The simplest and most intuitive way I have found so far for explaining which twin ages less in the Twin Paradox, is that it's the twin who's world-line is the longest (if it's the longest in one inertial reference frame, it's the longest in all inertial reference frames).

Now what I'm looking for is a more formal, mathematical way to express the comparative length of any 2 world lines, preferably in terms of variables that could be easily measured by each twin respectively. Variables like proper time, proper acceleration (or comparative velocity), ... and allowing both twins to travel freely in space between their point of separation and their point of reunion.

The solution I'm looking for would look something like this: The twin who ages less is the twin for whom the following mathematical formula [insert solution here] is smaller (or larger) than the same mathematical formula for his twin.

Intuitively I feel like the formula would have to be something like: [sum of each world-line segment's proper time multiplied with the same segment's speed relative to x] or [sum of the distance travelled in each world-line segment, distance as expressed by y] but I haven't been able to find a precise expression for this formula yet.

One solution that might work is: The twin who ages less is the twin who travelled a larger distance relative to an (imaginary) object traveling at a constant speed from their point of separation to their point of reunion. This distance could be calculated by each twin based on their relative speed to this imaginary object and the proper time they traveled at this speed. But: Can we do without the reference to this imaginary object and it's inertial reference frame ? Using only what can actually be observed by the twins ?

PS: here's an image to go along with my question. Intuitively I can "see" that the 2-segment journey is "longer" in euclidean space, so it must be "shorter" in terms of its proper time.

enter image description here

EDIT: Lots of great solutions in the answers below! My issue with the standard solution is it requires agreeing on a common shared coordinate system upon which to draw and measure the length of the world lines. If possible, I would like to find a criterion (for who ages less) based simply on what the twins observe between the moment of separation and the moment of reunion. What can they observe ? Their relative speed at separation, their changes of speed before reunion, and the proper times that passed for each (constant-speed) leg of their travels.

  • $\begingroup$ Search terms: "spacetime interval," "proper time." $\endgroup$
    – rob
    Sep 2, 2022 at 13:37
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    $\begingroup$ This is covered in detail in What is the proper way to explain the twin paradox? $\endgroup$ Sep 2, 2022 at 16:33
  • $\begingroup$ Response to your edit: You are asking who will be younger when they return to earth. The only thing that matters for that is how long their round-trip worldlines are. The length of a world-line is independent of coordinates. So each twin, can measure the length of his own worldline and his twin's worldline, using any coordinate system he cares to, and this determines who is younger at the end. $\endgroup$
    – WillO
    Sep 3, 2022 at 12:19
  • $\begingroup$ You already named the quantity you are looking for. It is the proper time. You can calculate the proper time for any time-like path in flat or curved spacetime, which is equal to the time, that a twin experiences along that path through spacetime. For the formulas, have a look at wikipedia. $\endgroup$ Sep 5, 2022 at 7:58

4 Answers 4


@Chris. In response to your update,

find a criterion (for who ages less) based simply on what the twins observe between the moment of separation and the moment of reunion. What can they observe ?

Each observer should count the receptions of birthday transmissions from the other observer, starting from the separation-event until the reunion-event. (If you wish, each transmission could be encoded with the birthday-number of the celebrant.)

When they reunite, each observer can reconcile what they celebrated-and-transmitted and what they received from the other. The result is Lorentz invariant.

Using a variation of your spacetime diagram (to allow nicer numbers to appear in the calculation), we have these spacetime diagrams. I didn't include the inertial observer as a celebrant or receiver, but you can sketch in the details for that case.

RRGP-robphy-count-A RRGP-robphy-count-B


My issue with the standard solution is it requires agreeing on a common shared coordinate system upon which to draw and measure the length of the world lines

The resulting count for each receiver are the same in any frame.
Here are the results from another frame.
(If you click on an image, then edit its url to remove to "m", you'll get the full quality image.)

