twinparadox

A Geometrical View of Time Dilation and the Twin Paradox

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Based on the amount of questions we receive on the topic at Physics Forums, there is a lot of confusion in the general public about how time dilation works and the resolution of the twin paradox. More often than not, the confusion is based upon misunderstandings about how time dilation works and an unawareness of the relativity of simultaneity and how time is treated within special relativity. The aim of this Insight is to make an attempt at explaining these issues from a geometrical viewpoint and to see that we have an analogous effect in the geometry which we are used to, although we do not find it strange.

A geometrical analogue

Consider the following two lines in normal two-dimensional space:

fig1

In order to describe the lines, we can introduce a coordinate system ##S##:

fig2

In this coordinate system, we may use the coordinate ##x## to parametrise the lines, we find that

$$y_1(x) = 0, \quad y_2(x) = kx,$$

where ##k## is some constant. The lengths of each line between ##x = 0## and ##x = x_0## are simply given by Pythagoras’ theorem ##\ell^2 = \Delta x^2 + \Delta y^2## and we find

$$\ell_1 = x_0, \quad \ell_2 = x_0\sqrt{1+k^2}.$$

Consequently we find that the ratio between these two lengths is ##\ell_2/\ell_1 = \sqrt{1+k^2}##. This is shown in the following figure:

fig3

However, there is nothing special about the coordinate system ##S##. In particular, the length of the lines between any two given points on the lines does not depend on the coordinate system chosen and we could just as well have chosen to introduce the coordinate system ##S’## according to:

fig4

In this new coordinate system, we can parametrise the lines using the ##x’## coordinate

$$y_1′(x’) = -kx’, \quad y_2′(x’) = 0.$$

For the distance between the points corresponding to ##x’ = 0## and ##x’=x_0’##, Pythagoras’ theorem now results in

$$\ell_1′ = x_0′ \sqrt{1+k^2}, \quad \ell_2′ = x_0’$$

and so the line ##\gamma_1## is longer than ##\gamma_2## for the same difference in the parameter ##x_0’##, unlike in the case when the lines were parametrised by ##x##, where we obtained the opposite result. Yet there is nothing strange going on here. If we select a point on ##\gamma_2## and draw the lines which have the same ##x## and ##x’## values as that point, it becomes clear that the points on ##\gamma_1## which share the same parameter value differs depending on whether we used ##x## or ##x’## for the parametrisation. We should therefore not be surprised to find that the length to these two different points is different. Let us call this effect the “relativity of the same ##x##-coordinate”. Naturally, this effect does not affect the actual length of any of the lines, but only what lengths are being compared. This is illustrated in the following figure:

fig5

The points ##B## and ##B’## have the same ##t## and ##t’## coordinate as ##A##, respectively, yet correspond to different points on the curve ##\gamma_1## and therefore the distance from the intersection along this curve is different.

Curves intersecting at two points

Now consider the following two curves:

fig6

Just from the figure, it should be clear that ##\gamma_2## is longer and we can verify this by introducing the coordinate system ##S## as above and computing the lengths of each of the curves, with the length of ##\gamma_2## being split into two equal contributions, one from each line segment.

fig7

We find that

$$\ell_1 = 2\ell, \quad \ell_2 = 2\ell \sqrt{1+k^2}.$$

We could also compute the length of the curves by looking at the ##S’## coordinate system:

fig8

By our previous discussion, the length of ##\gamma_2## up to the half-way point is given by

$$\ell_2’/2 = \ell’,$$

while the length of ##\gamma_1## to the same ##x’## coordinate is given by

$$\ell_1’/2 = \ell’ \sqrt{1+k^2}.$$

If we were not careful about the relativity of the same ##x##-coordinate, we might do the error of thinking that ##\ell_1’/2## would be half the length of the curve ##\gamma_1##. However, since we are aware of this, we know better than that and realise that if we multiplied the result by two, we would obtain a result which was too large.

