# Question about Special Relativity similar to twin clock experiment

This question is related to twin clock experiment but also involves a third clock. Here it goes:

Suppose the relative velocity between clock 2 and clock 3 is given by $$v=0.8c$$. This would mean that the relative time dilation is given by $$\gamma=f(0.8c)$$ where $$f$$ is time dilation function at $$v=0.8c$$.

Now imagine similar to twin clock experiment, clock 1 is the clock that stays at home while clock 2 and clock 3 travel at velocity $$v=0.5c$$ w.r.t to clock 1. Clock 2 and clock 3 travel in opposite direction so their relative velocity is $$v=0.8c$$. Clock 2 and clock 3 return to clock 1 after completing their journey. Clock 2 and clock 3 have aged less compared to clock 1 and the factor of their time dilation w.r.t clock 1 is given by $$\gamma=f(0.5c)$$. Clock 1 will say that time dilation experienced by clock 2 and clock 3 is same but clock 2 and clock 3 had a relative velocity and hence they should experience different time dilation w.r.t to each other.

So, my question is: is it not necessary to define a reference frame to arrive at correct time dilation experienced by moving bodies?

• Relative velocity $\neq$ time dilation in the second case, since neither clock 2 nor clock 3 is in an inertial reference frame (they both have to return to clock 1, which involves an acceleration). – probably_someone Oct 12 '18 at 8:45
• Please look up clock hypothesis. Also, clock 4 and clock 5 can be assumed to be moving opposite to clock 2 and clock 3, respectively. Clock 4 and clock 5 will sync time with clock 2 and 3, respectively, to do away with necessity of returning and hence accelerations. – user146021 Oct 12 '18 at 10:16
• Where are clocks 4 and 5 at the point in time in clock 1's frame when clocks 1, 2, and 3 are at the same position? – probably_someone Oct 12 '18 at 10:18
• Sorry, i didnt make it very clear before. clock 4 and 5 are assumed only to do away with stopping and returning motion of clock 2 and 3. It can be assumed that at the point of turning (start of return journey towards clock 1), clock 2 (going away from clock 1) encounter clock 4 (clock 4 is going towards clock 1), sync their time and no acceleration is involved. – user146021 Oct 12 '18 at 16:56

A spacetime diagram might help. I have drawn it on rotated graph paper to help us visualize the ticks better, and I have chosen speeds of (3/5)c for the travelers (instead of (1/2)c) so that we can more easily count on the rotated graph paper.

[It turns out that velocities with rational Doppler factors lead to easier arithmetic... (3/5)c has a Doppler factor of 2, whereas (1/2)c has a Doppler factor of $$\sqrt{3}$$. The relative speed of the outgoing twins turns out to be $$(15/17)$$, which has a Doppler factor of 4.]

With this spacetime diagram, one can tell a detailed story what each observer would measure... by visual observation of light signals, or assigning-coordinates-by-simultaneity, or both.

The bottom line is that: while the actual observations of the other traveler appear to be irregular because of their relative motions, they will be identical when they reunite and compare notes. Indeed, by your construction, their worldlines through spacetime are symmetrical.

I've located events that would be useful to describe what Bob [in BLUE] (the initially-forward traveler) would say about Carol [in GREEN] (the initially-backward traveler). Alice [in RED] is the inertial observer.

For example, Bob might watch what Carol does (because Carol will broadcast/stream her life).

• From O to Q, Bob will see Carol move in slow motion... since in 8 years of Bob's clock, Bob only sees 2 years of Carol's life from O to A. (Follow the light signals along the grid.)
• At Q, when Bob turns around, Bob will see Carol moving normally again... but delayed. During the first year after Q for Bob (the beginning of the ninth year after separation), Bob sees the start of Carol's third year after separation. In the six years from Q to F, Bob sees six years of Carol's life from A to R.
• At R, when Carol turns around, Bob will see Carol moving at high speed. In the last two years from F to reunion at Z, Bob will see the last eight years of Carol from R to Z.
• At the reunion event Z, Bob will have seen in his 16 years of travel, all 16 years of Carol... but in an irregular way.
.. And Carol will say the same thing about Bob.

Of course, Alice will tell a different story of how she watched Bob's 16 years pass irregularly in her 20 years of waiting for the reunion. However, Alice's description of Carol will be identical to her description of Bob.

Update: In response to the OP's questions, I have gone a little further to attempt to draw the "spacetime diagram" of Bob (the initially forward traveler).

• I draw the spacetime diagram by inbound-Bob [the inertial path from Q to Z].
(I draw this first because it is higher-up in the final diagram.)
• I draw the spacetime diagram by outbound-Bob [the inertial path from O to Q].
• Based on the simultaneity according to each above, I cut each diagram and splice them together to get a Frankensteined-spacetime diagram for Bob [ the noninertial path O to Q to Z].

The Frankensteined-spacetime diagram for Bob [ the noninertial path O to Q to Z]:

Note:
Event P on inertial-Alice's worldline is MISSING on noninertial-Bob's spacetime diagram.
Event X appears TWICE on noninertial-Bob's spacetime diagram. (update)
Events from B to C on Carol's worldline (in particular, R) are also MISSING.
Suppose Bob sent a "radar" signal to B. Bob would send it just before his first year and it would arrive just before the 4th year on Carol's worldline. However, Bob receives the echo just before his 10th year, although his "spacetime diagram" would predict he would get it just after his 15th year [with the timestamp of his transmission just before his first year]. But when Bob receives that predicted echo, it actually has a timestamp from just after his 6th year.
Quite peculiar... noninertial-Bob's Frankensteined-spacetime diagram is somehow flawed... there are weird features not found on an inertial-observer's spacetime diagram.

