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Section 52.15 Special Relativity Bootcamp

Subsection 52.15.1 Galilean Relativity

Subsection 52.15.2 Time Dilation

Subsection 52.15.3 Lorentz transformations

Subsection 52.15.4 Length Contraction

Subsection 52.15.5 Transformations of velocities

Subsection 52.15.6 The Relativisitc Doppler Effect

Subsection 52.15.7 Relativistic Momentum

Subsection 52.15.8 Relativistic Energy

Subsection 52.15.9 Miscellaneous Problems

Consider the relativistic form of \(\vec F = m \vec a\) given in the chapter. Show that if force \(\vec F\) is parallel to the velocity \(\vec u\text{,}\) then this equation becomes

\begin{equation*} F = m_0 \gamma^3 \dfrac{du}{dt},\ \ \textrm{where}\ \gamma = \dfrac{1}{\sqrt{1-u^2/c^2}}. \end{equation*}

Let \(\vec u = u \hat i\text{.}\)


See solution.


Let \(\vec u = u \hat i\text{.}\) Since \(\vec F \parallel \vec u\text{,}\) \(\vec F = F \hat i\text{.}\) Then

\begin{align*} F\hat i \amp = \gamma m_0 \vec a + m_0 u \hat i \gamma^3 \frac{u a_x }{c^2},\ \ a_x = \frac{du}{dt} \\ \amp = \gamma m_0( a_y \hat j + a_z \hat k) + \gamma m_0 a_x \hat i + m_0 u \hat i \gamma^3 \frac{u a_x }{c^2}, \end{align*}


\begin{align*} F \amp = \gamma m_0 a_x \left( 1 + \gamma^2\frac{u^2 }{c^2} \right) \\ \amp = \gamma m_0 (du/dt) \gamma^2 = \gamma^3 m_0 (du/dt) \end{align*}

In 1851 Hippolyte Fizeau carried out an experiment to measure the speed of light in moving water. When water is stationary the speed is

\begin{equation*} u = \frac{c}{n}, \end{equation*}

where \(n\) is the refractive index of water. Fizeau found that if water was flowing with speed \(v\) in the same direction as the direction of light his data for the speed of light would fit with the following formula. \(u = \frac{c}{n} + v - \frac{v}{n^2}.\) This formula is an approximate formula for an exact formula that you can derive based on velocity transformation law you have studied in this chapter.

(a) Show that the exact expression for the speed of light will be

\begin{equation*} u = \left( \frac{c+ n v}{nc + v}\right)\ c. \end{equation*}

(b) Expand the expression in part (a) for \(v/c \ll 1\) keeping only the linear term in \(v/c\) and show that you get the approximate formula agrees with Fizeau's experiment.


Use Maclauren series in \(v/nc\text{.}\)


Already given.


(a) We just add thwe velocity \(v\) of water to that of the velocity of light in the medium \(c/n\text{.}\) This gives

\begin{equation*} u = \frac{ v + \frac{c}{n} }{ 1 + (v)(c/n)/c^2 } = \left( \frac{c+ n v}{nc + v}\right)\ c. \end{equation*}

(b) Expand the denominator of the formula upto linear term in \(v/c\text{.}\)

\begin{equation*} \frac{1}{nc + v} = \frac{1}{nc} \frac{1}{1 + v/nc} \approx \frac{1}{nc}\left( 1 - v/nc \right) = \frac{1}{nc} -\frac{v}{n^2c^2}. \end{equation*}

Combining this with the rest of the terms we get the following when we drop quadratic term in \(v/c\text{.}\)

\begin{equation*} u \approx (c^2+ nc v) \left( \frac{1}{nc} -\frac{v}{n^2c^2} \right) = \frac{c}{n} + v - \frac{v}{n^2}. \end{equation*}

Cesium atomic clocks were flown around the world, once eastward and once westward, and their times were then compared with a reference clock at the U.S. Naval Observatory. The time difference in the clocks of the Eastward with Wesward flights cancels out the time dilation due to gravity so that we can obtain the time difference due to special relativity even at the typical aircraft speeds.

Let \(R\) be the radius of Earth, \(\omega\) the angular rotation speed of Earth, \(u\) the speed of flight with respect to the surface of Earth. Below we will calculate the difference in times due to motion if ground speed of the planes is \(u\text{.}\) We will assume that a clock based at the center of Earth is an inertial clock and the clocks at Earth's surface and in the planes move with uniform speed relative the clock at rest at the center of Earth.

