Suppose that a man departs from Earth to reach a planet that is ninety-nine light-years away with 0.9802 light speed, c. According to special relativity, it takes him 40 years (measured by his own clock) to reach the planet and then return to Earth. But for the Earth's clock, 202 years have passed. When he reaches the planet, he would see his own clock reading 20 years and the Earth's clock reading 3.96 years, but the people on Earth would see their own clock reading 101 years.

To make things simple, think of two rockets, each moving without change in velocity. The man rides the first going out and the second going out. Both of the rockets would meet at the planet, and it is time for him to change his rocket (change in frame). The time needed for him to return to earth is 3.96 years as well. So this just indicates that during the change of frame, at that instant 194.08 years pass when he looks at the Earth clock. In other words, when he is in the first rocket the moment he reaches the planet, he reads the Earth clock reading 3.96 years. But the instant he is in the second rocket, when he looks at the Earth clock, it reads 198.04 years!

Why does the change in the frame (suppose that no time is needed to change his frame) alter the reading of the Earth clock so suddenly? And how?

  • $\begingroup$ "suppose that no time is needed to change his frame".. right.. infinite acceleration. And you wonder why the clock changes so suddenly. $\endgroup$ Sep 18, 2021 at 1:36

4 Answers 4


It is a consequence of the relativity of simultaneity. When two observers move relative to each other, their time axes diverge and their respective planes of simultaneity become tilted relative to each other, the degree of tilt depending upon their relative speed of motion. When your imagined traveller hops from the outbound rocket to the inbound, her plane of simultaneity instantly switches from being tilted one way with respect to Earth's, to being tilted another way- it is that switch which causes the sudden change in the Earth's time coordinate in her frame of reference.

Indeed, you don't actually need observers- you can in a more abstract way simply consider three reference frames, each tilted relative to the others, and you will see that the time coordinate of a given event will change depending upon which reference frame you use to identify it.


In response to comments- the phenomena of SR reflect the interrelationships between space and time as expressed with reference to coordinate systems that are moving relative to each other. The phenomena do not depend upon observers and clocks- we mention observers and clocks to help us visualise the concepts.

The phenomenon of the twin paradox does not require travelling twins, and specifically it does not require acceleration.

Consider two events A and B which occur at different times at the same place in a given frame FX, and a third event C which occurs elsewhere at some time between the times of A and B in the frame FX.

In the frame FX, the time interval between events A and B is simply the sum of the time interval between events A and C and the time interval between the events C and B.

Consider the same events in the coordinates from two other frames FY and FZ, one of which is moving to the right of FX and the other moving to the left.

You will now find that the time interval between events A and B as measured in FX is greater than the sum of the time interval between A and C when measured in FY plus the time interval between C and B when measured in FZ.

In other words, the twin paradox effect is purely the result of transforming the intervals between events in reference frames moving relative to each other.


In other words, when he is in the first rocket the moment he reaches the planet, he reads the Earth clock reading 3.96 years. But the instant he is in the second rocket, when he looks at the Earth clock, it reads 198.04 years!

No, just before he turns around, he sees it reading about 2 years, and just after he turns around, he still sees it reading about 2 years.

The only way he can see Earth's clock is by light that travels from it to him. During the outward trip, that light is redshifted by a factor of $\sqrt{\frac{1+v/c}{1-v/c}} \approx 10$ (see Relativistic Doppler effect at Wikipedia). His own clock ticks off 20 years in that time, and he sees the Earth clock run at a tenth that speed, so it ticks off 2 years.

On the way back, the light from Earth is blueshifted by the same factor. His own clock ticks off another 20 years, and he sees the Earth clock run at ten times that speed, so it ticks off 200 years, for a total of 202.

If he could see Earth's clock "instantaneously in his own rest frame", then he would see what you describe, but that makes no sense. He would also see the Earth clock running backwards if he accelerated in the other direction. If Earth could see him instantaneously too, he could watch Earth's future and write it down, and they would see his written description of it before it happened. All of those absurdities are avoided because the speed of light is the maximum possible speed. And your problem is avoided too.

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    $\begingroup$ "see" never means "see" in SR, it means: measured by a comoving lattice of synchronized clocks. $\endgroup$
    – JEB
    Sep 8, 2021 at 0:09

When I face west, Los Angeles is 3000 miles in front of me (in my frame, of course). When I turn my body 90 degrees and face north, Los Angeles is 3000 miles to my left (in my new frame). Do you want to say that Los Angeles moved several thousand miles in the instant that I changed frames?

