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.

  • $\begingroup$ 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. $\endgroup$ Jul 24, 2018 at 16:00

1 Answer 1

  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.
  • $\begingroup$ 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? $\endgroup$ Jul 24, 2018 at 16:15
  • $\begingroup$ What else do you know about $T^{\mu \nu}$? Reread your equations and make sure you've not made any typos. $\endgroup$
    – gj255
    Jul 24, 2018 at 16:21
  • $\begingroup$ 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 $(\nabla_{\nu} K_{\mu})T^{\mu\nu}=0$ $\endgroup$ Jul 24, 2018 at 16:27
  • $\begingroup$ @Greg.Paul Bingo. $\endgroup$
    – gj255
    Jul 24, 2018 at 16:27
  • 1
    $\begingroup$ Regarding point (3) under debate here, I got nice answers to that exact question here. $\endgroup$
    – knzhou
    Jul 24, 2018 at 20:45

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