Subfield: Modern Physics

Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.

First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).

By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.

Some mathematics formulas that might be useful

- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
- $\tanh x=\frac{\sinh x}{\cosh x}$
- $1+\sinh ^{2} x=\cosh ^{2} x$
- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$



- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$


Part A. Single Accelerated Particle 

Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
Context question:
1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
Context answer:
\boxed{$a=\frac{F}{\gamma^{3} m}$}


Context question:
2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
Context answer:
\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}


Context question:
3.  Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.
Context answer:
\boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}


Context question:
4.  Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.
Context answer:
\boxed{证明题}


Context question:
5.  Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.
Context answer:
\boxed{$\beta=\tanh \frac{g \tau}{c}$}


Context question:
6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.
Context answer:
\boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$}


Extra Supplementary Reading Materials:

Part B. Flight Time 

The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A.
Context question:
1.  At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value?
Context answer:
\boxed{证明题}


Context question:
2.  Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value?
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Part C. Minkowski Diagram

In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame.

$$
\begin{aligned}
& x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\
& c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\
& x^{\prime}=\gamma(x-\beta c t), \\
& c t^{\prime}=\gamma(c t-\beta x) . \\
& \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}
\end{aligned}
$$

<img_4526>

Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S.

To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below.



<img_4417>

The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$.

$$
\begin{aligned}
\mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\
& =A C \cos \theta-A C \sin \theta \tan \theta \\
& =L \sqrt{1-\beta^{2}}
\end{aligned}
$$
Context question:
1.  Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame.
Context answer:
\boxed{$\sqrt{1-\beta^{2}} L$}


Context question:
2.  Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$.
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Part D. Two Accelerated Particles

For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part.
Context question:
1.  After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame.
Context answer:
\boxed{$\tau_{B}=\tau_{A}$}


Context question:
2.  Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that

$$
\sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c}
$$

where $C_{1}$ is a constant. Determine $C_{1}$.
Context answer:
\boxed{$C_{1}=\frac{g L}{c^{2}}$}


Question: 3.  The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is

$$
\frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)}
$$

where $C_{2}$ is a constant. Determine $C_{2}$.

Solution:
From the length contraction, we have

$$
L^{\prime} =\frac{x_{2}-x_{1}}{\gamma_{1}}
\tag{33}
$$

$$
\frac{d L^{\prime}}{d \tau_{1}} =\left(\frac{d x_{2}}{d \tau_{2}} \frac{d \tau_{2}}{d \tau_{1}}-\frac{d x_{1}}{d \tau_{1}}\right) \frac{1}{\gamma_{1}}-\frac{x_{2}-x_{1}}{\gamma_{1}^{2}} \frac{d \gamma_{1}}{d \tau_{1}}
\tag{34}
$$

Take derivative of eq.(30), eq.(31) and eq.(32), we get

$$
\frac{d x_{1}}{d \tau_{1}} =c \sinh \frac{g \tau_{1}}{c}
\tag{35}
$$

$$
\frac{d x_{2}}{d \tau_{2}} =c \sinh \frac{g \tau_{2}}{c}
\tag{36}
$$

$$
\frac{g L}{c^{2}} \cosh \frac{g \tau_{1}}{c} =\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)\left(\frac{d \tau_{2}}{d \tau_{1}}-1\right) .
\tag{37}
$$

The last equation can be rearrange to get

$$
\frac{d \tau_{2}}{d \tau_{1}}=\frac{\frac{g L}{c^{2}} \cosh \frac{g \tau_{1}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)}+1
\tag{38}
$$



From eq.(29), we have

$$
x_{2}-x_{1}=\frac{c\left(t_{2}-t_{1}\right)}{\beta_{1}}=\frac{c}{\tanh \frac{g \tau_{1}}{c}}\left(\frac{c}{g} \sinh \frac{g \tau_{2}}{c}-\frac{c}{g} \sinh \frac{g \tau_{1}}{c}\right)
\tag{39}
$$

Combining all these equations, we get

$$
\begin{aligned}
& \frac{d L_{1}}{d \tau_{1}}=\left(c \sinh \frac{g \tau_{2}}{c} \frac{\frac{g L}{c^{2}} \cosh \frac{g \tau_{1}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)}+c \sinh \frac{g \tau_{2}}{c}-c \sinh \frac{g \tau_{1}}{c}\right) \frac{1}{\cosh \frac{g \tau_{1}}{c}} \\
& -\frac{c^{2}}{g}\left(\sinh \frac{g \tau_{2}}{c}-\sinh \frac{g \tau_{1}}{c}\right) \frac{1}{\tanh \frac{g \tau_{1}}{c}} \frac{1}{\cosh ^{2} \frac{g \tau_{1}}{c}} \frac{g}{c} \sinh \frac{g \tau_{1}}{c} \\
& \frac{d L_{1}}{d \tau_{1}}=\frac{g L}{c} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)}
\end{aligned}
\tag{40}
$$

So $C_{2}=\frac{g L}{c}$.

Final answer: $C_{2}=\frac{g L}{c}$