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I am following along Chapter 2 of Takagi's Vacuum Noise and Stress Induced by Uniform Acceleration. For a free real scalar field $\phi$ the stress-energy tensor is: $$ T_{\mu\nu} = ( \partial_{\mu} \phi ) ( \partial_{\nu} \phi ) - g_{\mu\nu} \tfrac{1}{2} g^{\alpha\beta} ( \partial_{\alpha} \phi ) ( \partial_{\beta} \phi ) - \tfrac{1}{2} g_{\mu\nu} m^2 \phi^2 $$ For $K$ a timelike Killing vector of the spacetime, define: $$ H_{K} = - \int_{\Sigma} d^3\Sigma_{\nu}\ K^{\mu} T_{\mu}^{\ \nu} $$ where $\Sigma$ is a spacelike hypersurface and $d^3\Sigma_{\nu}$ the 3-volume 1-form over this surface. Then $H$ is a conserved charge and is independent of the choice of $\Sigma$ used to integrate it.

Takagi says that $K^{\mu} T_{\mu}^{\ \nu}$ is a conserved vector. So I have two questions:

1. Does $K^{\mu} T_{\mu}^{\ \nu}$ being a 'conserved vector' mean that it obeys $\partial_{\nu} K^{\mu} T_{\mu}^{\ \nu}= 0$? If this is true, how do I see this?

2. What does it mean that $H_K$ is a conserved charge? Does it mean $\mathcal{L}_{K} H_{K} = 0$ (Where $\mathcal{L}_{K}$ is the Lie derivative)? Normally you'd have $K = \frac{\partial}{\partial x^0}$ for ordinary Minkowski time and so I'd understand $H_{\partial_0}$ being conserved as the statement $\frac{\partial}{\partial x^0} H_{\partial_0} = 0$

EDIT: I've also read the following statement in DeWitt's A Global Approach to Quantum Field Theory: In a general stationary background $H_{K}$ is the only conserved charge that there is for this system. Why is this true? I know that in a general stationary spacetime there exists one global timelike Killing vector, but independent of this isn't it still true that $T_{\mu\nu}$ is a conserved current? To me it seems that there should still be four corresponding conserved charges, independent of whether the spacetime is stationary or not.

QuantumEyedea
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  • Just try to evaluate $\nabla_\mu (T^{\mu \nu} K_\nu) $, using that $\nabla_\mu T^{\mu \nu} = 0$ and the definition of a Killing vector. – TempAccount2020 Jul 24 '18 at 16:00

1 Answers1

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  1. $K_\mu T^{\mu \nu}$ being conserved means it has no covariant divergence:

    $$ \nabla_\nu (K_\mu T^{\mu \nu}) = 0 \,.$$

    To see this, expand using the product rule and apply energy-momentum conservation and Killing's equation. Note that $\partial_\nu (K_\mu T^{\mu \nu})$ is not a scalar.

