I was presented with the below problem in my Physics course.

Triplet A stays on the earth. Triplet B travels at speed $4 c / 5$ to a planet (a distance $L$ away from earth) and back. Triplet $\mathrm{C}$ travels out to the planet at speed $3 c / 4$, and then returns at the necessary speed to arrive back exactly when B does. [Triplet A,B,C: are of same age by birth.]

(a) How much does each triplet age during this process?

(b) Who is youngest?

The issue I am facing is that I am expected to solve this problem using Lorentz transformations from Special Relativity but wouldn't the spaceships undergo acceleration in their round trip when turning back and Special Relativity is only valid in inertial frames of reference.

I only have knowledge upto Lorentz transformation, I have referred to the Supplementary Topic B from Robert Resnick's Introduction to Special Relativity but it involves explanation using concepts I am not familiar with. Further research also revealed that the Twin paradox is not exactly a paradox by definition as it cannot be assumed that the relative motion of twins is symmetric because of the acceleration of the twin in the spaceship.

My doubt is whether the problem above as it is valid and can be explained using time dilation as explained by Lorentz transformation? If possible which frame of reference would be preferred while referring to the aging of twins and why?

  • $\begingroup$ The tiebreaker is due to acceleration, but the effect of time dilation is an SR effect. You can isolate the effect of the acceleration and show that, in a well-defined limit of almost negligible contribution from the acceleration region, the SR prediction of time dilation is correct. $\endgroup$ Oct 2 at 7:14
  • $\begingroup$ If we were to assume that the time dilation due to SR were close to the actual time dilation which frame of reference do we use? What I meant was from the frame of the twin in spaceship the twin on earth slows done but the same is reversed in the frame of reference of twin on earth. $\endgroup$
    – Klegzart
    Oct 2 at 7:29
  • 1
    $\begingroup$ There is only one frame of reference whereby we have one twin that is always having inertial motion. The other frames of reference we have acceleration, and that breaks the symmetry, i.e. the tiebreaker. We can only take the inertial frame as the basis to compare with. $\endgroup$ Oct 2 at 7:37
  • 1
    $\begingroup$ The standard simplifying assumption is that the velocity changes are instantaneous. $\endgroup$
    – WillO
    Oct 2 at 11:04

1 Answer 1


At the bottomline the question is related with the hyperbolic geometry of the 4-dimensional Minkowski space. But we go there step by step.

In the whole post the speed of light $c=1$.

Actually, in order to answer well the question one has to be familiar with the invariant line element in special relativity. I wonder if Lorentz transformations could be of real help. The invariant line element in special relativity --- i.e. in Minkowski space is (actually to simplify things we just assume that there is only one dimension in space, respectively choose that the coordinate system where the planet to be reached has only a non-zero x-coordinate (and of course a t-coordinate), but the y and z-coordinate are zero. Then we can neglect y and z-coordinates completely):

$$ ds^2 = dt^2 - dx^2$$

$ds$ is the invariant distance measure in (simplified) Minkowski space, analogue to the distance measure in Euclidean space: $ds^2 = dx^2 + dy^2$. This measure does not depend on the coordinate system which is chosen, and this is true for the line element in Minkowski space as well as that of the Euclidean space. In Minkowski space coordinate system changes in particular comprise Lorentz transformations, or even more concretely the line element does not change if it is measured from a moving system or from a system at rest.

However, the integrated line element $ \int ds $--- because we will consider finite distances and time ranges, not only infinitesimally small ones -- depends on the integation path taken.

In the following we will compare the integrated line element of each of the triplets and see what it will tell us.

We will choose a coordinate system as attached to the post.

Starting with triplet A:

He does not move, therefore along its "journey" along the time-axis $dx$=0. Therefore we find:

$$S_A=\int ds_A = \int dt =: T$$

This was the easiest case. Let's come to triplet B. We will use the same coordinate system as for triplet A --- our choice. The journey 1) to the planet + 2) return (from the planet) parts. Actually in both parts triplet B travels at $v=\pm 4/5 c \equiv \pm 4/5$ in our convention. Actually we do not care much about the sharp returning maneuver. But we take into account that the integration path for triplet B is different from triplet A (see graph attached):


$$S^1_B = \int ds_B = \int \sqrt{ dt^2 -dx^2} = \int dt \sqrt{1 -(\frac{dx}{dt})^2 } = \int dt \sqrt{1 -v^2 } $$


$$S^1_B = \int dt \sqrt{1 -(4/5)^2 } = \frac{T}{2} \cdot \frac{3}{5}$$

  1. Actually for the return trip it is exactly the same except that the sign of velocity changes sign which does not matter because the square of velocity enters in the formula for the line element. Therefore

$$S^2_B = \frac{T}{2} \cdot \frac{3}{5} \longrightarrow\text{(1)+(2):}\quad S_B =S^1_B + S^2_B = T\frac{3}{5} = 0.6T$$

For completeness I give also the result for triplet C. He only travels at $v =3/4c \equiv 3/4$. So he arrives at the planet much later (in the time coordinate of triplet A) than triplet B. As I've understood the text, he then returns "as fast as possible" to Earth in order to catch up fully with triplet B. We simply assume (other assumptions are possible which might yield another result, but for the pedagogy of the case it does not really matter): He travels on return with almost speed of light, i.e. $v \approx 1$ in order to make the catch up.

1.) We also assume that the time coordinate when he reaches the planet almost is $T$, i.e. almost the time coordinate when triplet B reaches Earth again.

$$S^1_C = \int dt \sqrt{1 -v^2} = \int dt \sqrt{1 -(3/4)^2} = T\frac{\sqrt{7}}{{4}}$$


$$S^2_C = \int dt \sqrt{1 -v^2} = \int dt \sqrt{1 -1^2} = 0$$

Therefore $$S_C = S^1_C + S^2_C =T\frac{\sqrt{7}}{{4}} \approx 0.66 T$$

Still a couple of comment have to be made: It actually turns out that the (integrated) line elements of triplet B and C are smaller than the one of triplet A although according to the graph attached to the post they look much longer than line element of A.

This is due to the hyperbolic geometry of the Minkowski space. The little "-" sign in the line element formula makes the difference.

enter image description here

Shlomo Sternberg formulated that in its book on "Curvature in Mathematics and Physics" (p.237) as: "The triangle unequality does not hold in Minkowski space".

However, the graph of course is a simplication based on Euclidean geometry.

Important Question:

What happens when we change the coordinate system chosen at the beginning. The answer is amazingly simple: As the line elements computed here are invariant, i.e. invariant upon coordinate transformation, in particular independent on whether the observer is a resting or moving frame, they do not change. It does not matter whether the coordinate system change is curvilinear (related with accelerations) or not.

The result is the same. This means: Actually there is no twin paradox !!

The variable $t$ is just a coordinate, it does not tell us of time evolved. The eigen time is given by the line element.


Not the answer you're looking for? Browse other questions tagged or ask your own question.