# 物理代写|广义相对论代写General relativity代考|PHYC90012

## 物理代写|广义相对论代写General relativity代考|Some Modified Theories of Gravity

In this chapter, we just provide an outline of some modified theories of gravity.

1. $f(R)$ theory of gravity: In the action of Einstein general theory of relativity, we use Lagrangian for geometry as $L_G=R$, where $R$ is a Ricci scalar. Now, it can be generalized by taking $L_G=f(R)$. This newly developed theory of gravity is known as $f(R)$ theory of gravity. For a purely phenomenological thought, $f(R)$ could be expanded in a power series with positive as well as negative powers of the curvature scalar as
$$f(R)=\ldots \ldots+\frac{\alpha_2}{R^2}+\frac{\alpha_1}{R}-2 \Lambda+R+\frac{R^2}{\beta_2}+\frac{R^3}{\beta_3}+\ldots \ldots \ldots$$
where the coefficients $\alpha_i$ and $\beta_i$ have the appropriate dimensions.
Following the same point of view, we can write the action for $f(R)$ gravity as
$$I=\int \sqrt{-g}\left[f(R)+2 k L_m\right] d^4 x .$$
Here, $L_m$ is the matter Lagrangian. Varying with respect to $g_{\mu v}$, we get modified Einstein field equation as
$$\Xi_{\mu \nu} \equiv F(R) R_{\mu \nu}-\frac{1}{2} g_{\mu v} f(R)-\left[\nabla_\mu \nabla_v-g_{\mu v} \square\right] F(R)=-k T_{\mu v} .$$

Here, $F(R)=\frac{d f R)}{d R}, \nabla_\mu$ is the covariant derivative with $\square F=\nabla^\mu \nabla_\mu F=\frac{1}{\sqrt{-g}} \partial_\mu\left(\sqrt{-g} g^{\mu \nu} \partial_v F\right)$. As usual $T_{\mu v}$ is the energy-momentum tensor of the matter fields. Conservation equation $\nabla^\mu T_{\mu v}=0$ implies $\nabla^\mu \Xi_{\mu v}=0$. Also trace of the above field equation in $f(R)$ gravity is
$$3 \square F(R)+F(R) R-2 f(R)=-k T,$$
where $T=g^{\mu v} T_{\mu v}$.
We can recover Einstein gravity without the cosmological constant by taking $f(R)=R$. This implies $F(R)=1$ with $\square F=0$ and we get $R=k T$. In general, the trace of field equation in $f(R)$ gravity indicates that the field equations of $f(R)$ theory will provide various different solutions than Einstein’s theory. For a simple example, one can notice here that the Birkhoff’s theorem does not hold good, which states that the Schwarzschild solution is the unique spherically symmetric vacuum solution. Note that for $T=0$ no longer implies that $R=0$, or is even constant.

## 物理代写|广义相对论代写General relativity代考|Proof of Poisson Equation

Let us consider a mass $M$ occupying a volume $V$, which is enclosed by a surface $S$. The gravitational flux passing through the elementary surface $d S$ is given by g.n $d S$, where $\mathbf{g}$ is gravitational vector field (also known as gravitational acceleration) and $\mathbf{n}$ is the unit outward normal vector to S. Now, the total gravitational flux through $\mathrm{S}$ is
$$\int_S \mathbf{g} \cdot \mathbf{n} d S=-G M \int_S \frac{\mathbf{e}_r \cdot \mathbf{n}}{r^2} d S$$
We know $\frac{e_r \mathrm{n}}{r^2} d S=\frac{\cos \theta d S}{r^2}$ is the elementary solid angle $d \Omega$ subtended at $M$ by the elementary surface $d S$, where $\mathbf{e}_r$ is the radial unit vector. Thus,
$$\int_S \mathbf{g} \cdot \mathbf{n} d S=-G M \int_S d \Omega=-4 \pi G M=-4 \pi G \int_V \rho(r) d V .$$
Applying Gauss divergence theorem to the left-hand side, we get
$$\int_V \nabla \cdot \mathbf{g} d V=-4 \pi G \int_V \rho(r) d V .$$
Thus, we get
$$\nabla \cdot \mathrm{g}=-4 \pi G \rho .$$

Using $\mathbf{g}=-\nabla \phi$ (gravity is a conservative force, therefore, it can be written as the gradient of a scalar potential $\phi$, known as the gravitational potential), we finally obtain
$$\nabla^2 \phi=4 \pi G \rho .$$
The gravitational field $(F)$ is proportional to the negative of $\nabla \phi$, i.e.,
$$F=-m \nabla \phi .$$
This is the force acting on a particle of mass $m$.
For a single mass $M$ that produces the potential $\phi$, then the solution of Poisson equation is given by
$$\phi=\frac{-G M}{r} .$$
The force acting on another particle with mass $m$ will be
$$F=G M m \nabla\left(\frac{1}{r}\right)=\frac{-G M m}{r^2} .$$
The ratio of gravitational force and electrical force between two electrons is given by
$$\frac{F_{\text {grav }}}{F_{\text {elec }}}=\frac{G m_e^2}{k_e e^2}=0.24 \times 10^{-42} .$$
Here, $k_e$ is Coulomb’s constant with $m_e$ and $e$ are mass and charge of the electron, respectively. This indicates that the gravitational force is very weak.

