# 物理代写|广义相对论代写General relativity代考|MATH4105

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## 物理代写|广义相对论代写General relativity代考|Schwarzschild Solution or Exterior Solution

The exact solution of the Einstein field equation in empty space was obtained by Schwarzschild in 1916, which describes the geometry of spacetime outside a spherically symmetric distribution of matter.

Consider the metric of the empty spacetime outside of a spherically symmetric distribution of matter of mass $M$ as
$$d s^2=e^v d t^2-e^\lambda d r^2-r^2 d \theta^2-r^2 \sin ^2 \theta d \phi^2$$
where $\lambda$ and $v$ are functions of $r$ alone.
A metric of this type is known as spherically symmetric.
Let us write
\begin{aligned} g_{11}=-e^\lambda ; \quad g_{22}=-r^2 ; \quad g_{33}-r^2 \sin ^2 \theta ; \quad g_{44}=e^v, \ \text { then, } \quad g^{11}=-e^{-\lambda} ; \quad g^{22}=-\frac{1}{r^2} ; \quad g^{33}=-\frac{1}{r^2 \sin ^2 \theta} ; \quad g^{44}=e^{-v} . \end{aligned}
The nonvanishing Christoffel symbols of second kind are
$$\begin{gathered} \Gamma_{11}^1=\frac{1}{2} \lambda^{\prime} ; \Gamma_{12}^2=\frac{1}{r} ; \Gamma_{31}^3=\frac{1}{r}, \Gamma_{22}^1=-r e^{-\lambda}, \Gamma_{33}^2=-\sin \theta \cos \theta, \ \Gamma_{23}^3=\cot \theta ; \Gamma_{33}^1=-r \sin ^2 \theta e^{-\lambda} ; \Gamma_{44}^1=\frac{1}{2} v^{\prime} e^{-\lambda+v} ; \Gamma_{14}^4=\frac{1}{2} v^{\prime} \ {\left[\Gamma_{a a}^a=\left[\ln \sqrt{\left|g_{a a}\right|}\right]a ; \Gamma{b a}^a=\left[\ln \sqrt{\left|g_{a a}\right|}\right]{, b} ; \Gamma{a a}^b=-\frac{1}{2 g_{b b}} g_{a a, b}\right],} \end{gathered}$$
where $A_{a b, c} \equiv \frac{\partial}{\partial x} A_{a b}$.
The complements of the Ricci tensor are
\begin{aligned} R_{11} & =-\frac{\lambda^{\prime}}{r}-\frac{1}{4} \lambda^{\prime} v^{\prime}+\frac{1}{2} v^{\prime \prime}+\frac{1}{4} v^{\prime 2}, \ R_{22}=\csc ^2 \theta R_{33} & =-1+e^{-\lambda}\left[1-\frac{1}{2} r \lambda^{\prime}+\frac{1}{2} r v^{\prime}\right], \ R_{44} & =e^{-\lambda+v}\left[\frac{1}{4} \lambda^{\prime} v^{\prime}-\frac{1}{2} v^{\prime \prime}-\frac{1}{r} v^{\prime}-\frac{1}{4} v^{\prime 2}\right], \ R_{\alpha \beta} & =0, \alpha \neq \beta . \end{aligned}

## 物理代写|广义相对论代写General relativity代考|A Classical Argument

The concept of a black hole, in a lucid sense, is that of a star whose gravitational field is very powerful, so that even light cannot seep to different areas. In the Newtonian regime, let us assume a particle of mass $m$ traveling radially from a spherically symmetric distribution of matter having total mass $M$, radius $R$, and uniform density $\rho$. Let us also assume that at a distance $r$ from the center, the particle possesses a velocity $v$, then the conservation of energy $E$ yields
$$E=K . E .+P . E .=\frac{1}{2} m v^2-\frac{G M m}{r} .$$
The escape velocity $v_0$ is the minimum velocity desired to take a body from the surface of the distribution of matter to infinity. This demands $v \rightarrow 0$ as $r \rightarrow \infty$. Hence from (i) we have $E=0$. Solving for $v$, we obtain, $v^2=\frac{2 G M}{r}$; therefore, the escape velocity is
$$v_0^2=\frac{2 G M}{R}$$
This indicates that when the particle’s radial velocity at the surface is less than $v_0$, it will ultimately come back to the surface of the distribution of matter due to the gravitational attraction. For the case of a light ray (with velocity $c$ ), which would just escape to infinity, one gets a relation between $c$ with mass and radius of the distribution as
$$c^2=\frac{2 G M}{R}$$
Thus, we have two possibilities for which light could no longer escape: either when the mass $M$ is increased with a fixed radius or the radius $R$ is decreased with fixed mass. Note that the limiting condition (iii) yields the radius $R$ as
$$R=\frac{2 G M}{c^2}$$
which is known as Schwarzschild radius.

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Schwarzschild Solution or Exterior Solution

$$g_{11}=-e^\lambda ; \quad g_{22}=-r^2 ; \quad g_{33}-r^2 \sin ^2 \theta ; \quad g_{44}=e$$

$$\Gamma_{11}^1=\frac{1}{2} \lambda^{\prime} ; \Gamma_{12}^2=\frac{1}{r} ; \Gamma_{31}^3=\frac{1}{r}, \Gamma_{22}^1=-r e^{-\lambda}, \Gamma_{33}^2=$$

Ricci 张量的补集是
$$R_{11}=-\frac{\lambda^{\prime}}{r}-\frac{1}{4} \lambda^{\prime} v^{\prime}+\frac{1}{2} v^{\prime \prime}+\frac{1}{4} v^{\prime 2}, R_{22}=\csc ^2 \theta R_3$$

## 物理代写|广义相对论代写General relativity代考|A Classical Argument

$$E=K . E .+P . E .=\frac{1}{2} m v^2-\frac{G M m}{r} .$$

$$v_0^2=\frac{2 G M}{R}$$

$$c^2=\frac{2 G M}{R}$$

$$R=\frac{2 G M}{c^2}$$

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