Question 39.5: Simultaneity and Time Dilation Revisited (A) Use the Lorentz......
Simultaneity and Time Dilation Revisited
(A) Use the Lorentz transformation equations in difference form to show that simultaneity is not an absolute concept.
(B) Use the Lorentz transformation equations in difference form to show that a moving clock is measured to run more slowly than a clock that is at rest with respect to an observer.
(A) Conceptualize Imagine two events that are simultaneous and separated in space as measured in the \mathrm{S}^{\prime} frame such that \Delta t^{\prime}=0 and \Delta x^{\prime} \neq 0. These measurements are made by an observer O^{\prime} who is moving with speed v relative to O.
Categorize The statement of the problem tells us to categorize this example as one involving the use of the Lorentz transformation.
Analyze From the expression for \Delta t given in Equation 39.14,
\left.\begin{array}{rl} \Delta x & =\gamma\left(\Delta x^{\prime}+v \Delta t^{\prime}\right) \\ \Delta t & =\gamma\left(\Delta t^{\prime}+\frac{v}{c^2} \Delta x^{\prime}\right) \end{array}\right\} S^{\prime} \to S (39.14)
find the time interval \Delta t measured by observer O :
\Delta t=\gamma\left(\Delta t^{\prime}+\frac{v}{c^{2}} \Delta x^{\prime}\right)=\gamma\left(0+\frac{v}{c^{2}} \Delta x^{\prime}\right)=\gamma \frac{v}{c^{2}} \Delta x^{\prime}
Finalize The time interval for the same two events as measured by O is nonzero, so the events do not appear to be simultaneous to O.
(B) Conceptualize Imagine that observer O^{\prime} carries a clock that he uses to measure a time interval \Delta t^{\prime}. He finds that two events occur at the same place in his reference frame \left(\Delta x^{\prime}=0\right) but at different times \left(\Delta t^{\prime} \neq 0\right). Observer O^{\prime} is moving with speed v relative to O.
Categorize The statement of the problem tells us to categorize this example as one involving the use of the Lorentz transformation.
Analyze From the expression for \Delta t given in Equation 39.14 , find the time interval \Delta t measured by observer O :
\Delta t=\gamma\left(\Delta t^{\prime}+\frac{v}{c^{2}} \Delta x^{\prime}\right)=\gamma\left[\Delta t^{\prime}+\frac{v}{c^{2}}(0)\right]=\gamma \Delta t^{\prime}
Finalize This result is the equation for time dilation found earlier (Eq. 39.7), where \Delta t^{\prime}=\Delta t_{p} is the proper time interval measured by the clock carried by observer O^{\prime}. Therefore, O measures the moving clock to run slow.
\Delta t={\frac{\Delta t_{\rho}}{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}}=\gamma\,\Delta t_{\rho} (39.7)