I'm asking this question to better understand how "abrupt" a frame of reference change is when a spaceship accelerates. Apologies if this is not the right forum, and for the unrealistic scenario.

Let's say Alice leaves Earth on a spaceship with an excellent camera (>50,000 fps), and starts going around it in an equilateral triangle inscribing the Solar System at 0.999999c, and comes back after 50 years. How much of Earth's history can she record?

I believe she would come back to Earth and see everything a little bit over 35,000 years older (0.999999c translates into Lorentz factor of 707.107, which times 50 years is ~35,355). But what percentage of these 35,355 years would she be able to actually record with her camera at 60 * 707.107 fps? Assuming the desired output when back on Earth is 60fps.

If I understand correctly, when she's not accelerating, Earth's clocks (and people) seem to move slower. It is only when she accelerates (when turning in the triangle) that Earth "abruptly" moves forward in her FoR. How abrupt is this, can she observe much of history continuously, or does she get only bits and pieces due to the "abrupt" nature of the change in frame of reference? What if she was going around in a circle instead?

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    $\begingroup$ This seems to be a different wording of the twin's paradox. Would an answer considering the twin's paradox scenario (where the traveling twin has the camera) suffice to answer your question? It is simpler to describe than the triangle but has the same essential physics. $\endgroup$ Commented Dec 28, 2022 at 19:25
  • $\begingroup$ @NíckolasAlves my current understanding is that in the twin's paradox case, the abrupt nature of acceleration very far away from Earth means that only a small percentage of history can be recorded. Similarly, in a triangle close to Earth, only bits a pieces can be recorded, although more than in the traveling twin scenario. And in a circle, everything can be recorded smoothly if the desired output is 60fps. If my understanding is incorrect, perhaps answering the twin paradox with camera question is sufficient. $\endgroup$
    – Bruno Fiss
    Commented Dec 28, 2022 at 19:28
  • $\begingroup$ There are quantum limits to how fast cameras can snap photos (the camera's photo would be processed by what is effectively a computer and you can't just have computers at 10^9999 gigahertz see: en.wikipedia.org/wiki/Bremermann%27s_limit ) so there is some weird science you can do here "a 1kg, object accelerating can only sample up to some threshold density" and depending on the acceleration pattern you could say things like "it is impossible to photograph events 1,2,3 unless the mass of the object is larger because of how close in time they are" $\endgroup$ Commented Dec 29, 2022 at 21:05

6 Answers 6


She could record all of it, in principle, with a slightly faster camera.

The relevant quantity here is the Doppler shift factor, not the time dilation factor.

The relativistic blueshift factor for a stationary emitter at the origin is $γ(1+β\cos θ)$, where $β=\frac{v}{c}$ and $θ$ is the angle between the receiver's velocity and a line between the emitter and receiver, which in your problem varies from 30° as Alice leaves a vertex to 150° as she arrives at one.

The camera therefore needs to support a variable frame rate in the range $γ(1 \pm β\frac{\sqrt3}{2})\cdot 60\text{ fps} \approx 5684\text{ – }79169\text{ fps}$. The frame rate will follow a sinusoidal sawtooth pattern, gradually falling from 79169 to 5684 along each side of the triangle and suddenly jumping back to 79169 at the vertices.

There is no time jump at the vertices. Nothing special happens during acceleration in special or general relativity. If you switch to a different coordinate system, the $t'$ in your new coordinate system will not match the $t$ in your old coordinate system, but that's because it's a different coordinate, not because something physically happened. The whole point of the principle of relativity is that you can use any coordinate system you want; you are not obliged to change coordinate systems when you accelerate. This problem, like the twin paradox, is most easily analyzed in the rest frame of the Earth, regardless of Alice's velocity, and there's no reason to use any other coordinate system.


Your triangular set-up is too complicated for a person as old and tired as I am to figure out without resorting to pencil and paper, so I will instead focus on the classic set-up of the twin paradox. The principles are exactly the same in both cases.

Suppose then that there is a huge clock on the Earth and the travelling twin has a special telescope that allows them to keep an eye on the clock throughout their journey. As the twin coasts away from Earth, the huge clock will appear to be running slow when the twin views it through the telescope. The apparent reduction will be the combined effect of time dilation and a classic Doppler red-shift. When the twin turns around, the clock will immediately appear to be running fast, the apparent reduction now being a combination of time dilation and a classic Doppler blue-shift. At no point will the time on the giant clock appear to the travelling twin to leap forward.

So, if the speeds and distances involved are such that 20 years pass on Earth during a round trip of 10 years by the travelling twin, then the twin will be able to record footage of all 20 years of Earth time, the first part of which will seem to be in slow motion and the rest (from the turn-around point) will be speeded up.

Of course, in reality the twin's view of the Earth would be screwed-up by the extent of the blue and red-shifting, by the fact that the Earth would be far too small to see for most of the trip, and by the fact that as the twin approach the Earth at near light speed, the change in the Earth's apparent size would probably be too fast for the eye to follow. Also, of course, accelerating to near the speed of light is utterly fanciful in any event.

