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# 物理代写| PlaneWave Solutions 相对论代考

## 物理代写

8.4 Plane Wave Solutions
We consider the simplest type of solutions and also the most important type of solutions, namely, the plane wave solutions-the surfaces of constant phase are planes or more precisely, hyperplanes in the background four dimensional Minkowski space. Astrophysically, since the sources are at a very large distance from the detectors, the waves would be plane to a great degree of accuracy.

We set the right hand side in Eq. (8.2.10), namely, the source term to zero, and so we have the equation:
$$\square \bar{h}{\mu \nu}=0 .$$ We now consider solutions of the form, $$\bar{h}{\mu v}=A_{\mu v} e^{-i k_{\alpha} x^{\alpha}}$$
where $A_{\mu \nu}$ is a constant symmetric matrix and $k_{\alpha}$ is a constant vector called the wave vector and signifies the direction of the wave in the 4 dimensional spacetime. Substituting this solution in the wave equation gives:
\begin{aligned} \square \bar{h}^{\mu v}=\eta^{\alpha \beta} \bar{h}{, \alpha \beta}^{\mu v} &=-\eta^{\alpha \beta} k{\alpha} k_{\beta} \bar{h}^{\mu v} \equiv 0 \ & \Longrightarrow \eta^{\alpha \beta} k_{\alpha} k_{\beta}=0 \end{aligned}
Or $k_{\alpha} k^{\alpha}=0$, which implies that the wave vector $k^{\alpha}$ is null. Thus, GW waves travel with the speed $c$ which is also the speed of light. This is the first result of GR.
We will further also show that the waves are transverse. The Lorenz gauge condition Eq. (8.2.5) gives,
Thus GW are transverse.
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8.4.1 The Transverse Traceless (TT) Gauge because we can still remain within the Lorenz gauge by further choosing $\xi^{\mu}$ such that $\square \xi^{\mu}=0$ (instead of using $\zeta^{\mu}$ we persist with $\xi^{\mu}$ without any cause for confusion). Let us choose $\xi_{\mu}=B_{\mu} e^{-i k_{\alpha} x^{\alpha}}$ then, Eq. (8.3.5) gives,
$$A_{\mu \nu}^{\prime}=A_{\mu \nu}+i B_{\mu} k_{v}+i B_{\nu} k_{\mu}-i \eta_{\mu \nu} B^{\alpha} k_{\alpha} .$$
We see that we automatically remain in the Lorenz guage because,
$$A_{\mu \nu}^{\prime} k^{\nu}=A_{\mu \nu} k^{\nu}+i B_{\mu} k_{\nu} k^{\nu}+i B_{v} k^{v} k_{\mu}-i \eta_{\mu \nu} k^{v} B^{\alpha} k_{\alpha} \equiv 0$$
because first two terms on the RHS of Eq. (8.4.6) are zero and the last two terms cancel out. This is no surprise because we are within the Lorenz gauges.
The additional gauge freedom allows us to make $A_{\mu v}$ tracefree, that is,
$$A_{\mu}^{\mu}=0$$
and further for any given timelike vector $U^{\mu}$ we are allowed to choose:
$$A_{\mu \nu} U^{\nu}=0 .$$
The equations (8.4.7) and (8.4.8) can be easily shown to hold in the new guage by explicitly solving for $B_{\mu}$ in terms of the old $A_{\mu \nu}$. See Exercise 8.1.

