# How can time dilation be symmetric?

Suppose we have two twins travelling away from each other, each twin moving at some speed $v$: Twin $A$ observes twin $B$’s time to be dilated so his clock runs faster than twin $B$’s clock. But twin $B$ observes twin $A$’s time to be dilated so his clock runs faster than twin $A$’s clock. Each twin thinks their clock is running faster. How can this be? Isn’t this a paradox?

• Seriously? Someone voted to close this as a homework-like question? Look who asked this question, and who wrote the top-rated answer. This is an ask-to-answer question, not a homework question. – David Hammen Feb 3 '18 at 22:36
• There is a similar paradox I realized years ago -- if someone leaves Earth on a ship going just less than the speed of light, travels out a ways, then heads back, he will find he's only aged a bit while centuries have gone by on Earth, and everyone he knew is dead. But couldn't you also say that he got aboard his spaceship which remained still while the rest of the universe moved in the opposite direction at near the speed of light then returned? In this case, the astronaut would have turned to dust while only a few years had gone by on Earth. I have no resolution for this paradox. – Jennifer Feb 4 '18 at 3:24
• @Jennifer: that's the standard twin paradox, and it isn't a paradox either. – John Rennie Feb 4 '18 at 5:14

The answer to this is that our twins, $A$ and $B$, are not measuring the same thing on their clocks. Since they are not measuring the same thing there is no paradox in the fact that each twin thinks their clock is running faster.

I’m going to try and give an intuitive feel for what is going on, and to do this I’ll use an analogy. This is going to seem a bit odd at first but bear with me and I hope everything will become clear.

Suppose I, Albert, and my two friends Bill and Charlie are all in cars driving at $1$ metre per second. I am driving due North, Bill is driving at an angle $\theta$ to my right and Charlie is driving at an angle $\theta$ to my left: Consider how fast we are travelling North, i.e. the component of our velocity in the North direction. I am travelling North at $1$ m/s while my friends are travelling North at $\cos\theta$ m/s, so my friends are travelling North more slowly that I am.

Now it turns out that our compasses have the odd feature that they show North as the direction in which our cars are travelling. That means both Bill and Charlie also consider themselves to be travelling North. Let’s have a look at the situation from Bill’s perspective: Bill considers himself to be travelling North at $1$ m/s while from his perspective I am travelling North more slowly, at $\cos(\theta)$, and Charlie is travelling North even more slowly, at $\cos(2\theta)$ m/s. And for completeness let’s show Charlie’s view: Like Bill, Charlie considers himself to be travelling North at $1$ m/s while he considers me to be travelling North more slowly, at $\cos(\theta)$, and Bill to be travelling North even more slowly, at $\cos(2\theta)$ m/s.

So all three of us think they are travelling North faster than the other two. Let me emphasise this because this is the key point in my argument:

Everyone thinks they are travelling North faster than everyone else

Now this isn’t rocket science. The reason we all think we are travelling North fastest is because we have different ideas of what direction North is in. But this is exactly what happens in special relativity if we replace the direction North in our diagrams by the time direction. And the reason everyone thinks everyone else’s time is dilated is because we all disagree about the direction of the time axis.

In special relativity we typically use spacetime diagrams with the time axis vertical and the $x$ axis horizontal (we omit the $y$ and $z$ axes because it’s hard to draw 4D graphs). I’ll get out of my car, so I’m not moving, then if I draw my spacetime diagram it looks like this: Although I’m no longer in the car I am still moving up the time axis because of course I’m moving through time at one second per second. So we have a diagram much like the one I started with except now the vertical direction is time not North, and I’m moving in the time direction not the North direction.

Bill and Charlie are moving away from me along the $x$ axis at speeds $+v$ and $-v$ just like the twins in the question: But, and this is the key point, what special relativity tells us is that for a moving observer the time and x axes are rotated relative to mine. Specifically, if the other observer is moving relative to me at a speed $v$ then their time axis is rotated by an angle $\theta$ given by:

$$\tan\theta = \frac{v}{c}$$

So if I draw Bill and Charlie’s time axes on my graph I get: Hopefully you can now see the point of my analogy. In Bill and Charlie’s rest frames they are stationary, so as far as they are concerned they are moving up the time axis at $1$ second per second just like me. But because their time axes are rotated relative to me I observe them to be moving in the time direction at less than $1$ second per second i.e. their time is dilated relative to mine.

