# 物理代写| Linearised Gravity 相对论代考

## 物理代写

8.2 Linearised Gravity
We choose a nearly Cartesian system so that we can write:
$$g_{\alpha \beta}=\eta_{\alpha \beta}+h_{\alpha \beta},$$
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S. Dhurandhar and S. Mitra, General Relativity and Gravitational Waves,
UNITEXT for Physics, https://doi.org/10.1007/978-3-030-92335-8_8
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8 Gravitational Waves
where $\left|h_{\alpha \beta}\right|<<1$ and $\eta_{\alpha \beta}$ is the Minkowski metric tensor with only nonzero diagonal components $(1,-1,-1,-1) .$ We can think of $h_{\alpha \beta}$ as a “field” over the Minkowski spacetime described by the metric $\eta_{\alpha \beta} .$ This is in the spirit of considering an electromagnetic field described by the electromagnetic vector potential $A_{\alpha}$ over the Minkowski spacetime. From astrophysical considerations, we typically find $h_{\alpha \beta} \sim 10^{-22}$ or less when the waves reach our detectors. Therefore, we are essentially justified in keeping only the first order terms in $h_{\alpha \beta} .$ It is possible that near to the source, $h_{\alpha \beta}$ may not be so small and higher orders are required, but we will not be dealing with such situations in this book.

We start with linearising the Einstein tensor which consists of the Ricci tensor and the scalar curvature which are themselves contractions of the Riemann tensor. Therefore, we first start with the Riemann tensor. Note that,
$$R_{\alpha \mu \beta v}=\frac{1}{2}\left(g_{\alpha v, \mu \beta}+g_{\mu \beta, \alpha v}-g_{\alpha \beta, \mu v}-g_{\mu v, \alpha \beta}\right)+\Gamma^{2} \text { terms }$$
Given the weak field limit we find that $\Gamma \sim o(h)$ and so the product of Gamma terms are of the second and higher order in $h_{\alpha \beta}$ and so can be dropped in this approximation. Also the expression in $g_{\alpha \beta}$ between the parenthesis reduces to the expression with $g$ s replaced by $h$ secause the $\eta_{i k}$ are constants whose derivatives are zero. Thus in the weak field limit,
$$R_{\alpha \mu \beta v} \simeq \frac{1}{2}\left(h_{\alpha v, \mu \beta}+h_{\mu \beta, \alpha v}-h_{\alpha \beta, \mu \nu}-h_{\mu v, \alpha \beta}\right) .$$
In order to proceed further, that is, compute the Ricci tensor and the scalar curvature, it helps to go over to a new variable,
$$\bar{h}{\alpha \beta}=h{\alpha \beta}-\frac{1}{2} h \eta_{\alpha \beta},$$
where $h=h_{\mu}^{\mu}$, is the trace of $h_{\alpha \beta}$. Note that the indices are lowered and raised by the metric tensors $\eta_{\alpha \beta}$ and $\eta^{\alpha \beta}$, because we are keeping terms only to the first order in $h_{\alpha \beta}$. The quantity $\bar{h}{\alpha \beta}$ is called the trace reverse of $h{\alpha \beta}$. By making infinitismal coordinate transformations of the form $x^{\prime \mu}=x^{\mu}+\xi^{\mu}$ where $\xi^{\mu}$ are infinitismal we can choose a coordinate system in which the divergence of $\bar{h}{\alpha \beta}$ vanishes, that is, $$\bar{h}{, \beta}^{\alpha \beta}=0$$
This is called the Lorenz gauge in analogy with the Lorenz gauge in electrodynamics (see Eq. (1.7.17)). It can be shown that such a coordinate system can always we found if the relevant functions are well behaved-continuous, differentiable etc. We show this in the next section. Moreover, just as in electrodynamics, this choice of gauge does not exhaust all freedom, because we can still remain in the gauge by making further transformations where the functions $\xi^{i}$ satisfy $\square \xi^{i}=0$. So it is really a family

## 物理代考

8.2 线性重力

$$g_{\alpha \beta}=\eta_{\alpha \beta}+h_{\alpha \beta}，$$
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(1) 作者，获得 Springer Nature Switzerland AG 2022 的独家许可
S. Dhurandhar 和 S. Mitra，广义相对论和引力波，
UNITEXT 物理，https://doi.org/10.1007/978-3-030-92335-8_8
132
8 引力波

$$R_{\alpha \mu \beta v}=\frac{1}{2}\left(g_{\alpha v, \mu \beta}+g_{\mu \beta, \alpha v}-g_{\alpha \beta, \mu v}-g_{\mu v, \alpha \beta}\right)+\Gamma^{2} \text { terms }$$

$$R_{\alpha \mu \beta v} \simeq \frac{1}{2}\left(h_{\alpha v, \mu \beta}+h_{\mu \beta, \alpha v}-h_{\ alpha \beta, \mu \nu}-h_{\mu v, \alpha \beta}\right) 。$$

$$\bar{h}{\alpha \beta}=h{\alpha \beta}-\frac{1}{2} h \eta_{\alpha \beta}，$$

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