Application to relativity

So how does this connect to relativity and time dilation? In its foundation, relativity is a theory about the geometry of space-time, which may be described using a coordinate system with four coordinates ##(t,x,y,z)##, where the first coordinate is the time coordinate and the last three are spatial coordinates. It is important to realise that, just as in the case we just described, these are just coordinates which, by themselves, do not have a direct physical interpretation. For the sake of brevity, let us consider only one spatial coordinate ##x## as the generalisation is straight-forward.

In relativistic space-time, geometry works a bit differently as compared to the geometry which we are used to working with. In particular, this manifests itself by a change in Pythagoras’ theorem, which now comes with a minus sign

$$c^2 \tau^2 = c^2 \Delta t^2 – \Delta x^2$$

as a direct consequence of the speed of light being equal in all frames. Here, ##\tau## is the proper time along a straight line in space-time with ##\Delta t## being the difference between the enpoint time-coordinates and ##\Delta x## being the difference between the endpoint space-coordinates. This also changes what kind of coordinate transformations we can do in order for the proper time to be the same regardless of the coordinates. Instead of rotating both coordinate axes in the same direction as happens for rotations, the coordinate axes are rotated by the same angle in opposite directions.

fig9

In relativity, the proper time defined above is the only time which will be measured by any observer. A curve in space-time is called a world line and will generally describe the position of the observer as a function of the time. The proper time of a world line, i.e., the “length” of the world line with the modified Pythagorean theorem, is the time a clock following that world line will measure, the time coordinate ##t## has no particular meaning other than that it happens to numerically be the same as the time measured by an observer for which the spatial coordinates are not changing. This is completely analogous with our previous example, where the value of the ##x##-coordinate was equal to the length of a line for which the ##y##-coordinate was not changing.

We can now do the exact same exercise as we did using regular geometry. Consider the following two world lines described in the coordinate system ##S##:

fig10

The first world line ##\gamma_1## has a constant ##x##-component while the second world line describes an observer travelling with velocity ##v##. The world lines may be parametrised as

$$x_1(t) = 0, \quad x_2(t) = vt.$$

We can now compute the proper time from the time coordinate taking values ##t=0## to ##t=t_0## for both world lines using the modified Pythagorean theorem and find that

$$\tau_1 = t_0, \quad \tau_2 = t_0\sqrt{1-\frac{v^2}{c^2}}$$

or, equivalently,

$$\frac{\tau_1}{\tau_2} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \equiv \gamma,$$

where ##\gamma## is the famous gamma-factor of special relativity. In other words, the proper time elapsed for an observer at rest between two time coordinates will differ from the proper time elapsed for a moving observer between the same time coordinates by a factor ##\gamma##. This is the essence of time-dilation but, just as for the example we started with, it crucially depends on the coordinate system.

Taking a different coordinate system ##S’## where ##\gamma_2## describes a world line with constant ##x’##

fig11

the exact same argumentation will now lead us to conclude

$$\frac{\tau_1′}{\tau_2′} = \sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}.$$

This is the exact opposite relation as that we found in the coordinate system ##S##, just as was the case in our first example! The resolution of this apparent mismatch is also completely analogous. We should not be surprised to find a different relation between the times elapsed between different ##t## coordinates and that elapsed between different ##t’## coordinates simply because the point on the curve ##\gamma_1## with the same ##t## coordinate as the point ##A## on ##\gamma_2## will not be the same as the point on ##\gamma_1## with the same ##t’## coordinate as ##A##:

fig12

This effect is called the “relativity of simultaneity” and simply states that two events which are simultaneous in one set of coordinates are generally not simultaneous in a different set of coordinates (simultaneous simply refers to having the same time-coordinate). As such, there is really nothing strange about the proper times to ##B## and ##C## along ##\gamma_1## being different, since they are simply not the same point.

