Perturbation Theory II

It is a fact that any small, but otherwise arbitrary, transformation of coordinates is given by choice of vector field \(\xi\), whose component functions w.r.t. the original chart, \(\xi^m_{(x)}\), are small. Then the change incurred by the metric components under such a small, but otherwise arbitrary, transformation is given by¹ \[\Delta_{\xi}g_{ab} = \big(\mathcal{L}_{\xi}g\big)_{ab},\] where \(\mathcal{L}\) is the Lie derivative. In other words, if the \(\delta g_{ab}\) takes this form it is a ‘fake’ perturbation, as described at the end of last lecture. But of course we write \[g_{ab} + \delta g_{ab},\] and so we must study how \(\delta g_{ab}\) itself changes if we choose a small, arbitrary \(\xi\). If it is a fake perturbation then we know the change in the metric components is of the above form, but \(\delta g_{ab}\) is the change in the metric, and so taking a further change \(\xi\) will simply give us \[\big(\Delta_{\xi} \delta g\big)_{ab} = \big(\mathcal{L}_{\xi}g\big)_{ab},\] where we note that there is no \(\delta\) on the right-hand side. The aim is to find which components are fake and then work out which, if any, combinations of components cancel on the right-hand side giving us a real perturbation.

25.1 Calculate \(\Delta_{\xi}a\), \(\Delta_{\xi}c\), \(\Delta_{\xi}b_{\alpha}\) & \(\Delta_{\xi}e_{\alpha\beta}\)

Even though we did the decomposition in the last lecture to the capital letters (\(A,B_{\alpha}\), etc.), it is actually cleverer to work out the change of the lower case letters first.

Our chart is \((U,x)\) where \(g = dx^0\otimes dx^0 - \gamma_{\alpha\beta}dx^{\alpha}\otimes dx^{\beta}\). We can decompose the vector field \(\xi\) in this chart as \[\xi = T\partial_0 +L^{\beta}\partial_{\beta},\] where \(T\) and \(L^{\mu}\) are scalar fields that can depend on all the chart coordinates, including \(x^0\). We then calculate the change of \(\delta g_{ab}\) incurred by \(\xi\) by considering \(T\) and \(L^{\mu}\). We have \[\begin{split} \big(\Delta_{\xi}\delta g\big)_{ab} & = \big(\mathcal{L}_{\xi}g\big)_{ab} \\ & = \xi^mg_{ab,m} + {\xi^m}_{,a}g_{mb} + {\xi^m}_{,b}g_{am} \\ & = Tg_{ab,0} + L^{\mu}g_{ab,\mu} + T_{,a}g_{0b} + {L^{\mu}}_{,a}g_{\mu b} + T_{,b}g_{a0} + {L^{\mu}}_{,b}g_{a\mu}. \end{split}\] We now consider each of the components separately. Using a dot to indicate a derivative w.r.t. \(x^0\), we have: \[\big(\Delta_{\xi}\delta g\big)_{00} = 0 + 0 + \dot{T} + 0 + \dot{T} + 0 = 2\dot{T},\] where we have used the fact that \(g_{00}=1\) and \(g_{0\alpha}=0\).

Next we have \[\big(\Delta_{\xi} \delta\big)_{0\beta} = 0 + 0 + 0 - \dot{L}^{\mu} \gamma_{\mu\beta} + T_{,\beta} + 0 = -\dot{L}^{\mu}\gamma_{\mu\beta} + T_{,\beta},\] where we have used \(g_{0\beta}=0\), \(g_{00}=1\) and \(g_{\alpha\beta} = -\gamma_{\alpha\beta}\). We can then use the fact that \(\gamma_{\alpha\beta}\) is independent of \(x^0\) to give us \[\big(\Delta_{\xi} \delta\big)_{0\beta} = T_{,\beta} - \big(L^{\mu}\gamma_{\mu\beta}\big)_{,0} = D_{\beta}T - \dot{L}_{\beta},\] where \(D_{\beta}\) is the Levi-Civita covariant derivative using \(\gamma\).

Finally we have \[\big(\Delta_{\xi}\delta g\big)_{\alpha\beta} = 0 - L^{\mu} \gamma_{\alpha\beta,\mu} + 0 - {L^{\mu}}_{,\alpha}\gamma_{\mu\beta} + 0 - {L^{\mu}}_{,\beta} \gamma_{\alpha\mu}.\] We then use \[{L^{\mu}}_{,\alpha}\gamma_{\mu\beta} = \big(L^{\mu}\gamma_{\mu\beta}\big)_{\alpha} - L^{\mu} \gamma_{\mu\beta,\alpha}\] to give \[\big(\Delta_{\xi}\delta g\big)_{\alpha\beta} = -2\bigg[ L_{(\alpha,\beta)} -\frac{1}{2}L^{\mu}\big( \gamma_{\mu\beta,\alpha} + \gamma_{\mu\alpha,\beta} - \gamma_{\alpha\beta,\mu}\big) \bigg] = -2 D_{(\alpha}L_{\beta)},\] where again \(D_{\alpha}\) is the Levi-Civita covariant derivative using \(\gamma\).

