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

## 物理代写|广义相对论代写General relativity代考|Motion of Test Particle

Let us consider the motion of a massive test particle or a massless particle, i.e., photon in Schwarzschild spacetime. It is known that all massive particles move along time-like geodesics, whereas photons move along null geodesics. We shall consider geodesics of test particles (either time-like or null) in Schwarzschild spacetime.
Let us take the Lagrangian in the following form as (with $p$ is an affine parameter)
$$L=\left(1-\frac{2 m}{r}\right)\left(\frac{d t}{d p}\right)^2-\left(1-\frac{2 m}{r}\right)^{-1}\left(\frac{d r}{d p}\right)^2-r^2\left(\frac{d \theta}{d p}\right)^2-r^2 \sin ^2 \theta\left(\frac{d \phi}{d p}\right)^2 .$$
We know
$$\delta \int d s=0 \Rightarrow \int \delta\left(\frac{d s}{d p}\right) d p=0 \Rightarrow \delta \int L d p=0,$$
where
$$L=\left(\frac{d s}{d p}\right)=\left(g_{\mu \gamma} \frac{d x^\mu}{d p} \frac{d x^\gamma}{d p}\right)^{\frac{1}{2}}$$
$\Rightarrow$ Euler-Lagrangian equation
$$\frac{d}{d p}\left(\frac{\partial L}{\partial\left(\frac{d x^*}{d p}\right)}\right)-\frac{\partial L}{\partial x^\mu}=0 .$$
Thus, the corresponding Euler-Lagrange’s equations are
\begin{aligned} &\frac{d}{d p}\left(\frac{\partial L}{\partial r^1}\right)-\frac{\partial L}{\partial r}=0 \ &\frac{d}{d p}\left(\frac{\partial L}{\partial \theta^1}\right)-\frac{\partial L}{\partial \theta}=0, \text { etc. } \end{aligned}
$[” \cdots$ ” implies differentiation with respect to $p]$

## 物理代写|广义相对论代写General relativity代考|The precession of the perihelion motion of mercury

The point on the orbit nearest to the sun is called perihelion. Observations indicate that the perihelion of mercury’s orbit shows a precession of $5599.74 \pm .41$ arc sec per century. Among this $5557.18 \pm .85$ arc sec can be explained by Newtonian gravitational theory. It is due to the influences of other planets. Therefore, an amount $\approx 43.03^{\prime \prime}$ per century is still unsolved. The unaccounted perihelion motion of $42.56 \pm .94 \operatorname{arc~sec}$ per century can be explained by general theory of relativity.

Consider the motion of a planet. Since mass of a planet is very small compared to the sun, we can regard the planet as a test particle moving in the field of the sun (see Fig. 17). Again, the sun is at least, to a high degree of approximation, a spherical body. So the geometry outside the sun is given by Schwarzschild metric and the planet will move in geodesics (massive particle) in this field. The relativistic equation of planet’s orbit is
$$\frac{d^2 U}{d \phi^2}+U=\frac{m}{h^2}+3 m U^2 .$$
[we have considered planet as a test particle (massive) in the gravitational field of sun and for this $\epsilon=1]$

## 物理代写|广义相对论代写广义相对论代考|试验粒子运动

$$L=\left(1-\frac{2 m}{r}\right)\left(\frac{d t}{d p}\right)^2-\left(1-\frac{2 m}{r}\right)^{-1}\left(\frac{d r}{d p}\right)^2-r^2\left(\frac{d \theta}{d p}\right)^2-r^2 \sin ^2 \theta\left(\frac{d \phi}{d p}\right)^2 .$$

$$\delta \int d s=0 \Rightarrow \int \delta\left(\frac{d s}{d p}\right) d p=0 \Rightarrow \delta \int L d p=0,$$

$$L=\left(\frac{d s}{d p}\right)=\left(g_{\mu \gamma} \frac{d x^\mu}{d p} \frac{d x^\gamma}{d p}\right)^{\frac{1}{2}}$$
$\Rightarrow$欧拉-拉格朗日方程
$$\frac{d}{d p}\left(\frac{\partial L}{\partial\left(\frac{d x^*}{d p}\right)}\right)-\frac{\partial L}{\partial x^\mu}=0 .$$

\begin{aligned} &\frac{d}{d p}\left(\frac{\partial L}{\partial r^1}\right)-\frac{\partial L}{\partial r}=0 \ &\frac{d}{d p}\left(\frac{\partial L}{\partial \theta^1}\right)-\frac{\partial L}{\partial \theta}=0, \text { etc. } \end{aligned}
$[” \cdots$暗示了对$p]$ 的微分

## 物理代写|广义相对论代写广义相对论代考|水星近日点运动的进动

$$\frac{d^2 U}{d \phi^2}+U=\frac{m}{h^2}+3 m U^2 .$$
[我们认为行星是太阳引力场中的一个测试粒子(质量)，为此$\epsilon=1]$

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