How do we find the energy momentum tensor as Noether charge for translations in curved spaces. This should still exist since the action is still an integral over space such that it it invariant under translations right?

Attempt at a solution

Our Lagrangian will be of the form: $\mathcal{L} = \mathcal{L}[\phi, \partial_\mu \phi]$ and the corresponding Euler Lagrange equations are:

$$\frac {\partial \mathcal{L}}{\partial \phi} - \nabla_\mu \frac{\partial \mathcal{L}}{\partial \partial_\mu \phi} = 0$$

1)To obtain the Noether charge we must demand that the on-shell variation of $\mathcal{L}$ is a surface term indeed, we find that:

$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi}\delta \phi + \partial_\mu(\frac{\partial \mathcal{L}}{\partial \partial_\mu \phi}\delta \phi) - \partial_\mu(\frac{\partial \mathcal{L}}{\partial \partial_\mu \phi})\delta \phi$$ The two partial derivatives in the second and third terms can be changed into covariant derivatives since the additional christoffel symbols will cancel out. After doing so we find that the first and third terms cancel due to the equations of motion such that we end up with:

$$\delta \mathcal{L} = \nabla_\mu(\frac{\partial \mathcal{L}}{\partial \partial_\mu \phi}\delta \phi)$$

2)We must also study the variation of the Lagragian due to the variation of the fields these changes are:

$x^\mu \rightarrow x^\mu + \epsilon^\mu$ such that $ \phi \rightarrow \phi + \epsilon^\mu \partial_\mu \phi$ therefore we find that(for a free scalar):

$$\delta \mathcal{L} = -\partial_\mu \phi \partial^\mu \delta \phi$$ $$=-\partial_\mu \phi \partial^\mu(\epsilon^\kappa \partial_\kappa \phi)$$ $$=\epsilon^\kappa \partial_\kappa(-1/2 \partial_\mu \phi \partial^\mu \phi) = \epsilon^\kappa \partial_\kappa(\mathcal{L})$$

3)The next step is typically to state that $j_\mu$ = (1) - (2) is conserved $(\nabla_\mu J^\mu = 0)$ but I cannot find any way to rewrite (2) such that it contains only covariant derivatives. I think that I am mistaken somwhere in my derivation of point 2 but I cannot figure it out...

Any help would be greatly appreciated ! :)


3 Answers 3


Noether's current of local translation symmetry (=diffeomorphism invariance) in curved spacetime is indeed the stress-energy tensor. To derive it, remember that infinitesimal coordinate transformation, $x^m\rightarrow x^m+\epsilon^m$, induces the transformation of metric tensor, $g^{mn}\rightarrow g^{mn}+\nabla^m\epsilon^n +\nabla^n\epsilon^m$ ($\nabla^m$ = covariant derivative).

Vary the action w.r.t. metric and its derivative. $$ \delta S=\delta\int d^4x\sqrt{-g}\mathcal{L}=\int d^4x\left[ \frac{\partial\sqrt{-g}\mathcal{L}}{\partial g^{mn}}\delta g^{mn}+\frac{\partial\sqrt{-g}\mathcal{L}}{\partial(\partial_p g^{mn})}\delta(\partial_p g^{mn}) \right].\tag{1} $$

Integrate by parts the second term in $(1)$, then $$ \delta S=\int d^4x\left[ \frac{\partial\sqrt{-g}\mathcal{L}}{\partial g^{mn}}-\partial_p\left(\frac{\partial\sqrt{-g}\mathcal{L}}{\partial(\partial_p g^{mn})}\right) \right]\delta g^{mn}= \tag{2} $$ $$ =\int d^4x\sqrt{-g}\;T_{mn}\nabla^m\epsilon^n \tag{3} $$

where in the last step I used $\delta g^{mn}=\nabla^m\epsilon^n +\nabla^n\epsilon^m$ and defined stress-energy tensor $T_{mn}$ as $$ \frac{1}{2}\sqrt{-g}\;T_{mn}\equiv \frac{\partial\sqrt{-g}\mathcal{L}}{\partial g^{mn}}-\partial_p\left(\frac{\partial\sqrt{-g}\mathcal{L}}{\partial(\partial_p g^{mn})}\right). $$

Integrating by parts $(3)$, we finally have $$ \delta S=\int d^4x\sqrt{-g}\;\epsilon^n\nabla^m T_{mn}=0.\tag{4} $$

So there are 4 covariantly conserved currents $(T_{m})_{n}$ associated with 4 translations $\epsilon^n$.