The twin paradox

The twin paradox is the analogue of our second example with two curves starting and ending at the same points. In the case of two twins, where one leaves to travel to a faraway land and later returns, the world lines in the coordinate system ##S## would look like:

fig13

Just applying the modified Pythagorean theorem it becomes clear that the proper times along the world lines are given by

$$\tau_1 = 2\tau, \quad \tau_2 = 2\tau \sqrt{1-\frac{v^2}{c^2}} \quad \Longrightarrow \quad \frac{\tau_1}{\tau_2} = \gamma$$

and therefore more proper time will pass for the twin staying behind, i.e., the red curve. Going to the coordinate system ##S’##, we can compute the relation between the proper time ##\tau’## taken for the travelling twin to reach the turnaround point and the proper time elapsed for the staying twin during the same ##t’## interval as

$$\frac{\tau_1′}2 = \tau’ \sqrt{1-\frac{v^2}{c^2}}, \quad \frac{\tau_2′}2 = \tau’.$$

However, just as in our example above, the relativity of simultaneity tells us that ##\tau_1’## is not equal to half the proper time elapsed for the staying twin until the reunion:

fig14

Instead, the change of geometry indicates that we are not counting part of the world line if we make this assumption and therefore underestimate the elapsed proper time. If taken properly into accout, we would find that the time elapsed for the staying twin is longer also as computed by the travelling observer. This resolves the twin paradox and shows that it only appears due to not taking the relativity of simultaneity properly into account.

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Associate professor in theoretical astroparticle physics. He did his thesis on phenomenological neutrino physics and is currently also working with different aspects of dark matter as well as physics beyond the Standard Model. Currently teaching courses in mathematical methods for physicists, special relativity, and quantum field theory. A member at Physics Forums since 2014.
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  1. greswd
    greswd says:

    I think many people are interested in the travelling twin's perspective.This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin's frame.And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).

  2. Dale
    Dale says:

    Thanks Orodruin. This is definitely worth referencing.

    The geometrical view does help clear up a lot of the confusing aspects of SR, but it seems to be a big leap for new students. I wonder which is more difficult for most students: understanding the geometrical view or understanding the relativity of simultaneity.

  3. Orodruin
    Orodruin says:

    The geometrical view does help clear up a lot of the confusing aspects of SR, but it seems to be a big leap for new students.

    I find the geometrical comparison to Euclidean space helps students and it is how I start the presentation of the subject in my SR class, by introducing Lorentz transforms as the hyperbolic rotations leaving the Minkowski line element invariant. Admittedly, these are students who are in their fourth year of university and they may be more ready for it. I had the idea of writing this Insight three days ago when I was giving the corresponding lecture, although I have tried to make it slightly more accessible than the level I present it on in class.

  4. SlowThinker
    SlowThinker says:

    Perhaps you’re going a bit fast near the end.

    Just applying the modified Pythagorean theorem it becomes clear that the proper times along the world lines are given by
    $$tau_1=2tau, tau_2=2tausqrt{1-frac{v^2}{c^2}} implies frac{tau_1}{tau_2}=gamma$$
    and therefore less proper time will pass for the twin staying behind, i.e., the red curve.

    It seems the twin staying behind should have more proper time, since ##frac{tau_1}{tau_2}=gammageq1##.

  5. greswd
    greswd says:

    What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

    \""

    As described in this paper: [URL]http://arxiv.org/abs/gr-qc/0104077v2[/URL]

  6. greswd
    greswd says:

    What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

    \""

    As described in this paper: http://arxiv.org/pdf/gr-qc/0104077v2.pdf%5B/quote%5DI think many people are interested in the travelling twin’s perspective.

    This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin’s frame.

    And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).

  7. Orodruin
    Orodruin says:

    I think many people are interested in the travelling twin’s perspective.

    This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin’s frame.

    And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).

    This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

    Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.

  8. greswd
    greswd says:

    This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

    Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.

    By “same frame”, I mean same inertial reference frame. Both being at rest w.r.t. each other.

    I’m not sure what you mean by “special status”, I don’t think I mentioned that any frame had a special status.

    The “time compression” may be viewed as a form of gravitational blueshift. This and the “same frame” effect are described as being due to “shock like scale discontinuity”.

    Speaking of additional pedagogical value, I was just curious about the travelling twin’s perspective. I think many people have never seen such a diagram before too.

  9. Nugatory
    Nugatory says:

    This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time

    By “same frame”, I mean same inertial reference frame. Both being at rest w.r.t. each other.

    Whether the twins are at rest relative to one another has nothing to do with whether they are in the same frame. Both twins are always in all frames all the time.

