# Twin Paradox, calculating spacetime intervals from both perspectives

I've very recently started to try to understand special relativity. I'm want to get a decent understanding of the twin paradox. I'll post what I've done so far and highlight what's gone wrong for me.

The situation is that Alice and Bob both begin at point $x_1$ in Alice's coordinate system $(x,t)$ (we have orientated the axis so that y and z will not matter). Bob then instantaneously moves off with velocity $v$ in the positive $x$ direction. Bob's coordinate system is now $(x',t')=(\gamma (x-vt), \gamma (t-v\frac{x}{c^2}))$. In Alice's coordinates, Bob reaches point $x_2$ after having moved distance $d$, then instantly turns around and travels back to Alice at $-v$. I want to show that Alice's space time interval $\Delta s_A^2$ is greater than Bob's $\Delta s_B^2$, as each path's space time interval is proportional to the proper time passed along that path.

According to Alice: $\Delta s_A^2=-c^2\Delta t^2 + \Delta x_A^2$ where $\Delta t=\dfrac{2d}{v}$ and $\Delta x_A=0$ as she didn't move, but the time elapsed is the time it took bob to travel the distance and then come back. Thus $\Delta s_A^2=-\dfrac{4c^2d^2}{v^2}$

Still according to Alice, Bob's $\Delta s_B^2=-c^2\Delta t^2+\Delta x_B^2$, where the change in time is the same, but now bob has moved distance $d$ twice, thus $\Delta x_B=2d$ ($\Delta x$ refers to total distance traveled rather than displacement which is 0 in this case). Now we have $\Delta s_B^2=-\dfrac{4c^2d^2}{v^2}+4d^2$.

The |size| of Bob's spacetime interval now definitely smaller than Alice's, and this would be all OK, except that when I do the calculations in Bob's frame, they don't agree. This is contradictory to the fact that $\Delta s^2$ is conserved under Lorentz transformations.

According to Bob: $\Delta {s'}_A^2=-c^2\Delta t'^2 + \Delta {x'}_A^2$ where $\Delta t'=\dfrac{2d'}{v}$. I'm not sure if I'm right in saying that $v$, the relative velocity of the two frames, is the only velocity upon which both Alice and Bob will agree on, other than the speed of light. Anyway, $d=\gamma d'$ as lengths contract by $\gamma$ i.e. $d'$ is smaller than $d$ by factor $\gamma$. Also, $\Delta {x'}_A=2d'$, so we have $\Delta {s'}_A^2=-\dfrac{4c^2d'^2}{v^2}+4d'^2=-\dfrac{4c^2d^2}{\gamma ^2 v^2}+\dfrac{4d^2}{\gamma ^2}=4d^2\dfrac{v^2-c^2}{\gamma^2 v^2}\neq \Delta s_A^2$

Although I haven't included them here, my calculations for $\Delta {s'}_B^2$ and $\Delta s_B^2$ agree. Sorry for the ultra-long post, but any help would we well appreciated!

• Thanks for the reply! I've spent a while trying to rearrange my result for $\Delta {s'}_B^2$ with gamma explicitly substituted in, but to no avail. Do you mean that they are in fact equal and that I've just not got the correct expression? – James Machin Feb 11 '14 at 13:31
• oh sorry, yes indeed I meant to say $\Delta {s'}_A^2$, my bad! I'm using the definition $\gamma=\sqrt{\dfrac{1}{1-\frac{v^2}{c^2}}}$ and unless I'm being extremely stupid (which is entirely possible) using that defn it won't rearrange. Thanks again for the reply by the way – James Machin Feb 11 '14 at 14:41
• I'm going to remove my previous comments because they are incorrect and might confuse future readers. – Wouter Feb 11 '14 at 16:08

You can't calculate Bob's proper time using the Lorentz transformations because Bob does not travel in an inertial frame. The frame he sets out in is not the same as the frame he returns in because it's travelling in a different direction. You say Bob instantly turns around and travels back to Alice at −v, and what this means is that at the halfway stage Bob must accelerate to change his velocity from $v$ to $-v$. You have to take this acceleration into account when calculating the proper time for Bob's trip.

You'll hear people say you need General Relativity to treat acceleration, but this isn't so. The procedure for treating acceleration in SR is described in Gravitation by Misner, Thorne and Wheeler chapter 6. If you just want the equations have a look at John Baez's article on the relativistic rocket.

• Oh ok I see, thank you! so if I had gotten identical answers to the calculations in both reference frames, I'd have done something wrong? – James Machin Feb 13 '14 at 19:12
• @JamesMachin, there are three reference frames, not two. This is the point John made when we wrote: "The frame he sets out in is not the same as the frame he returns in". – Alfred Centauri Feb 13 '14 at 22:47
• You can pick, say, Bob's initial frame as inertial and calculate all three intervals (Alice's, Bob's out bound, Bob's return) in that frame. The work is harder, but you get the same result. – dmckee --- ex-moderator kitten May 17 '14 at 16:21

John has already provided a correct answer containing pretty much all the necessary information, but I'll just try to show the resolution of the paradox and visually point out where things go wrong for Bob. So let's draw a spacetime diagram of the situation. For simplicity we set $c = 1$ in what follows.