Bearing in mind my analogy, to find out what Bill observes we rotate everything to the left to make Bill’s time axis vertical, and now Bill considers himself to be moving up the time axis fastest. Likewise we rotate to the right to make Charlie’s time axis vertical, and we find that Charlie considers himself to be moving up the time axis fastest.

And this answers our question. All three of us think we are moving through time fastest, and the other two people’s time is dilated, because when we measure time we are all measuring time in a different direction. Our clocks differ because we are measuring different things.

• @JohnRennie: You described the perception very well. What about reality? When twins on earth with synchronized clocks, now, one stays on earth, the other makes a fast and long, quick round trip and then compare the clocks. It is expected that the trip making clock can be years behind the one left on earth. Doesn't this tell us that the ticking of clock was actually slowed down due to initial acceleration, then sped up again during retardation at far end. And same repeated on the return journey. This tells me whatever acceleration does to clocks, retardation in same direction reverses it. – kpv Jan 30 '18 at 20:46
• @kpv: The round-trip involves acceleration, which introduces an objective asymmetry. The above-described twin paradox doesn't involve acceleration, only transformations. – AtmosphericPrisonEscape Jan 31 '18 at 0:42
• – Chris Jan 31 '18 at 2:23
• Great explanation! Question though, and I might be way off base here (feel free to tell me ;)). If your traveling example was 1D instead of 2D, this wouldn't apply (as everyone would be stuck moving in the same direction, or the opposite direction). So does your example highlight the fact that time (by itself) is more than a single dimension? – Flater Feb 2 '18 at 9:33
• @Acccumulation that's true. However anyone who understands the maths at that level understands there is no paradox anyway. I was looking for a way to explain the situation to people without a knowledge of SR. I have considered extending the answer to add a rigorous discussion as an appendix, but I think that exists in existing answers already. – John Rennie Feb 3 '18 at 4:58

This effect (B is slower from p.o.v of A and vice versa) does not seem very mysterious and can be observed even in a very simple model. The effect is the direct consequence of Einstein – synchronization of clocks in an observer’s frame of reference.

To demonstrate that, let us consider the behavior of objects that, although slow-moving, nonetheless act in accordance with the laws of the special theory of relativity. Fig. 1. The ship on the left is at rest on the water surface. A shuttle moves at a velocity of $V$ from a ship to the bottom and back. The ship on the right is moving at a velocity of $v$ along the water body surface. The speed of movement of the shuttle equals $V$, the shuttle’s horizontal velocity component equals $v$, and the vertical component, $V_Z$, equals $V \sqrt{1-(v/V)^2}$

Let’s imagine the surface of a flat-bottomed water body with a depth of $h$, filled with still water. A ship equipped with a pendulum clock and with instruments that operate based on signals generated by this clock (in time with this clock) is located on the water body surface. A high-speed shuttle that is in continuous motion along a plumb line (relative to a given ship) between the ship and the bottom performs the function of the clock’s pendulum. Each shuttle trip to the bottom and back requires a time of $Δt = 2h/V_Z$ , where $V_Z$ – rate of descent and ascent of the underwater shuttle, and is accompanied by a change in the clock reading. The shuttle moves at a constant speed of V relative to the water, and if the ship is at rest, the shuttle moves perpendicular to the bottom, and the rate of the shuttle’s descent and ascent, $V_Z$, equals $V$. The time, $Δt$, of a shuttle trip to the bottom and back equals $2h/V$. The $V$ velocity value exceeds the ship’s speed of $v$; i.e., the condition $v < V$ is satisfied.

