The action of a massless scalar field in curved spacetime is given by:

\begin{equation} S(\phi)=\int d^{4}x \sqrt{-g}\left(g^{\mu\nu}\phi_{,\mu}\phi_{,\nu}\right) \end{equation}

Now the action can be rewritten using the tetrad formalism as:

\begin{equation} S(\phi)=\int d^{4}x e\left(\eta^{ab}\phi_{,a}\phi_{,b}\right) \end{equation}

where $e$ is the determinant of the veilbein $e^{a}_{\mu}$ and we have used the identity $\phi_{,a}=e^{\mu}_{a}\phi_{,\mu}$.

Is it correct to assume that the equation of motion can be given by:

$$\partial_{a}\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_{a}\phi\right)}\right) - \frac{\partial\mathcal{L}}{\partial\phi} = 0 $$ where $\mathcal{L}$ is the Lagrangian density?

In which case we have explicitly:

$$e \eta^{ab}\phi_{,ab}+\eta^{ab}\phi_{,b}e_{,a}=0$$

  • $\begingroup$ Have you tried doing the minimization in the tetrad formalism and seeing how it compares? $\endgroup$
    – Kyle Kanos
    Commented Jul 16, 2015 at 20:18
  • 2
    $\begingroup$ Hi yess. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Commented Jul 16, 2015 at 20:49

1 Answer 1


The action of a massless scalar field is given by:

\begin{eqnarray} S(\phi)&=&\int{\cal{L}}dt\\ &=&\int d^{4}x \sqrt{-g}\left(g^{\mu\nu}\phi_{,\mu}\phi_{,\nu}\right) \end{eqnarray}

Now choosing a tetrad, i.e., a basis of one form at each spacetime point $\{e^{a}=e^{a}_{\mu}dx^{\mu}\}$ we can rewrite the action as:

\begin{eqnarray} S(\phi)&=&\int{\cal{L}}dt\\ &=&\int d^{4}x e\left(e^{\mu}_{a}e^{\nu}_{b}\eta^{ab}\phi_{,\mu}\phi_{,\nu}\right) \end{eqnarray}

where $e$ is the determinant of $e^{a}_{\mu}$ which is equal to $\sqrt{-g}$ and we have use the identity $e_{a}^{\mu}e_{b}^{\nu}\eta^{ab}=g^{\mu\nu}$.

Now the equations of motion are given by the Euler-Lagrange equations: \begin{equation} \frac{\delta \mathcal{S}}{\delta\phi}=\frac{\partial\mathcal{L}}{\partial\phi} -\partial_\mu \left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)=0. \end{equation}

So we have

\begin{eqnarray} \frac{\partial\mathcal{L}}{\partial\phi}&=&0\\ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}&=& e e_{a}^{\mu}e_{b}^{\nu}\eta^{ab}\phi_{,\nu} \end{eqnarray}

where the equation the motion takes the form:

\begin{equation} \partial_{\mu}\left(e e_{a}^{\mu}e_{b}^{\nu}\eta^{ab}\phi_{,\nu}\right)=0 \end{equation}

Now taking into account that $\partial_{b}=e^{\nu}_{b}\partial_{\nu} $ gives:

\begin{equation} \partial_{\mu}\left(e e_{a}^{\mu}\eta^{ab}\phi_{,b}\right)=0 \end{equation}

Then the equation of motion takes the form:

\begin{eqnarray} e\eta^{ab}\partial_{a}\phi_{,b}+e\eta^{ab}\phi_{,b}\partial_{\mu}(e_{a}^{\mu})+\eta^{ab}\phi_{,b}\partial_{a}e=0\\ \end{eqnarray}


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