If we have two objects (clocks) in (empty) space and they separate (the distance between them increases) at relativistic speeds then get back together (just for easier comparison of their times), what determines which clock would advance more (tick faster) and which less?

In other words: Let's name one object A, the other object B. In case one, object A stands still and object B moves away (then returns). In second case object B is still and B moves.

Does the above make sense? If there is nothing else around, can we claim, that one object is standing still and the other moves?

Update: This is basically the twin paradox. Does it make any change if the moving object does not turn around? Just stops at some distance? Or never stops?

  • 1
    $\begingroup$ This is basically a question about The Twin "Paradox", for which several answers exist on this (and other) sites, e.g. here. In essence, the object who experienced the least acceleration ages faster. $\endgroup$
    – pela
    May 8, 2015 at 13:45
  • $\begingroup$ Acceleration can always be discerned by inertial forces that are local to the accelerated observer. $\endgroup$
    – CuriousOne
    May 8, 2015 at 13:48
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    $\begingroup$ Any observer moving along a geodesic experiences the maximum amount of proper time between two events. The more an observer deviates from following a geodesic between the two events, the less proper time they experience. That determines which clock advances more when they meet back up $\endgroup$
    – Jim
    May 8, 2015 at 14:08

3 Answers 3


Beginners to special relativity generally learn the equation for time dilation:

$$ t' = \frac{t}{\gamma} \tag{1} $$

where $\gamma$ is the Lorentz factor:

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

For any $v > 0$ the Lorentz factor is greater than one and hence $t' < t$, which is what we mean by time dilation.

And this is all very well, but equation (1) is a special case that applies only for uniform unaccelerated motion, and it's absolutely no use for understanding the twin paradox. Indeed, the twin paradox only arises because people mistakenly apply equation (1) when it isn't appropriate. To understand the twin paradox properly we have to go a bit deeper into how special relativity works.

Special relativity is a geometrical theory of spacetime, just like general relativity in fact though considerably simpler. The basic equation for special relativity is the Minkowski metric:

$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 \tag{2} $$

Suppose I set up a coordinate system $(t, x, y, z)$ with myself stationary at the origin and the time $t$ measured with my clock and the distances $x$, $y$ and $z$ measured with my rulers. What the Minkowski metric says is that if I observe an object to move a distance $(dx, dy, dz)$ in a time $dt$ then the proper time $d\tau$ can be calculated using equation (2).

Why would we do this? Well firstly the proper time is an invariant i.e. every observer in any coordinate frame moving, rotating, accelerating or whatever will calculate the same value for the proper time $d\tau$. Secondly the proper time is just the time shown on a clock carried by our moving object.

The invariance of the proper time is an assumption and actually it's the key assumption in special relativity. You just have to accept that's how SR works. However we can easily prove the second point i.e. that the proper time is the time shown on the moving object's clock. In the moving object's rest frame it is stationary so $dx = dy = dz = 0$ and therefore equation (2) simplifies to:

$$ c^2d\tau^2 = c^2dt^2 $$

or just:

$$ d\tau = dt $$

So in the object's rest frame the proper time $d\tau$ is equal to the time measured by the object's clock $dt$. But we've just said that $d\tau$ is an invariant and is the same for all observers. That means for all observers the proper time is the time shown on the moving objects clock.

The point of all this is that we now have a way to calculate time dilation because we can use equation (2) to calculate the proper time. Let's do this for the simple case of uniform unaccelerated motion and check we get equation (1). We'll assume the object passes us at time zero with velocity $v$, and for convenience we'll assume it's moving along the $x$ axis so $dy = dz = 0$. The equation (2) becomes:

$$ c^2d\tau^2 = c^2dt^2 - dx^2 $$

But if the object is moving at velocity $v$ this means that $dx/dt = v$ because that's what we mean by velocity. So $dx = vdt$, and substituting this into the equation about we get:

$$ c^2d\tau^2 = c^2dt^2 - v^2dt^2 $$

and a quick rearrangement gives:

$$ d\tau^2 = dt^2(1 - v^2/c^2) $$


$$ d\tau = \frac{dt}{\frac{1}{\sqrt{1 - v^2/c^2}}} $$

which is just equation (1).

This is a good point to stop and summarise where we've got to. We've stated that the Minkowski metric, equation (2) above, is the core equation in special relativity, and that it can be used to calculate the proper time $d\tau$, which is just the time measured by the moving object's clock. And this is what you asked, i.e. how do we calculate how much time has elapsed for the moving object. So in principle at least we've answered your question.

