I've been reading up on how the linearised Einstein field equations can be derived from a field theoretic perspective, by constructing the most general Lagrangian for a massless spin-2 field, $h_{\mu\nu}$, compatible with locality and Lorentz invariance. This is given by $$\mathcal{L}=a\partial_{\lambda}h^{\mu\nu}\partial^{\lambda}h_{\mu\nu}+b \partial_{\mu}h^{\mu\lambda}\partial^{\nu}h_{\lambda\nu}+c\partial_{\mu}h\partial_{\nu}h^{\mu\nu}+d\partial_{\mu}h\partial^{\mu}h$$ Now, the requirement of Lorentz invariance requires that $\mathcal{L}$ is invariant under the transformation $$h_{\mu\nu}\rightarrow h_{\mu\nu}+2\partial_{(\mu}\xi_{\nu)}$$ where $\xi^{\mu}$ is a generic 4-vector. This transformation corresponds exactly to a gauge transformation of the spin-2 field.

In principle, one should be able to determine the forms of the coefficients $a$, $b$ $c$ and $d$. However, it seems impossible to do so without relaxing the condition of strict Lorentz invariance of $\mathcal{L}$ to requiring that it is invariant up to a boundary term. What does this mean for the gauge invariance of the theory? If it is to correspond to GR then the theory should be invariant under these linearised diffeomorphisms, right?

What am I missing? If someone could enlighten me on this it would be much appreciated.

  • $\begingroup$ "...the requirement of Lorentz invariance requires..." no, that transformation corresponds to diffeomorphism invariance. Your Lagrangian is Lorentz invariant regardless of this latter invariance. $\endgroup$ Nov 6, 2017 at 21:30
  • $\begingroup$ @AccidentalFourierTransform If one studies how the spin-2 field transforms under Lorentz transformations though, one finds that $U(\Lambda)h^{\mu\nu}U^{-1}(\Lambda)=(\Lambda^{-1})^{\mu}_{\;\;\sigma}(\Lambda^{-1})^{\nu}_{\;\;\rho}h^{\sigma\rho} +2\partial^{(\mu}\xi_{_{\Lambda}}^{\nu)}$, so isn't such a transformation needed to negate the non-covariant part of the transformation? $\endgroup$
    – Will
    Nov 6, 2017 at 21:42
  • $\begingroup$ Oh, now I see what you mean. You are going Weinberg style, aren't you? E.g., $h^{0\mu}=\partial_i h^{i\mu}=0$, etc. $\endgroup$ Nov 6, 2017 at 21:56
  • $\begingroup$ @AccidentalFourierTransform I've actually been following these notes: google.co.uk/url?sa=t&source=web&rct=j&url=http://… but this may be equivalent to Weinberg. I didn't intend to gauge fix $h_{\mu\nu}$. Does the spin-2 field only transform like this under Lorentz transformations if it is gauge fixed? $\endgroup$
    – Will
    Nov 6, 2017 at 22:01
  • $\begingroup$ Yep, that document has "Weinberg" written all over it. See e.g. the references. $\endgroup$ Nov 6, 2017 at 22:07

1 Answer 1


First of all, as soon as one neglects curvature by resorting to a linear perturbation around the Minkowski metric, one is then forced to use the tenets of classical field theory. One of these is that the fields (in this case the 10 independent $h$ functions) are Schwartz test functions, i.e functions whose modulus and spacetime differentials of all orders go fast to 0, as the „radius of Minkowski spacetime" is pushed to $+\infty$. In this simple perspective, there is no discussion of "Lorentz invariance of the Lagrangian (density)". The true invariance is imposed upon the Lagrangian action, i.e. to the integral on the flat spacetime continuum of the Lagrangian density. So there's no "relaxing" needed, it is all there in the standard assumptions of classical field theory.

Feynman in his GR lectures explains how one derives the a,b,c,d factors. That is an elegant method reminding me of the Pauli and Fierz one. On the other hand, one can linearize by brute force the Hilbert-Einstein action and get the 4 coefficients. This works, too.

  • $\begingroup$ I guess I'm just used to the spin-1 case of electromgnetism in which the Lagrangian density is already gauge invariant before considering the action. $\endgroup$
    – Will
    Nov 6, 2017 at 21:49
  • $\begingroup$ In classical field theory functions are not in general of the Schwartz type. Or rather, there is no need to assume them to be. For our purposes, it suffices to take $h\in C^2_c(\mathbb R^4)$. $\endgroup$ Nov 6, 2017 at 22:00
  • $\begingroup$ Note also that OP is taking $h$ to be the gauge-fixed field (Coulomb? DeDonder?), where $h^{0\mu}=\partial_i h^{i\mu}=h^\mu{}_\mu=0$ , so the field has only two independent components. $\endgroup$ Nov 6, 2017 at 22:03
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    $\begingroup$ Well, this is a too stringent assumption. Spacetime is in principle infinite, so there is a reason to expect that fields "stretch" to infinity. $\endgroup$
    – DanielC
    Nov 6, 2017 at 22:03

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