Question

You stand at the center of your $100m$ spaceship and watch Anna's identical ship pass at $0.6c$. At $t=0$ on your wristwatch, Anna, at the center of her ship, is directly across you and her wristwatch also reads $0$.

(a) A friend on your ship,$24m$ from you in a direction towards the tail of the ship, looks at a clock directly across from him on Anna's ship. What does it read?

(b) Your friend now steps onto Anna's ship. By this very act he moves from a frame where Anna is one age to a frame where she is another. What is the difference in these ages? Explain.

(c) Answer parts (a) and (b) for a friend $24m$ from you but in a direction toward the front of Anna's passing ship.

(d) What happens to the reading on a clock when you accelerate toward it? Away from it?

Short answer

(a) The clock directly across the friend at $x=-24\text{\hspace{0.17em}m}$ reads $60\text{\hspace{0.17em}ns}$.

(b) When the friend shifted to Anna's ship, her age, according to the friend, jumped forward by $60\text{\hspace{0.17em}ns}$.

(c) The clock directly across the friend at $x=24\text{\hspace{0.17em}m}$ reads $-60\text{\hspace{0.17em}ns}$. When the friend shifted to Anna's ship, her age, according to the friend, jumped backward by $60\text{\hspace{0.17em}ns}$.

(d) Clock readings jump up when the observer accelerates toward it and jumps down when the observer accelerates away from it.

Step-by-step solution

Given data

Speed of Anna with respect to the stationary observer is,

$v=0.6c$

Lorentz transformation

Thetime interval $t^{\text{'}}$ measured by a moving observer at velocity$v$ is related to the time interval $t$ of two events measured by a stationary observer and the space interval $x$ of the two events measured by the stationary observer as,

$t^{\text{'}}=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}$ .....(I)

Here $c$ is the speed of light in vacuum.

Step 3:Determining time in Anna's spaceship

(a)

When the clock at the mid-point of the spaceship reads $t=0$,the time at$24\text{\hspace{0.17em}m}$behind the ship also reads$t=0$. The time at the same location at Anna's ship can be obtained from equation (I) as,

$\begin{array}{c}{t}^{\text{'}}=\frac{0-\frac{0.6c}{c^2}(-24\text{\hspace{0.17em}m})}{\sqrt{1-{0.6}^2}}\\ =6\times {10}^{-8}\text{\hspace{0.17em}s}\\ =60\text{\hspace{0.17em}ns}\end{array}$

Thus, the clock reads $60\text{\hspace{0.17em}ns}$.

Determining Anna's age jump if the friend shifts to Anna's ship

(b)

The clock at $x=0$ and $x=-24\text{\hspace{0.17em}m}$ both read $t=0$. Suppose the friend shifts to Anna's ship, and the clock there reads $t^{\text{'}}=60\text{\hspace{0.17em}ns}$, which should also be the reading on the clock with Anna as they are synchronized.

Thus, Anna's age jumped forward by $60\text{\hspace{0.17em}ns}$.

Determining reading on the clock and Anna's age jump if the friend is toward the front of the ship

(c)

The friend is now at $x=24\text{\hspace{0.17em}m}$. The time at the same location at Anna's ship can be obtained from equation (I) as,

$\begin{array}{c}{t}^{\text{'}}=\frac{0-\frac{0.6c}{c^2}(+24\text{\hspace{0.17em}m})}{\sqrt{1-{0.6}^2}}\\ =-6\times {10}^{-8}\text{\hspace{0.17em}s}\\ =-60\text{\hspace{0.17em}ns}\end{array}$

Thus the clock reads $-60\text{\hspace{0.17em}ns}$, which will also be the jump in Anna's age.

Determining the change in clock reading if the observer accelerates towards it or away from it

(d)

In the first case, the friend accelerates towards Anna when he jumps ship. Initially, Anna was moving away from him, and after the jump, Anna was at rest with respect to him. So the friend accelerated toward Anna. Here the clock reading jumped up.

In the second case, initially, Anna was moving toward the friend, and after the jump, Anna was at rest with respect to the friend. So the friend accelerated away from Anna. Here the clock reading jumped backward.

Thus, clock readings jump up when the observer accelerates toward it and jumps down when the observer accelerates away from it.