But can we do the calculation for an accelerating observer like the twin? And the answer is that yes we can. The maths will get harder by the same basic principle applies. We'll do it for a simplified case. Suppose at time zero our object starts accelerating at a constant (in our frame) acceleration $a$. Then in our frame the distance is given by the equation we all learned in school:

$$ x = \tfrac{1}{2}at^2 $$

and differentiating gives:

$$ dx = atdt $$

and as before we substitute this into the Minkowski metric to get:

$$ c^2d\tau^2 = c^2dt^2 - a^2t^2dt^2 $$


$$ d\tau = dt\sqrt{1 - \frac{a^2}{c^2}t^2} $$

Integrating this turns out to be rather sticky, so we'll assume that $at \ll c$ in which case we can approximate the square root as;

$$ d\tau \approx dt\left(1 - \frac{a^2}{2c^2}t^2\right) $$

And we can integrate this to get the elapsed time for the accelerating object:

$$ \tau \approx t\left(1 - \frac{a^2}{6c^2}t^2\right) $$

And we find that $\tau \lt t$ so the time for the accelerating object is dilated - the accelerating twin ages less than us.

  • $\begingroup$ Hmm, I think this answer got away from me a bit. I thought it would be an opportunity to give a simple explanation of how time dilation works, but I'm not sure it's all that simple. Oh well, I'll leave the answer here in case anyone fins it useful. $\endgroup$ May 8, 2015 at 17:41

"What would determine which clock ticks faster?"

If both clocks experience a uniform change in motion they will show the same time when they convene. Acceleration/deceleration changes a reference frame's time dilation factor. If these two clocks experienced different changes in motion (ex. one accelerates/decelerates differently than the other) the difference in time they exhibit will be based on that asymmetry.


Whichever clock switches it's rest frame in order to meet with the other clock will be the one with a lower count (younger twin).

If both clocks switch their rest frame symmetrically, then both will be of the same age.

Meaning, if an observer who was at rest with those clocks before they started accelerating, sees those clocks move away from him at Vrel_a in the first phase. Then sees both clocks accelerate back towards him at the same Vrel_b, when they meet at the observer eventually, then both their counters will be in sync.

To understand what is going on. Draw an x,t diagram using the perspective of an observer traveling along with one of the two clocks.

The observer considers himself at rest, so his and the clock's(he is at rest to) worldline, would go straight up logically. The other clock would either be a diagonal to the left or right. Mark a few events on this diagonal. Each event has one x and one t coordinate E1(x,t) E2(x,t) etc.

Now have your clock+observer accelerate, hence switch the rest frame.

Draw a second x,t diagram now, and use the formulas of SR to map those events inside the new rest frame. You will notice that if event E1 for example was at x=5 (ls), t=10s that in the new rest frame, this event won't be at the same place anymore. It space and time shifted. So if for example 10 seconds into the experiment the clock at x=5 ls t=10 displays 8 seconds (remember, you see it tick slower because of time dilation), when you switch the rest frame, that event E1 won't be anymore at t= 10s on your diagram, but below. Meaning Event E1 for you now happened "earlier" in the past, in the classical sense. If we were to use the classical notion of present, which is two events happening "at the same time" which is a parallel line to the x axis in your diagram.

The clock you will see(as in calculate) now, which is on the line that is parallel to the x-axis and which crosses the t-axis at 10 second, hence simultaneously to you being at x=0 t= 10s, will be a clock with a higher counter.

So the observer who is at rest with the clock, accelerated towards the other clock and switched his rest frame. Events that formerly were simultaneous to him, are not anymore. In fact, all events E1, E2,..., En happening at a given (x1,t1),...,(xn,tn) coordinate, require a transformation in his new rest frame, having different coordinates now.

And while he is moving now back towards the other clock, and keeps seeing(as in calculate) it as ticking slower, the instance of the clock he is moving towards, in the classical sense of considering the clock that is simultaneous to him (on the same line parallel to the x axis), is one with a higher counter than his own clock displays.

The slower ticking does not make up for this time shift until they meet again and results in the observer reaching the clock with a higher counter even though it was ticking slower before and after the acceleration all the way through this experiment. Seen (as in calculated) by the observer. The exception being the acceleration phase, where if he was omnipresent would see the instance of the clock which is simultaneous to him, speed up and the counter shoot up.

edit: Einstein in some of his thought-experiments was using an infinite army of observers at rest with meters and synced clocks which would then report every event and mark down the x,t coordinate. He was using the classical sense of simultaneously even though simultaneous does not have much of a meaning anymore in relativity. It's basically giving you only a basis which allows you to assign coordinates to events.

Unfortunately, many have mixed up their explanations. Using time dilation and simultaneity which is based on the basis or methodology of measuring this scenario the way i did it, while at the same time argue with different methodologies of how to describe this scenario.

edit2: To fully understand SR, you will have to derive it yourself by taking the two postulates and mapping them in mathematical form, then get the formulas out of this derivation. Especially the formulas for how to transform coordinates of events from one rest frame into another are very important. Otherwise, when someone tells you to take a given formula and you magically get the proper time or whatever else, while you might be able to solve some problems, you did not really understand SR.


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