When I am at the end of my outbound journey, the earth clock reads 3.96 right now (in my frame, of course). When I change my velocity by 180 degrees, the earth clock reads 3.96 about 980 years ago (in my new frame). Do you want to say that earth clocks jumped forward almost 1000 years in the instant that I changed frames?

Los Angeles is what it is, and does not care what frame I use to describe it. If I change frames (and turning around is changing frames in space), I will have a new and different way to describe the same old location in space of the same old Los Angeles. An earth clock reading 3.96 is what it is and does not care what frame I use to describe it. If I change frames (and switching from an outbound rocket to an inbound rocket is changing frames), I will have a new and different way to describe the position in spacetime of the same old reading on the same old earth clock.

Again: "The clock reads 3.96 right now" and "The clock was reading 3.96 980 years ago" are perfectly analogous to "Los Angeles is 3000 miles ahead" and "Los Angeles is 3000 miles to my left". Each of these four statements is true in some frames and not in others. If you combine a statement that is true only in Frame A with a statement that is true only in Frame B, you are of course going to appear to get nonsense (like "Los Angeles moved several thousand miles in an instant" or "those clocks jumped 980 years in an instant"). But really, there's nothing to worry about.

Original post ends here.

This addendum is in response to a comment from which I will quote, because comments have a way of disappearing:

I still have a doubt. If I am correct, the event E mentioned above refers to "Earth clock reads 3.96 in return ship frame". If so...
Los Angeles is a location in space. Frames on earth --- like lines of latitude/longitude, or the grid I impose when I face north and define "forward, back, left, right", or the grid I impose when I face northeast and do the same thing --- give different descriptions of that location (like "3000 miles to the left" or "2121 miles to the left and 2121 miles backward"). But the location is what it is. "Los Angeles in the latitude/longitude frame" is not a location; it is a description of a location. There is no such location as "Los Angeles in the latitude/longitude frame".


"Earth clock reads 3.96" is an event in spacetime. Frames on spacetime --- like the grids imposed by travelers on various spaceships --- give different descriptions of that event (by assigning it time and location coordinates). But the event is what it is. "Earth clock reads 3.96 in return ship frame" is not an event; it is a pair of coordinates that describe an event. There is no such event as "Earth clock reads 3.96 in return ship frame".

  • $\begingroup$ I had to do a double take, but that's what $\Lambda^{\mu}_{\ \ \nu}$ says. Luckily, I have a python script that does the 3-frame twin paradox, with all notable events recorded. $\endgroup$
    – JEB
    Sep 8, 2021 at 0:11
  • $\begingroup$ @Eden: Let $E$ be the event "Earth clock reads 3.96" and let $F$ be the event "traveler arrives at star and immediately switches ships". In earthbound coordinates, $E$ is at $(x=0,t=3.96)$ and $F$ is at $(x=99,t=101)$. So the events are separated by $(\Delta x=99, \Delta t=97.04)$. Once the traveler boards the return ship, his velocity is $v=-.9802$. To find the time difference between $E$ and $F$ in the traveler's frame, apply the Lorentz transformation $\Delta t' = (\Delta t - v \Delta x)/\sqrt{1-v^2}$ and you will get about 980. $\endgroup$
    – WillO
    Sep 8, 2021 at 2:10
  • $\begingroup$ @Eden: There's also a much more geometric (and in my opinion much more illuminating) way to see this, which I will try to write up and post tomorrow if I find time. Meanwhile, I feel like I already posted this once on a similar question years ago (with different numbers, but same idea). I will try to find that and post the link if I can. $\endgroup$
    – WillO
    Sep 8, 2021 at 2:12
  • $\begingroup$ @WillO Yeah I have just tried using Lorentz Transformation (the same way you wrote in the comment section) and I found the exact answer as you wrote. So I deleted my comment. At first, I used the formula of time dilation, but since the Earth is moving relative to the first rocket (not stationary), \Delta x does not equal zero. And I think that is the reason why the formula of time dilation cannot be used. Am I correct? $\endgroup$
    – Eden
    Sep 8, 2021 at 2:22
  • $\begingroup$ @Eden: Yes, exactly. $\endgroup$
    – WillO
    Sep 8, 2021 at 2:23

The key here is in the accelerations involved. You can't just say that the man simply departs the Earth at 0.9802c. If his frame of reference immediately before the point of departure is coincident with the Earth's frame of reference (or, more precisely, they are not moving relative to each other) then he will undergo a significant acceleration to reach 0.9802c relative to the Earth. Thus he will experience relativistic effects as a result of accelerating away from the Earth. Once he reaches the distant planet (or more precisely, before he reaches the planet) he will also have to accelerate to achieve a velocity of -0.9802c relative to the Earth to return, undergoing further relativistic effects. And then, approaching the Earth, he will have to accelerate once again to achieve a velocity of 0c relative to the Earth, undergoing further relativistic effects.