RRGP-robphy-count-A2 RRGP-robphy-count-B2


original answer begins here

Inspired by your spacetime diagram, I show the following visualizations, which are related to the formula given by @Dale when applied to piecewise-inertial worldlines.

From my answer to Equivalence of two definitions of proper time in special relativity


This spacetime diagram has the spacetime-paths of light-signals in standard-issue longitudinal light-clocks. On a (1+1)D-spacetime diagram, the area enclosed by "one tick" of an inertial light-clock is the same, independent of the velocity of the light-clock and independent of the inertial observer drawing the spacetime diagram. (Lorentz invariance preserves the area (since the determinant of the boost equals 1) and the directions-of-the-diamond-edges (since the lightlike directions of the diamonds are eigenvectors of the boost).)

By drawing the spacetime on "rotated graph paper", it is easy to construct equal-area light-clock diamonds for velocities with rational Doppler-factors $k=\sqrt{\frac{1+v}{1-v}}$ (like $v=(3/5)c$, $v=(4/5)c$, and $v=(5/13)c$ but not $v=(1/2)c$ or $v=0.99c$), which lead to triangles with a set of sides that are Pythagorean triples and thus most are calculations are fractions.

Alternatively, one can draw the causal-diamond of an inertial worldline segment.
This method works well for velocities that are rational... so that the diamond-areas can be computed by counting. From my poster https://www.aapt.org/docdirectory/meetingpresentations/WM18/FG07-Salgado-RelativityRotatedGraphPaper-CalculatingWithCausalDiamonds.pdf


You can play with this idea on my GeoGebra visualization
"Visualizing Segments with Light-Clock Diamonds (robphy)" https://www.geogebra.org/m/HYD7hB9v#material/VrQgQq9R
from my "Relativity on Rotated Graph Paper (robphy) - MAA2016" https://www.geogebra.org/m/HYD7hB9v#

All of this is based on my article
“Relativity on Rotated Graph Paper”,
Am. J. Phys. 84, 344 (2016);

Based on your diagram, we have
RRGP-robphy-example and RRGP-robphy-example-boost

The diagram on the right uses the clock-diamonds of an observer moving with velocity $v=(3/5)c$ (so $k=2$). Observe that the number of clock-diamonds in a causal diamond is unchanged, as expected since the area in terms of the number of clock diamonds is the invariant square interval of the timelike diagonal of the causal diamond.

By counting the number of light-clock diamonds in the causal diamonds shown, we have:

  • proper-time along ABCD is $AB+BC+CD=\sqrt{15}+\sqrt{12}+\sqrt{15}=11.2100683076$
    where $v_{AB}=(-1/4)c$, $v_{BC}=(2/4)c$, and $v_{CD}=(-1/4)c$
  • proper-time along APD is $AP+PD=\sqrt{20}+\sqrt{20}=8.94427191$,
    where $v_{AP}=(4/6)c$ and $v_{PD}=(-4/6)c$.
  • proper-time along AD is $AD=\sqrt{144}=12$ ; AD is an inertial worldline.
    for any intermediate event M on the line segment AD, we have $AD=AM+MD$.
    By contrast, $AD \neq AP+PD$ since P is not on the line segment AD; APD is a non-inertial worldline.

The twin who ages less is the twin for whom the following mathematical formula: the sum of "the square-root of the areas of the causal diamonds along piecewise-inertial segments" (i.e. the proper-time elapsed along that piecewise-inertial worldline) is smaller than the analogous quantity for his twin.

The area of a diamond in light-cone coordinates is $UV$, where $U=t+x$ and $V=t-x$.

