Question

Figure 37-18 shows two clocks in stationary frame $\mathit{S}$(they are synchronized in that frame) and one clock in moving frame $\mathit{S}\mathbf{\text{'}}$.Clocks ${\mathit{C}}_{\mathbf{1}}$and $\mathit{C}{\mathbf{\text{'}}}_{\mathbf{1}}$read zero when they pass each other. When clocks ${\mathit{C}}_{\mathbf{1}}$and ${\mathit{C}}_{\mathbf{2}}$pass each other, (a) which clock has the smaller reading and (b) which clock measures a proper time?

Short answer

(a) The clock$C{\text{'}}_1$ measure the smaller reading.

(b) The clock$C{\text{'}}_1$ measure the proper time.

Step-by-step solution

Write the given data from the question.

Clocks$C_1$ and $C{\text{'}}_1$read zero when they pass each other.

The frame$S$ is stationary while frame$S\text{'}$ is moving with the velocity$v$ .

Determine the clock which read the smaller reading between and .

(a)

According to the special theory of the relativity, the clock which is in the motion reads the smaller time with respect to the clock which is stationary.

Therefore when the clock $C{\text{'}}_1$pass the clock$C_2$ , then clock$C{\text{'}}_1$ measure the smaller time compare to the clock $C_2$because clock $C{\text{'}}_1$is in the motion and clock$C_2$ is stationary.

Hence the clock$C{\text{'}}_1$ measure the smaller reading.

Determine the clock which measure the proper time between C'1and C2

.

(b)

According to the special theory of the relativity, the clock which is in the motion measure the proper time with respect to the clock which is stationary.

Therefore when the clock$C{\text{'}}_1$ pass the clock$C_2$ , then clock $C{\text{'}}_1$measure the proper time compare to the clock$C_2$ because clock$C{\text{'}}_1$ is in the motion and clock$C_2$ is stationary.

Hence the clock $C{\text{'}}_1$measure the proper time.