(a) Let a clock be at rest in a frame \(S'\) wich is moving with speed \(v\) with respect to another frame \(S\text{.}\) The, show that if \(v \ll c\text{,}\)

\begin{equation*} \Delta t_{S} = \gamma \Delta t_{S'} \approx \Delta t_{S'} \left( 1 + \frac{1}{2}\ \frac{v^2}{c^2}\right). \end{equation*}


\begin{equation*} \Delta t_{S'} \approx \Delta t_{S} \left( 1 - \frac{1}{2}\ \frac{v^2}{c^2}\right). \end{equation*}

(b) Use the simplified formula for the time dilation to find the times elapsed \(t_s\) for the clock at the surface of Earth if the time elapsed in the clock at the center of the Earth is \(t_0\text{.}\)

(c) Find time \(t_a\) elapsed in a clock in the air plane if the time elapsed in the clock at the center of the Earth is \(t_0\text{.}\) Assume the airplane is going towards East with speed given by \(R\omega + u\text{,}\) which amounts to ignoring the altitude of the plane. Also, note the change in the formula if the plane is going towards West.

(d) Find the formula for difference in the times in planes going East and going West.

(e) Use your relation to find the time difference between the Earth-based clock and the flying clock for a 48-hour trip as observed in the Earth-based clock. Use speed of the flight to be \(u = 232 \text{ m/s}\) and ignore the altitude of the plane compared to the radius of Earth.


Follow the instructions in the problem.


(e) \(4.13\times 10^{-10}\text{ s}\text{.}\)


(a) From the observation that clock in its rest frame goes slower by \(\gamma\) we have

\begin{equation*} \Delta t_{S} = \gamma \Delta t_{S'}, \end{equation*}

where we can expand \(\gamma\) in powers of \((v/c)^2\text{.}\)

\begin{equation*} \gamma = \left( 1 - (v/c)^2 \right)^{-1/2} = 1 - (-1/2) (v/c)^2 + ... \end{equation*}

Dropping higher powers gives the answer. Similarly,

\begin{equation*} \frac{1}{\gamma} \approx \left( 1 +(1/2) (v/c)^2 \right)^{-1} \approx 1 - \frac{1}{2}\frac{v^2}{c^2}. \end{equation*}

(b) We place a clock at surface of Earth. So, \(t_s\) is the time in the \(S'\) frame of (a). Hence, with \(v = \omega R\) since moving in a circle of radius \(R\) with angular speed \(\omega\text{.}\)

\begin{equation*} t_s = \frac{t_0}{\gamma} = t_0 \left( 1 - \frac{1}{2}\frac{\omega^2 R^2}{c^2} \right). \end{equation*}

(c) Similar to (b), with speed \(v=\omega R + u\text{,}\) we get time in the easward moving plan

\begin{equation*} t_{ae} = t_0 \left( 1 - \frac{1}{2}\frac{(\omega R + u)^2}{c^2} \right). \end{equation*}

For flight in the opposite direction \(u\) will change sign with respec to \(\omega R\text{.}\) Therefore,

\begin{equation*} t_{aw} = t_0 \left( 1 - \frac{1}{2}\frac{(\omega R - u)^2}{c^2} \right). \end{equation*}

(d) The difference will be

\begin{equation*} t_{aw} - t_{ae} = \frac{2 \omega R u}{c^2}\; t_0. \end{equation*}

(e) Now, we use the numerical values. From \(\omega= 2\pi/24\text{hr}\) and \(R=6.37\text{ km}\text{,}\) we get \(t_0\) from \(t_s= 48\text{ hr}\text{.}\) First we have

\begin{equation*} \frac{\omega R}{c} = \frac{2\pi \times 6.37 \text{ km}}{24\text{hr} \times 3\times 10^5\times 3600 \text{ km/hr}} = 1.544\times 10^{-9}. \end{equation*}

Then, we calculate

\begin{align*} t_0 \amp = t_s \left( 1 + \frac{1}{2} (\omega R/c)^2 \right) \\ \amp = 48\text{ hr}\, \left( 1 + \frac{1}{2} (0.02)^2 \right) = 48.0096\text{ hr}. \end{align*}

Now, we evaluate the difference

\begin{align*} t_{aw} - t_{ae} \amp = \frac{2 \omega R u}{c^2}\; t_0 \\ \amp = 2\times 1.5432\times 10^{-9} \frac{u}{c}\; t_0 = 4.13\times 10^{-10}\text{ s}. \end{align*}