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物理代写| Einstein’s Field Equations 相对论代考

物理代写|Einstein’s Field Equations 相对论代考

物理代写

5.1 Einstein’s Field Equations
The curvature of spacetime manifests itself as gravitation. But since gravity is caused by matter distribution, one must connect the curvature to the matter distribution. This is exactly what Einstein’s equations do. They connect the curvature tensor to a tensor which describes the matter distrbution. But there are several conditions that must be satisfied by the field equations. First of all the equations should be tensor equations so that they are true for all observers or hold in all coordinate systems. Secondly, Einstein’s theory must reduce to Newton’s theory to a very good approximation in the limit of weak fields and slow motion of sources.
Newton’s equation for the Newtonian gravitational potential $\phi$ is given by:
$$
\nabla^{2} \phi=4 \pi G \rho,
$$
where $\rho$ is the mass density and $G$ the Newton’s constant of gravitation $\simeq 6.67 \times$ $10^{-8}$ in cgs units. Clearly Newton’s equation is not Lorentz covariant in SR. This can be seen as follows: for an observer moving with uniform velocity $v$, the mass transforms as $m \longrightarrow m \gamma$, while the volume shrinks as $V \longrightarrow V / \gamma$ because of Lorentz contraction and thus the $\rho \longrightarrow \rho \gamma^{2}$. Since there are two factors of $\gamma$ this indicates that the $\rho$ must be a component of a second rank tensor. Also in SR mass and energy are interconvertible and so $\rho$ or $\rho c^{2}$ should appear in the equations. Moreover, energy is just one component of the 4-momentum vector and so momentum must also appear in the Einstein’s equations and thus must also be a source of gravitation and affect the curvature. The entity that fits the bill is the energy momentum tensor $T_{i k}$-infact the zero-zero component of $T_{i k}$ is $T_{00}=\rho c^{2}$. We will physically describe this tensor in the next section. This is the right hand side of Einstein’s equations. The left hand in the next section. This is the right hand side of Einstein’s equations. The left hand side of the equations describe the curvature. The tensor describing the curvature must Thus the Riemann tensor by itself cannot be used but its contraction, namely, the
(C) The Author(s), under exclusive license to Springer Nature Switzerland AG 2022
87 S. Dhurandhar and S. Mitra, General Relativity and Gravitational Waves,
UNITEXT for Physics, https://doi.org/10.1007/978-3-030-92335-8_5
88
5 Einstein’s Equations
Ricci tensor $R_{i k}$ is a good candidate. But even this will not do by itself because the energy-momentum tensor $T_{i k}$ has vanishing divergence:
$$
T_{; k}^{i k}=0 .
$$
This equation implies the conservation of energy and monentum. But $R^{i k}$ is not divergence free. But this can be easily remedied by subtracting its divergence. Thus we arrive at the Einstein tensor:
$$
G_{i k}=R_{i k}-\frac{1}{2} R g_{i k},
$$
where $R$ is the Ricci scalar. By the Bianchi identities, as shown in Eq. (4.5.26) of the last chapter the divergence of this tensor vanishes. Thus we have the Einstein’s equations:
$$
R_{i k}-\frac{1}{2} R g_{i k}=\kappa T_{i k},
$$
where $\kappa$ is a constant that must be determined. This is achieved by going to the weak field limit and then comparing with Newton’s equation Eq. (5.1.1). The energymomentum tensor can be chosen to have dimensions of either pressure or mass density – one is related to the other by the factor of $c^{2}$. We make the former choice. Then the constant $\kappa$ becomes $8 \pi G / c^{4}$. Thus the Einstein’s equations are:
$$
G_{i k}=R_{i k}-\frac{1}{2} R g_{i k}=\frac{8 \pi G}{c^{4}} T_{i k} .
$$
equations reduce to the Newton’s equation Eq. (5.1.1).
The Einstein tensor $G_{i k}$ can also be obtained via the action principle. The re action is called the Einstein-Hilbert action and is given by:
$$
\mathfrak{A}=\int_{\mathcal{V}} R \sqrt{-g} d^{4} x
$$
This action is varied with respect to $g^{i k}$ with the boundary conditions that $\delta g$ its first derivatives vanish on the boundary of $\mathcal{V}$. This approach is followed in $v$ other books and we refer the interested reader to these (Narlikar 2010). Here not further elaborate.

The Einstein’s equations are second order coupled nonlinear partial diffe equations for the metric components $g_{i k}$. Since the Riemann tensor involves th ond order derivatives of the metric, the equations are also of second order. are also nonlinear as they involve products of $g_{i k}$ and their first and second derivatives with respect to the coordinates. The physical meaning of this is gravitational field also produces a gravitational field since it carries energy. I trast, the Newton’s equation is linear and so also are the Maxwell’s equatic
$5.1$ Einstein’s Field Equations
electromagnetism. The implication is that in GR we cannot superpose soluti we could conveniently do in Newton’s theory of gravitation and also in electro netism. This makes solutions hard to obtain. Also the equations are coupled. V make a counting argument about the number of equations: From the symmetry Einstein tensor, we have 10 equations (symmetric matrix in 4 dimensions) in functions $g_{i k}$. However, the Bianchi identities or vanishing of the divergence Einstein tensor, puts 4 conditions on the 10 equations. This brings down the $\mathrm{n}$ of independent equations to 6 . But this is balanced by the fact that there are $4 \mathrm{~d}$ of freedom in choice of coordinates. Thus there are really 6 independent $g_{i k}$ mathematically the problem is well posed.

