Linearized Einstein equations

Let us consider small perturbations on Minkowski background, such that in some frame the metric can be presented in the form \begin{equation} \label{WFL} g_{\mu\nu}(x)=\eta_{\mu\nu}+h_{\mu\nu}(x), \qquad |h_{\mu\nu}(x)|\ll 1. \end{equation} We can also consider perturbations on the background of other exact solutions of the Einstein equations by replacing $\eta_{\mu\nu}$ with the corresponding $g_{\mu\nu}^{(0)}$. Thus cosmological perturbations are naturally studied in the Friedmanninan background. The linearized Einstein equations are obtained in the first order by $h_{\mu\nu}$, discarding quadratic terms. On Minkowski background the zero-order terms for the curvature tensor and its contractions vanish, so from the Einstein's equation the stress-energy tensor in the considered region must also be small (if non-zero) and $\sim h$. The constraints this places on matter will be considered in more detail in the next section.

Problem 1: Inverse metric

Show that on Minkowski background the inverse metric is \[g^{\mu\nu}(x)=\eta^{\mu\nu}-h^{\mu\nu}(x)+O(h^2), \quad\text{where}\quad h^{\mu\nu}\equiv\eta^{\mu\rho}\eta^{\nu\sigma} h_{\rho\sigma},\] and we agree to use $\eta$ for raising and lowering of the indices.

solution

Let \[g^{\mu\nu}=\eta^{\mu\nu}+\tilde{h}^{\mu\nu}.\] Then by definition \[\delta^{\mu}_{\nu}=g^{\mu\lambda}g_{\lambda\nu} =(\eta^{\mu\lambda}+\tilde{h}^{\mu\lambda}+O(h^2)) (\eta_{\lambda\nu}+h_{\lambda\nu}+O(h^2)) =\delta^{\mu}_{\nu}+h^{\mu}_{\nu}+\tilde{h}^{\mu}_{\nu}+O(h^2),\] so \[\tilde{h}^{\mu}_{\nu}=-h^{\mu}_{\nu}.\]

Problem 2: Raising indices

Show$^*$ that using the background metric $g_{\mu\nu}^{(0)}$ to raise and lower indices instead of the true metric $g_{\mu\nu}$ only makes difference in the next order by $h$.
Consider for definiteness a second rank tensor $A_{\mu\nu}$: \begin{align*}A_{\mu\nu} =g_{\mu\rho}g_{\nu\sigma}A^{\rho\sigma} =g_{\mu\rho}^{(0)}g_{\nu\sigma}^{(0)}A^{\rho\sigma} +O(hA). \end{align*}

$^*$ Is it really a problem at all?

Problem 3: Linearized curvature tensors

Derive the curvature, Ricci and Einstein tensors in the first order by $h_{\mu\nu}$.

