For Noether theorem with only internal symmetry, I've found there has been a very clear proof. But I still struggle with the proof of coordinate transformation. Because there are so many different proofs which are not consistent with each other.

  1. The first question is what's the definition of "an action has some spacetime symmetry"?

Because for an action $$S[g_{\mu\nu}(x), \phi_A(x)]=\int \sqrt{-g} d^4x \mathcal{L}(\phi_A(x),\partial_\mu\phi_A(x),g_{\mu\nu}(x)) $$

where $A$ is index of the field, e.g spinor field, vector field etc.

If I transform $\partial_\mu$, $\phi_A(x)$, $g_{\mu\nu}$, $\sqrt{-g} $, $d^4x$ simultaneously according to the change of $x^\mu$ , the action $S$ and Lagrangian $\mathcal{L}$ are certainly always unchanged under any coordinate transformation. But certainly not any coordinate transformation correspends to conservative currents. So I firstly need to know how to define the symmetry of coordinate transformation of an action? I haven't yet found any QFT textbook gave me an explicit definition. But I think that the definition of spacetime symmetries of action should give me the isometries of the background metric $g_{\mu\nu}(x)$. While up to now I don't know how to define?

Now I found that there are about two kinds of proof in different QFT textbooks:

The first kind proof (e.g Greiner p40 )

For coordinate transformation: $$x_\mu\rightarrow x_\mu'(x)$$ $$\phi_A(x)\rightarrow \phi_A^{\prime}(x')$$ $$\partial_\mu \phi_A (x)\rightarrow \partial'_\mu \phi'_A (x')$$ $$\delta x^\mu= x^{\prime \mu}-x^\mu$$ $$\delta \phi_A(x)= \phi_A^{\prime}(x') -\phi_A(x)$$ $$\delta (\partial_\mu\phi_A(x))= \partial'_\mu\phi_A^{\prime}(x') -\partial_\mu\phi_A(x)\neq \partial_\mu \delta \phi_A(x) \quad !$$

He defines that the coordinate transformation is a symmtry of action if $$\int_{\Omega'} d^4 x' \mathcal{L}'(x')-\int_{\Omega}d^4x \mathcal{L}(x)=0\tag{2.45}$$

So in this step he keeps the $\sqrt{-g}=\sqrt{-\eta}=1$ unchanged before and after transformation.

In eq(2.49),we see that there is no variation of Lagrangian with respect to $g_{\mu\nu}$, so he also keep the metric components in Lagrangian unchanged. So in his derivation the Lagragian is not in fact a scalar w.r.t. general coordinate transformation.

That means e.g scalar field $$\mathcal{L}=\eta_{\mu\nu}\partial^{\mu}\phi(x)\partial^{\nu}\phi(x) =(\partial_t\phi(x))^2-(\nabla\phi(x))^2\\\rightarrow \eta_{\mu\nu}\partial^{\prime\mu}\phi'(x')\partial^{\prime\nu}\phi'(x')=(\partial'_t\phi'(x'))^2-(\nabla'\phi'(x'))^2 $$

It's obvious that the above $\mathcal{L}$ is Lorentz-invariant scalar, but not a scalar in arbitary coordinate transformation.

$\textbf{Remarks}$: Whenever the background is Minkovski or curved spacetime with dynamics of metric or not, after the coordinate transformation, the metric tensor itself, i.e. $\eta=\eta_{\mu\nu}\frac{\partial}{\partial_{\mu}}\frac{\partial}{\partial_{\nu}}$ is not changed, but the components, i.e. $\eta_{\mu\nu}$, will change, unless the transformation is isometry.

The second proof (Schwartz p35, Peskin p19)

There is an assumption that they view $\mathcal{L}$ as a scalar, i.e. $\mathcal{L}(x)=\mathcal{L}'(x')$, that requires the metric components $\eta_{\mu\nu}$ change according to coordinate transformation (which is different in the first proof ). But they use the noether's current fomula derived in the condition of internel symmetry. And in the case of internel symmetry, $d^4x$, $\sqrt{-g}$ are unchanged and $\delta\partial_\mu \phi(x)= \partial_{\mu}\delta\phi(x)$, which do not hold in case of coordinate transformation. (Is it in fact totally wrong in Peskin's derivation?)

What's more, you cannot even use Peskin's method to derive the conservative currents of rotation, because they view $\mathcal{L}(x)=\mathcal{L}'(x')$ as scalar, so under Lorentz transformation $\delta\mathcal{L}(x)=(\partial_\mu \mathcal{L}) \delta \omega^{\mu}_{\nu}x^{\nu}$ that is not a total derivative term.

