# 物理代写| TThe Perihelion Shift in the Orbit of Mercury 相对论代考

## 物理代写

7.3 The Perihelion Shift in the Orbit of Mercury
In Newton’s theory a test particle orbit is a conic section-an ellipse, a hyperbola or a parabola. If the orbit is bound as in the case of planetary motion, then the planet must describe an ellipse (in the absence of other perturbing forces). It was found that the orbit of Mercury is a precessing ellipse. Surely, there are other planets orbiting the Sun which are responsible in part for the precession, but not all of the precession could be explained away. The discrepancy not accounted for is minute-about 43 arc seconds per century. Although the discrepancy is minute, it cannot be denied. It was a great triumph for GR that it could exactly explain this discrepancy and GR became a theory to be reckoned with. Figure $7.2$ shows a precessing elliptical orbit with the sun (orange disc) at the focus. The orbit is not closed. The precession has been grossly exaggerated for clarity. In this subsection we will establish this result.
We begin with Eqs. (6.4.9) and (6.4.10), then divide one by the other to obtain the following equation:
$$\frac{1}{r^{4}}\left(\frac{d r}{d \phi}\right)^{2}=\frac{E^{2}-1}{L^{2}}+\frac{2 m}{r}\left(\frac{1}{r^{2}}+\frac{1}{L^{2}}\right)-\frac{1}{r^{2}}$$
Changing the variable to $u=1 / r$ as before results in,
$$\left(\frac{d u}{d \phi}\right)^{2}+u^{2}=\frac{E^{2}-1}{L^{2}}+2 m u\left(u^{2}+\frac{1}{L^{2}}\right)$$
This is however somewhat an inconvenient equation. A simpler equation is obtained on differentiating this equation:
$$u^{\prime \prime}+u=\frac{m}{L^{2}}+3 m u^{2}$$
where in the above equation we have used a short hand notation by denoting differentiation with respect to $\phi$ with a prime.

Let us now pause and understand the magnitude of the terms in Eq. (7.3.3) for Mercury’s orbit. If one ignores the second term in this equation on the RHS, this equation describes just the Newtonian orbit. The quantity $L$-angular momentum per unit rest mass at infinity-is just the Newtnonian angular momentum $L_{N} / c$. This is evident from Eq. (6.4.9) in which the differentiation is with respect to $s$ which
Fig. 7.2 The figure depicts a precessing elliptical orbit with the sun (orange disc) at the focus. The orbit is not closed. The precession has been grossly exaggerated for clarity
is the proper time multiplied by $c$ or $d s=c d \tau$. But since the velocity of Mercury is small compared with $c, d s \approx c d t$. This brings in the factor $c$ or $L_{N}=c L$. The Newtonian equation is,
$$u^{\prime \prime}+u=\frac{m}{L^{2}}=\frac{G M}{L_{N}^{2}} \equiv \frac{1}{l}$$
where $l$ is the latus rectum. This equation has the solution:
$$u=\frac{1}{l}(1+e \cos \phi)$$
Eq. (7.3.5) is the Newtonian solution for the orbit. $e$ is the eccentricity of Mercury’s orbit and it is reasonably small so that the orbit can be considered to be roughly circular. Then $l \sim$ radius of the orbit $\sim 6 \times 10^{7} \mathrm{~km}$. Now going back to Eq. (7.3.3) we can see that the second term on the RHS is $\sim 3 m / l^{2}$ because $u \sim 1 / l$ while the first term is $1 / l$. Thus the second term is smaller by a factor of $3 m / l \sim 10^{-7}$ as compared to the first. Recall that the Sun’s mass $m \sim 1.5 \mathrm{~km}$. It is therefore a small perturbation of the orbit and therefore it has an extremely small effect. The essential effect we are interested in is the precession of the elliptical orbit.
7 Classical Tests of General Relativity
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Let us therefore write the total solution as $u=u_{0}+u_{1}$, where $u_{0}$ is the Newtonian solution given in Eq. (7.3.5). Then to the first order $u_{1}$ satisfies the equation:
\begin{aligned} u_{1}^{\prime \prime}+u_{1} &=3 m u_{0}^{2} \ &=\frac{3 m}{l^{2}}\left(1+2 e \cos \phi+e^{2} \cos ^{2} \phi\right) \end{aligned}
We remark that the perturbation scheme can be carried out in a systematic way just as we did for the case of deflection of light by the Sun. We could have defined $v=l u$ and written the full equation as $v^{\prime \prime}+v=1+\epsilon v^{2}$, where $\epsilon=3 m / l$, then separated out the equation at orders 0 and 1 of $\epsilon$. Carrying out these steps, we would have obtained identical results.

There are now three terms on the RHS of Eq. (7.3.6). The first and third term, that is the constant term and the $\cos ^{2} \phi$ term only produce a bounded oscillatory solution. It is the term in $\cos \phi$ which is responsible for a secular increase in phase and hence for precession of the orbit. Therefore, taking only this term into account, we find that the solution is,
$$u_{1}=\frac{3 m e}{l^{2}} \phi \sin \phi$$
The above solution can be easily verified by direct substitution. The full solution is,
\begin{aligned} u &=\frac{1}{l}(1+e \cos \phi)+\frac{3 m e}{l^{2}} \phi \sin \phi \ &=\frac{1}{l}[1+e(\cos \phi+\delta \sin \phi)] \end{aligned}
where,
$$\delta=\frac{3 m}{l} \phi$$

## 物理代考

7.3 水星轨道的近日点位移

$$\frac{1}{r^{4}}\left(\frac{dr}{d \phi}\right)^{2}=\frac{E^{2}-1}{L^{2} }+\frac{2 m}{r}\left(\frac{1}{r^{2}}+\frac{1}{L^{2}}\right)-\frac{1}{r ^{2}}$$

$$\left(\frac{du}{d \phi}\right)^{2}+u^{2}=\frac{E^{2}-1}{L^{2}}+2 mu\left (u^{2}+\frac{1}{L^{2}}\right)$$

$$u^{\prime \prime}+u=\frac{m}{L^{2}}+3 m u^{2}$$

$$u^{\prime \prime}+u=\frac{m}{L^{2}}=\frac{G M}{L_{N}^{2}} \equiv \frac{1}{l}$$

$$u=\frac{1}{l}(1+e \cos \phi)$$

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$$\开始{对齐} u_{1}^{\prime \prime}+u_{1} &=3 m u_{0}^{2} \ &=\frac{3 m}{l^{2}}\left(1+2 e \cos \phi+e^{2} \cos ^{2} \phi\right) \end{对齐}$$

$$u_{1}=\frac{3 m e}{l^{2}} \phi \sin \phi$$

$$\开始{对齐} u &=\frac{1}{l}(1+e \cos \phi)+\frac{3 m e}{l^{2}} \phi \sin \phi \ &=\frac{1}{l}[1+e(\cos \phi+\delta \sin \phi)] \end{对齐}$$

$$\delta=\frac{3 m}{l} \phi$$

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