  10. PAllen
    PAllen says:

    This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

    Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.

    Note that radar coordinates are exactly Minkowski coordinates if the defining world line is eternally inertial. They are asymptotically the same as Fermi-Normal coordinates near the defining world line. They have the specific feature that as long as the defining world line is inertial for all proper time before some event, and also inertial for all proper time after some event, no matter how complex the motion in between, and however ‘long’ such complex motion occurs, they give a complete coordinate chart of Minkowski space (unlike Fermi-Normal coordinates). Thus, they can be seen as a generalization of Minkowski coordinates to describe the ‘experience’ of a non-inertial observer, having several nice properties (including also they are built from more achievable measurements than positing very long rulers).

    All the same, one must never attach ‘more reality’ to a description in radar coordinates than in some other coordinates.

  11. Dale
    Dale says:

    What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

    As described in this paper: [URL]http://arxiv.org/abs/gr-qc/0104077v2[/URL]

    I like it. It is one of my favorites.

    That said, I wouldn’t worry too much about details like the part of the trip where the distance is constant. It is a non-inertial frame so distance and speed don’t have their ordinary meanings anyway. The point is mostly that the travelling twin’s “perspective” is not symmetrical to the home twin.

  12. Orodruin
    Orodruin says:

    I like it. It is one of my favorites.

    That said, I wouldn’t worry too much about details like the part of the trip where the distance is constant. It is a non-inertial frame so distance and speed don’t have their ordinary meanings anyway. The point is mostly that the travelling twin’s “perspective” is not symmetrical to the home twin.

    But exactly here lies the problem. I am not arguing that the coordinates are abad idea, I just question the pedagogical value to people who are just learning relativity. I do not see the point of introducing non-inertial coordinate systems on top of the struggles they already have.

  13. Dale
    Dale says:

    But exactly here lies the problem. I am not arguing that the coordinates are abad idea, I just question the pedagogical value to people who are just learning relativity. I do not see the point of introducing non-inertial coordinate systems on top of the struggles they already have.

    I agree. Pedagogically, for a novice I would encourage them to stick with inertial frames. Then if they persist use the paper in the spirit of “your question does have an answer, here it is, come back to it when you are ready”.

  14. the_emi_guy
    the_emi_guy says:

    Seems to me that SR cannot be applied to a problem that involves three inertial frames and only two
    observers. Isn’t an observer moving from one inertial frame to another is out of the scope of SR?
    Seems that any solution would require a footnote indicating that
    the presumably infinite acceleration experienced by the “travelling” twin is assumed to have no implication.

    On the other hand, allowing three observers, one in each frame, is a valid SR problem, and its solution is a trivial
    application of time dilation and removes all of the complexity of the two observer problem.

    Observers:
    Observer1 (Earthbound observer) is on earth.
    Observer2 (outbound observer) passes earth traveling toward Alpha Centauri (and never comes back)
    Observer3 (inbound observer) is traveling from Alpha Centauri and first passes outbound observer, then later passes earth.

    Events:
    Earth passes his time to outbound observer when they meet.
    Outbound observer passes his time to inbound observer when they meet.

    Analysis:
    Outbound observer’s point of view (exclusively): Earthbound clock is running slow (simple time dilation)
    thus inbound observer’s clock is set ahead of Earth’s clock when outbound and inbound meet.

    Now continuing along with outbound observer (*not* switching inertial frames). Earth’s clock is still running slow, but the
    inbound observer’s clock is running even slower since this relative velocity is larger.
    Thus, by the time that the outbound observer sees the inbound observer reach earth, he will see that the inbound observer’s clock will have gone from being ahead of Earth’s to being behind Earth’s by simple application of time dilation applied to this larger relative velocity.

    This can be worked using simple SR time dilation from any of the three inertial frames to obtain identical results.

  15. Orodruin
    Orodruin says:

    Seems to me that SR cannot be applied to a problem that involves three inertial frames and only two
    observers. Isn’t an observer moving from one inertial frame to another is out of the scope of SR?

    No. It is perfectly within the scope of SR. You do not need an actual observer to define an inertial frame in SR.

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