If a ship is proceeding at a speed of $v$, the clock tick rate and the operating speed of the instruments on the ships are decreased. This occurs due to the fact that when a ship is moving at a speed of $v$, the ascent and descent rate, $V_Z$, of a shuttle making trips in the water between a ship and the bottom of the water body according to the hypotenuses of right triangles happens to equal $V \sqrt{1-(v/V)^2}$ . Time on the ship in motion, which can be called simulated time, $t'$, passes more slowly than time, $t$, on the ship at rest also by $1 \sqrt{1-(v/V)^2}$ times. Thus, the more rapidly a ship proceeds through the water, the less often the pendulum “swings” and the more slowly the operations of the instruments located on this ship are performed, the operating speed of which is proportional to the shuttle pendulum frequency.

It is easy to simulate time dilation using ships of this type.

Let us assume that two ships at rest are located on a water surface at some distance from one another. Let’s imagine that the ships are equipped with speedboats that, like the shuttles, run at a speed of $V$, but only on the water surface. Let us assume that the ship instruments synchronize the clocks using a speedboat to transmit the information, which runs from one ship to the other and back. If the instruments have information that the speed of the boat relative to ships in opposite directions are equal, then using the boat, the instruments synchronize the clocks, as is done using a light signal in the special theory of relativity.

Having synchronized the clocks, the instruments on the ships at rest can compare their clock rate to that of a ship that is moving past them along the line that connects them. Taking the clock readings of the ship in motion at the locations of the ships at rest and comparing them to the readings of the synchronized clocks on their own ships, the instruments record time dilation of the moving ship $1 \sqrt{1-(v/V)^2}$ times.

Now imagine two ships under way one after another at a speed of $v$. Let’s assume that the first ship moves past a ship at rest at some point in time, then the second ship also moves past the ship at rest at some later point in time. Comparing the clock readings of the ship at rest with those of the previously synchronized clocks of their own ships, the instruments of the ships in motion detect a difference in the rate of their clock and that of the clock on the ship in motion. The result of a comparison of the clock on the ship at rest and the clocks on the ships in motion will depend upon the clock synchronization technique.

If the instruments on the ships in motion are able to measure the speed, $v$, of their ships, or if they have information concerning the fact that their ships are moving at a speed of $v$, then by synchronizing their clocks using a boat moving between the ships, they take into account the disparity of the speed of the speedboat they are using relative to their ships in the direction and opposite the direction of their movement. By synchronizing the clocks in this manner, they obtain a true result, according to which time on the ship at rest passes $1\sqrt{1-(v/V)^2}$ times more quickly than their own time.

However, this result can be just the opposite if the instruments on the ships in motion have no information concerning the movement of their ships and no other means of communication between the ships other than a speedboat. The truth of the matter is that by sending a boat that carries the requisite information from ship to ship, the instruments can only record the fact of the movement of the ships relative to one another. Basic calculations reveal that the instruments have no way of determining which ship is in motion and which ship is at rest relative to the water. If the instruments use false information concerning the repose of their ships, then mistaking their ships in motion relative to the water for ships at rest, they mistake the ship at rest in the water for a ship in motion relative to them. Here, they use the false condition of the equality of the boat’s speed relative to their ships in the direction of their movement and opposite it.

In this instance, by synchronizing the clocks using the Einstein technique, the instruments on the ships in motion, strange as it may seem, record a false time dilation on the ship at rest in the water, which in their estimation is moving relative to them.

Some references:

Dorling, J. „Length Contraction and Clock Synchronization: The Empirical Equivalence of the Einsteinian and Lorentzian Theories“, The British Journal for the Philosophy of Science, 19, pp. 67-9

Chapter 3.5.5 The reciprocity of the Lorentz transformation https://www.mpiwg-berlin.mpg.de/litserv/diss/janssen_diss/Chapter3.pdf

Simulation of Kinematics of Special Theory of Relativity by means of classical mechanics https://arxiv.org/abs/1201.1828

Here are some spacetime diagrams that display the symmetry of time dilation.
These diagrams underlie the various analogies one can use to motivate the symmetry.

First, we draw spacetime diagrams on rotated graph paper so that we can more easily visualize the ticks along the inertial observer worldlines.

