Figure 37-17 shows two clocks in stationary frame $\mathit{S}\mathbf{\text{'}}$(they are synchronized in that frame) and one clock in moving frame $\mathit{S}$. 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 $\mathit{C}{\mathbf{\text{'}}}_{\mathbf{2}}$pass each other, (a) which clock has the smaller reading and (b) which clock measures a proper time?

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

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

The expression to calculate the recorded time is given as follows.

$\mathit{t}\mathbf{=}\frac{{\mathbf{t}}_{\mathbf{0}}}{\sqrt{\mathbf{1}\mathbf{-}\mathbf{\beta }}}$ …… (i)

Here, is the speed parameter.

The expression to calculate the speed parameter is given as follows,

role="math" localid="1663126785804" $\mathit{\beta }\mathbf{=}\frac{\mathbf{v}}{\mathbf{c}}$

Here,$\mathit{v}$ is the relative speed of the frame $\mathit{S}\mathbf{\text{'}}$with respect to frame $\mathit{S}$and $\mathit{c}$ is the speed of the light.

(a)

Since the frame$S$ is moving with the velocity$v$ , the recorded time is given by,

$t=t_0\sqrt{1-{\beta }^2}$

Substitute $v/c$for$\beta$ into above equation.

$t=\frac{t_0}{1-{\left(v/c\right)}^2}$

Since the due to effect of time dilation, the moving clock read the smaller value.

Hence the clock $c_1$read the smaller value.

(b)

For the two events occur at the same location with respect to the inertial reference frame, the time interval measure between two events is known as the proper time and proper time interval.

The value of Lorentz factor is always greater than one, therefore time interval measure by the clock with the velocity is longer than the time interval measured by the clock which is at the rest.

Hence the clock$C_1$ measure the proper time.