It is known that the Hamiltonian action principle for general relativity (with cosmological constant) is \begin{align} I_H\left[\gamma_{ij},\pi^{ij}\right]=\frac{1}{16\pi G}\int dt\;\int d^{n-1}x\;\left(\pi^{ij}\dot{\gamma}_{ij}-N\mathcal{H}-N^{i}\mathcal{H}_i\right), \end{align} where $\gamma_{ij}$ and $\pi^{ij}$ are the spatial metric and its canonical momenta, respectively, and \begin{align} \mathcal{H}&=\frac{1}{\sqrt{\gamma}}\left(\pi_{ij}\pi^{ij}-\frac{1}{n-2}\pi^2\right)-\sqrt{\gamma}\left(^{(n-1)}R-2\Lambda\right),\\ \label{eq: Shift definition} \mathcal{H}_i&=-2\pi^j_{\;\;i|j}, \end{align} where $\pi^j_{\;\;i|j}$ is the spatial covariant derivative with index $j$ of $\pi^{j}_i$. I am trying to reproduce the following Hamiltonian equation of motion \begin{align} \dot{\pi}^{ij}&=\frac{1}{2}\gamma^{-1/2}N\gamma^{ij}\left(\pi_{kl}\pi^{kl}-\frac{1}{n-2}\right)-2\gamma^{-1/2}N\left(\pi^{ik}\pi_{k}^{j}-\frac{1}{n-2}\pi\pi^{ij}\right)-N\gamma^{1/2}\left(R^{ij}-\frac{1}{2}\gamma^{ij}\left(R-2\Lambda\right)\right)\\ &\hspace{2cm}+\gamma^{1/2}\left(N^{|ij}-\gamma^{ij}\Box N\right)+{\color{red}{\gamma^{1/2}\left(\gamma^{-1/2}\pi^{ij}N^k\right)_{|k}}}-2\pi^{k(i}N^{j)}_{\;\;\,|k}. \end{align} Except for the red term, I obtained the remaining terms by taking the variation of the Hamiltonian action with respect to $\gamma^{ij}$. I suspect that the red term appears from $\mathcal{H}_i$, however, I am unable to obtain it.

  • $\begingroup$ Did you vary the action with respect to $\gamma^{ij}$ or $\gamma_{ij}$? You say the former but the position of your indices on $\dot{\pi}^{ij}$ indicates the latter. They're not quite the same thing and it's important to be consistent. $\endgroup$ Commented Jun 23, 2021 at 15:54
  • $\begingroup$ I haven't carefully checked, but to a lazy eye, it appears that your factors of $\sqrt{\gamma}$ aren't right, and note that the red term differs from $\frac{\delta}{\delta_{\gamma_{ij}}} \left(\sqrt{\gamma}H_{a}N^{a}\right)$ by constant factors $\endgroup$ Commented Jun 23, 2021 at 15:58

1 Answer 1


Note that when you vary the spatial metric, you must also vary the spatial covariant derivative so that it remains compatible with the spatial metric. Specifically, if $D_a$ is the spatial covariant derivative, we have \begin{align*} \delta(D_i \pi^{j} {}_k) &= \delta( \gamma_{kl} D_i \pi^{jl}) \\ &= (\delta \gamma_{kl}) D_i \pi^{jl} + \gamma_{kl} \delta C^{j} {}_{im} \pi^{ml} + \gamma_{kl}\delta C^l {}_{im} \pi^{jm} + \gamma_{kl} D_i \delta\pi^{jl} \end{align*} where $$ \delta C^{j} {}_{im} = \frac{1}{2} \gamma^{jn} \left[ D_i \delta\gamma_{mn} + D_m \delta\gamma_{in} - D_n \delta \gamma_{mn} \right]. $$ You would then perform an integration by parts on the resulting expression & relabel indices to obtain something of the form $\delta\gamma_{ij}$ contracted with some other tensor; this should yield the given term.

  • $\begingroup$ Let me know if you need more detail here; I have to run to lunch now, but I can return later to fill in more of the gaps. $\endgroup$ Commented Jun 23, 2021 at 16:07
  • $\begingroup$ Maybe this is a stupid question, however, if I perform the variation I obtained: \begin{align} \delta(D_i \pi^j_k)&=\delta\left(\gamma_{kl}D_i \pi^{jl}\right)\\ &=\delta \gamma_{kl}D_i \pi^{jl}+\gamma_{kl}\delta\left(\partial_i \pi^{jl}+\Gamma^j_{im}\pi^{ml}+\Gamma^l_{im}\pi^{jm}\right)\\ &=\delta \gamma_{kl}D_i \pi^{jl}+\gamma_{kl}\delta \Gamma^j_{im}\pi^{ml}+\gamma_{kl}\delta\Gamma^l_{im}\pi^{jm}+\gamma_{kl}D_i \delta\pi^{jl}, \end{align} which differs from yours, up to a metric factor, right? $\endgroup$ Commented Jun 24, 2021 at 0:56
  • $\begingroup$ I have another, maybe stupid, one: Why I cannot perform \begin{align} N^i\delta\left(D_i \pi^j_k\right)&=N^i\delta\left(\gamma_{kl}D_i\pi^{jl}\right)\\ &=N^i\delta\left( \gamma_{kl}D_i\pi^{jl}+\gamma_{kl} D_i\delta \pi^{jl}\right) \end{align} and then proceeding to integrate by parts? Thanks! $\endgroup$ Commented Jun 24, 2021 at 1:01
  • $\begingroup$ @SoniaLlambias: I forgot the metric terms in the last step of my first equation; thanks for pointing that out to me. I'll correct this shortly. Concerning your second comment, I'm not sure exactly where your $\delta$s are being applied in the second step; with the parentheses where they are, it seems like you're taking the variation of a variation on the last term. Can you clarify? $\endgroup$ Commented Jun 24, 2021 at 13:35

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