Since $A_{\mu \nu}$ can be considered as a symmetric matrix, it has at most 10 independent components. The Lorenz gauge condition $A_{\mu v} k^{\nu}=0$, Eqs. (8.4.7) and (8.4.8) constitute 8 independent conditions on these 10 components of $A_{\mu v}$. Therefore, $A_{\mu v}$ has only 2 independent components and these are the physical degrees of freedom of the GW-the two polarisations. The above conditions fix the gauge and is called the transverse traceless gauge or the TT gauge for short. The $A_{\mu \nu}$ matrix is denoted by $A_{\mu \nu}^{T T}$ in this gauge and the corresponding $h_{\mu \nu}$ by $h_{\mu \nu}^{T T}$.
8.4 Plane Wave Solutions
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$$A_{\mu v}^{T T}=\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \ 0 & A_{x x} & A_{x y} & 0 \ 0 & A_{x y} & -A_{x x} & 0 \ 0 & 0 & 0 & 0 \end{array}\right)$$
It is customary to write $A_{x x}=A_{+}$and $A_{x y}=A_{\times}$for reasons that will become clear later in the text. We therefore write $A_{x x}=A_{+}$and $A_{x y}=A_{\times}$and the corresponding $h_{+}=A_{+} e^{-i k_{\alpha} x^{\alpha}}$ and $h_{\times}=A_{\times} e^{-i k_{\alpha} x^{\alpha}} .$ We can associate polarisation tensors corresponding to each polarisation (just as there are polarisation vectors for the electromagnetic wave):
$$e_{\mu v}^{+}=\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 \end{array}\right), \quad e_{\mu v}^{\times}=\left(\begin{array}{llll} 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 \end{array}\right)$$
These are the plus $(+)$ and cross $(x)$ polarisation tensors. Each of the polarisation tensors describe linear polarisation again denoted by $+$ and $\times$. The nomenclature will become clear in the next section when we consider the effect of each polarisation separately on a ring of test particles. The general wave is a linear combination of the two polarisations and can be written in terms of the polarisation tensors:
$$h_{\mu v}^{T T}=h_{+} e_{\mu v}^{+}+h_{\times} e_{\mu v}^{\times} .$$
Here we have polarisation tensors because we have a tensor wave. The polarisations can have a phase difference between them which is realised by the fact that the $A_{+}$ and $A_{\times}$can be complex quantities and so the phases are be accommodated in the phases of the complex numbers.

The polarisation tensors are orthogonal in the sense that their scalar product is zero:
$$e_{\mu v}^{+} e^{\times \mu v}=0$$
In the above discussion we have restricted ourselves to a specific frequency $\omega$ of the wave. These results can be easily generalised to a wave coming from a fixed direction, as a general function of $t$. The result follows on taking the inverse Fourier transform with respect to $\omega$. Then the GW amplitudes $h_{+, \times}$become functions of $t$, that is, now we have $h_{+}(t), h_{\times}(t), h_{\mu \nu}^{T T}(t)$ etc. Much of the above discussion follows (Schutz 1995). For further reading we refer the reader to Misner et al. (1973) and Schutz (1995).

## 物理代考

8.4 平面波解决方案

$$\square \bar{h}{\mu \nu}=0 。$$ 我们现在考虑以下形式的解决方案， $$\bar{h}{\mu v}=A_{\mu v} e^{-i k_{\alpha} x^{\alpha}}$$

$$\开始{对齐} \square \bar{h}^{\mu v}=\eta^{\alpha \beta} \bar{h}{, \alpha \beta}^{\mu v} &=-\eta^{\阿尔法 \beta} k{\alpha} k_{\beta} \bar{h}^{\mu v} \equiv 0 \ & \Longrightarrow \eta^{\alpha \beta} k_{\alpha} k_{\beta}=0 \end{对齐}$$

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8.4.1 横向无迹（TT）规范，因为我们仍然可以通过进一步选择 $\xi^{\mu}$ 使得 $\square \xi^{\mu}=0$（而不是使用$\zeta^{\mu}$ 我们坚持使用 $\xi^{\mu}$ 没有任何混淆）。让我们选择 $\xi_{\mu}=B_{\mu} e^{-i k_{\alpha} x^{\alpha}}$ 然后，等式。 (8.3.5) 给出，
$$A_{\mu \nu}^{\prime}=A_{\mu \nu}+i B_{\mu} k_{v}+i B_{\nu} k_{\mu}-i \eta_{\mu \nu} B^{\alpha} k_{\alpha} 。$$

$$A_{\mu \nu}^{\prime} k^{\nu}=A_{\mu \nu} k^{\nu}+i B_{\mu} k_{\nu} k^{\nu} +i B_{v} k^{v} k_{\mu}-i \eta_{\mu \nu} k^{v} B^{\alpha} k_{\alpha} \equiv 0$$

$$A_{\mu}^{\mu}=0$$

$$A_{\mu \nu} U^{\nu}=0 。$$

8.4 平面波解决方案
137
$$A_{\mu v}^{T T}=\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \ 0 & A_{x x} & A_{x y} & 0 \ 0 & A_{x y} & -A_{x x} & 0 \ 0 & 0 & 0 & 0 \end{数组}\右）$$

$$e_{\mu v}^{+}=\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 \end{array}\right), \quad e_{\mu v}^{\times}=\left(\begin{array}{llll} 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 \end{数组}\右）$$

$$h_{\mu v}^{T T}=h_{+} e_{\mu v}^{+}+h_{\times} e_{\mu v}^{\times} 。$$

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