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
光在引力场中弯曲。在存在引力场的情况下，它似乎不会直线运动。让我们考虑一个质量为 $M$ 的中心质量，其外部有一个球对称的引力场。然后这个引力场由史瓦西度量来表示。我们将认为该领域是弱的。通常我们会想到光线掠过太阳表面的问题。这是广义相对论的第一个经典测试，在历史上是广义相对论的重要胜利。探险队是由一个数字测试结果非常好地证实了 GR 的预测——根据 GR 的预测，发现偏转角为 1.75 美元角秒。我们现在将继续在下面获得这个结果。
我们从方程式开始。 （6.4.23）从上一章开始。为了方便起见，我们在下面重写这个方程：
$$
\frac{d^{2} u}{d \phi^{2}}+u=3 m u^{2} 。
$$
由于 $3 m u^{2}$ 项，这个方程是非线性的，因此不容易精确求解。但是我们不需要确切的解决方案；一个近似解就足够了，因为非线性项在某种意义上是“小”的，我们很快就会看到。考虑一条光线，它从离中心质量很远的地方开始，掠过中心质量。我们选择坐标使得质量在原点$O$；光线从 $x \longrightarrow-\infty$ 开始，几乎平行于 $x$ 轴，冲击参数为 $b$，并经过原点朝向 $x \longrightarrow \infty$。情况如图 7.1 所示。如果没有质量，光线将沿着直线 $y=b$ 从 $-\infty$ 到 $\infty$，平行于 $x$-轴，如果它开始平行于 $x$ -轴。在极坐标中，这
图 7.1 一条光线从 $x \longrightarrow-\infty$ 向位于原点的具有冲击参数 $b$ 的中心质量 $M$ 传播，并继续朝向 $x \longrightarrow \infty$。光线被原点附近的质量 $M$ 偏转。光线发生 $\Delta \phi \sim 4 m / b$ 的偏转。为了清楚起见，图中的偏转角 $\Delta \phi$ 被严重夸大了
7.2 中心质量对光的偏转
123
方程变为$r \sin \phi=b$。但是因为有一个中心质量，当光线接近原点时，它会向质量弯曲，并且它几乎会沿一条直线向稍微不同的方向继续前进。因此，光线将偏离其原始路径。这是粗略的画面，现在我们将继续计算。
通过设置$v=b u$ 可以方便地转换为无量纲坐标。然后我们可以写Rq。 (7.2.1) 在 $v$ 方面为：
$$
\frac{d^{2} v}{d \phi^{2}}+v=\epsilon v^{2}
$$
其中 $\epsilon=3 \mathrm{~m} / b$。请注意，$\epsilon$ 是无量纲的，对于所考虑的情况，它很小。对于掠过太阳表面的光线的具体情况，$b$ 可以取为太阳的半径，$m \sim 1.5 \mathrm{~km}$ 是太阳的质量，以长度为单位.因此$\epsilon$ 是$10^{-6}$ 的几倍，远小于unity。现在我们准备在小参数 $\epsilon$ 中按顺序执行逼近。我们只去除$\epsilon$ 中的第一个订单，并相应地写出Eq。 (7.2.2) 用这种形式的 $v$ 我们得到：
$$
\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}+\epsilon \frac{d^{2} v_{1}}{d \phi^{2 }}+\epsilon v_{1}=\epsilon\left(v_{0}+\epsilon v_{1}\right)^{2}
$$
$$
\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}+\epsilon \frac{d^{2} v_{1}}{d \phi^{2 }}+\epsilon v_{1}=\epsilon\left(v_{0}+\epsilon v_{1}\right)^{2}
$$
我们现在可以以 $\epsilon$ 的不同阶数分离出方程。我们以 0 阶和 1 阶写出方程：
$$
\开始{对齐}
&\frac{d^{2} v_{0}}{d \phi^{2}}+v_{0}=0 \
&\frac{d^{2} v_{1}}{d \phi^{2}}+v_{1}=v_{0}^{2}
\end{对齐}
$$
选择边界条件的 $v_{0}$ 方程的解是 $v_{0}=\sin \phi$。这对应于$r \sin \phi=b$，即$y=b$。这是问题的零阶解。我们现在将这个解代入第二个方程并得到：
$$
\frac{d^{2} v_{1}}{d \phi^{2}}+v_{1}=\sin ^{2} \phi
$$
这个方程的解是：
$$
v_{1}=\frac{2}{3}-\frac{1}{3} \sin ^{2} \phi
$$
然后给出替换 $\epsilon, v=\sin \phi+2 m / b-(m / b) \sin ^{2} \phi .$ 我们需要这条曲线的渐近线，当 $v \rightarrow 0$。删除最后一项，因为它具有更高的小阶，我们发现渐近线由下式给出
124
广义相对论的 7 个经典检验
$\sin \phi=-2 m / b$ 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。
我们注意到仅使用
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