  • Using the left diagram, Diamond AP has $U=10$ and $V=2$. Note that $t=(U+V)/2=6$ and $x=(U-V)/2=4$. The area is $UV=20$: $$(10)(2)=(t+x)(t-x)=t^2-x^2=(6^2-4^2)=36-16=20$$
  • Using the right diagram, Diamond AP has $U'=5$ and $V'=4$. Note that $t'=(U'+V')/2=(9/5)$ and $x=(U-V)/2=(1/5)$. The area is $U'V'=20$: $$(5)(4)=(t'+x')(t'-x')=t'^2-x'^2=(\left(\frac{9}{5}\right)^2-\left(\frac{1}{5}\right)^2)=\frac{81}{25}-\frac{1}{25}=20$$
  • a general diamond has area $$UV=(\Delta t)^2-(\Delta x)^2=\left(1 - \left( \frac{\Delta x}{\Delta t} \right)^2 \right) (\Delta t)^2$$ so that its square-root (the proper-time) is $$\sqrt{UV}=\sqrt{\left(1 - \left( \frac{\Delta x}{\Delta t} \right)^2 \right) (\Delta t)^2} = \sqrt{1 - \left( \frac{\Delta x}{\Delta t} \right)^2 }\Delta t $$ in agreement with @Dale's formula.

Using, for example, desmos.com (by copying the $\TeX$-format below from "Show Math as" to desmos),
with your worldline ABCD
$t_{AB}=\int_{0}^{4}\sqrt{1-\left(\frac{-1}{4}\right)^{2}}dt $ $\qquad=\int_{0}^{4}\sqrt{\frac{15}{16}} dt = \sqrt{\frac{15}{16}}(4-0)= 3.87298334621 = \sqrt{15}$
$t_{BC}=\int_{4}^{8}\sqrt{1-\left(-\frac{2}{4}\right)^{2}}dt $ $\qquad=\int_{4}^{8}\sqrt{\frac{12}{16}} dt= 3.46410161514 = \sqrt{12} $
$t_{CD}=\int_{8}^{12}\sqrt{1-\left(\frac{1}{4}\right)^{2}}dt $ $\qquad= 3.87298334621 = \sqrt{15} $
as we got from counting light-clock diamonds.

  • 1
    $\begingroup$ Wow, this is nice. I can only do one +1, otherwise I would do another. I like the big causal diamonds better than counting lots of small ones. I haven't seen that before $\endgroup$
    – Dale
    Sep 2, 2022 at 15:30
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    $\begingroup$ @Dale Thanks. After my method (for hand drawings and counting) was characterized as being limited by my restriction to rational Doppler factors, I looked at my method and saw that I can weaken it to rational velocities as long as one is happy with the abstraction of calculating the square root of an integer and not need the clock-diamonds (of its ticking light-clock) drawn along the diagonal of the causal diamond. $\endgroup$
    – robphy
    Sep 2, 2022 at 17:01

The twin who ages less is the twin for whom the following mathematical formula $$ \tau = \int \sqrt{1-\frac{v(t)^2}{c^2}}dt$$ is smaller than the same mathematical formula for his twin.

In this formula $v(t)$ is the velocity of the twin in question. This formula only works for an inertial frame in flat spacetime (no significant gravity), but the twin can accelerate or move arbitrarily.

My issue with the standard solution is it requires agreeing on a common shared coordinate system upon which to draw and measure the length of the world lines

This is not correct. There does not need to be any agreement on a common shared coordinate system. The above formula will work in any inertial frame. Further, if some twin wishes to use a non-inertial frame then they can use this equation: $$\tau=\int \sqrt{-g_{\mu \nu} dx^\mu dx^\nu}$$

  • $\begingroup$ Since relativity is self consistent, when working in Minkowski space, there is no apparent paradox. So you are not solving the twin paradox, your erasing it. $\endgroup$
    – JEB
    Sep 2, 2022 at 3:54
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    $\begingroup$ @JEB of course. As with most "paradoxes" the twin paradox is not paradoxical. When the twins are reunited, all observers agree on which twin is the younger $\endgroup$
    – Tristan
    Sep 2, 2022 at 10:28
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    $\begingroup$ Perhaps it's worth mentioning that this is the formula for the proper time (interval) so that anyone can look into it further if desired. $\endgroup$
    – Fato39
    Sep 2, 2022 at 11:05

If it’s a piecewise straight world line, for each segment of it add the “proper time”

$$ \Delta \tau = c^{-1} \sqrt{c^2 \Delta t^2 - \Delta r^2}. $$

If it’s a smooth curve, the formula above is still valid for infinitesimal intervals and the sum becomes an integral.

