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A Space Station S has two ships parked at it: P - PuttPutt and Q - Quick. S, P and Q synchronize their clocks. At noon P and Q take off in the same direction. P quickly accelerates to $c/4$ and Q quickly accelerates to $c/2$.

Both P and Q are given a time to turn around and return to S. This time is calculated such that both P and Q will arrive at S concurrently.

On the way out, once P and Q arrive at their cruising velocity, S observes their clocks. It notes that Q's clock is moving slower than their own. $$\gamma_q=\sqrt{\frac{1}{1-\frac{(c/2)^2}{c}}}=\sqrt{\frac{4}{3}}\approx1.1547$$

S measures each second on Q being $1.1547s$. The same analysis for P shows each second measuring $\sqrt{\frac{16}{15}}\approx1.0328s$.

In Q's reference frame (once cruising velocity is achieved):

  • Q1: How fast are S and P moving?
  • Q2: Does Q see S's clocks as moving slower or faster?
  • Q3: Does Q see P's clocks as moving at $\gamma_p$ or at $\gamma_q/\gamma_p$?

In P's reference frame:

  • Q4: How fast are S and Q moving?
  • Q5: Does P see a difference in S and Q's clocks?

Once S, P and Q are reunited. What is the order of the clocks? Which one is earliest, which one is latest and which one is in between?

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Q1: How fast are S and P moving (in Q's frame)?

Any two observers will always agree on their relative velocity. So if $S$ sees $Q$ moving at $c/2$ then $Q$ sees $S$ moving at $-c/2$.

Calculating the velocity of $P$ in $Q$'s frame is harder. You could just use the formula for relativistic addition of velocities to get:

$$ v'_p = \frac{v_p - v_q}{1 - \frac{v_pv_q}{c^2}} \tag{1} $$

but this brings no understanding with it, so let's use the Lorentz transformations:

$$\begin{align} t' &= \gamma \left( t - \frac{vx}{c^2}\right) \\ x' &= \gamma \left( x- vt \right) \end{align}$$

Start in $S$'s frame. In this frame $P$ is moving with velocity $v_p$ so in a time $t$ $P$ moves from the point $(0, 0)$ to $(t, v_pt)$. Well use the usual convention that the origins of all frames coincide at time $t = 0$, so the point $(0, 0)$ is the same in all frames. We just have to calculate where the point $(t, v_pt)$ is in $Q$'s frame. The velocity of $Q$ is $v_q$, so just plugging the values into the Lorentz transformations we get:

$$\begin{align} t' &= \gamma \left( t - \frac{v_qv_pt}{c^2}\right) \\ x' &= \gamma \left( v_pt - v_qt \right) \end{align}$$

The velocity of $P$ in $Q$'s frame is simply $x'/t'$ so:

$$\begin{align} v'_p &= \frac{\gamma(v_pt - v_qt)}{\gamma(t - \frac{v_qv_pt}{c^2})} \\ &= \frac{v_p - v_q}{1 - \frac{v_qv_p}{c^2}} \end{align}$$

which is the same as equation (1).

Q2: Does Q see S's clocks as moving slower or faster?

Every observer sees the clocks of a moving observer running slow. So $S$ sees $Q$'s clocks running slow by a factor of $\gamma$, but $Q$ sees $S$'s clocks running slow by the same factor of $\gamma$. This is the origin of the famous twin paradox.

Q3: Does Q see P's clocks as moving at $\gamma_p$ or at $\gamma_q/\gamma_p$?

Neither.

If you take equation (1) and feed in $v_p = 0.25c$ and $v_q = 0.5c$ you get the velocity of $P$ in $Q$'s frame as:

$$ v'_p \approx 0.286 c $$

If you work out the corresponding value of $\gamma$ you get:

$$ \gamma'_p \approx 0.958 $$

Q4: How fast are S and Q moving?

In Q1 we worked out that $Q$ sees $P$ moving at about $0.286c$, so $P$ sees $Q$ moving at $0.286c$.

Likewise $P$ sees $S$ moving at $0.25c$.

Q5: Does P see a difference in S and Q's clocks?

Yes, because in $P$'s frame $S$ and $Q$ are moving at different speeds.

Once S, P and Q are reunited. What is the order of the clocks? Which one is earliest, which one is latest and which one is in between?

To do the calculation would be hard because you'd have to calculate the time dilation in the acceleration phases. Fortunately if you just want the relative order that's easy because you can get it from the following thought experiment:

We send $P$ and $Q$ off, and they both return a time $t$ later. We know that time dilation means their elapsed times, $t_p$ and $t_q$, will both be less than $t$. Suppose we make the velocity of $P$ very small, then $t_p \approx t$, and if we take the limit of $P$ not moving at all we'd have $t_p \approx t$. As we increase the velocity of $P$ the time $t_p$ will start falling smoothly with increasing velocity. The exact form of the function $t_p(v_p)$ will be complicated, but we expect it to fall smoothly towards zero as $v_p \rightarrow c$.

So because $v_q > v_p$ we know $t_q < t_p$, so $S$'s clock shows the most elapsed time and $Q$'s clock shows the least elapsed time.

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  • $\begingroup$ Thanks for this answer. It will take me a bit to digest it. $\endgroup$ – aepryus Aug 5 '14 at 11:43
  • $\begingroup$ Again thanks for the answer. Two follow up questions: 1. This analysis says that P doesn't see Q and S as equidistant from P while S does. Length contraction is constant along the direction of motion. How can the relative distance of things be a function of inertial frame? 2: While Q is traveling it sees S's clock moving the slowest. When they all meet again Q than sees its clock as the slowest. This transition occurs gradually or instantly? How would a plot of S vs Q's clock look in Qs reference frame? $\endgroup$ – aepryus Aug 7 '14 at 7:28
  • $\begingroup$ 1. suppose a ship passes me at 0.99c, and as it passes the ship fires a missile travelling at 0.99c. Then the observers on the ship see me and the missile as equidistant. But from my perspective the ship is travelling at 0.99c and the missile at some speed between 0.99c and c. So in my frame the distances are not equal. That is basically what's happening here. 2. You have to solve the equations of accelerated motion for that. See this question for a beginners guide. $\endgroup$ – John Rennie Aug 7 '14 at 8:38

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