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    $\begingroup$ Just a comment to tickle everyone's imagination: suppose there are two giant (synchronized) clocks near Earth, satellites that maintain their position on the Earth-twin line, one slightly in front of the Earth, one just behind the planet. Now that's the interesting scenario. For the traveling twin, they will show different times. And the twin will see somewhat different times of history (according to the Earh-local record) on the different (near and far) sides of the Earth :) $\endgroup$ Commented Dec 29, 2022 at 20:45
  • $\begingroup$ @FilipMilovanović, is there something interesting about this beyond the expected speed-of-light delay? $\endgroup$
    – Mark
    Commented Dec 30, 2022 at 23:53
  • $\begingroup$ @Mark: Yes, the space-faring twin doesn't see exactly the same history as the earthbound twin, even when compensating for any signal delay. It's because the simultaneity hyperplanes differ for each observer. As you go across the Earth, the points/events that are simultaneous for the astronaut become increasingly separated in time for the earthbound observer - going from one end of the Earth to the other along the line that connects the twins (parallel to the motion). But, if I'm not mistaken, the effect is tiny at the Earth-diameter scale. $\endgroup$ Commented Jan 6, 2023 at 23:47

I'll consider a similar problem in the setup of the twin's paradox, since it is easier to analyze and allows me to write the expressions explicitly.

Consider the traveling twin departs from Earth at a constant velocity $v$ (as measured in the Earth frame) and travels away from Earth for time $\tau_0$ (as measured in his watch). In the Earth frame, his trajectory for $\tau < \tau_0$ is $x(t) = v t$. For $\tau < \tau_0$, his proper time $\tau$ is related to time at Earth ($x = 0$) $t$ through $$\tau = \gamma (t - v x(t)) = \gamma(1 - v^2) t = \sqrt{1 - v^2} t.$$

At $\tau = \tau_0$, he is at $x = v t_0 = \gamma v \tau_0$ and starts his journey back. We can assume the acceleration happens infinitely fast, so that his velocity is simply discontinuous. Fun fact: the details of the acceleration are not so relevant. The most important fact about the paradox is just that the traveling twin is non-inertial. Hence, for $\tau > \tau_0$ his trajectory is given by $x_r(t) = v (2 t_0 - t)$. The subscript $r$ stands for "return"). Using this, we find that for $\tau > \tau_0$ the relation between $\tau$ and $t$ is $$\tau = \gamma(t + v x_r(t)) = 2 \gamma v^2 t_0 + \gamma (1 - v^2) t = 2 \gamma v^2 t_0 + \sqrt{1 - v^2} t.$$

Notice that in both cases $\tau$ is running slower than $t$, since $\sqrt{1 - v^2}$. However, this does not refer to what the traveling twin sees. It refers to how two different clocks attribute time to the same event.

Now to get to your question, let us find the relation between $\tau$ and the retarded time $u = t-x$. Notice that a light beam that leaves Earth in the direction of the traveling twin moves in a way that keeps $u$ constant, and hence this is a convenient variable to see how the traveling twin receives light from Earth. Furthermore, at the Earth, $u = t$. Hence, finding $u(\tau)$ means finding which instant on Earth the traveling twin is seeing in his camera. I should notice that I'm working with $u$ because I'm picking the traveling twin to be moving to the right (positive $v$). Otherwise, we'd want to work with $t+x$ instead (the light rays moving to the left do not keep $u$ constant, but keep $t+x$ constant).

Using our previous expressions for $x(t)$ and $x_r(t)$, we can see that for $\tau < \tau_0$ we have \begin{align} u &= t - x(t), \\ &= (1 - v) t. \end{align} Therefore, for $\tau < \tau_0$, $t = \frac{u}{1-v}$ and $$\tau = \frac{\sqrt{1 - v^2}}{1 - v} u,$$ implying $$u = \frac{1 - v}{\sqrt{1 - v^2}} \tau.$$

Similarly, for $\tau > \tau_0$ we get \begin{align} u &= t - x_r(t), \\ &= (1 + v) t - 2 v t_0. \end{align} This implies $t = \frac{u + 2 v t_0}{1+v}$ and hence, for $\tau > \tau_0$, $$\tau = \frac{\sqrt{1 - v^2}}{1+v} (u + 2 v t_0).$$ Solving for $u$, $$u = \frac{1+v}{\sqrt{1 - v^2}} \tau - 2 v t_0.$$

Notice that $t_0 = \gamma \tau_0$. Bringing everything together, we find that at proper time $\tau$ the traveling twin sees (literally) Earth at time $$u(\tau) = \begin{cases} \frac{1 - v}{\sqrt{1 - v^2}} \tau, \text{ if } \tau < \tau_0, \\ \frac{1+v}{\sqrt{1 - v^2}} \tau - \frac{2 v \tau_0}{\sqrt{1 - v^2}}, \text{ if } \tau > \tau_0. \end{cases}$$

A few comments are in place:

  1. This function is continuous at $\tau_0$. There is no "time skip" when the ship turns around.
  2. At $\tau = 2 \tau_0$, when the ship gets back to earth, one has $u(2 \tau_0) = 2 t_0$, which is precisely the time the Earth measures it takes for the traveling twin to come back.

