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I'm familiar with the Lagrangian for linearized gravity about a flat background, $$ \mathcal{L} = \frac{1}{2}[(\partial_\mu h^{\mu\nu} \partial_\nu h - \partial_\mu h^{\rho \sigma} \partial_\rho h^\mu … I am interested in the Lagrangian for linearized gravity around a curved background. My background metric is diagonal and the only source is a cosmological constant. …

Action for linearized gravity is well-known, see for example David Tong: Lectures on General Relativity: $\mathbf{The\;Fierz-Pauli\;Action}$ The linearised equations of motion can be derived from … (At linear order, the expansion of the Lagrangian is equal to the linearised Ricci scalar $(5.4)$ which is a total derivative.) $$-200-$$ This action is invariant under diffeomorphism transformation …

Following a derivation for the action in linearized gravity I have reached the point where I have the following expression $$S = \frac{1}{16 \pi G} \int d^{4}x \sqrt{-g} R.$$ I know that $$R = \partial … Combining the $\frac{h_{\nu}^{\nu}}{2}$ term with the two terms in R gives the two middle terms in the final Lagrangian density. …

The Einstein-Hilbert Lagrangian is: $$\mathcal{L}_{EH}=\sqrt{-g} R$$ where $g={\rm Det}[g_{\mu\nu}]$ and $R$ is the Ricci scalar. … So I really don't see why the first order term in the Einstein-Hilbert Lagrangian should vanish? …

I'm doing some field theory, on linearized gravity backgrounds by treating $h_{\mu\nu}$ as a massless spin-2 field. I can't seem to find the ${\cal O}(h^3)$ terms in the Lagrangian anywhere. … Are there any known references that just lists the next to leading order terms in the above Lagrangian? …

My attempt: I thought the only straight forward way to do this was by showing that the Lagrangian density of the theory is lorentz invariant. … According to Carroll the Lagrangian density is $$ \mathcal{L} = \frac{1}{2} \left[ (\partial_\mu {h^{\mu \nu }} )(\partial_\nu h) - (\partial_\mu {h^{\rho \sigma}} )(\partial_\rho {h^{\mu }}_{ \sigma} …

-1, -1, -1). $$ I have the question about lagrangian of fields in this case: in the book it is written in a form $$ L_{f} = -\frac{1}{32 \pi G}\left( \frac{1}{2}(\partial_{\alpha}h_{\nu \beta}) \partial … alpha} \bar {h}_{\mu \alpha })\partial_{\beta}\bar {h}^{\mu \beta}\right), \qquad (1) $$ where $$ \bar {h}^{\mu \nu} = h^{\mu \nu} - \frac{1}{2}\eta^{\mu \nu}h^{\alpha}_{\alpha} $$ It is argued that this lagrangian …

Let's say we consider Einstein's Lagrangian from GR. In linearized gravity, we would expand the Ricci scalar to quadratic order in the perturbation parameter to find the propagator. …

In linearized gravity it is assumed that the gravitational field couples to the stress tensor and not directly to the field $\phi$, with the form $\kappa h_{\mu\nu}T^{\mu\nu}$, as seen here. … In the linearized case, why do we not take the coupling directly with $\phi$ as $\kappa h\phi ^2$, where $h=h^{\mu}_{\mu}=\eta^{\mu\nu}h_{\mu\nu}$ instead? …

I have heard that non-linearity of Einstein's field equations has to do with the fact that gravity self-couples. What does non-linearity have to do with self-coupling? …

He mentions that a straightforward generalization will work for any real, bosonic, linear field provided that it has a well-posed initial value problem; it is derivable from a Lagrangian. … What I find particularly surprising in these remarks is that linearized gravity is described by a spin $s = 2$ field. …

Quantum electrodynamics based upon Euler-Heisenberg or Born-Infeld Lagrangians predict photons to move according to an effective metric which is dependent on the background electromagnetic field. … In other words, photon trajectories are curved in presence of electromagnetic fields, meaning that an effective gravity is acting upon. …

Expressing the metric as $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, assuming $h_{\mu \nu} \ll 1$ we can write the Einstein Hilbert action to leading order in $h_{\mu \nu}$ and quantize the linearized … (This should be especially true if we take the action to be the Kretschmann scalar, which seems to be equivalent to the Yang Mills Lagrangian. $$ \mathrm{Tr}(F_{\mu \nu} F^{\mu \nu}) \leftrightarrow R^ …

I'm familiar with the Lagrangian point for two bodies, but is there a similar concept for 3+ bodies? I tried googling this, but wasn't sure exactly what to google, so got nothing useful. … experiences more gravity, not less. …

I have heard that the Lagrangian that is used for quantum gravity but produces a theory that requires infinite constants to be renormalized. … This Lagrangian is supposed to be a linearization of the Lagrangian from GR? What do the infinite constants represent? …