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If one consider the Maxwell action as $$S=-\int \mathrm{d^{4}}x\! \ \frac{1}{4}F_{ab}F^{ab} \,$$ one find the usual Maxwell equation $$\partial_{a}F^{ab}=0$$ Then one can simply arrive the following the Maxwell on-shell action $$-\int \mathrm{d^{4}}x\! \ \frac{1}{2}\partial_{a}(A_{b}F^{ab}) \,$$

Now my question is for Einstein Hilbert action. What is the expression of on-shell Einstein Hilbert action $$S=\int \mathrm{d^{4}}x\! \ R \,$$ I know how to find Einstein equation from variational principle, which is given as $$R_{ab}-\frac{1}{2}g_{ab}R=0$$

How to write on-shell Einstein Hilbert action with above equation?

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  • $\begingroup$ Comment to the question (v2): There are several notions of on-shell actions. Consider including a definition or reference for clarity. $\endgroup$ – Qmechanic Jul 2 '14 at 11:25
  • $\begingroup$ By on-shell action, I mean the action - the equation of motion. For clarity I have included the example with Maxwell action. $\endgroup$ – mattrodric Jul 2 '14 at 11:31
  • $\begingroup$ I want a similar form for On-shell Einstein Hilbert action just like the-shell Maxwell action$$-\int \mathrm{d^{4}}x\! \ \frac{1}{2}\partial_{a}(A_{b}F^{ab}) \,$$ $\endgroup$ – mattrodric Jul 2 '14 at 11:35
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The action you're considering yields Einstein's equations in vacuum, so $R=0$ (this follows immediately from contracting Einstein's equations). Therefore the action vanishes on shell.

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  • $\begingroup$ That is true. But my question is more general. Suppose one consider Einstein-Hilbert-Maxwell action. Then there will be two equations of motion, one for metric and one for gauge field. The what would be the form of on-shell action then? $\endgroup$ – mattrodric Jul 2 '14 at 11:42
  • $\begingroup$ Then the equations of motion will be coupled: due to the Maxwell term it's not a vacuum solution anymore. In this case the EE become $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=\kappa T_{\mu\nu}$ and the Maxwell equations become $D_{\mu}F^{\mu\nu}=0$. If for example we consider only electric fields in four dimensions, a solution is given by the Reissner Nordstrom metric together with $A_0=-Q/r+\Phi$, where $Q$ is the black hole charge and $\Phi$ is constant. As it turns out, the EM tensor is traceless in this specific case so that $R=0$ again. $\endgroup$ – ScroogeMcDuck Jul 2 '14 at 11:52
  • $\begingroup$ In fact a more general result holds: the EM tensor for Maxwell fields is $T_{\mu\nu}=F_{\mu\rho}F_{\nu}^{\rho}-\frac{1}{4}F_{\rho\sigma}F^{\rho\sigma}$, so clearly $T_{\mu}^{\mu}=0$ in four dimensions. Hence $R=0$ for the Einstein-Hilbert-Maxwell action (but only in 4 dimensions!), so that you're only left with the on-shell Maxwell part. $\endgroup$ – ScroogeMcDuck Jul 2 '14 at 12:00

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