Hence, let us get to your comments.

If I understand correctly, when she's not accelerating, Earth's clocks (and people) seem to move slower.

Turns out this is wrong! When the traveling twin is coming back home, he sees the Earth running. Think of it as in the Doppler effect: he'd see the peaks of a wave coming toward him passing by him faster. Notice we are this time taking into account the actual optical effect, not just changes of coordinates (which would be a Lorentz boost).

It is only when she accelerates (when turning in the triangle) that Earth "abruptly" moves forward in her FoR.

This also turns out to be a misconception! The twin paradox is often misunderstood as if it needed General Relativity due to acceleration of the traveling twin. However, the traveling twin needs not even to accelerate for a finite time. All of the analysis we carried out has used only Lorentz boosts.

How abrupt is this, can she observe much of history continuously, or does she get only bits and pieces due to the "abrupt" nature of the change in frame of reference? What if she was going around in a circle instead?

There is nothing really abrupt, except that when the traveling twin is coming back they see the Earth running faster. The twin gets to see the whole history of the Earth, without any cuts. While I worked out the simplest possible trajectory, acceleration wouldn't change anything. This can be seen with a spacetime diagram: draw the Earth at rest at $x=0$ and any causal trajectory for the traveling twin. You can then draw the light rays leaving Earth and see that all of them intersect the traveling twin's trajectory once, and only once.


Assuming she slightly rounds off those corners (to avoid infinite acceleration), Alice records 35,000 years of history in 50 years. Therefore if Bob (who stays home on earth) is recording that history at 60 frames per second, Alice is recording it at 42,000 frames per second on average, leading to at least some segments with very low video quality.

  • $\begingroup$ @xac : Thank you for the edit. $\endgroup$
    – WillO
    Commented Dec 29, 2022 at 13:17

how "abrupt" [is] a frame of reference change

The time-dilation effects caused by Alice's acceleration would modify her frame of reference instantly compared to the reference frame of an observer on the Earth. There is no reason to suspect it would delay or be quantized into discrete steps — certainly not on any scale that an ordinary camera could detect.

I suspect this intuition comes from observing cameras, which quantize their record of reality into discrete frames. They manage this even though there's nothing discrete about the ordinary passage of time. So, how do cameras generally account for this?

Each frame of the video is a "time lapse" of all the light that entered the camera during the time the shutter was open. All the light which touches the image sensor (or the film) during one frame gets added together as part of that frame.

And, if you record motion that happens faster than the camera's FPS, then the footage will be blurry; if it's significantly faster, it becomes what we traditionally think of as a time lapse.

So, if you want to know the minimum FPS needed (for any application) you need to know the fastest motion (from the camera's frame of reference) that you want to track, and set a reasonable FPS from there.

There are still many ways you would need to post-process the video in order to make it match anything resembling normal footage, as there are many other effects going on here which are tangential to your question.


I think that the heart of the OP question is: if time on Earth, under Alice's perspective, is always running slowly, how it is possible that her clock shows a shorter time when back home? It seems that something must happen during the acceleration period.

As shown in other answers, there is nothing special with that sharp points from the point of view of perception. A signal at each second send by the earth can be received with an interval greater or smaller than a second, depending on the Doppler effect and time dilation together.

But there is something that changes abruptly at the sharp point: it is the calculated time on the earth correspondent to each time on the ship. It is different from what is seen at the earth at the same moment, because that image was sent before, and the light had to travel until the ship. That is why it can only be calculated, not perceived.

One way to understand what happens is to suppose 2 ships, one receding and another approaching the earth with the same magnitude of velocity and crossing in a given point. If we take this moment as $x = x' = t = t'= 0$ for both ships, the time on earth is calculated as:$t_e = vx$ and $t_e = -vx'$ (taking c = 1). $x$, that is equal to $x'$ for $t = t' = 0$, is the distance to the earth at that moment, and $v$ is the relative velocity of the earth (positive for the receding ship and negative for the other). The ships don't agree about what is the time at the earth at that exact moment. Note that no experience can tell who is right, because the speed of light is finite, and nobody can see earth's clock at $t = t' = 0$. Any vision corresponds to earlier times.

When a ship suddenly changes velocity from receding to approaching, the calculated time on earth increases at once by $2vx$ because it is equivalent to change from a receding ship to an approaching one. It is like a slice of time was erased and the earth was suddenly in the future. That is why when arriving here, Alice is younger that Bob, even if during all trip (except at the sharp point) she calculates the corresponding Bob's time intervals as smaller than hers.


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