Using the expression for \(\delta g\) in terms of \(a,b_{\alpha},c\) and \(e_{\alpha\beta}\) from last lecture, we conclude that \[\begin{split} \Delta_{\xi}(2a) & = 2\dot{T} \\ \Delta_{\xi}(-b_{\alpha}) & = D_{\alpha}T - \dot{L}_{\alpha}\\ \Delta_{\xi} \big(-2c\gamma_{\alpha\beta} -e_{\alpha\beta}\big) & = -2D_{(\alpha}L_{\beta)}. \end{split}\] The first two expressions clearly tell us that \[\Delta_{\xi} a = \dot{T}, \qquad\text{and}\qquad \Delta_{\xi} b_{\alpha} = \dot{L}_{\alpha} - D_{\alpha}T,\] but the third expression needs a little work. We contract both sides with \(\gamma^{\alpha\beta}\) and use \(\gamma^{\alpha\beta}\gamma_{\alpha\beta}=3\) along with the fact that \(D\) is \(\gamma\)-metric compatible (so we can ‘take \(\gamma\) inside it’), giving \[6\Delta_{\xi}c + \Delta_{\xi}\gamma^{\alpha\beta}e_{\alpha\beta} = 2D_{\alpha}\big(\gamma^{\alpha\beta}L_{\beta}\big) \qquad \implies \qquad \Delta_{\xi} c = \frac{1}{3}D_{\alpha}L^{\alpha},\] where we used the trace-free condition of \(e_{\alpha\beta}\). From this it follows that \[\Delta_{\xi} e_{\alpha\beta} = 2D_{(\alpha}L_{\beta)} - \frac{2}{3}D_{\mu}L^{\mu}\gamma_{\alpha\beta}.\]

25.2 Scalar Perturbations

We now wan to consider the decomposed fields, and we start with the scalar perturbations.

We do not prove this claim here but just use it to simplify the following.

Using the above claim and the results from last lecture we see that \[\begin{split} \Delta_{\xi} A & = \dot{T}, \\ \Delta_{\xi}B & = \dot{L}-T \\ \Delta_{\xi} C & = \frac{1}{3}\Delta L \\ \Delta_{\xi} E & = L, \end{split}\] where on the right-hand side of the \(C\) equation \(\Delta\) is the Laplacian.

We can now use these results to show the results quoted at the end of last lecture, namely \[\Psi := A + \dot{B} - \Ddot{E}, \qquad\text{and}\qquad \Phi := C - \frac{1}{3}\Delta E\] are gauge invariants.

For convenience only, and precisely because the \(\Phi\) and \(\Psi\) are gauge invariants, we are free to pick any \((T,L^{\alpha})\) we want. Such a choice is known as a gauge, and the process gauge fixing. Again should be familiar from electromagnetism. We decide to use the gauge \[T := B - \dot{E}, \qquad\text{and}\qquad L := -E.\] We choose this gauge as then \[\begin{split} \Delta_{\xi} E = L \qquad \implies \qquad E & = 0 \\ \Delta_{\xi} B = \dot{L} - T \qquad \implies \qquad B & = 0, \end{split}\] and so \[\Psi = A, \qquad\text{and}\qquad \Phi = C.\] This is known as the longitudinal gauge. Working in this gauge, we finally get the left-hand side for the scalar perturbations for Einsteins equations, \(\delta G_{ab} = T_{ab}\), as \[\begin{split} \delta G_{00} & = 2\Delta \Phi \\ \delta G_{0\alpha} & = 2 D_{\alpha} \dot{\Phi} \\ \delta G_{\alpha\beta} & = \bigg[2\Ddot{\Phi} - \frac{2}{3}\Delta(\Psi+\Phi)\bigg]\gamma_{\alpha\beta} + \bigg[ D_{(\alpha}D_{\beta)} - \frac{1}{3}\gamma_{\alpha\beta} \Delta\bigg](\Psi+\Phi) \end{split}\]

25.3 Vector Perturbations

Again, we do not prove this claim but just use it.

The results here are \[\Delta_{\xi} B_{\alpha} = \dot{L}_{\alpha}, \qquad\text{and}\qquad \Delta_{\xi}E_{\alpha} = L_{\alpha},\] and so the only gauge invariant quantities are \[\Theta := B_{\alpha} - \dot{E}.\]

We then use the vector gauge, which sets \(E=0\), and obtain \(\Theta_{\alpha} = B_{\alpha}\). We then obtain the left-hand side of the vector perturbations of Einsteins equations, \[\delta G_{0\alpha} = \frac{1}{2}\Delta \Theta_{\alpha}, \qquad\text{and}\qquad \delta G_{\alpha\beta} = D_{(\alpha} \dot{\Theta}_{\beta)}.\]

25.4 Tensor Perturbations

With this claim it follows trivially that \[\Delta_{xi} E_{\alpha\beta} =0,\] and so all \(E_{\alpha\beta}\) are gauge invariants. The left-hand side equations are \[\delta G_{\alpha\beta} = - \ddot{E}_{\alpha\beta} + \Delta E_{\alpha\beta}.\]