In our example,
our observers have relative velocity of $v/c=\tanh\theta=(6/10)$,
and the corresponding time-dilation factor is $\gamma=\cosh\theta=(10/8)$,
where $\theta$ is the Minkowski-angle [the "rapidity"] between the timelike-worldlines.

We have drawn the diagram from Alice's frame. Alice regards P and P' as simultaneous, whereas Bob (traveling with velocity (6/10)c with respect to Alice) regards Q and Q' as simultaneous.
Note that:

• $\triangle OPP'$ is a Minkowski-right triangle, where $OP$ is Minkowski-perpendicular to $PP'$.
$$\cosh\theta=\gamma=\frac{OP}{OP'}=\frac{10}{8}$$
• $\triangle OQQ'$ is a [similar] Minkowski-right triangle, where $OQ$ is Minkowski-perpendicular to $QQ'$.
$$\cosh\theta=\gamma=\frac{OQ}{OQ'}=\frac{10}{8}$$ In my diagram, the "light-clock diamonds" have lightlike edges and equal area. In addition, the diagonals of the "light-clock diamonds" are Minkowski-perpendicular to each other.

By drawing in the hyperbolas with center at the meeting event $O$, one can see that $PP'$ is tangent to that hyperbola at the event $P$, where "radius vector" $OP$ meets the hyperbola. Similarly, $QQ'$ is tangent to that hyperbola at the event $Q$, where "radius vector" $OQ$ meets the hyperbola. To see that this "tangent is perpendicular to radius" construction is analogous to the Euclidean construction (and to see the Galilean analogue), play around with my visualization https://www.desmos.com/calculator/wm9jmrqnw2 by tuning the E-parameter.
(In this visualization, time runs to the right [like the standard position-vs-time graphs].)

The first diagram is based on Fig. 17 in my paper "Relativity on Rotated Graph Paper" [American Journal of Physics 84, 344 (2016)] http://dx.doi.org/10.1119/1.4943251

I'm assuming the ships take off simultaneously from earth, with all three clocks set to 0. Events connected by blue lines are simultaneous according to the observer on earth. Events connected by red lines are simultaneous according to the observer on ship A. Events connected by green lines are simultaneous according to the observer on ship B:

(Note: these times are approximations; to make this fully realistic, I'd have to show events happening at times like 1:47, which I'll have rounded off to 2:00.) The earthbound observer says things like this:

I see by my clock that it's now 4:00. At this moment both ship clocks say 3:00. They are running slow.

Or

I see by my clock that it's now 8:00. At this moment both ship clocks say 6:00. They are running slow.

The captain on Ship A says things like:

I see by my clock that it is now 4:00. At this moment, the earth clock says 3:00. It is running slow. Also at this moment, the B-clock says 2:00. It is running even slower.

Or:

I see by my clock that it is now 8:00. At this moment, the earth clock says 6:00. It is running slow. Also at this moment, the B-clock says 4:00. It is running even slower.

The captain on Ship B says things like:

I see by my clock that it is now 4:00. At this moment, the earth clock says 3:00. It is running slow. Also at this moment, the A-clock says 2:00. It is running even slower.

Or:

I see by my clock that it is now 8:00. At this moment, the earth clock says 6:00. It is running slow. Also at this moment, the A-clock says 4:00. It is running even slower.

Where's the alleged paradox?

This is often expressed as "Each of the twins thinks his clock is moving faster". However a more precise way to say it would be "Each of the twins thinks his clock is moving faster when observed in his own coordinate system."

The difference is important in that if the travelers understand relativity, they will know that their observation only applies into their own coordinate system. They can also calculate and agree with what the other traveler thinks, so they are not disagreeing.

An analogy can be made with movement. When traveler A looks out of his window and sees distance to B's ship increasing, he might think "I'm moving and he is staying where he is". But B can think exactly the same. And yet, both of them understand that their observations are not in conflict, because movement is always relative. Another example from Wikipedia:

While this seems self-contradictory, a similar oddity occurs in everyday life. If person A sees person B, person B will appear small to them; at the same time, person A will appear small to person B. Being familiar with the effects of perspective, there is no contradiction or paradox in this situation.

