77
$\begingroup$

The paradox in the twin paradox is that the situation appears symmetrical so each twin should think the other has aged less, which is of course impossible.

There are a thousand explanations out there for why this doesn't happen, but they all end up saying something vague like it's because one twin is accelerating or you need general relativity to understand it.

Please will someone give a simple and definitive explanation for why both twins agree on which twin is younger when they meet for the second time?

$\endgroup$
64
+50
$\begingroup$

Introduction

This is the third (and last) in a series of posts explaining time dilation, and it is going to assume you’ve read the preceding posts What is time dilation really? and What is time, does it flow, and if so what defines its direction?. Much of what follows won’t make sense unless you’re familiar with the topics discussed in the previous two questions. This is also going to be the hardest of the three posts by quite some way, but it just isn’t possible to gain a real understanding of the twin paradox without exploring some hard ideas. You have been warned!

In what follows I’m going to assume I am the stationary twin i.e. I remain on Earth while you go zooming off on your return trip in your spaceship. Remember that when you see me or my it refers to the stationary twin and you and your refer to the accelerating twin.

So as not to keep you in suspense, I'm going to explain that the asymmetry arises because the geometry of spacetime looks different for the two twins. To calculate the elapsed time we need a function called the metric, and in the coordinate system of an accelerating observer the metric looks different from normal flat spacetime. When we take this into account both twins agree about their respective ages.

My version of events

In the question on time dilation I explained what we mean by time dilation and how we calculate it. In particular I showed this spacetime diagram:

Figure 1

Figure 1

This shows our two trajectories through spacetime using my rest coordinates i.e. the coordinates in which I remain stationary at the origin. In these coordinates I remain at $x=0$ and simply travel up the time axis from the starting point $A$ to the finishing point $B$ as shown by the black arrow. You go hurtling away along the $x$ axis from point $A$, then stop, reverse and scream back to meet me again at point $B$, as shown by the red arrows. So the red line shows your trajectory through spacetime as measured using my coordinates.

From the time dilation question we know that the elapsed time shown by a clock carried by an observer, $\Delta\tau$, is related to the length of the observer’s trajectory, $\Delta s$, by:

$$ \Delta s^2 = -c^2 \Delta\tau^2 $$

And we know that the length $\Delta s$ is calculated using a function called the metric. In flat spacetime this function is the Minkowski metric, and it tells that if you move a distance $\mathrm dx$ along the $x$ axis, $\mathrm dy$ along the $y$ axis and $\mathrm dz$ along the $z$ axis in a time $\mathrm dt$ then the total distance you have moved in spacetime is given by the Minkowski metric:

$$\mathrm ds^2 = -c^2\mathrm dt^2 + \mathrm dx^2 +\mathrm dy^2 +\mathrm dz^2 $$

Since it’s hard to draw 4D graphs it’s usual to assume all motion is on the $x$ axis, so $\mathrm dy =\mathrm dz = 0$, in which case the metric simplifies to:

$$\mathrm ds^2 = -c^2\mathrm d\tau^2 = -c^2\mathrm dt^2 + \mathrm dx^2 \tag{1} $$

To calculate the length of the red curve we use the cunning trick of noting that velocity is defined by $v = \mathrm dx/\mathrm dt$ so $\mathrm dx = v\,\mathrm dt$, and if we take equation (1) and substitute for $\mathrm dx$ we end up with:

$$ \mathrm d\tau = \sqrt{1 - \frac{v^2(t)}{c^2}}\,\mathrm dt $$

So the elapsed time $\tau_{AB}$ is given by the integral:

$$ \tau_{AB} = \int_{t_A}^{t_B} \, \sqrt{1 - \frac{v^2(t)}{c^2}} \,\mathrm dt \tag{2} $$

where $v(t)$ is your velocity as a function of time. The exact form of $v(t)$ will depend on how you choose to accelerate, but since $v^2$ is always positive that means the term inside the square root is always less than or equal to one:

$$ 1 - \frac{v^2(t)}{c^2} \le 1 $$

And therefore the integral from $t_A$ to $t_B$ must be less than or equal to $t_B-t_A$. This means your elapsed time $\tau_{AB}$ must be less than my elapsed time $t_{AB}$ i.e. when we meet again you have aged less than I have.

So far so good, but the paradox is that we could draw the spacetime diagram in figure 1 using your coordinates, i.e. the coordinates in which you are at rest, to give something like:

Figure 2

Figure 2

In these coordinates you remain stationary so your trajectory shown by the red line goes straight up your time axis, while my trajectory shown by the black line heads off in the $-x$ direction before returning. If we use the same argument as above we would conclude that I should have aged less than you, but we can’t both have aged less than each other.

And that’s the paradox.

Your version of events

The resolution to the paradox turns out to be very simple. When I calculated the length of your trajectory in the previous section I used the Minkowski metric, equation (1), and after some algebra ended up with the equation for your path length in equation (2):

$$ \Delta t_\text{you} = \int_{t_A}^{t_B} \, \sqrt{1 - \frac{v^2(t)}{c^2}}\,\mathrm dt $$

The resolution to the paradox is simply that in your rest frame the metric is not the Minkowski metric, and therefore the equation you have to use to calculate my path length is not the same as equation (2):

$$ \Delta t_\text{me} \ne \int_{t’_A}^{t’_B} \, \sqrt{1 - \frac{v’^2(t)}{c^2}}\,\mathrm dt’ $$

and that’s why when you calculate my path length we both agree that my path length is longer than yours i.e. we both agree that I age more than you do.

