I have trouble calculating equations of motion for scalar and Yang-Mills field in tetrad formalism. Action for the gravitational field is given by: $$S= \int \epsilon_{abcd} e^a \wedge e^b \wedge R^{cd},$$ and action for scalar and Yang-Mills field is: $$S = \int |e| \left ( g^{\mu\nu} \partial_\mu \Phi \partial_\nu \Phi + g^{\mu\nu}g^{\rho\sigma}\,\mathrm{tr}\,(F_{\mu\rho}F_{\nu\sigma}\right))d^4 x \,,$$ where $g_{\mu\nu} = e_\mu^a e_\nu^b \eta_{ab}$ and $\partial_{\mu}\Phi=\nabla_{\mu}\Phi=e_{\mu}^a \nabla_a \Phi$. I also have that: \begin{equation} \frac{1}{4!}\epsilon_{\mu \nu \rho \sigma} d x^\mu \wedge d x^\nu \wedge d x^\rho \wedge d x^\sigma = d^4 x \end{equation} and \begin{equation} \epsilon^{\mu \nu \rho \sigma}= |e| \epsilon^{a b c d} {e^\mu}_{a} {e^\nu}_{b} {e^\rho}_{c} {e^\sigma}_{d}. \end{equation} I can write scalar part as: $$ S=\int |e|^2 \eta^{mn}\nabla_m \Phi \nabla_n \Phi \epsilon_{abcd}e^a \wedge e^b \wedge e^c \wedge e^d, $$ and taking $\frac{\delta S}{\delta \Phi}$ variation and using partial integration I would get equation of motion for scalar field. I know we have $\nabla_a \eta^{mn}= 0$, but what are $\nabla_a e_b$ and $\nabla_a|e|$? I'm not very familiar with tetrad formalism and any help would be very much appreciated.

Also, is there any smarter way to write down Yang-Mills action to simplify calculation of $\frac{\partial S}{\partial A_a}$, where $$F_{\mu \nu}^a=\partial_\mu A_\nu^a-\partial_\nu A_\mu^a+\frac{1}{2}f^{abc}A^a_{[\mu} A^b_{\nu]}?$$


1 Answer 1


The action for the gravitational field is simply the Einstein Hilbert action expressed in the language of differential forms, i.e: $$S=\int \mathcal R^{cd}\wedge \star e_{ab}$$ where $\star e_{ab}\equiv \star (e_a\wedge e_b)$ is just a shortcut for brevity. Variation wrt the vielbein results into the standard EFE. You can either choose first or second order formulation to do the variations (standard exterior calculus). If you choose the first order approach, then you will also vary wrt the connection $\omega$ to get the compatibility condition, i.e you will consider $\mathcal R^{ab}(\omega)$ and proceed. Otherwise, (second order) you will have to find an expression that relates the variation of $\omega$ with the variation of the vielbein by introducing a new (for simplicity) torsionless connection $\tilde \omega$ and parameterising the difference of the latter from the Levi-Civita connection $\omega$. Use $\tilde \omega=\omega+\delta \omega$, $\tilde e^a=e^a+\delta e^a$ and $\tilde e_a=e_a-\delta e_a$, keep perturbations first order and use all assumptions plus the tetrad postulate. I suggest you do it a la Palatini, since it is by far easier in this case. For your aid: $$\mathcal R^{ab}=\mathcal D\omega^{ab}=\dfrac{1}{2}R^{ab}{}_{cd}e^c\wedge e^d\quad \star e_{a_1\cdots a_q}=\dfrac{1}{(D-q)!}\epsilon_{a_1\cdots a_q a_{q+1}\cdots a_D}e^{a_{q+1}\cdots a_D}$$

where $D$ is the dimension of the manifold and $\mathcal D$ the covariant exterior derivative, defined by: $$\mathcal D\phi^a_b=\mathrm d\phi^a_b+\omega^a{}_c\wedge \phi^c_b-(-1)^p\phi^a_c\wedge\omega^c{}_b $$ for a $p$-form $\phi^a_b$. Use also the fact that $\epsilon$ is fully antisymmetric and $$\epsilon_{a_1\cdots a_q a_{q+1}\cdots a_D}\epsilon^{b_1\cdots b_q a_{q+1}\cdots a_D}=(D-q)!\delta^{b_1\cdots b_q}_{a_1\cdots a_q}$$ where $\delta^{b_1\cdots b_q}_{a_1\cdots a_q}$ is the gKd. Furthermore, the no torsion condition generally translates to $\mathcal D e^a=0$ and hence, $\mathcal D(\star e_{a_1\cdots a_q})=0$. Finally, $e^{a_1\cdots a_D}=\epsilon^{a_1\cdots a_D}e^{1\cdots D}=\epsilon^{a_1\cdots a_D}\star 1$. These are almost all relations you will need and if you are stuck remember to always use integration by parts, wedge and Levi-Civita symbol antisymmetries and Stokes theorem to drop the boundary terms.

Your kinetic terms is of the form $-2X\star 1={\phi}_{;}{}^a\phi_{;a}\star 1$, so I think there is no need to even use exterior algebra here, just use $\eta_{ab}$ for the indices and you get $-2\delta_\phi X=2(\delta \phi)_;{}^a\phi_{;a}$. Do integration by parts, Stokes and then the result is $\delta\phi\Box\phi\star 1$. Remember that $\phi,\delta\phi,\Box\phi$ are all 0-forms (scalars), that is why I'm dropping the wedge. This term however also contributes to the vielbein eom, but that is something you can calculate on your own. You can show the equivalence between $-2X\star 1$ and $\phi_{;}{}^a\mathrm d\phi \wedge \star e_a$. For your aid I'm noting here that $\mathcal D\phi=\mathrm d\phi$ and $\delta_e \mathrm d\phi=0$. Also, it may come handy that $\star 1=|e|\mathrm d^4x$ and $\star e_{a}=i_{e_a}\star 1$, where $i_v$ is the interior product wrt the vector field $v$ (google it).


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