Q20P As in Fig. 37-9, reference frame... [FREE SOLUTION]

Chapter 37: Q20P (page 1146)

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

Short Answer

Expert verified

The value of $鈭� t^{'}$ is $6.3 脳 10^{- 7} s$ .

Step by step solution

The Lorentz transformation

For two events the value of $鈭� t$ is given by $鈭� t = \frac{鈭� t^{'} + \frac{尾鈭� x^{'}}{c}}{\sqrt{1 - 尾^{2}}}$ or $鈭� t = \frac{1}{\sqrt{1 - 尾^{2}}} . 鈭� t^{'} + \frac{1}{\sqrt{1 - 尾^{2}}} . \frac{尾鈭� x^{'}}{c}$ .

The calculation

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

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

$鈭� t^{'} = 鈭� t (\sqrt{1 - 尾^{2}}) = 2 脳 10^{- 6} 脳 (\sqrt{1 - {(0.949)}^{2}}) = 6.3 脳 10^{- 7} s$

Thus, the value of $鈭� t^{'}$ is $6.3 脳 10^{- 7} s$ .

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Short Answer

Step by step solution

The Lorentz transformation

The calculation

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