The path Alice takes is simply a vertical line in the spacetime diagram (from her frame of reference): she is at rest for a period of time $\Delta t$ and so passes through event $B$ on her path from $A$ to $C$. Bob, however, takes the scenic route and first goes out to $B'$, travelling at a speed $v<c$. Then he moves on to $C$ from $B'$, with the same speed. The spatial distance Bob travels between $A$ and $B'$ is $\Delta x =\frac{1}{2}v\Delta t$.

We use proper time to measure interval lengths for timelike paths: $\Delta\tau^2 = -\Delta s^2$. It's now obvious how long it takes Alice to travel to her midway point:

$$\Delta\tau_{AB} = \frac{1}{2}\Delta t$$

while it takes Bob

\begin{align} \Delta\tau_{AB'} &= \sqrt{\frac{1}{4}(\Delta t)^2-(\Delta x)^2} \\ &= \sqrt{\frac{1}{4}(\Delta t)^2-\left(\frac{1}{2}v\Delta t\right)^2} \\ &= \frac{1}{2}\sqrt{1-v^2}\Delta t \end{align}

to get to his midway point, where we have filled in our earlier expression for $\Delta x$.

Of course this means $\Delta\tau_{ABC} = \Delta t$ and $\Delta\tau_{AB'C} = \sqrt{1-v^2}\Delta t < \Delta\tau_{ABC}$.

As John mentions and as you can see from the spacetime diagram, Bob needs to accelerate at some point to return to the same spatial location as Alice (only straight lines on a Minkowski spacetime diagram are unaccelerated and Bob's path has a bend). This constitutes a change in inertial frame, but just looking at it from Alice's frame should convince you that a non-straight path through spacetime always has a shorter proper time than a straight path. Why Bob doesn't find that Alice experienced a shorter proper time is precisely because his frame of reference changes halfway through his journey. You can use the references John provided to check this.

• Let me pick up on a point Wouter makes as this is an essential point (and I wish I had mentioned it :-). The Principle of Extremal Aging states: The path a free object takes between two events in spacetime is the path for which the time lapse between these events, recorded on the object's wristwatch, is an extremum. A straight line in spacetime, as followed by Alice, maximises the proper time. Any observer like Bob who travels from Alice's start point to her end point on a different trajectory has a lower (i.e. shorter) proper time. – John Rennie Feb 11 '14 at 16:21
• Thank you very much for the detailed answer! I've read through it fully now and I think I understand it. About the Principle of Extremal aging; the way I understand it is that every person is the center of their own reference frame, so their delta x,y,z is always 0 in their coordinates. So the proper time is equal to the time measured passing by that person. I don't understand what a geodesic is in spacetime though! – James Machin Feb 13 '14 at 19:16
• @JamesMachin, if you plot Alice's worldline through the initial and final events, you'll see that it is a straight line; a geodesic in flat (uncurved) spacetime. If you plot Bob's worldline through the same two events, you'll see that it has a 'kink'. According to the Principle of Extremal Aging, the straight path is the path of greatest proper time. Since Bob's path is not straight, the proper time along that path is less than along Alice's straight path. – Alfred Centauri Feb 13 '14 at 22:51

There are just 3 events that need be considered here: (a) the initial event that Alice and Bob are co-located, (b) the event that Bob turns around and (c) the final event that Alice and Bob are again co-located.

If, according to Alice, Bob's speed on both legs is $v$, and the distance to the turnaround point is $r$, then:

$$\Delta s^2_{ac} = (c2r/v)^2$$

$$\Delta s^2_{ab} = \Delta s^2_{bc} = (cr/v)^2 - r^2 = r^2[(c/v)^2 - 1)]$$

The proper time for Alice is then:

$$\tau_A = 2r/v$$

and the proper time for Bob is:

$$\tau_B = 2r/v\sqrt{1 - \frac{v^2}{c^2}} = \frac{\tau_A}{\gamma}$$

So, clearly, Bob's proper time along his path from (a) to (c) is less than Alice's proper time on her path.

• Thank you for the reply! I have already done the calculations in Alice's frame of reference and got the correct answers! I was just wondering why there was a discrepancy between the calculations done in Alice's and Bob's frame. – James Machin Feb 13 '14 at 19:20
• @JamesMachin, keep in mind that Bob doesn't have a frame; the outbound frame and the inbound frame are two different (inertial) frames. Bob changes frames, at the turn-around event and that must be taken into account in your calculations. In essence, at the turn-around, events that are simultaneous in Bob's outbound frame are no longer simultaneous - the inbound frame and outbound frames do not agree on what events are simultaneous. In a certain sense, Bob 'sees' Alice's age instantaneously jump at the turnaround. – Alfred Centauri Feb 13 '14 at 21:02