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{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}


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.

Solution:
Consider the following systems, a frame $S^{\prime}$ is moving with respect to another frame $S$, with velocity $u$ in the $x$ direction. If a particle is moving in the $\mathrm{S}^{\prime}$ frame with velocity $v^{\prime}$ also in $x$ direction, then the particle velocity in the $\mathrm{S}$ frame is given by

$$
v=\frac{u+v^{\prime}}{1+\frac{u v^{\prime}}{c^{2}}}
\tag{7}
$$



If the particles velocity changes with respect to the $S$ ' frame, then the velocity in the $S$ frame is also change according to

$$
\begin{aligned}
d v & =\frac{d v^{\prime}}{1+\frac{u v^{\prime}}{c^{2}}}-\frac{u+v^{\prime}}{\left(1+\frac{u v^{\prime}}{c^{2}}\right)^{2}} \frac{u d v^{\prime}}{c^{2}} \\
d v & =\frac{1}{\gamma^{2}} \frac{d v^{\prime}}{\left(1+\frac{u v^{\prime}}{c^{2}}\right)^{2}}
\end{aligned}
\tag{8}
$$

The time in the $\mathrm{S}^{\prime}$ frame is $t^{\prime}$, so the time in the $\mathrm{S}$ frame is given by

$$
t=\gamma\left(t^{\prime}+\frac{u x^{\prime}}{c^{2}}\right)
\tag{9}
$$

so the time change in the $S^{\prime}$ frame will give a time change in the $\mathrm{S}$ frame as follow

$$
d t=\gamma d t^{\prime}\left(1+\frac{u v^{\prime}}{c^{2}}\right)
\tag{10}
$$

The acceleration in the $\mathrm{S}$ frame is given by

$$
a=\frac{d v}{d t}=\frac{a^{\prime}}{\gamma^{3}} \frac{1}{\left(1+\frac{u v^{\prime}}{c^{2}}\right)^{3}}
\tag{11}
$$

If the $\mathrm{S}^{\prime}$ frame is the proper frame, then by definition the velocity $v^{\prime}=0$. Substitute this to the last equation, we get

$$
a=\frac{a^{\prime}}{\gamma^{3}}
\tag{12}
$$

Combining Eq.(3) and Eq.(12), we get

$$
a^{\prime}=\frac{F}{m} \equiv g
\tag{13}
$$