontal) transformation $\delta x$.

However in eq. (7.3.31) he changes perspective and think of it as

  • (ii) an infinitesimal vertical transformation $\delta_0\psi=\epsilon^{\mu}\partial_{\mu}\psi$ in the $\psi$ target space.

If the action $I_V[\psi]=\int_V \! d^4x {\cal L}$ is integrated over a spacetime region $V\subseteq \mathbb{R}^4$, it turns out that a purely horizontal transformation (i) only produces boundary terms along the boundary $\partial V$ whose contributions get cancelled in the corresponding full Noether current.

In contrast, to successfully apply Noether's theorem, it is important that the infinitesimal transformation contains a vertical component (ii).

In the second method (ii), Weinberg considers a purely vertical transformation $\delta_0\psi=\epsilon^{\mu}\partial_{\mu}\psi$ with no horizontal component $\delta x=0$. This answers OP's main question about why the spacetime integration measure $d^4x$ is not changed in eq. (7.3.31).

Note that Weinberg's transformation (ii) is only a quasi-symmetry rather than a strict symmetry.

All this is explained in more details in my Phys.SE answer here for the case of point mechanics.


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