4
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ Have you tried doing the minimization in the tetrad formalism and seeing how it compares? $\endgroup$ – Kyle Kanos Jul 16 '15 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 Jul 16 '15 at 20:49
2
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.