  1. It seems that these two derivations are based on different prerequistes or different defitions of spacetime symmetry of action, although at the end they get the same Noether current of translation. Greiner's way can calculate the current of rotational symmetry, and Peskin's way cannot. Am I right?

  2. Is there some rigorous derivation of Noether's current of spacetime symmetry? I thinks there may be some clear proof in qft in curved spacetime such that the effect of metric can be seen explicitly.

  • 1
    $\begingroup$ Possible duplicate of: physics.stackexchange.com/q/138202 See, in particular, John Rennie's comment. See also page 126 of these notes damtp.cam.ac.uk/user/tong/qft/qft.pdf regarding local gauge symmetries and whether they correspond to conservation laws via Noether's Theorem. $\endgroup$ Apr 19, 2017 at 4:16
  • $\begingroup$ @joshphysics Thanks for your comment. Is my argument right that there are some loopholes of derivation in QFT's textbooks? $\endgroup$
    – 346699
    Apr 19, 2017 at 4:21
  • $\begingroup$ I'm sorry to say that I don't have enough time at the moment to read the question carefully enough to say. $\endgroup$ Apr 19, 2017 at 4:27

1 Answer 1


This post (v13) asks many questions. Let us here make some general remarks, which OP hopefully will find useful.

  1. A lot of confusion seems to stem from the fact that many physicists (see e.g. Refs. 1-3) consider the Lagrangian density ${\cal L}$, which is the integrand in the action functional $$S~=~\int d^4x ~{\cal L} \tag{A}$$ without a factor $\sqrt{|g|}$. The Lagrangian density ${\cal L}$ is not a scalar. As the name suggests, it transforms as a density $${\cal L}\quad\longrightarrow\quad{\cal L}^{\prime}~=~\frac{{\cal L}}{J}, \qquad J ~:=~\det(\frac{\partial x^{\prime \nu}}{\partial x^{\mu}}), \tag{B}$$ under general coordinate transformations $$ x^{\mu} \quad\longrightarrow\quad x^{\prime \nu}~= ~f^{\nu}(x).\tag{C}$$ E.g. in this notation the Lagrangian density for a scalar field $\phi$ reads $${\cal L}~=~\pm \frac{1}{2}~\sqrt{|g|}~\partial_{\mu}\phi ~g^{\mu\nu}~\partial_{\nu}\phi \tag{D} $$ with Minkowski signature $(\pm ,\mp,\mp,\mp)$, respectively.

  2. On one hand, in QFT, the metric tensor $g_{\mu\nu}$ is a fixed background (usually taken to be the Minkowski metric $\eta_{\mu\nu}$), which we don't vary. On the other hand, in GR, the metric tensor $g_{\mu\nu}$ is a dynamical field, which we vary.

  3. Full diffeomorphism invariance is in the realm of Noether's second theorem.

  4. However OP seems (for now) mostly interested in Noether's first theorem in the context of both vertical and horizontal variations. [For terminology, see e.g. my Phys.SE answer here.]

    • Refs. 2-3, joshphysics's Phys.SE answer and JamalS's Phys.SE answer discuss only vertical variations.

    • Ref. 1 discusses both vertical and horizontal variations.

    • However, Ref. 1 with eq. (2.45) only discusses strict symmetries. Noether's theorem may be generalized to quasi-symmetries, cf. my Phys.SE answers here & here. This generalization leads to additional terms in the Noether current.

  5. Noether's theorems are discussed in numerous textbooks, but often with various shortcomings, presumably because the authors are only interested in special classes of applications. See also this Phys.SE post.


  1. W. Greiner & J. Reinhardt, Field quantization; p. 40.

  2. M.D. Schwartz, QFT and the Standard Model; p. 35.

  3. M.E. Peskin & D.V. Schroeder, An Intro to QFT; p. 19.

  • $\begingroup$ Comment on 2: After the coordinate transformation, the metric tensor itself, i.e. $\eta=\eta_{\mu\nu}\frac{\partial}{\partial_{\mu}}\frac{\partial}{\partial_{\nu}}$ is not changed, but the components, i.e. $\eta_{\mu\nu}$, will still change, unless the transformation is isometry. $\endgroup$
    – 346699
    Apr 19, 2017 at 14:55
  • $\begingroup$ $\uparrow$ Right. $\endgroup$
    – Qmechanic
    Apr 19, 2017 at 15:05
  • $\begingroup$ So if I make general coordinate transformation, the components $\eta_{\mu\nu}$ should vary. $\endgroup$
    – 346699
    Apr 19, 2017 at 15:11
  • $\begingroup$ Assuming that the metric tensor itself is not varied, the change of its components is a 'passive' effect of the coordinate transformation. $\endgroup$
    – Qmechanic
    Apr 19, 2017 at 15:24

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