Another important part of the different coordinate systems is that there is no direct way to simultaneously measure the clock times when they are not next to each other. Because light speed is the maximum speed of any information, what you see of the other clock is delayed more and more as it goes further away.

However, if traveler B decides to turn his ship around and catch up with A, the situation changes. Traveler B's coordinate system now changes as his speed is changing. This breaks the symmetry. By the time when B catches up with A, both of them will observe that B's clock is lagging behind A's clock.

This phenomenon follows directly from the principle of time dilation of special relativity:

Proper time = time before time dilation

Observed coordinate time = time after time dilation

That means in this case: When each twin observes his own clock, the observed coordinate time is the proper time (time dilation factor 1, that means absence of any time dilation). When he observes some other clock moving at a relative velocity with respect to himself, the time dilation is not one, it is bigger than one, that means that there is some time dilation.

One way to understand relativity is to think of space time as being described by a geometry, in which $x^2+y^2=z^2$, with $x$ and $y$ being two legs of a right triangle, and $z$ as the hypotenuse, is replaced with $x^2-y^2=z^2$ with $x$ representing the distance in space between two events in spacetime, $y$ being the distance in time between two events in spacetime, and $z$ being the distance in spacetime between two events.

The distance in time between two events, if those two events are connected by a world line of an object that is in an inertial reference frame, is the proper time of the world line. So the proper time of a world line that is in an inertial reference frame can be expressed using the equation $\tau^2=-\left(\frac{\Delta_x}{c}\right)^2+{\Delta_t}^2$, with $\tau$ being the proper time, $\Delta_x$ being the objects displacement in space, $c$ being the speed of light, and $\Delta_t$ being the objects displacement in time.

If twin A and B are in inertial reference frames, and B is moving relative to A, then you could draw a right triangle with one of the legs representing the proper time of twin A, the other leg representing the displacement in space of twin B relative to A from the initial time for A to the final time for A and the hypotenuse being the proper time for B so ${\tau_b}^2=-\left(\frac{\Delta_{x_a}}{c}\right)^2+\tau_a^2$. The direction of an objects proper time is also the direction of that objects time axis so A and B also disagree on which direction is time and that is how they can both say that it is the other whose clock has slowed down. While different observers can disagree on the displacement in space, and the displacement in time they can agree on the distance in spacetime between two events.

(This answer was originally posted for a newer question, asked on 2018 Nov 10, that was later marked as a duplicate question and linked to this one, so I relocated my answer here.)

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.

Everything is clear once you remember that simultaneity is relative.

Let's look at a simple case where frame F' is moving to the right with velocity v with respect to frame F, in another words, F is moving at -v with respect to F'.

The Lorentz transformation in one direction is given by \begin{align} x' &= \gamma (x-vt) \\ t' &= \gamma (t - \frac{v}{c^2}x) \end{align}

If you fix $$x'=0$$, then $$t'$$ would be the time of the clock inside F'. Simple substitution shows \begin{align} x=vt &\Rightarrow t'=\gamma(t-\frac{v^2}{c^2}t)\\ &\Rightarrow t'=\frac{t}{\gamma} \end{align}

We usually denote this $$t'$$ with $$\tau$$ and call it the proper time because the "clock" is fixed to $$x'=0$$ in frame F'. We can see the time dilation in this case.

However, at time $$\tau$$ in frame F', we can ask the question of what does the observer at frame F' see at this moment on the clock sitting at the origin frame F? The key thing here is that at this moment means different things to F' and to F due to the relativity of simultaneity. In the Minkowski diagram, it is tilted.

If we fix $$t'=\tau$$ for some value of $$\tau$$, we can figure out what $$t$$ is at $$x=0$$. \begin{align} \tau&=\gamma(t-0)\\ \frac{\tau}{\gamma}&= t \end{align} Here, we can denote $$t$$ with something else such as $$\overline{t}$$ meaning it is what the F' observer sees on the clock at the origin of F. You can see that both observers think that there is a slowdown and at the root it works because simultaneity is relative.

## protected by Qmechanic♦Jan 30 '18 at 22:15

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