solution

All the tensors considered are constructed of derivatives of the metric tensor, and thus are of the first order by $h$. Partial derivative is tensorial in the zeroth order by $h$: $\partial_{\mu}=\nabla_{\mu}+O(h)$. Also, as shown above, if we use $\eta_{\mu\nu}$ to lower and raise indices, the result will differ from the exact expression in the next order by $h$. Therefore in the assumed approximation we can neglect the difference between the partial and covariant derivatives and freely put the metric in and out of action of differential operators. Using the notation \begin{align*} h&\equiv h^{\mu}_{\mu};\\ \partial^{\lambda}& \equiv \eta^{\lambda\mu}\partial_{\mu};\\ \square&\equiv\eta^{\mu\nu}\partial_{\mu}\partial_{\nu} \equiv \partial^{\mu}\partial_{\mu}, \end{align*} where $\square$ is the usual flat D'Alembertian, we get \begin{align*} \Gamma^{\lambda}_{\mu\nu} &=\tfrac12 g^{\lambda\rho} (\partial_{\mu}h_{\rho\nu}+\partial_{\nu}h_{\rho\mu} -\partial_{\rho}h_{\mu\nu})=\\ &=\tfrac12 (\partial_{\mu}h^{\lambda}_{\nu} +\partial_{\nu}h^{\lambda}_{\mu} -\partial^{\lambda}h_{\mu\nu})+O(h^2);\\ {R^{\mu}}_{\nu\rho\sigma} &=\partial_{\rho}\Gamma^{\mu}_{\nu\sigma} -\partial_{\sigma}\Gamma^{\mu}_{\nu\rho}+O(h^2)=\\ &=\tfrac{1}{2}\big( \partial_{\rho}\partial_{\nu}h^{\mu}_{\sigma} +\underline{\partial_{\rho}\partial_{\sigma}h^{\mu}_{\nu}} -\partial_{\rho}\partial^{\mu}h_{\nu\sigma} -\partial_{\sigma}\partial_{\nu}h^{\mu}_{\rho} -\underline{\partial_{\sigma}\partial_{\rho}h^{\mu}_{\nu}} +\partial_{\sigma}\partial^{\mu}h_{\nu\rho}\big)+O(h^2)=\\ &=\tfrac{1}{2}\big( \partial_{\rho}\partial_{\nu}h^{\mu}_{\sigma} -\partial_{\rho}\partial^{\mu}h_{\nu\sigma} -\partial_{\sigma}\partial_{\nu}h^{\mu}_{\rho} +\partial_{\sigma}\partial^{\mu}h_{\nu\rho}\big) +O(h^2);\\ R_{\nu\sigma} &=\tfrac{1}{2}\big( \partial_{\nu}\partial_{\mu}h^{\mu}_{\sigma} -\square h_{\nu\sigma} -\partial_{\sigma}\partial_{\nu}h +\partial_{\sigma}\partial_{\mu}h^{\mu}_{\nu}\big) +O(h^2);\\ R&\equiv \eta^{\nu\sigma}R_{\nu\sigma}= \partial_{\nu}\partial_{\mu}h^{\mu\nu} -\square h +O(h^2);\\ G_{\mu\nu} &\equiv R_{\mu\nu}-\tfrac{1}{2}R\eta_{\mu\nu}=\\ &=\tfrac{1}{2}\big[ \partial_{\mu}\partial_{\lambda}h^{\lambda}_{\nu} +\partial_{\nu}\partial_{\lambda}h^{\lambda}_{\mu} -\partial_{\mu}\partial_{\nu}h -\square h_{\mu\nu} +\eta_{\mu\nu}(\square h - \partial_{\rho}\partial_{\sigma}h^{\rho\sigma}) \big]. \end{align*}

Problem 4: Trace-reversed perturbation

Write the Einstein's tensor in terms of the trace-reversed metric perturbation \[\bar{h}_{\mu\nu} =h_{\mu\nu}-\frac{1}{2}h\;\eta_{\mu\nu}.\]

solution

Taking the trace we get \[\bar{h}=h-2h=-h \quad\text{and}\quad h_{\mu\nu}=\bar{h}_{\mu\nu} -\frac{1}{2}\bar{h}\;\eta_{\mu\nu},\] so now it is clear why $\bar{h}_{\mu\nu}$ can be named the "trace-reversed" metric. Then \begin{align*} G_{\mu\nu}&=\tfrac{1}{2}\big[ \partial_{\mu}\partial_{\lambda}\bar{h}^{\lambda}_{\nu} -\underline{\tfrac{1}{2}\partial_{\mu}\partial_{\nu}\bar{h}} +\partial_{\nu}\partial_{\lambda}\bar{h}^{\lambda}_{\mu} -\underline{\tfrac{1}{2}\partial_{\mu}\partial_{\nu}\bar{h}} +\underline{\partial_{\mu}\partial_{\nu}\bar{h}} -\square \bar{h}_{\mu\nu} +\tfrac{1}{2}\eta_{\mu\nu}\square \bar{h} +\\ &\qquad+\eta_{\mu\nu}(-\square \bar{h} -\partial_{\rho}\partial_{\sigma}\bar{h}^{\rho\sigma} +\tfrac{1}{2}\square \bar{h}) \big]=\\ &=\tfrac{1}{2}\big[ \partial_{\mu}\partial_{\lambda}\bar{h}^{\lambda}_{\nu} +\partial_{\nu}\partial_{\lambda}\bar{h}^{\lambda}_{\mu} -\square\bar{h}_{\mu\nu} -\eta_{\mu\nu} \partial_{\rho}\partial_{\sigma}\bar{h}^{\rho\sigma} \big]. \end{align*}