In your alternative, where you posit two inertial rockets you're still not considering the accelerations the man has to undergo to move between the frames of reference. Actually, it's impossible to move instantaneously between two frames of reference that are moving relative to each other - there must always be acceleration involved.

[EDIT] The answer provided by Marco Ocram is not really an answer. An observer cannot move from one inertial frame of reference to another instantaneously. Certainly an observer in one frame, knowing the relative velocities of the other frames, can calculate what an observer in any other frame would see, but that's different to what s/he would observe directly. You simply can't get around the fact that to change your frame of reference involves acceleration. 'Suppose[ing] that no time is needed to change his frame' is proposing a completely different theory of relativity, in which case it's whatever you want it to be and you'd have to test this new theory. It's certainly not Einstein's General Relativity.

[EDIT2] With regard to Marco Ocram's comment about two observers synchronsing clocks: You need to consider what each of the observers observes. Call them Observer-Out (for the one travelling away from the Earth) and Observer-In (for the one travelling towards the Earth). First, I assume by 'synchronise clocks' you mean observe the time displayed on each other's clock at a certain point when each, from their own frame, considers the other to be passing. Go back a bit and think about what Observer-In would observe when s/he observes Observer-Out being beside the Earth - the times on the Earth's clock and on Observer-Out's clock. (Note, from the Earth's frame of refence Observer-In would be 180 light years away when making that observation.) Now think about what what will happen to Observer-In's observations of each of those clocks and remember that Observer-In's velocity relative to the Earth is significantly different to his/her velocity relative to Observer-Out. Think about what that would mean as Observer-In and Observer-Out approached and passed each other at the hypothetical planet. Then think about the further observations Observer-In would make on Observer-Out's clock as they flew apart, with Observer-In continuing towards the Earth and Observer-Out continuing to fly 90 light years past the hypothetical planet.

Anyway, apart from all that, the original question was really about the Twin Paradox, which is about why the clock on a ship that travels away from and then back to the Earth shows a different time when it returns, and the resolution of that paradox is in the fact that the travelling clock is not always in an inertial frame - i.e. it undergoes accelerations at various points in the journey. And, most importantly, the travelling observer will not instantly see a change in the Earth's clock, but, rather, will see the rate of change in the Earth's clock change as s/he undergoes those various accelerations.

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    $\begingroup$ The thought experiment can be performed without any accelerations at the turning point, by simply having separate observers for each leg of the journey who synchronise clocks as they pass each other. $\endgroup$ Sep 8, 2021 at 7:35
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    $\begingroup$ Thanks, but accelerations are beside the point. The principles of SR don't even need observers or clocks- they apply at an abstract level to frames of reference. Simply put, if you have two events occurring at different times in one place in one frame of reference, then the time interval between them will be longer than the time interval between the same two events and an intermediate one elsewhere when expressed in the coordinate systems of two other frames moving relative to the first. $\endgroup$ Sep 9, 2021 at 21:48
  • $\begingroup$ The whole point about these fictitious scenarios is that they are thought experiments, designed to tease out the underlying principles, like questions about frictionless surfaces, or perfectly rigid shapes, or strings of zero mass. $\endgroup$ Sep 9, 2021 at 21:50
  • $\begingroup$ Clearly a real astronaut couldn't switch between frames in that way- indeed it is laughable to suppose a human could be accelerated to 0.98c using existing technologies, but that's entirely beside the point. $\endgroup$ Sep 9, 2021 at 21:51
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    $\begingroup$ Take note, there is no paradox between two inertial frames. Each frame will observe a time dilation effect in the other that is consistent with the equations of Special Relativity, and the effect will be the same. The paradox, supposedly, occurs when one observer returns to the other's frame of reference - i.e. one of the frames becomes non-inertial at some points. $\endgroup$ Sep 9, 2021 at 22:05

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