Question

As in Fig. 37-9, reference frame ${\mathit{S}}^{\mathbf{\text{'}}}$passes reference $\mathit{S}$frame with a certain velocity. Events 1 and 2 are to have a certain temporal separation $\mathbf{∆}{\mathit{t}}^{\mathbf{\text{'}}}$according to the ${\mathit{S}}^{\mathbf{\text{'}}}$observer. However, their spatial separation $\mathbf{∆}{\mathit{x}}^{\mathbf{\text{'}}}$according to that observer has not been set yet. Figure 37-24 gives their temporal separation $\mathbf{∆}{\mathit{x}}^{\mathbf{\text{'}}}$according to the observer as a function of for a range of values. The vertical axis scale is set by $\mathbf{∆}{\mathbf{t}}_{\mathbf{a}}\mathbf{=}\mathbf{6}\mathbf{\mu s}$. What is $\mathbf{∆}{\mathbf{t}}^{\mathbf{\text{'}}}$?

Short answer

The value of $∆t^{\text{'}}$is $6.3\times {10}^{-7}s$.

Step-by-step solution

The Lorentz transformation

For two events the value of $∆t$is given by $∆t=\frac{∆t^{\text{'}}+{\displaystyle \frac{\beta ∆x^{\text{'}}}{c}}}{\sqrt{1-{\beta }^2}}$or$∆t=\frac{1}{\sqrt{1-{\beta }^2}}.∆t^{\text{'}}+\frac{1}{\sqrt{1-{\beta }^2}}.\frac{\beta ∆x^{\text{'}}}{c}$ .

The calculation

Here, in the above equation the coefficient of $∆x^{\text{'}}$is considered as the slope of the graph. Here, the slope of the graph is $\frac{6-2\mu s}{400m}=0.01$.

So, the expression for$∆t^{\text{'}}=∆t\left(\sqrt{1-{\beta }^2}\right)$ becomes . From the graph, we can interpret the value of$\beta$ as $0.949$. So, the value of $∆t^{\text{'}}$can be calculated as follows:

$$\begin{gathered} ∆t^{\text{'}}=∆t\left(\sqrt{1-{\beta }^2}\right) \\ =2\times {10}^{-6}\times \left(\sqrt{1-{\left(0.949\right)}^2}\right) \\ =6.3\times {10}^{-7}s \end{gathered}$$

Thus, the value of$∆t^{\text{'}}$ is $6.3\times {10}^{-7}s$.