# 物理代写| Shapiro Time Delay 相对论代考

## 物理代写

7.5 Shapiro Time Delay
An electromagnetic wave such as light or a radio wave experiences an extra timedelay if the path of the light passes close to a gravitating object. This was first pointed out by Shapiro. For example consider say Venus and Earth on opposite sides of the Sun and the null ray from one to the other passes close to the surface of the Sun. Then the null ray takes a little longer to travel the distance as compared to the flat space situation. If a radar beam is sent from Earth to Venus and it reflects of its surface, the total extra time taken by the ray is $\sim 200 \mu \mathrm{sec}$. We will now carry out the analysis below.

We again use the variational principle given by Eq. (6.4.3) to obtain the photon orbits. The first integrals of the orbits are given in Eq. (6.4.20). Here we need the $\dot{r}$ and $\dot{t}$ equations which we must divide to yield the result:
$$\left(\frac{\dot{r}}{c \dot{t}}\right)^{2} \equiv\left(\frac{d r}{c d t}\right)^{2}=\left(1-\frac{2 m}{r}\right)^{2}\left[1-\frac{b^{2}}{r^{2}}\left(1-\frac{2 m}{r}\right)\right]$$
where as before we have put $L / E \equiv b$-the impact parameter which is also a constant of motion. In our situation because the radar ray grazes the surface of the Sun the value of $b$ can be taken to be the radius of the Sun or $b$ is approximately $10^{6} \mathrm{~km}$. Since $2 \mathrm{~m} \sim 3 \mathrm{~km}$ we have $b>>2 \mathrm{~m}$. Further, since the planets are at a distance $\sim 10^{8} \mathrm{~km}$, this distance, say $r_{1}$, is much larger than the radius of the Sun and so $r_{1}>>b$. We must keep these numbers in mind while performing the integrals so that suitable approximations can be made to simplify the integrals.
$7.5$ Shapiro Time Delay
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The bounce of the ray occurs approximately at $r \approx b$. This is the minimum distance the ray gets near to the Sun. It is convenient to switch to dimensionless variables. Let $x=r / b, T=c t / b$ and $\epsilon=2 m / b$. In these variables Eq. (7.5.1) becomes,
$$\left(\frac{d x}{d T}\right)^{2}=\left(1-\frac{\epsilon}{x}\right)^{2}\left[1-\frac{1}{x^{2}}\left(1-\frac{\epsilon}{x}\right)\right] .$$
We now take the square root of this equation and then take its reciprocal. The result is:
$$\frac{d T}{d x}=\left(1-\frac{\epsilon}{x}\right)^{-1}\left[1-\frac{1}{x^{2}}\left(1-\frac{\epsilon}{x}\right)\right]^{-\frac{1}{2}} \approx\left(1+\frac{\epsilon}{x}\right)\left(1-\frac{1}{x^{2}}\right)^{-\frac{1}{2}}$$
where we have used the binomial theorem and kept terms upto the first order in $\epsilon$. We now integrate the above equaion from $b$ to $r_{1}$ or in our dimensionless variables from 1 to $x_{1}=r_{1} / b$. We obtain:
\begin{aligned} T &=\int_{1}^{x_{1}} \frac{x d x}{\sqrt{x^{2}-1}}+\epsilon \times \int_{1}^{x_{1}} \frac{d x}{\sqrt{x^{2}-1}} \ & \equiv T_{\text {Newtonian }}+T_{\text {Shapiro }} \end{aligned}
The first integral is just $\sqrt{x_{1}^{2}-1}$ which in terms of $r$ becomes:
$$\sqrt{r_{1}^{2}-b^{2}} \equiv c \times t_{\text {Newtonian }}$$
which is just $c$ times the time taken by the photon to travel from $r=r_{1}$ to $b$ in flat space, in the absence of gravity. This quantity we identify with the Newtonian value. The second integral gives,
$$T_{\text {Shapiro }}=\epsilon \ln \left(x_{1}+\sqrt{x_{1}^{2}-1}\right)$$
This expression gives the time-delay for the photon path from $r_{1}$ to $b$ :
$$t_{\text {Shapiro }}=2 m \ln \left(\frac{r_{1}+\sqrt{r_{1}^{2}-b^{2}}}{b}\right) \approx 2 m \ln \left(\frac{2 r_{1}}{b}\right)$$
We may now compute the full Shapiro delay of the radar pulse travelling from $r_{1} \longrightarrow b \longrightarrow r_{2}$ and back the same way, where $r_{1}$ and $r_{2}$ are the radial coordinates of the planets. Then using Eq. (7.5.7) from $r_{1}$ to $b$ and again from $b$ to $r_{2}$ and doubling the result for the return path, we obtain,
130
7 Classical Tests of General Relativity
$t_{\text {Shapiro }} \approx 4 m \ln \left(\frac{4 r_{1} r_{2}}{b^{2}}\right) .$
Taking the values $r_{1} \sim 1.5 \times 10^{8} \mathrm{~km}$ (Earth-Sun distance), $r_{2} \sim 1.08 \times 10^{8} \mathrm{~km}$ (Venus-Sun distance) and $b \sim 0.7 \times 10^{6} \mathrm{~km}$, the radius of the Sun, the above equation results in $t_{\text {Shapiro }} \sim 2.4 \times 10^{-4}$ seconds.

## 物理代考

7.5 夏皮罗时间延迟

$$\left(\frac{\dot{r}}{c \dot{t}}\right)^{2} \equiv\left(\frac{dr}{cdt}\right)^{2}=\left (1-\frac{2 m}{r}\right)^{2}\left[1-\frac{b^{2}}{r^{2}}\left(1-\frac{2 m }{r}\右）\右]$$

$7.5$ 夏皮罗时间延迟
129

$$\left(\frac{dx}{d T}\right)^{2}=\left(1-\frac{\epsilon}{x}\right)^{2}\left[1-\frac{1 }{x^{2}}\left(1-\frac{\epsilon}{x}\right)\right] 。$$

$$\frac{d T}{dx}=\left(1-\frac{\epsilon}{x}\right)^{-1}\left[1-\frac{1}{x^{2}}\左(1-\frac{\epsilon}{x}\right)\right]^{-\frac{1}{2}} \approx\left(1+\frac{\epsilon}{x}\right) \left(1-\frac{1}{x^{2}}\right)^{-\frac{1}{2}}$$

$$\开始{对齐} T &=\int_{1}^{x_{1}} \frac{xdx}{\sqrt{x^{2}-1}}+\epsilon \times \int_{1}^{x_{1}} \frac{dx}{\sqrt{x^{2}-1}} \ & \equiv T_{\text {牛顿}}+T_{\text {夏皮罗}} \end{对齐}$$

$$\sqrt{r_{1}^{2}-b^{2}} \equiv c \times t_{\text {牛顿}}$$

$$T_{\text {夏皮罗}}=\epsilon \ln \left(x_{1}+\sqrt{x_{1}^{2}-1}\right)$$

$$t_{\text {夏皮罗}}=2 m \ln \left(\frac{r_{1}+\sqrt{r_{1}^{2}-b^{2}}}{b}\right) \大约 2 m \ln \left(\frac{2 r_{1}}{b}\right)$$

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$t_{\text {Shapiro }} \约 4 m \ln \left(\frac{4 r_{1} r_{2}}{b^{2}}\right) .$

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