  2. $H_K$ is not some field defined through spacetime so Lie derivatives aren't really appropriate here. The statement that $H_K$ is independent of $\Sigma$ can be proved as follows: for any $\Sigma, \Sigma'$ consider the volume in spacetime bounded by these two surfaces along with the timelike boundary at spatial infinity. Then integrate $ \nabla_\nu (K_\mu T^{\mu \nu})$ over this volume and apply the divergence theorem, assuming that $T_{\mu \nu}$ vanishes sufficiently quickly at spatial infinity. If we choose $\Sigma = \Sigma_t$ to be a surface of constant $t$, then this result can be specialised to $$ \frac{\mathrm{d}}{\mathrm{d}t} H_K(\Sigma_t) = 0 \,,$$ which is what is meant by $H_K$ being conserved.
  3. The conservation law $\nabla_\mu T^{\mu \nu} = 0$ is a little different from a conservation law of the form $\nabla_\mu J^\mu = 0$. The presence of an extra free index in the former case means we cannot apply the usual (covariant) divergence theorem to conclude the existence of a conserved charge.
gj255
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  • Thanks very much! Killing's equation is $\nabla_{\nu} K_{\mu} + \nabla_{\mu} K_{\nu} = 0$. This along with $\nabla_{\nu} T^{\mu\nu}=0$ seems to tell me that $\nabla_{\nu} ( K_{\mu} T^{\mu\nu})=(\nabla_{\nu} K_{\mu}) T^{\mu\nu}+K_{\mu}(\nabla_{\nu} T^{\mu\nu})=-\nabla_{\mu} K_{\nu} \neq 0$. What am I missing here? – QuantumEyedea Jul 24 '18 at 16:15
  • What else do you know about $T^{\mu \nu}$? Reread your equations and make sure you've not made any typos. – gj255 Jul 24 '18 at 16:21
  • Ah I think I got it, we need that $T$ is symmetric. The above equation implies $(\nabla_{\nu} K_{\mu})T^{\mu\nu} = - (\nabla_{\mu} K_{\nu}) T^{\mu\nu}$. Swapping indices on the RHS and then using $T^{\nu\mu}=T^{\mu\nu}$ gives $(\nabla_{\nu} K_{\mu})T^{\mu\nu} = - (\nabla_{\nu} K_{\mu}) T^{\mu\nu}$ and so $(\nabla_{\nu} K_{\mu})T^{\mu\nu}=0$ – QuantumEyedea Jul 24 '18 at 16:27
  • @Greg.Paul Bingo. – gj255 Jul 24 '18 at 16:27
  • Could you expand on your last point 3.? I don't understand. Why is the conservation law different when you have the extra index? I think I recall in the derivation of $T$ via Noether's theorem (from translational invariance) that you need to peel a shift vector $s$ off of $s_{\mu}T^{\mu\nu}$ to define $ T^{\mu\nu}$, even though it is the overall quantity $J^{\nu} \equiv s_{\mu}T^{\mu\nu}$ that is the actual conserved current. Does it have something to do with this? – QuantumEyedea Jul 24 '18 at 17:34
  • @Greg.Paul I think that's only how things work in flat spacetime. In curved spacetime there aren't necessarily any global translational symmetries, but $T^{\mu \nu}$ is still conserved in the sense that $\nabla_\mu T^{\mu \nu} = 0$. But since $T$ is a rank-2 tensor, the covariant derivative acts on it differently to how it acts on vectors. – gj255 Jul 24 '18 at 18:10
  • @gj255 Regarding 3rd item. $\nabla_\mu T^{\mu\nu} = 0$ implies $P_\mu \equiv \int dV T_{0\mu}$ is conserved. So, the conserved charge is the four-momentum. – Oktay Doğangün Jul 24 '18 at 18:59
  • @Oktay_Doğangün I think only when we have $\mathcal{K} = \frac{\partial}{\partial x^0}$, where $x^0$ is ordinary Minkowski time. The argument being made here is that if you have arbitrary coordinates you can't define the $P_{\mu}$ this way... – QuantumEyedea Jul 24 '18 at 19:25
  • @OktayDoğangün That's not true. Consider for instance that the total energy of radiation decreases in an expanding (say de Sitter for finiteness) universe. – gj255 Jul 24 '18 at 19:54
  • @gj255 These equations here are strictly local, there can not be global conservation law for spacetime symmetries in GR (curved spacetime), but these are not global considerations. As a matter of fact, $\nabla_\mu T^{\mu\nu} (x)=0$ is also local, one can not generalize it to distinct time slices (except some special cases: SR, eternal cosmologies with vanishing $\Lambda$, etc). The symmetry that implies vanishing divergence of T is diffeomorphism invariance, which is actually "local translation invariance". – Oktay Doğangün Jul 24 '18 at 20:16
  • @OktayDoğangün Yes, precisely. You cannot integrate $\nabla_\mu T^{\mu \nu} = 0$ to generate some globally conserved charge. This is in contrast to currents $J$ which satisfy $\nabla_\mu J^\mu = 0$. This is exactly the content of the third point in my answer. – gj255 Jul 24 '18 at 20:33
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    Regarding point (3) under debate here, I got nice answers to that exact question here. – knzhou Jul 24 '18 at 20:45
  • Indeed, you can not integrate over the spacetime, but you can integrate over a timeslice volume where you get the definition of P. Otherwise you could not predict, for example, that binary pulsars lose their energy in the form of gravitational waves. By the way, currents are also considered as local conservation laws, but of course in QFT they also apply globally (due to SR). – Oktay Doğangün Jul 24 '18 at 21:07
  • @OktayDoğangün There are subtleties with defining the energy in the gravitational field, which is of relevance for binary pulsars, of course, but this is not what I'm talking about. I'm talking about the energy-momentum of matter in some fixed background. In that case, there is a conserved charge $Q$ corresponding to any conserved current $J^\mu$, and there is not necessarily a conserved charge $P$ corresponding to $T^{\mu \nu}$. Try integrating $\nabla_\mu T^{\mu 0}$ over some 'timeslice volume' – you won't get a conserved energy! – gj255 Jul 24 '18 at 21:33