# General Relativity as a Special Relativistic Field Theory

In this question, I want to consider only the classical case. I have seen the statement that general relativity can be considered as a spin-2 field living on a Minkowski background. In that case, you start with some inertial coordinate system, and you consider a propagating spin-2 field $h_{\mu \nu}$. You impose the gauge invariance $\delta h_{\mu \nu} = \partial_{\mu} \xi_{\nu} + \partial_{\nu} \xi_{\mu}$. Then you can show that even if you start with only linearized terms in the action, you have to introduce all higher order terms in the Einstein-Hilbert action. I have two questions:

1-) Using the gauge invariance, one can show that $h_{\mu \nu}$ must couple to a conserved energy-momentum tensor $t_{\mu \nu}$. That is, you define $t_{\mu \nu}$ by $\delta S \sim \int t^{\mu \nu}\delta h_{\mu \nu}$ , and the gauge invariance condition gives you $\partial _\mu t^{\mu \nu} = 0$. However, at the end, we need to identify $\xi^\mu$ with diffeomorphisms, so under a gauge transformation, $t^{\mu \nu}$ should also vary. However, in the literature, as far as I can see, the converse is assumed. Why?

EDIT: More properly, in the reference I have added below, he couples $h_{\mu \nu}$ to matter energy-momentum tensor $T_{\mu \nu}: \int h_{\mu \nu}T^{\mu \nu}$. Then he uses the convergence of matter energy-momentum tensor (this is assumed at the beginning, then you show that it can't be) to show that the coupling term in the action is invariant under gauge transformations. However he assumes that, $\delta T_{\mu \nu}=0$ under a gauge transformation.

2-) One can show in special relativity that a particle can't go faster than the speed of light because of causality. However, in general relativity, "coordinate velocity" $\left| \frac{dx^i}{dx^0}\right|$ can be larger than the speed of light. This should also be true in the formalism where you consider general relativity as a special relativistic field theory. However, this would then contradict the proof which shows that if the action is invariant under global Lorentz symmetry, then nothing can go faster than the speed of light. That is, even if you include $h_{\mu \nu}$, the action is still Lorentz invariant, and it is a special relativistic theory. What am I missing ?

Reference: For first question, see "Gravitational Waves Volume 1: Theory and Experiments" by Michele Maggiore, page 77, especially (2.109).

• What is "the converse"? Also, conservation of the gauge-coupling current is a consequence of the global version of a symmetry (i.e. translation in this case), not the gauge symmetry. – ACuriousMind Feb 22 '16 at 19:40
• @ACuriousMind They assume that $t_{\mu \nu}$ doesn't change under a gauge transformation. About your second comment, if you assume that action is invariant (up to a boundary term, lets say) than you can show the conservation of current using local gauge symmetry. Can you show what is wrong ? Perhaps you have a different definition of current than I have used in question. Precisely speaking, symmetric part of $t^{\mu \nu}$ is conserved. Thanks. – atlaspbody Feb 22 '16 at 19:50
• One can usually derive all conservation laws that are "from the gauge symmetry" also from the global symmetry + the matter equations of motions. Also, can you give a reference where it is assumed that $t^{\mu\nu}$ doesn't change under a gauge transformation? (It has two free indices, of course it changes) – ACuriousMind Feb 22 '16 at 19:53
• @ACuriousMind I understand your point, but in that case I think you can derive it as an off-shell conservation law. Actually, I am not sure if they argue like this, but in any case, my actual question is still there. I will add a reference in a moment. – atlaspbody Feb 22 '16 at 20:40
• @ACuriousMind The reference I added doesn't use the same argument to show that divergence of $t_{\mu \nu}$ is 0. Rather, he couples $h_{\mu \nu}$ to matter energy-momentum tensor, and argue that since matter energy momentum tensor is divergenceless coupling term $\int h_{\mu \nu} T^{\mu \nu}_matter$ should be invariant under gauge transformations. But he assume that under a gauge transformation $\delta T^{\mu \nu}=0$. – atlaspbody Feb 22 '16 at 20:47