However, in practice obtaining exact analytic solutions to Einstein’s equ is extremely difficult. One must usually impose symmetry conditions so th problem becomes tractable. But this has the disadvantage of leading us away realistic astrophysical situations. Nevertheless we have the Schwarzschild so for a spherically symmetric empty spacetime with a central point mass and th responding uncharged, non-spinning black hole. We also have the Kerr so of the uncharged rotating black hole. There are also solutions for charged ro black holes-the Kerr-Newman solutions. But even in the simple case of th body problem, even considering point masses, there is no exact solution in GR must stitch together approximate solution (post-Newtonian) to a numerical so of the orbits. Astrophysically, however, this solution has attained enormous i of the orbits. Astrophysically, however, this solution has attained enormous tance with the construction of laser interferometric gravitational wave observa as neutron stars or black holes is among the promising gravitational wave so for these observatories.

物理代写| Einstein’s Field Equations 相对论代考

物理代考

5.1 爱因斯坦场方程
时空的曲率表现为万有引力。但是由于重力是由物质分布引起的,因此必须将曲率与物质分布联系起来。这正是爱因斯坦方程所做的。他们将曲率张量连接到描述物质分布的张量。但是场方程必须满足几个条件。首先,方程应该是张量方程,这样它们对所有观察者都是正确的,或者在所有坐标系中都成立。其次,爱因斯坦的理论必须简化为牛顿的理论,在弱场和源慢速运动的极限下得到很好的近似。
牛顿引力势 $\phi$ 的牛顿方程由下式给出:
$$
\nabla^{2} \phi=4 \pi G \rho,
$$
其中 $\rho$ 是质量密度,$G$ 是牛顿引力常数 $\simeq 6.67 \times$ $10^{-8}$,单位为 cgs。显然,牛顿方程在 SR 中不是洛伦兹协变的。这可以看作如下:对于匀速运动 $v$ 的观察者,质量变换为 $m \longrightarrow m \gamma$,而由于洛伦兹收缩和因此 $\rho \longrightarrow \rho \gamma^{2}$。由于$\gamma$ 有两个因子,这表明$\rho$ 必须是二阶张量的一个分量。同样在 SR 中,质量和能量是可相互转换的,因此 $\rho$ 或 $\rho c^{2}$ 应该出现在方程中。此外,能量只是 4 动量矢量的一个组成部分,因此动量也必须出现在爱因斯坦方程中,因此也必须是引力源并影响曲率。符合要求的实体是能量动量张量 $T_{ik}$——事实上 $T_{ik}$ 的零零分量是 $T_{00}=\rho c^{2}$。我们将在下一节中物理地描述这个张量。这是爱因斯坦方程的右手边。下一节中的左手。这是爱因斯坦方程的右手边。方程的左侧描述曲率。描述曲率的张量必须因此不能单独使用黎曼张量,只能使用它的收缩,即
(C) 作者,获得 Springer Nature Switzerland AG 2022 的独家许可
87 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_5
88
5 爱因斯坦方程
Ricci 张量 $R_{ik}$ 是一个很好的候选者。但即使这样也不会自行完成,因为能量-动量张量 $T_{ik}$ 具有消失的散度:
$$
T_{; k}^{i k}=0 。
$$
这个方程意味着能量守恒和动量守恒。但是 $R^{i k}$ 不是无散的。但这可以通过减去其散度来轻松解决。因此我们得到了爱因斯坦张量:
$$
G_{i k}=R_{i k}-\frac{1}{2} R g_{i k},
$$
其中 $R$ 是 Ricci 标量。通过 Bianchi 恒等式,如方程式所示。上一章的 (4.5.26) 这个张量的散度消失了。因此我们有爱因斯坦方程:
$$
R_{i k}-\frac{1}{2} R g_{i k}=\kappa T_{i k},
$$
其中$\kappa$ 是一个必须确定的常数。这是通过达到弱场极限然后与牛顿方程 Eq 进行比较来实现的。 (5.1.1)。能量动量张量可以选择为具有压力或质量密度的维度——一个与另一个相关的因子为$c^{2}$。我们做出前一种选择。则常数 $\kappa$ 变为 $8 \pi G / c^{4}$。因此爱因斯坦方程为:
$$
G_{i k}=R_{i k}-\frac{1}{2} R g_{i k}=\frac{8 \pi G}{c^{4}} T_{i k} 。
$$
方程简化为牛顿方程(5.1.1)。
爱因斯坦张量$G_{ik}$也可以通过作用原理得到。该反应称为爱因斯坦-希尔伯特作用,由下式给出:
$$
\mathfrak{A}=\int_{\mathcal{V}} R \sqrt{-g} d^{4} x
$$
这个动作相对于 $g^{ik}$ 是变化的,边界条件是 $\delta g$ 它的一阶导数在 $\mathcal{V}$ 的边界上消失。 $v$ 其他书籍也采用了这种方法,我们将感兴趣的读者推荐给这些书籍(Narlikar 2010)。这里不再赘述。

爱因斯坦方程是度量分量$g_{ik}$ 的二阶耦合非线性偏微分方程。由于黎曼张量涉及度量的三阶导数,所以方程也是二阶的。也是非线性的,因为它们涉及 $g_{ik}$ 的乘积及其相对于坐标的一阶和二阶导数。其物理意义是引力场也产生引力场,因为它携带能量。我认为,牛顿方程是线性的,麦克斯韦方程也是线性的
$5.1$ 爱因斯坦的场方程
电磁学。这意味着在 GR 中我们不能叠加我们在牛顿万有引力理论和电学中可以方便地做的解。这使得解决方案难以获得。方程也是耦合的。 V 数 a

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上至高压线,下至发电机,只要用到电的地方就有相对论效应存在!相对论是关于时空和引力的理论,主要由爱因斯坦创立,相对论的提出给物理学带来了革命性的变化,被誉为现代物理性最伟大的基础理论。

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