This is general. For your particular case the second term in $(1)$ will vanish.

Update: $\nabla^m T_{mn}=0$ means that the stress-energy tensor is, generally speaking, only conserved if we take sufficiently small patch of spacetime (i.e. locally), where curvature is negligible. In that case Christoffel symbols would vanish and $\nabla^m T_{mn}$ would reduce to a proper conservation law $\partial^m T_{mn}=0$. Proper conservation is also possible whenever there are Killing vectors, as emphasized by Jerry Schirmer and Qmechanic.

  • $\begingroup$ Why doesn't the translation have any effect on the fields themselves ? And also, I am a bit confused now since other answers imply that translation symmetry is lost and still you seem to apply it ? $\endgroup$
    – gertian
    May 10, 2017 at 12:48
  • $\begingroup$ Translations do effect other fields and generate corresponding terms, but those terms will cancel due to the equations of motion, so it is sufficient to consider variation w.r.t. metric. As for the second question, I updated the answer. $\endgroup$
    – Kosm
    May 11, 2017 at 1:15

You don't have translation symmetry in general. You need to have a translational Killing vector of the metric in order to do this. If you assume this, then $L_{v}\sqrt{|g|} = 0$ (I'm sorry, neither \pounds \textsterling nor \mathsterling works in our version of MathJax, I mean lie derivative with respect to the translational killing vector), and the usual proof works, otherwise, the inhomogenity of the metric kills the noether charge.

  • $\begingroup$ The metric that I'm interested in(Rindler) does have a killing vecetor $\partial_t$ but how does this help me to do the usual proof ? ps: I think that \mathcal{} works $\endgroup$
    – gertian
    May 10, 2017 at 12:39

OP is asking how to apply Noether's first theorem:

How do we find the energy-momentum tensor as Noether charge for translations in curved spaces?

  1. The problem is that by assuming a generic curved spacetime manifold ($M,g$), we have lost the translational symmetry that we enjoyed in Minkowski space $\mathbb{R}^{3,1}$. There's no free lunch. As said in Jerry Schirmer's answer, we need a Killing symmetry to construct an on-shell conserved charge/quantity.

  2. However, we may still define the symmetric Hilbert stress-energy-momentum (SEM) tensor $T^{\mu\nu}$. Diffeomorphism invariance leads via either Noether's first or second theorem (Ref. 1 & Kosm's answer use the second theorem) to $$\nabla_{\mu} T^{\mu\nu}~\stackrel{m}{\approx}~0 \tag{A}$$ for an arbitrary metric $g_{\mu\nu}$, cf. my Phys.SE answer here. [Here the $\stackrel{m}{\approx}$ symbol means equality modulo matter eom. The connection $\nabla$ is the Levi-Civita connection.]

  3. It's important to realize that there are generically no on-shell conserved charges/quantities associated with eq. (A), cf. this Phys.SE post. We still need a Killing symmetry.


  1. R.M. Wald, GR, Appendix E.1.
  • $\begingroup$ Thank you for the broad answer with many relevant links! I now understand that I should not be trying to prove Noethers theory but rather construct $T^{\mu\nu}$ such that it is covariantly constant and symmetric and then contract it with killing vectors. One question remains: is there any "well known" derivation for $T^{\mu\nu}$ as the noether charge of diffeomorphism because I fail to find it (or is Kosm's answer exactly that ?) ps: I know what $T^{\mu\nu}$ is by construction, i just want to link it to a symmetry... $\endgroup$
    – gertian
    May 10, 2017 at 13:23
  • $\begingroup$ No, cf. point 3. $\endgroup$
    – Qmechanic
    May 11, 2017 at 8:09

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