Chapter 2, Problem 17P

(a)

To determine

Time at which signal will be received at $x=3\text{ m},y=5\text{ m},z=10\text{ m}$ from origin.

Answer to Problem 17P

At $t=3.86\times {10}^{-8}\text{ s}$ signal will be received at $x=3\text{ m},y=5\text{ m},z=10\text{ m}$.

Explanation of Solution

Formula used:

The equation of wave fronts observed in system K at time t at x, y, and z.

$x^2+y^2+z^2=c^2{t}^2$

Where, x, y, z are the coordinates of the t, is the time and c is the velocity of the light.

Calculation:

Here given data are as follows:

$t=3.86\times {10}^{-5}\text{ s}$ and $x=3\text{ m},y=5\text{ m},z=10\text{ m}$

Substituting the value of the coordinates and time in equation $x^2+y^2+z^2=c^2{t}^2$

$\begin{array}{c}{\left(3\text{ m}\right)}^2+{\left(5\text{ m}\right)}^2+{\left(10\text{ m}\right)}^2={\left(3\times {10}^8\text{ m / s}\right)}^2{t}^2\\ t^2=\frac{{\left(3\text{ m}\right)}^2+{\left(5\text{ m}\right)}^2+{\left(10\text{ m}\right)}^2}{{\left(3\times {10}^8\text{ m / s}\right)}^2}\\ t=\sqrt{\frac{{\left(3\text{ m}\right)}^2+{\left(5\text{ m}\right)}^2+{\left(10\text{ m}\right)}^2}{{\left(3\times {10}^8\text{ m / s}\right)}^2}}\\ t=\sqrt{\frac{9+{25}^2+100}{{\left(3\times {10}^8\right)}^2}}\text{ s}\end{array}$

After further simplification,

$\begin{array}{c}t=\sqrt{\frac{9+{25}^2+100}{{\left(3\times {10}^8\right)}^2}}\text{ s }\\ =3.\text{86}\times {10}^{-8}\text{ s}\end{array}$        (I)

Conclusion:

Therefore, time taken by the signal to reach at $x=3\text{ m},y=5\text{ m},z=10\text{ m}$ from origin is $t=3.86\times {10}^{-8}\text{ s}$.

(b)

To determine

The coordinates of the system Kʹ at time $t=3.86\times {10}^{-8}\text{ s}$, when speed of the system Kʹ is 0.8c along the x-axis of the system K.

Answer to Problem 17P

Coordinates of the system Kʹ at time $t=3.86\times {10}^{-8}\text{ s}$ are $x\prime =-10.4\text{ m},y\prime =5\text{ m, }z\prime =10\text{ m, and }t\prime =51.0\times {10}^{-9}\text{ s}$.

Explanation of Solution

Formula used:

With the use of the Lorentz transformation equations:

$\begin{array}{c}x\prime =\gamma \left(x-vt\right)\\ y\prime =y\\ z\prime =z\\ t\prime =\gamma \left(t-\left(vx/c^2\right)\right)\end{array}$        (II)

Where, x, y, and z are the coordinates of the system K at time t, v is the speed of the system Kʹ and $x\prime ,y\prime ,z\prime$ are the coordinates of the system Kʹ at time $t=3.86\times {10}^{-8}\text{ s}$.

Where γ is relativistic factor which can be obtained by the formula:

$\begin{array}{c}\gamma =\frac{1}{\sqrt{1-v^2/c^2}}\\ =\frac{1}{\sqrt{1-{\left(0.8c\right)}^2/c^2}}\\ =\frac{1}{\sqrt{1-0.64}}\\ =\frac{1}{0.6}\\ =\frac{5}{3}\end{array}$        (III)

Calculation:

The motion of the system Kʹ is only along x- direction, therefore, $y\prime =y=5\text{ m}$ and $z\prime =z=10\text{ m}$

Now, from equation (I), (II), and (III) calculate value of xʹ and tʹ by substituting the values of x, t, v, and $\gamma$.

$\begin{array}{c}x\prime =\gamma \left(x-vt\right)\\ =\frac{5}{3}\left(3\text{ m}-0.8\times 3\times {10}^8\text{ m / s}\times 3.86\times {10}^{-8}\text{ s}\right)\\ =\frac{5}{3}\left(3-9.264\right)\text{ m}\\ =-10.4\text{ m}\end{array}$        (IV)

Now calculate the value of the tʹ

$\begin{array}{c}t\prime =\gamma \left(t-\left(vx/c^2\right)\right)\\ =\frac{5}{3}\left(3.86\times {10}^{-8}\text{ s}-\left(\frac{0.8c\times 3\text{ m}}{c^2}\right)\right)\\ =\frac{5}{3}\left(3.86\times {10}^{-8}\text{ s}-\left(\frac{0.8\times 3\text{ m}}{c}\right)\right)\\ =\frac{5}{3}\left(3.86\times {10}^{-8}\text{ s}-\left(\frac{0.8\times 3\text{ m}}{3\times {10}^8\text{ s}}\right)\right)\end{array}$

After further simplification

$\begin{array}{c}t\prime =\frac{5}{3}\left(3.86\times {10}^{-8}\text{ s}-\left(\frac{0.8\times 3\text{ m}}{3\times {10}^8\text{ s}}\right)\right)\\ =\frac{5}{3}\left(3.86\times {10}^{-8}\text{ s}-0.8\times {10}^{-8}\text{ s}\right)\\ =\frac{5}{3}\left(3.06\times {10}^{-8}\text{ s}\right)\\ =51\times {10}^{-9}\text{ s}\end{array}$        (V)

Conclusion:

Therefore, the coordinates of the system Kʹ at time $t=3.86\times {10}^{-8}\text{ s}$ are $x\prime =-10.4\text{ m},y\prime =5\text{ m, }z\prime =10\text{ m, }$ and $t\prime =51.0\times {10}^{-9}\text{ s}$.

(c)

To determine

Speed of the light in system Kʹ.

Answer to Problem 17P

The speed of the light in the system Kʹ $2.994\times {10}^8\text{ m }/\text{s}$, which is almost equal to c.

Explanation of Solution

The equation of wave fronts observed in system Kʹ at time tʹ at xʹ, yʹ, and zʹ.

$x{\prime }^2+y{\prime }^2+z{\prime }^2=c{\prime }^{2}t{\prime }^2$

Where, xʹ, yʹ, and zʹ are the coordinates of the system Kʹ, tʹ, is the time and cʹ is the velocity of the light measured in system Kʹ.

Calculation:

Substitute the values $x\prime =-10.4\text{ m},y\prime =5\text{ m, }z\prime =10\text{ m, }$ and $t\prime =51.0\times {10}^{-9}\text{ s}$ in equation $x{\prime }^2+y{\prime }^2+z{\prime }^2=c{\prime }^{2}t{\prime }^2$ to calculate the value of the tʹ.

$\begin{array}{c}{\left(10.4\text{ m}\right)}^2+{\left(5\text{ m}\right)}^2+{\left(10\text{ m}\right)}^2={\left(51\times {10}^{-9}\text{ s}\right)}^{2}c{\prime }^2\\ c{\prime }^2=\frac{{\left(10.4\text{ m}\right)}^2+{\left(5\text{ m}\right)}^2+{\left(10\text{ m}\right)}^2}{{\left(51\times {10}^{-9}\text{ s}\right)}^2}\\ c\prime =\sqrt{\frac{{\left(10.4\text{ m}\right)}^2+{\left(5\text{ m}\right)}^2+{\left(10\text{ m}\right)}^2}{{\left(51\times {10}^{-9}\text{ s}\right)}^2}}\\ c\prime =2.994\times {10}^8\text{ m / s}\end{array}$

Which is almost equal to $c=3\times {10}^8\text{ m / s}$.

Conclusion:

Therefore, speed of the light in both system is $c=3\times {10}^8\text{ m / s}$.