The twin for which the resulting sum / integral is larger will age more. The value of the sum / integral gives you the time experienced by the twin.

The value of the sum / integral is a spacetime invariant, it doesn’t depend on the reference frame. It’s values computed for different frames will all coincide.

  • $\begingroup$ What is "r" in this formula ? $\endgroup$
    – Chris
    Sep 2, 2022 at 16:02
  • $\begingroup$ I like the idea of a formula that is reference frame independent! $\endgroup$
    – Chris
    Sep 2, 2022 at 16:10
  • $\begingroup$ @Chris $\Delta r$ is the spatial displacement. If a twin traveled 1 light year in 10 years (as measured on Earth), $\Delta r$ will be 1 light year and $\Delta t$ will be 10 years; and $\Delta \tau$ will be the time experienced by the twin (slightly less than 10 years). $\endgroup$ Sep 3, 2022 at 1:04
  • $\begingroup$ @Dale gives an equivalent formula in their answer. The equivalence is easy to see, because $\Delta r / \Delta t = v$ by definition (and also their answer is written in terms of an integral rather than a sum). $\endgroup$ Sep 3, 2022 at 1:05

Imagine twins who do fast space travel in opposite directions. Whatever one of them does, the other does the same in the opposite direction. Finally they are reunited at the spot they left, both at the same position with the same velocity. Which do you think will be older?

They will be the same age.

The one who is younger will be the one who has experienced more acceleration. When you accelerate your time slows down, and when you accelerate in the opposite direction your time slows down the same amount.

So make your graph show not positions, or velocities but accelerations independent of direction.

I'm an amateur at this and I could be wrong.

Oops! Your original problem calls for instantaneous velocity changes, which in the real world is impossible. When you change instantaneously from one direction and speed to a different direction and speed, you subject the twin not to 1g of acceleration, or 100g of acceleration but infinite acceleration.

So I don't know how to calculate the acceleration in this problem.

The experiment can never be done to find out which twin ages faster. It is not physics. It would not be surprising if it results in paradoxes.

  • 1
    $\begingroup$ Thanks @JThomas ! I agree with everything you said. Now we just need to get it into a coherent mathematical formula ! $\endgroup$
    – Chris
    Sep 2, 2022 at 7:42
  • $\begingroup$ At least two people disagree, so .... $\endgroup$
    – J Thomas
    Sep 2, 2022 at 11:16
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    $\begingroup$ The infinite acceleration might be impractical but it presents absolutely no theoretical obstruction. Just integrate piecewise as in the answer from @prof.legoslav. The business about acceleration affecting clocks is completely wrong. $\endgroup$
    – WillO
    Sep 2, 2022 at 12:17
  • $\begingroup$ @JThomas the problem can be simplified by having clocks fly past each other and synchronizing their time when they do. Then instead of acceleration you have relative speed change. Both could be observed and measured by a local observer. So I would be happy with either approach for my question. $\endgroup$
    – Chris
    Sep 2, 2022 at 15:27
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    $\begingroup$ According to the clock postulate, the proper time only depends on the speed, not the acceleration. That is, the proper time is just the arc-length of the worldline, and it's unaffected by the curvature of the worldline. Please see math.ucr.edu/home/baez/physics/Relativity/SR/clock.html We also have numerous posts about the clock postulate on Physics.SE. $\endgroup$
    – PM 2Ring
    Sep 3, 2022 at 7:02

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