• 1/2 Yes, it is very easy to understand this way, thank you. But the doubt that i have is on much different level. Here you have Alice, the state of which is known and the space time is kind of fixed to her frame...? Case 1: I want to know what if there is no Alice, i.e. just Bob and Carol who have same relative velocity as you assumed. How will the space-time diagram will change then? – user146021 Oct 13 '18 at 10:56
• 2/2 Case 2: Suppose we also have Alice but this time Bob is home clock (the home clock in twin experiment). Alice and Carol move such that at all given instants, Alice is midway between Carol and Bob. You see, case 2 is similar to the original question but this time i am viewing it from frame of Bob. How will the space time diagram change in case 2? I suppose this time the space time diagram will be drawn attached to the frame of Bob (instead of Alice as in your answer), and we will arrive at Bob ageing differently (different time dilation) w.r.t Carol. – user146021 Oct 13 '18 at 10:59
• Since Bob is non-inertial, his "spacetime diagram (using simultaneity)" will have problems...no one-to-one correspondence of events on this "diagram" with events in spacetime (as would appear on Alice's diagram). With the turnaround at Q, Bob's "diagram" will miss events from B to C along Carol's worldline.. and will double-count some events on the forward side of Q. So, Bob cannot be considered equivalent to an inertial observer. Analyzing what happens using his "diagram" will be more complicated. While info from Carol's birthday parties arrives, some of those events are not on his diagram. – robphy Oct 13 '18 at 15:00
• wow thanks! i need some time to carefully understand the updated part of the answer. In the mean time, what do you mean by Bob's spacetime diagram is flawed? Is it something to be expected (because he is non inertial) or is it something else? – user146021 Oct 13 '18 at 18:01
• Carefully go through that last paragraph. Use inertial-Alice's diagram to determine what actually happens. (Actually you can use any inertial-observer's diagram... say inbound-Bob.) Compare it with what noninertial-Bob predicts using his Frankensteined-spacetime "diagram". (I'm going to update the diagrams with an event X to the right of Q.) – robphy Oct 13 '18 at 18:15

It would be most useful if you kept a firm point. Time a clock reads is its proper time, i.e. the length in spacetime of its path.

You defined paths of clocks 2 and 3 in frame where clock 1 is stationary as exactly symmetrical, so of same length. Morover, consider that the length of a spacetime path in invariant, i.e. it is the same as computed in whichever (inertial) frame.

Thus you may be certain that a computation in other frames, where 2 and 3 appear not to have symmetrical paths, would notwithstanding give equal results.

• Can you simplify it a bit? – user146021 Oct 12 '18 at 20:03
• I would be glad to help you, if you explained what do you mean by "simplify". Perhaps the idea of a length in spacetime is novel to you? Referring to @robphy diagram, can you see two symmetrical paths? Whichever meaning we give to their lengths, it is quite obvious they are equal. Or perhaps you don't see the connection between proper time and path length? – Elio Fabri Oct 13 '18 at 13:08

$$\def\ns#1#2{#1_{\rm#2}} \def\xA{\ns xA} \def\tA{\ns tA} \def\xB{\ns xB} \def\tB{\ns tB}$$ I'll try to answer your new questions.

Here you have Alice, the state of which is known and the space time is kind of fixed to her frame...?

Not at all. it's better to think of spacetime as independent of particular observers and their frames. Maybe an analogy with Earth's surface helps.

Since more than a century international agreements exist as to which geographical coordinates to use: latitude and Greenwich longitude. It was not so before. French used Paris meridian as origin of longitudes, Italy (I suppose) Rome's meridian, and so on. (Latitudes were uniformly reckoned, since poles and equator are privileged references.)

The same, and more varied, happens in spacetime. Every reference frame is as good as any other, thus spacetime coordinates used in different frames are equally valid. Alice will use $$(\xA,\tA)$$ coordinates, Bob $$(\xB,\tB)$$ (but with an important proviso, see later) etc.

The proviso is that only inertial frames are allowed and this causes a difficulty with Bob and Carol, who change their speeds halfway. Two different frames must be introduced, both for Bob and for Carol, and a junction must be defined within each couple.

Those spacetime diagrams would be very awkward, as a consequence of what I said before. So much so, that it would be better to invent a "virtual" Alice with its frame. We are always allowed to do such things, since physically relevant results, relating to measurable quantities, are to be computable in whichever frame we like. In physicists jargon, are invariant.

2/2 Case 2: Suppose we also have Alice but this time Bob is home clock (the home clock in twin experiment).

You already know the answer: this is not allowed. Or better $$\dots$$ contrary to a diffused opinion, it is possible to use accelerated frames in SR. But at the price of considerable complication in reasoning and calculations. You ought to model Bob's motion wrt to an inertial frame, then to introduce special coordinates for Bob's accelerated frame, and so on. Out of reach for beginners, as well as useless.

In my opinion, for this and other reasons, it is better not to think in terms of time dilation (and to leave apart Lorentz transformations, seldom needed in elementary problems). This is why I suggested to work with spacetime paths and their lengths. An approach leading you to think of the physical situation and the quantities requested, instead of relying on rote formulas, prone to frequent misuse.