Difference between revisions of "Linearized Einstein equations"
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[[Category:Weak field limit and gravitational waves|2]] | [[Category:Weak field limit and gravitational waves|2]] | ||
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+ | |||
+ | 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} | ||
+ | %Actually, we can always do this in a small enough neighborhood of any event, and even along the given worldline, due to the principle of equivalence: locally space-time is always Minkowski. Therefore, when we talk about the weak field limit, we usually imply something stronger: that the imposed condition holds in a region of spacetime of scale $L$ much larger than the characteristic curvature radius of the perturbation\footnote{UNCLEAR!}: | ||
+ | %\[R_{curv} \ll L.\] | ||
+ | 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. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | <div id="gw11"></div> | ||
+ | <div style="border: 1px solid #AAA; padding:5px;"> | ||
+ | === 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. | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;">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 | ||
+ | \[\tilde{h}^{\mu}_{\nu}=-h^{\mu}_{\nu}.\]</p> | ||
+ | </div> | ||
+ | </div></div> | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | <div id="gw12"></div> | ||
+ | <div style="border: 1px solid #AAA; padding:5px;"> | ||
+ | === 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 $h$. | ||
+ | <br/> | ||
+ | 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*} | ||
+ | <br/> | ||
+ | |||
+ | $^*$ Is it really a problem at all | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;"></p> | ||
+ | </div> | ||
+ | </div></div> | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | <div id="gw13"></div> | ||
+ | <div style="border: 1px solid #AAA; padding:5px;"> | ||
+ | === Problem 3: Linearized curvature tensors=== | ||
+ | Derive the curvature, Ricci and Einstein tensors in the first order by $h_{\mu\nu}$. | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;">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*}</p> | ||
+ | </div> | ||
+ | </div></div> | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | <div id="gw14"></div> | ||
+ | <div style="border: 1px solid #AAA; padding:5px;"> | ||
+ | === Problem 4: Trace-reversed perturbation=== | ||
+ | Write the Einstein's tensor in terms of the trace-reversed metric perturbation | ||
+ | \[\bar{h}_{\mu\nu} | ||
+ | =h_{\mu\nu}-\frac{1}{2}h\;\eta_{\mu\nu}.\] | ||
+ | <div class="NavFrame collapsed"> | ||
+ | <div class="NavHead">solution</div> | ||
+ | <div style="width:100%;" class="NavContent"> | ||
+ | <p style="text-align: left;">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*}</p> | ||
+ | </div> | ||
+ | </div></div> |
Revision as of 13:03, 26 December 2012
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} %Actually, we can always do this in a small enough neighborhood of any event, and even along the given worldline, due to the principle of equivalence: locally space-time is always Minkowski. Therefore, when we talk about the weak field limit, we usually imply something stronger: that the imposed condition holds in a region of spacetime of scale $L$ much larger than the characteristic curvature radius of the perturbation\footnote{UNCLEAR!}: %\[R_{curv} \ll L.\] 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.
Contents
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.
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 $h_{\mu\nu}$.
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 \[\bar{h}_{\mu\nu} =h_{\mu\nu}-\frac{1}{2}h\;\eta_{\mu\nu}.\]
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*}