# 物理代写|宇宙学代写cosmology代考|PHYS3080

## 物理代写|宇宙学代写cosmology代考|The Einstein equations for scalar perturbations

We are now ready to embark on our computation of the Einstein equations at linear order in perturbations. To begin, we will focus on scalar perturbations and continue to work in conformal-Newtonian gauge, so that our starting point is
\begin{aligned} &g_{00}(\boldsymbol{x}, t)=-1-2 \Psi(\boldsymbol{x}, t), \ &g_{0 i}(\boldsymbol{x}, t)=0, \ &g_{i j}(\boldsymbol{x}, t)=a^2(t) \delta_{i j}[1+2 \Phi(\boldsymbol{x}, t)] . \end{aligned}
Evaluating the left-hand side of the Einstein equation (3.1) requires three steps:

• Compute the Christoffel symbol, $\Gamma^\mu{ }_{\alpha \beta}$, for the perturbed metric of Eq. (6.20); we have already done this in Sect 3.3.1.
• From these, form the Ricci tensor, $R_{\mu v}$, using Eq. (3.3).
• Contract the Ricci tensor to form the Ricci scalar, $R \equiv g^{\mu v} R_{\mu v}$.
We will also immediately switch to Fourier space, exchanging spatial derivatives with powers of $i k$. We need two independent equations for the two variables $\Phi, \Psi$. Given that we are dealing with scalar perturbations, we can already anticipate that the 00 component as well as the scalar component of the $i j$ Einstein equations will be useful.

## 物理代写|宇宙学代写cosmology代考|Ricci tensor

The Ricci tensor is most easily expressed in terms of the Christoffel symbol we derived in Sect. 3.3.1. First, consider the time-time component of Eq. (3.3):
$$R_{00}=\Gamma^\alpha{ }{00, \alpha}-\Gamma^\alpha{ }{0 \alpha, 0}+\Gamma^\alpha{ }{\beta \alpha} \Gamma^\beta{ }{00}-\Gamma^\alpha{ }{\beta 0} \Gamma^\beta{ }{0 \alpha} .$$
All of these terms contribute at first order. One simplification comes from considering the $\alpha=0$ part of all these terms. The first and second terms are equal and opposite to each other as are the last two. So the sum over the index $\alpha$ contributes only when $\alpha$ is spatial. Let us consider each of the terms one by one.

• The first is
$$\Gamma^i{ }_{00, i}=-\frac{k^2}{a^2} \Psi,$$
after translating the first line of Eq. (3.56) into Fourier space.
• The second term in Eq. (6.21) is
$$-\Gamma^i 0 i, 0=-3\left(\frac{\ddot{a}}{a}-H^2+\Phi_{, 00}\right)$$
using the second line of Eq. (3.56). The factor of 3 in front comes from the implicit sum in $\delta_{i i}$.
• The next term is $\Gamma^i{ }{i \beta} \Gamma^\beta{ }{00}$. Note that $\Gamma^\beta{ }{00}$ is first order no matter what $\beta$ is, so we need keep only the zeroth-order part of $\Gamma^i{ }{i \beta}$. However, the last line of Eq. (3.56) shows that $\Gamma^i{ }{i \beta}$ is first-order unless $\beta=0$. So to first-order, \begin{aligned} \Gamma^i{ }{i \beta} \Gamma^\beta{ }{00} &=\Gamma^i{ }{i 0} \Gamma^0{ }{00} \ &=3 H \Psi{, 0} • \end{aligned} •
• Finally the last term is $-\Gamma^i{ }{\beta 0} \Gamma^\beta{ }{0 i}$. In this case, if $\beta=0$ both $\Gamma$ are first-order, so their product is second-order and can be neglected. Therefore, only spatial $\beta$ need to be considered, leading to
• • \begin{aligned} • -\Gamma^i{ }{\beta 0} \Gamma^\beta{ }{0 i} &=-\Gamma^i{ }{j 0} \Gamma^j 0 i \ &=-3\left(H^2+2 H \Phi{, 0}\right) • \end{aligned} •

## 物理代写|宇宙学代写cosmology代考|标量摄动的爱因斯坦方程

\begin{aligned} &g_{00}(\boldsymbol{x}, t)=-1-2 \Psi(\boldsymbol{x}, t), \ &g_{0 i}(\boldsymbol{x}, t)=0, \ &g_{i j}(\boldsymbol{x}, t)=a^2(t) \delta_{i j}[1+2 \Phi(\boldsymbol{x}, t)] . \end{aligned}

## 物理代写|宇宙学代写cosmology代考|里奇张量

$$R_{00}=\Gamma^\alpha{ }{00, \alpha}-\Gamma^\alpha{ }{0 \alpha, 0}+\Gamma^\alpha{ }{\beta \alpha} \Gamma^\beta{ }{00}-\Gamma^\alpha{ }{\beta 0} \Gamma^\beta{ }{0 \alpha} .$$

$$\Gamma^i{ }{00, i}=-\frac{k^2}{a^2} \Psi,$$
。Eq.(6.21)中的第二项是使用Eq.(3.56)的第二行
$$-\Gamma^i 0 i, 0=-3\left(\frac{\ddot{a}}{a}-H^2+\Phi{, 00}\right)$$
。前面的因子3来自$\delta_{i i}$中的隐式和下一项是$\Gamma^i{ }{i \beta} \Gamma^\beta{ }{00}$。注意，无论$\beta$是什么，$\Gamma^\beta{ }{00}$都是一阶的，所以我们只需要保留$\Gamma^i{ }{i \beta}$的零阶部分。然而，式(3.56)的最后一行表明，$\Gamma^i{ }{i \beta}$是一阶的，除非$\beta=0$。一阶，\begin{aligned} \Gamma^i{ }{i \beta} \Gamma^\beta{ }{00} &=\Gamma^i{ }{i 0} \Gamma^0{ }{00} \ &=3 H \Psi{, 0}\end{aligned} 最后一项是$-\Gamma^i{ }{\beta 0} \Gamma^\beta{ }{0 i}$。在这种情况下，如果$\beta=0$和$\Gamma$都是一阶的，那么它们的乘积是二阶的，可以忽略。因此，只需要考虑空间$\beta$，导致 \begin{aligned}-\Gamma^i{ }{\beta 0} \Gamma^\beta{ }{0 i} &=-\Gamma^i{ }{j 0} \Gamma^j 0 i \ &=-3\left(H^2+2 H \Phi{, 0}\right)\end{aligned}

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