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My problem is about finding the spatial components of canonical momenta $\pi_{i j}=\frac{\partial \mathcal{L}}{\partial \dot{h}_{i j}}$ corresponding to the Fierz-Pauli Lagrangian. I am using the metric convention $\eta_{\mu \nu}=(-,+,+,+,...)$. In this Lagrangian I have terms proportional to $\partial_{\mu}h^{\sigma \nu}$ and these terms will definitely make a contribution to the momenta. To find that contribution, I should first lower the indices. Since my field $h_{\mu \nu}$ is symmetric on its indices, I do the contraction in the following way:

$$ h^{\mu \nu} = \frac{1}{2}\left( \eta^{\mu \alpha}\eta^{\nu \beta} + \eta^{\mu \beta}\eta^{\nu \alpha} \right)h_{\alpha \beta}. $$

I have a big problem in evaluating two terms:

\begin{align} (1) & = \partial_{\mu}h_{\nu \lambda} \partial^{\nu}h^{\mu \lambda}, \\ (2) & = \partial_{\mu}h^{\mu \nu}\partial_{\nu}h. \end{align}

Let me take only the second one for simplicity. Here $h=h^{\mu}{}_{\mu} = \eta^{\mu \nu}h_{\mu \nu}$. So I have

\begin{align} (2) & = \partial_{\mu}h^{\mu \nu}\partial_{\nu}h = \frac{1}{2}\eta^{\sigma \lambda}\left( \eta^{\mu \alpha}\eta^{\nu \beta} + \eta^{\mu \beta}\eta^{\nu \alpha} \right)\partial_{\mu}h_{\alpha \beta}\partial_{\nu}h_{\sigma \lambda} \\ & = \frac{1}{2}\eta^{\sigma \lambda}\eta^{\mu \alpha}\eta^{\nu \beta}\partial_{\mu}h_{\alpha \beta}\partial_{\nu}h_{\sigma \lambda} + \frac{1}{2}\eta^{\sigma \lambda}\eta^{\mu \beta}\eta^{\nu \alpha}\partial_{\mu}h_{\alpha \beta}\partial_{\nu}h_{\sigma \lambda}. \end{align}

Summation over $\mu$ $(\mu = \{0,k\})$ gives

\begin{align} (2) = \frac{1}{2}\eta^{\sigma \lambda}\eta^{0 \alpha}\eta^{\nu \beta}\partial_{0}h_{\alpha \beta}\partial_{\nu}h_{\sigma \lambda} + \frac{1}{2}\eta^{\sigma \lambda}\eta^{k \alpha}\eta^{\nu \beta}\partial_{k}h_{\alpha \beta}\partial_{\nu}h_{\sigma \lambda} \\ + \, \frac{1}{2}\eta^{\sigma \lambda}\eta^{0 \beta}\eta^{\nu \alpha}\partial_{0}h_{\alpha \beta}\partial_{\nu}h_{\sigma \lambda} + \frac{1}{2}\eta^{\sigma \lambda}\eta^{k \beta}\eta^{\nu \alpha}\partial_{k}h_{\alpha \beta}\partial_{\nu}h_{\sigma \lambda}. \end{align}

Now, after contraction over $\alpha$ in the first line and contraction over $\beta$ in the second line I get

\begin{align} (2) = -\frac{1}{2}\eta^{\sigma \lambda}\eta^{\nu \beta}\partial_{0}h_{0 \beta}\partial_{\nu}h_{\sigma \lambda} + \frac{1}{2}\eta^{\sigma \lambda}\eta^{\nu \beta}\partial_{k}h_{k \beta}\partial_{\nu}h_{\sigma \lambda} \\ - \, \frac{1}{2}\eta^{\sigma \lambda}\eta^{\nu \alpha}\partial_{0}h_{\alpha 0}\partial_{\nu}h_{\sigma \lambda} + \frac{1}{2}\eta^{\sigma \lambda}\eta^{\nu \alpha}\partial_{k}h_{\alpha k}\partial_{\nu}h_{\sigma \lambda}. \end{align}

To get a result for canonical momenta I should only keep the $\partial_0 h_{ij}$ terms. Hence I take $\nu = 0$ and don't bother the spatial indices, i.e; I automatically drop the $\nu = k$ terms. In that case

\begin{align} (2) = -\frac{1}{2}\eta^{\sigma \lambda}\eta^{0 \beta}\partial_{0}h_{0 \beta}\partial_{0}h_{\sigma \lambda} + \frac{1}{2}\eta^{\sigma \lambda}\eta^{0 \beta}\partial_{k}h_{k \beta}\partial_{0}h_{\sigma \lambda} \\ - \, \frac{1}{2}\eta^{\sigma \lambda}\eta^{0 \alpha}\partial_{0}h_{\alpha 0}\partial_{0}h_{\sigma \lambda} + \frac{1}{2}\eta^{\sigma \lambda}\eta^{0 \alpha}\partial_{k}h_{\alpha k}\partial_{0}h_{\sigma \lambda}. \end{align}

Contraction over $\beta$ in the first line and contraction over $\alpha$ in the second line gives

\begin{align} (2) = \frac{1}{2}\eta^{\sigma \lambda}\partial_{0}h_{0 0}\partial_{0}h_{\sigma \lambda} - \frac{1}{2}\eta^{\sigma \lambda}\partial_{k}h_{k 0}\partial_{0}h_{\sigma \lambda} \\ + \, \frac{1}{2}\eta^{\sigma \lambda}\partial_{0}h_{0 0}\partial_{0}h_{\sigma \lambda} - \frac{1}{2}\eta^{\sigma \lambda}\partial_{k}h_{0 k}\partial_{0}h_{\sigma \lambda}. \end{align}

Here, although I am not sure, I assume theory has a gauge $\partial_0 h_{00}=0$ so first terms in both of the lines drop. Lastly I take $\sigma = i, \; \lambda = j$ and I obtain

\begin{align} (2) & = - \frac{1}{2}\delta_{i j}\partial_{k}h_{k 0}\partial_{0}h_{i j} - \frac{1}{2}\delta_{i j}\partial_{k}h_{0 k}\partial_{0}h_{i j} \\ & = - \delta_{i j} \partial_{k}h_{0 k} \dot{h}_{i j}, \end{align}

by using $\eta^{i j}=\delta^{i j} = \delta_{i j}$. All these calculations seems correct to me but somehow I am making a mistake at some point. The correct result should be

$$ (2) = 2\delta_{i j} \partial_{k}h_{0 k} \dot{h}_{i j}, $$

so that I have

$$ \frac{\partial (2)}{\partial \dot{h}_{i j}} = 2\delta_{i j} \partial_{k}h_{0 k}, $$

which is what is found in the article I am studying. Thus I have a sign mistake and also a missing factor of $2$. I also have a similar problem with the first term $(1)$. Can anybody show the point I am missing and guide me on that?


EDIT: Today I have learned that instead of ignoring $\partial_0 h_{00}$ terms, it would be better to use $$h=\eta^{\mu \nu}h_{\mu \nu} = \eta^{00}h_{00} + \eta^{ij}h_{ij}= -h_{00} + \eta^{ij}h_{ij},$$ so that $$ h_{00}= \eta^{ij}h_{ij} - h.$$ The $h$ term will possibly make a contribution to some other term in canonical momenta as I am told. But I am still not able to find the correct result.

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