Chapter 36, Problem 24P

(II) Reference frame S′ moves at speed v = 0.92c in the +x direction with respect to reference frame S. The origins of S and S′ overlap at t = t′ = 0. An object is stationary in S′ at position x′ = 100 m. What is the position of the object in S when the clock in S reads 1.00 μs according to the (a) Galilean and (b) Lorentz transformation equations?

24.    (a) The Galilean tansformation in given in Eq. 36-4.

$x=x^{\prime }+v{t}^{\prime }=x^{\prime }+vt=100\text{ m}+(0.92)(3.00\times {10}^8\text{ m/s})(1.00\times {10}^{-6}\text{s})=\boxed{376\text{ m}}$

(b) The Lorentz transformation is given in Eq. 36-6. Note that we are given t, the clock reading in frame S.

$\begin{array}{l}t=\gamma \left(t^{\prime }+\frac{v{x}^{\prime }}{c^2}\right)\to t^{\prime }=\frac{t}{\gamma }-\frac{v{x}^{\prime }}{c^2}\\ x=\gamma (x^{\prime }+v{t}^{\prime })=\gamma \left[x^{\prime }+v\left(\frac{t}{\gamma }-\frac{v{x}^{\prime }}{c^2}\right)\right]=\gamma \left[x^{\prime }+\frac{v}{c}\left(\frac{ct}{\gamma }-\frac{v{x}^{\prime }}{c}\right)\right]\\ =\frac{1}{\sqrt{1-{0.92}^2}}\left[(100\text{ m})+(0.92)(\sqrt{1-{0.92}^2}(3.00\times {10}^8\text{ m/s})(1.00\times {10}^{-6}\text{ s})-0.92(100\text{ m}))\right]\\ =\boxed{316\text{ m}}\end{array}$