## 物理代写|广义相对论代写General relativity代考|Some Modified Theories of Gravity

1. $f(R)$ 引力理论: 在爱因斯坦广义相对论的作用下，我们使用拉格朗日几何作为 $L_G=R$ ，在哪里 $R$ 是 Ricci 标量。现在，它可以概括为 $L_G=f(R)$. 这种新发展的引力理论被称为 $f(R)$ 引力理论。对于纯粹 的现象学思想， $f(R)$ 可以在曲率标量的正募和负募的募级数中展开为
$$f(R)=\ldots \ldots+\frac{\alpha_2}{R^2}+\frac{\alpha_1}{R}-2 \Lambda+R+\frac{R^2}{\beta_2}+\frac{R^3}{\beta_3}+\ldots \ldots . .$$
其中系数 $\alpha_i$ 和 $\beta_i$ 有合适的尺寸。
遵循相同的观点，我们可以将动作写成 $f(R)$ 重力为
$$I=\int \sqrt{-g}\left[f(R)+2 k L_m\right] d^4 x .$$
这里， $L_m$ 是拉格朗日问题。变化相对于 $g_{\mu v}$ ，我们得到修改后的爱因斯坦场方程为
$$\Xi_{\mu \nu} \equiv F(R) R_{\mu \nu}-\frac{1}{2} g_{\mu v} f(R)-\left[\nabla_\mu \nabla_v-g_{\mu v} \square\right] F(R)=-k T_{\mu v} .$$
这里， $F(R)=\frac{d f R)}{d R}, \nabla_\mu$ 是协变导数 $\square F=\nabla^\mu \nabla_\mu F=\frac{1}{\sqrt{-g}} \partial_\mu\left(\sqrt{-g} g^{\mu \nu} \partial_v F\right)$. 照常 $T_{\mu v}$ 是物质场的能 量动量张量。守恒方程 $\nabla^\mu T_{\mu v}=0$ 暗示 $\nabla^\mu \Xi_{\mu v}=0$. 上面的场方程也可以在 $f(R)$ 重力是
$$3 \square F(R)+F(R) R-2 f(R)=-k T,$$
在哪里 $T=g^{\mu v} T_{\mu v}$.
我们可以在没有宇宙学常数的情况下恢复爰因斯坦引力 $f(R)=R$. 这意味着 $F(R)=1$ 和 $\square F=0$ 我们得 到 $R=k T$. 一般来说，场方程的迹 $f(R)$ 重力表示场方程 $f(R)$ 理论将提供与爱因斯坦理论不同的各种解决 方案。举一个简单的例子，这里可以注意到 Birkhoff 定理并不成立，它表明 Schwarzschild 解是唯一的球 对称真空解。请注意，对于 $T=0$ 不再暗示 $R=0$ ，甚至是常数。

## 物理代写|广义相对论代写General relativity代考|Proof of Poisson Equation

$$\int_S \mathbf{g} \cdot \mathbf{n} d S=-G M \int_S \frac{\mathbf{e}r \cdot \mathbf{n}}{r^2} d S$$ 我们知道 $\frac{e_r \mathrm{n}}{r^2} d S=\frac{\cos \theta d S}{r^2}$ 是基本立体角 $d \Omega$ 对向于 $M$ 由基本面 $d S$ ，在哪里 $\mathbf{e}_r$ 是径向单位向量。因此， $$\int_S \mathbf{g} \cdot \mathbf{n} d S=-G M \int_S d \Omega=-4 \pi G M=-4 \pi G \int_V \rho(r) d V$$ 将高斯散度定理应用于左侧，我们得到 $$\int_V \nabla \cdot \mathbf{g} d V=-4 \pi G \int_V \rho(r) d V .$$ 因此，我们得到 $$\nabla \cdot \mathrm{g}=-4 \pi G \rho .$$ 使用 $\mathrm{g}=-\nabla \phi$ (引力是一种保守的力，因此，它可以写成标量势的梯度 $\phi$ ，称为引力势)，我们最終得 到 $$\nabla^2 \phi=4 \pi G \rho .$$ 引力场 $(F)$ 与负数成正比 $\nabla \phi$ ，那是， $$F=-m \nabla \phi .$$ 这是作用在有质量的粒子上的力 $m$. 对于单个质量 $M$ 产生潜力的 $\phi$ ，则泊松方程的解由下式给出 $$\phi=\frac{-G M}{r} .$$ 作用在另一个有质量的粒子上的力 $m$ 将会 $$F=G M m \nabla\left(\frac{1}{r}\right)=\frac{-G M m}{r^2} .$$ 两个电子之间的引力和电力之比由下式给出 $$\frac{F{\text {grav }}}{F_{\text {elec }}}=\frac{G m_e^2}{k_e e^2}=0.24 \times 10^{-42} .$$

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