19th Ave New York, NY 95822, USA

# 物理代写| The Deflection of Light by a Central Mass 相对论代考

## 物理代写

7.2 The Deflection of Light by a Central Mass
Light is bent in a gravitational field. It does not appear to travel in straight lines in the presence of a gravitational field. Let us consider a central mass with mass $M$ having in its exterior a spherically symmetric gravitational field. Then this gravitational field is represented by the Schwarzschild metric. We will consider the field to be weak. Typically we have in mind the problem of a light ray grazing the surface of the Sun. This is the first of the classical tests of general relativity and was historically an important triumph for general relativity. Expeditions were led by no less a figure results of the tests bore out the predictions of GR extremely well-the deflection angle was found to be $1.75$ arc seconds in accordance with the predictions of GR. We will now proceed to obtain this result below.

We start with Eq. (6.4.23) from the last chapter. We rewrite this equation here for convenience below:
$$\frac{d^{2} u}{d \phi^{2}}+u=3 m u^{2} .$$
This equation is nonlinear because of the $3 m u^{2}$ term and therefore not easy to solve exactly. But we do not require the exact solution; an approximate solution suffices, because the nonlinear term is “small” in a sense we will see soon. Consider a light ray which starts from a large distance from the central mass and grazes past the central mass. We choose coordinates so that the mass is at the origin $O$; the light ray starts from $x \longrightarrow-\infty$ almost parallel to the $x$-axis with an impact parameter $b$ and goes past the origin towards $x \longrightarrow \infty$. The situation is shown in Fig. 7.1. If the mass were absent, the light ray would travel along a straight line $y=b$ from $-\infty$ to $\infty$ parallel to the $x$-axis, if it had started parallel to the $x$-axis. In polar coordinates this
Fig. 7.1 A light ray travels from $x \longrightarrow-\infty$ towards a central mass $M$ located at the origin with impact parameter $b$ and continues towards $x \longrightarrow \infty$. The light ray is deflected by the mass $M$ near the origin. The light ray suffers a deflection of $\Delta \phi \sim 4 m / b$. The deflection angle $\Delta \phi$ is grossly exaggerated in the figure for clarity
7.2 The Deflection of Light by a Central Mass
123
equation becomes $r \sin \phi=b$. But because there is a central mass, the light ray, as it approaches the origin will bend towards the mass and it will continue almost in a straight line in a slightly different direction. Thus the light ray will be deflected from its original path. This is the rough picture and now we will proceed with the calculation.

It is convenient to transform to dimensionless coordinates by setting $v=b u$. Then we can write Rq. (7.2.1) in terms of $v$ as:
$$\frac{d^{2} v}{d \phi^{2}}+v=\epsilon v^{2}$$
where $\epsilon=3 \mathrm{~m} / b$. Note that $\epsilon$ is dimensionless and for the situation under consideration it is small. For the specific case of a light ray grazing the surface of the Sun, $b$ can be taken to be the radius of the Sun and $m \sim 1.5 \mathrm{~km}$, the mass of the Sun in length units. Therefore $\epsilon$ is few times $10^{-6}$, much smaller than unity. Now we are ready to carry out the approximation order by order in the small parameter $\epsilon$. We weed only go the first order in $\epsilon$ and accordingly writing Eq. (7.2.2) with this form of $v$ we obtain:
$$\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}+\epsilon \frac{d^{2} v_{1}}{d \phi^{2}}+\epsilon v_{1}=\epsilon\left(v_{0}+\epsilon v_{1}\right)^{2}$$
$$\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}+\epsilon \frac{d^{2} v_{1}}{d \phi^{2}}+\epsilon v_{1}=\epsilon\left(v_{0}+\epsilon v_{1}\right)^{2}$$
We can now separate out the equations at various orders of $\epsilon$. We write out the equations at order 0 and order 1 :
\begin{aligned} &\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}=0 \ &\frac{d^{2} v_{1}}{d \phi^{2}}+v_{1}=v_{0}^{2} \end{aligned}
The solution to the $v_{0}$ equation with boundary conditions as chosen is $v_{0}=\sin \phi$. This corresponds to $r \sin \phi=b$ which is $y=b$. This is the zero’th order solution to the problem. We now substitute this solution into the second equation and obtain:
$$\frac{d^{2} v_{1}}{d \phi^{2}}+v_{1}=\sin ^{2} \phi$$
The solution to this equation is:
$$v_{1}=\frac{2}{3}-\frac{1}{3} \sin ^{2} \phi$$
which then gives on substituting for $\epsilon, v=\sin \phi+2 m / b-(m / b) \sin ^{2} \phi .$ We need the asymptotes to this curve which are obtained when $v \rightarrow 0$. Dropping the last term as it is of higher order of smallness, we find that the asymptotes are given by
124
7 Classical Tests of General Relativity
$\sin \phi=-2 m / b$ which gives two solutions $\phi=-2 m / b, \pi+2 m / b$. Thus the deflection angle is $\Delta \phi=4 m / b$. If one puts in the values corresponding to the Sun’s mass and radius, we find $\Delta \phi \sim 1.75^{\prime \prime}$ in agreement with observations.

We remark that using only the equivalence principle for solving this problem gives half this value. This problem has been worked out in Exercise $2.3 .2$ of Chapter $2 .$

## 物理代考

.2 中心质量对光的偏转

$$\frac{d^{2} u}{d \phi^{2}}+u=3 m u^{2} 。$$

7.2 中心质量对光的偏转
123

$$\frac{d^{2} v}{d \phi^{2}}+v=\epsilon v^{2}$$

$$\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}+\epsilon \frac{d^{2} v_{1}}{d \phi^{2 }}+\epsilon v_{1}=\epsilon\left(v_{0}+\epsilon v_{1}\right)^{2}$$
$$\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}+\epsilon \frac{d^{2} v_{1}}{d \phi^{2 }}+\epsilon v_{1}=\epsilon\left(v_{0}+\epsilon v_{1}\right)^{2}$$

$$\开始{对齐} &\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}=0 \ &\frac{d^{2} v_{1}}{d \phi^{2}}+v_{1}=v_{0}^{2} \end{对齐}$$

$$\frac{d^{2} v_{1}}{d \phi^{2}}+v_{1}=\sin ^{2} \phi$$

$$v_{1}=\frac{2}{3}-\frac{1}{3} \sin ^{2} \phi$$

124

$\sin \phi=-2 m / b$ 给出两个解 $\phi=-2 m / b, \pi+2 m / b$。因此偏转角为$\Delta \phi=4 m / b$。如果输入与太阳质量和半径相对应的值，我们会发现 $\Delta \phi \sim 1.75^{\prime \prime}$ 与观察结果一致。

Matlab代写