- #1
ergospherical
- 977
- 1,278
In the context of cosmology, you can perturb around the FRW background, conventionally:$$g = a^2(\tau)[(1+2A)d\tau^2 - 2B_a dx^a d\tau -(\delta_{ab} + h_{ab}) dx^a dx^b]$$with ##a,b## being spatial indices only (1,2,3). You can do gauge transformations ##\tilde{x} = x + \xi## of the coordinates. These are typically split into ##\xi^0 \equiv T## and ##\xi^a = \partial^a L + \hat{L}^a##. You find the following first order transformations,\begin{align*}
\tilde{A} &= A - T' - \mathcal{H} T \\
\tilde{B}_a &= B_a + \partial_a T - L_a' \\
\tilde{h}_{ab} &= h_{ab} - 2\partial_{(a} L_{b)} - 2\mathcal{H} T \delta_{ab}
\end{align*}The problem is specifically, is starting from the synchronous gauge, where$$g = a^2(\tau)[d\tau^2 - (\delta_{ab} + h_{ab}) dx^a dx^b]$$with ##h_{ab} = \tfrac{1}{3}\delta_{ab} h + (\hat{k}_a \hat{k}_b - \tfrac{1}{3} \delta_{ab}) h_S## and where ##h, h_S## are functions. And then converting to the Newtonian gauge, where$$\tilde{g} = a^2(\tilde{\tau})[(1+2\Phi)d\tilde{\tau}^2 - (1-2\Psi) \delta_{ab} d\tilde{x}^a d\tilde{x}^b]$$There are supposed to be transformations of the form
$$\tilde{\tau} = \tau + \sum_k T_k(\tau) e^{i\mathbf{k} \cdot \mathbf{x}}, \quad \tilde{\mathbf{x}} = \mathbf{x} + \sum_k L_k (\tau) i\hat{\mathbf{k}} e^{i\mathbf{k} \cdot \mathbf{x}}$$The problem is to find ##T_k## and ##L_k## in terms of ##h_S## and ##h##. I can re-write the second one of these with ##\xi^j = \partial^j \left[ \sum_k L_k \tfrac{1}{k} e^{i\mathbf{k} \cdot \mathbf{x}} \right]##, in other words my two transformation functions are\begin{align*}
T &= \sum_k T_k e^{i\mathbf{k} \cdot \mathbf{x}} \\
L &= \sum_k L_k \frac{1}{k} e^{i\mathbf{k} \cdot \mathbf{x}}
\end{align*}I must be missing something obvious about how to determine the forms of ##T_k## and ##L_k##. For example, in the synchronous gauge then ##A=0## and we get\begin{align*}
\tilde{A} = 0 - T' - \mathcal{H} T = - \sum_{k} \left[ (T_k' + \mathcal{H} T_k) e^{i\mathbf{k} \cdot \mathbf{x}} \right]
\end{align*}where ##\tilde{A} = \Phi##. I'm supposed to find, apparently, that ##T_k = \tfrac{h_S'}{2k^2}## and ##L_k = \tfrac{h_S}{2k}##. Would someone be able to point me in the right direction?
\tilde{A} &= A - T' - \mathcal{H} T \\
\tilde{B}_a &= B_a + \partial_a T - L_a' \\
\tilde{h}_{ab} &= h_{ab} - 2\partial_{(a} L_{b)} - 2\mathcal{H} T \delta_{ab}
\end{align*}The problem is specifically, is starting from the synchronous gauge, where$$g = a^2(\tau)[d\tau^2 - (\delta_{ab} + h_{ab}) dx^a dx^b]$$with ##h_{ab} = \tfrac{1}{3}\delta_{ab} h + (\hat{k}_a \hat{k}_b - \tfrac{1}{3} \delta_{ab}) h_S## and where ##h, h_S## are functions. And then converting to the Newtonian gauge, where$$\tilde{g} = a^2(\tilde{\tau})[(1+2\Phi)d\tilde{\tau}^2 - (1-2\Psi) \delta_{ab} d\tilde{x}^a d\tilde{x}^b]$$There are supposed to be transformations of the form
$$\tilde{\tau} = \tau + \sum_k T_k(\tau) e^{i\mathbf{k} \cdot \mathbf{x}}, \quad \tilde{\mathbf{x}} = \mathbf{x} + \sum_k L_k (\tau) i\hat{\mathbf{k}} e^{i\mathbf{k} \cdot \mathbf{x}}$$The problem is to find ##T_k## and ##L_k## in terms of ##h_S## and ##h##. I can re-write the second one of these with ##\xi^j = \partial^j \left[ \sum_k L_k \tfrac{1}{k} e^{i\mathbf{k} \cdot \mathbf{x}} \right]##, in other words my two transformation functions are\begin{align*}
T &= \sum_k T_k e^{i\mathbf{k} \cdot \mathbf{x}} \\
L &= \sum_k L_k \frac{1}{k} e^{i\mathbf{k} \cdot \mathbf{x}}
\end{align*}I must be missing something obvious about how to determine the forms of ##T_k## and ##L_k##. For example, in the synchronous gauge then ##A=0## and we get\begin{align*}
\tilde{A} = 0 - T' - \mathcal{H} T = - \sum_{k} \left[ (T_k' + \mathcal{H} T_k) e^{i\mathbf{k} \cdot \mathbf{x}} \right]
\end{align*}where ##\tilde{A} = \Phi##. I'm supposed to find, apparently, that ##T_k = \tfrac{h_S'}{2k^2}## and ##L_k = \tfrac{h_S}{2k}##. Would someone be able to point me in the right direction?
Last edited: