Question

A clock moves along an $\mathbf{x}$axis at a speed of$\mathbf{0}\mathbf{.600}\mathbf{c}$ and reads zero as it passes the origin of the axis. (a) Calculate the clock’s Lorentz factor. (b) What time does the clock read as it passes $\mathbf{x}\mathbf{=}\mathbf{180}\mathbf{\text{\hspace{0.17em}m}}$?

Short answer

1. The clock’s Lorentz factor is $1.25$.
2. The clock reads $8.00\times {10}^{-7}\text{s}$ as it passes $\mathrm{x}=180\text{\hspace{0.17em}m}$.

Step-by-step solution

The Lorentz factor

The formula of the Lorentz factor is $\mathbf{\gamma }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{\beta }}^{\mathbf{2}}}}$, where$\mathbf{\beta }$ is the speed parameter.

Find Lorentz factor

(a)

Here, given that $\mathrm{v}=0.600\mathrm{c}$. So, the speed parameter becomes $\mathrm{\beta }=\frac{\mathrm{v}}{\mathrm{c}}=0.600$.

The Lorentz factor can be calculated as follows:

$\begin{array}{c}\mathrm{\gamma }=\frac{1}{\sqrt{1-{\mathrm{\beta }}^2}}\\ =\frac{1}{\sqrt{1-{\left(0.600\right)}^2}}\\ =1.25\end{array}$

Thus, the value of Lorentz factor of the clock is $1.25$.

The reading in the clock

(b)

The time reading in the clock to travel from the origin to $x=180\text{\hspace{0.17em}m}$in the unprimed frame is given by:

$\begin{array}{c}\mathrm{t}=\frac{\mathrm{x}}{\mathrm{v}}\\ =\frac{180}{0.600(3.00\times {10}^8\text{\hspace{0.17em}m/s})}\\ =1.00\times {10}^{-6}\text{\hspace{0.17em}s}\end{array}$

Here, the reading is taken by the clock itself between two events. So, the reading on the clock at $\mathrm{x}=180$is:

$\begin{array}{c}{\mathrm{t}}^{\text{'}}=\frac{\mathrm{t}}{\mathrm{\gamma }}\\ =\frac{1.00\times {10}^{-6}}{1.25}\\ =8.00\times {10}^{-7}\text{\hspace{0.17em}s}\end{array}$

Thus, the reading on the clock is $8.00\times {10}^{-7}\text{\hspace{0.17em}s}$.