Question

A meterstick is glued to the wall with its 100 cm end farther to the right, in the positive direction. It has a clock at its center and one on each end. You walk by the meterstick in the positive direction at speed v . (a) When you reach the center clock, it reads 0. What do the other two read at this instant in your frame. (b) You instantly reverse direction. The clock at the center is still reading 0 and so is yours. What do the others read? (c) How does this relate to the twin paradox?

Short answer

(a) The time of the two clocks when you are moving right are $\frac{v}{c^2}\left(0.5\right)\sec \mathrm{and}-\frac{\mathrm{v}}{{\mathrm{c}}^2}\left(0.5\right)\sec$.

(b) The time of the two clocks when you are moving left are $-\frac{v}{c^2}\left(0.5\right)\sec \mathrm{and}\frac{\mathrm{v}}{{\mathrm{c}}^2}\left(0.5\right)\sec$.

(c) By changing the direction, the reference frame is changed.

Step-by-step solution

Given data

The position of the event, x = 100 m

Determine the formulas to calculate the time on other two clocks when you moving right andthe time on other two clocks when you moving left.

7lThe Lorentz transformation is defined the relative speed between the inertial and non-inertial frame that move at a constant velocity. This transformation between the two frame is linear transformation in which mapping occurs between two frames.

The expression to convert the time t from the frame S to frame S' is given as follows.

$\mathit{t}\mathbf{\text{'}}\mathbf{=}{\mathit{y}}_{\mathbf{v}}\left(t-\frac{v}{c^2}x\right)$ ...(i)

Here, $\mathit{t}\mathbf{\text{'}}$is the time in frame $\mathit{S}\mathbf{\text{'}}$, is the time in frame $\mathit{S}\mathbf{}\mathbf{,}{\mathit{y}}_{\mathbf{v}}$, is the Lorentz factor for the velocity, c is the speed of the light.

(a) Calculate the clock time of other two ends clock in respect of your frame.

Since reading of the in your clock is 0 as you pass centre of the ruler that means the time in the frame S' is zero.

Calculate the time of the clock in Centaury,

Substitute 0 for t' into equation (i).

$$\begin{gathered} 0=y_v\left(t-\frac{v}{c^2}x\right) \\ 0=t-\frac{v}{c^2}x \\ t-\frac{v}{c^2}x \end{gathered}$$ ...(ii)

The time on the clock on 100 cm end, +50 cm in front of you and for the time clock on 0 cm , -50 cm behind you.

For the frame in front of you,

The position of the event, x = + 50 cm.

Substitute + 50 cm for x into equation (ii)

role="math" localid="1660044962156" $$\begin{gathered} {\mathrm{t}}_{0\mathrm{cm}}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\frac{50}{100} \\ {t}_{0\mathrm{cm}}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\left(0.5\right)\sec \end{gathered}$$

For the frame in front of you,

The position of the event, x = - 50 cm.

Substitute -50 cm for x into equation (ii)

role="math" localid="1660045157787" $$\begin{gathered} t_{0cm}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(-50\times {10}^{-2}\right) \\ {t}_{0cm}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(-\frac{50}{100}\right) \\ {t}_{0\mathrm{cm}}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(0.5\right)\sec \end{gathered}$$

Hence the time of the two clocks when you are moving right are $\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(0.5\right)\sec$and $\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(0.5\right)\sec$.

 Step 4: (b) Calculate the clock time of other two ends clock in respect of your frame when you reverse the direction.

The time on the clocks of the other two end when the direction is reversed. The time on the clock on 100 cm end, -50 cm in behind you and for the time clock on 0 cm,+ 50 cm in front of you.

For the frame in front of you,

The position of the event, x = - 50 cm.

Substitute -50 cm for x into equation (ii)

$$\begin{gathered} t_{100cm}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(-50\times {10}^{-2}\right) \\ {t}_{100cm}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(\frac{50}{100}\right) \\ {t}_{100\mathrm{cm}}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(0.50\right)\sec \end{gathered}$$

For the frame in front of you,

The position of the event, x = + 50 cm.

Substitute + 50 for x into equation (ii)

$$\begin{gathered} t_{0cm}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(50\times {10}^{-2}\right) \\ {t}_{0cm}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(\frac{50}{100}\right) \\ {t}_{0\mathrm{cm}}=\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{x}\left(0.50\right)\sec \end{gathered}$$

Hence the time of the two clocks when you are moving left are role="math" localid="1660045360017" $-\frac{\mathrm{v}}{{\mathrm{c}}^2}\left(0.5\right)\sec$and $\frac{\mathrm{v}}{{\mathrm{c}}^2}\left(0.5\right)\sec$.

(c) Relation to twin paradox.

It is similar to the twin paradox, because by changing the direction, the reference frame is changed and you will the different time for the same positions. If you are accelerated in the reverse direction then you would be out of the reference frame and your clock would be not valid.