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This question already has an answer here:

I couldn't figure this out...

I understand that if Alice is moving relatively to Bob at a high speed, Alice's time elapses slower than Bob's. So when Alice is back to Bob and compares the clock, Alice's shows 10:01 but Bob's shows 10:10, for instance. But if we use Alice as the reference, Bob is moving. So Bob's clock should show 10:01... But there is only one outcome, isn't it? I must have messed up something, but I just couldn't find out what.

Help will be much appreciated.

P.S. I've searched around here and found something called "twin paradox". Is that what this is about? If yes, I can go read more about how people solve twin paradox. Thanks.

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marked as duplicate by John Rennie, peterh, Dale, Aaron Stevens, ZeroTheHero Nov 16 '18 at 0:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Yes, this is basically the twin paradox. Go read up on it. $\endgroup$ – Chris Nov 10 '18 at 16:15
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Like Chris's comment said, this is basically the twin paradox.

There are lots of sources that describe the twin paradox, but I still felt compelled to add this answer to help clarify one thing: there are two different kinds of situation that have both been described by the name "twin paradox". One is symmetric, and one is not.

First situation

Consider two objects that meet each other twice. Each object can record the proper time between these two meetings according to its own internal clock. If $\tau_A$ is the elapsed proper time between meetings according to object $A$, and $\tau_B$ is the elapsed proper time between meetings according to object $B$, then they can have $\tau_A=\tau_B$, but typically they will have $\tau_A\neq \tau_B$. In the typical case $\tau_A\neq \tau_B$, the situation is not symmetric. One of the two objects will age less than the other one, and both objects will agree about which one of them has aged less. For example, in flat (Minkowski) spacetime, suppose that:

  • Object $A$ remains in free-fall (that is, weightless) between the two meetings.

  • Object $B$ undergoes constant acceleration (in the sense that it has constant weight) between the two meetings.

In this case, object $B$ ages less between meetings than object $A$ does, and both objects agree about this. This situation is not symmetric.

Second situation

Now consider two objects flying past each other with constant velocities. They do not meet twice; they just keep on going after passing each other once. Each object has its own internal clock, and each object is able to observe (see) the other object's internal clock. In this situation, both of the following statements are true:

  • Object $A$ sees object $B$'s clock running more slowly than its own clock.

  • Object $B$ sees object $A$'s clock running more slowly than its own clock.

This is symmetric. The two objects are behaving symmetrically, so their observations of each other's clocks are necessarily also symmetric.

The second situation is more complicated, because in order for each object to observe the other object's clock, some kind of signal must travel from each object to the other. For example, each object could continually broadcast the time according to its own internal clock, using some kind of radio signal for the broadcast. Most importantly, this situation involves more than just the two objects; it also involves the radio signals that travel from one object to the other. This is why the second situation is more complicated.

Both of the situations described above, the first one and the second one, are described by special relativity using precisley the same principles. The principles are the same, but the situations are different. The second situation is symmetric, and the first one is not.


Appendix

For convenience, this appendix summarizes how the "principles" mentioned in the last paragraph can be expressed mathematically. In flat spacetime (which is the arena of special relativity), we can choose a coordinate system $t,x,y,z$ in which the proper-time increment $d\tau$ is given by $$ d\tau^2 = dt^2 - \frac{dx^2+dy^2+dz^2}{c^2} \tag{1} $$ whre $c$ is the vacuum speed of light and $dt,dx,dy,dz$ are the coordinate increments along any infinitesimal piece of the object's worldline. Equation (1) makes sense only when the right-hand side is non-negative, which is another principle: the worldline of a physical object must be such that the right-hand side of (1) is non-negative. Another principle gives a recipe for converting the proper-time equation into an equation that describes the motion of freely-falling objects. Applied to equation (1), this recipe says that the worldline of a freely-falling object is such that $x,y,z$ are all proportional to $t$. For a freely-falling massless entity, like a pulse of light, the worldline is such that the right-hand side of (1) is zero. This is consistent with calling the constant $c$ the "speed of light." Using these principles, we can analyze both of the types of situation that were described above. The principles about the motion of massless freely-falling entities are used, for example, to determine how light or radio signals propagate from one object to the other in the second type of scenario.

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  • $\begingroup$ @WillO How is that not the standard use of proper time? It's a clear and not confusing answer $\endgroup$ – Run like hell Nov 12 '18 at 12:53
  • $\begingroup$ dan and @runlikehell : i was completely wrong; somehow i read "proper time" as "interval". My comment was silly and I'm deleting it. Many apologies. $\endgroup$ – WillO Nov 12 '18 at 15:33
  • $\begingroup$ @WillO No worries! I do make plenty of mistakes, so I'm glad people speak up when something I wrote doesn't look right. I deleted my comment (from the other post, too) to wipe the slate clean. I may also add an appendix to help give a little more substance to my wordy answer. $\endgroup$ – Chiral Anomaly Nov 13 '18 at 0:41

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