So what is your metric?

The form of your metric will depend on exactly how you accelerate, and in general will not be a simple function. However there is a special case that is reasonably simple, and that is what I’m going to assume for the rest of this answer. I’ll assume that your acceleration (or rather deceleration) is constant, so your motion consists of the following:

  • at time zero you pass me with some positive velocity $v$ and constant deceleration $a$ - constant deceleration means you are accelerating towards me and in the opposite direction to your velocity

  • the constant deceleration eventually slows you to a stop at some distance $x$ away from me

  • you maintain the constant deceleration and now you start moving back towards me i.e. your velocity becomes negative

  • eventually you pass me again now with a velocity of $-v$

For motion at constant acceleration your metric is a function called the Rindler metric:

$$\mathrm ds^2 = -\left(1 + \frac{a\,x}{c^2} \right)^2 c^2\mathrm dt^2 +\mathrm dx^2 \tag{3} $$

For now I won’t attempt to justify this (I may do so in an appendix) I’ll just make a few comments on it before showing how to use it to calculate the trajectory length.

The Rindler metric doesn’t look completely different to the Minkowski metric that I used before. Indeed at the point $A$, where we part company, the value of $x$ is zero for both of us, and if we set $x=0$ the Rindler metric reduces to:

$$\mathrm ds^2 = -c^2\mathrm dt^2 +\mathrm dx^2 $$

which is just the Minkowski metric. Likewise if we take the acceleration $a$ to zero, the equation (3) just reduces to the Minkowski metric. However when $a \ne 0$ and $x \ne 0$ the two metrics are different, and the further $a$ and $x$ are from zero the more different the metrics become.

OK let’s attempt the calculation

Now we can calculate my elapsed time in your rest frame using the correct metric i.e. the Rindler metric. Let’s remind ourselves of the spacetime diagram:

Calculation

In your frame I pass you at time zero with a negative velocity, and I head off to negative $x$ before turning round to come back. What is perhaps not obvious is that the acceleration $a$ is negative. This is because $a$ is your acceleration. In the diagram above my acceleration relative to you is obviously positive so your acceleration relative to me must be negative.

We start as before by writing down the metric:

$$\mathrm ds^2 = -c^2\mathrm d\tau^2 = -\left(1 + \frac{a}{c^2}x \right)^2 c^2 \mathrm dt^2 +\mathrm dx^2 $$

And we use the same trick of substituting $\mathrm dx = v(x)\mathrm dt$. After rearranging we end up with:

$$ \Delta t_\text{me} = \int_{t_A}^{t_B} \, \sqrt{\left(1 + \frac{a\,x(t)}{c^2}\right)^2 - \frac{v^2(t)}{c^2}}\,\mathrm dt \tag{4} $$

This is actually pretty similar to the equation (2) that I used to calculate your elapsed time, apart from that extra term $a\,x(t)/c^2$. But it’s that extra term that makes the difference. To see why consider the leftmost point on my trajectory in figure 2. At this point my velocity is zero so the term in the square root becomes:

$$ 1 + \frac{a\,x(t)}{c^2} $$

But the product $a\,x(t)$ is positive, which means $1+ax(t)/c^2 \gt 1$ and therefore at this point $\mathrm d\tau \gt\mathrm dt$. Doing the integration in this region gives my elapsed time as greater than your elapsed time.

And this is the key to understanding the twin paradox. When you use equation (4) to calculate the length of my trajectory you’re going to find that my elapsed time is greater than your elapsed time, which is exactly what I found when I did the calculation in my frame. The resolution to the paradox is that the metric you use to do the calculation is not the same as the metric I use to do the calculation.

$\endgroup$
  • $\begingroup$ Also, notice that in the formulae above, when calculating the elapsed time for the accelerated observer, the integral must be taken upon the path $\gamma$ in the space-time (not on the real axis), hence the argument being smaller (bigger, respectively) than 1 doesn't necessarily mean that $$\int_{t_A}^{t_B} f(t;v) dt \leq \int_{t_A}^{t_B} dt = t_B -t_A$$ (as the last equality should instead by the length of the arc path $\gamma$ between $A-B$ which is in general not the difference of the extrema). $\endgroup$ – gented Nov 11 '16 at 22:20
  • 3
    $\begingroup$ This post was nominated for the best self-answer of 2016. It's an excellent post. I'd note, however, that it's probably worth saying, right at the outside, that the resolution of the paradox is that only one observer is in an inertial frame. It would help to say that before getting into the nitty-gritty of proper time integrals and metrics, IMHO. I'm all for mathematically complete and detailed answers on this site, but I'm also just suggesting that a little up-front intuition is good too. $\endgroup$ – DanielSank Jan 19 '17 at 18:59
  • $\begingroup$ But doesn't this break isotropic symmetry? If the twin went in the other direction and came back, according to this argument, in his reference frame, Minowski metric would apply. But then the whole argument would be reversed. Can anyone please point out what I'm missing? $\endgroup$ – Kalpak Gupta Jan 20 '17 at 6:38
  • 1
    $\begingroup$ @KalpakGupta One of the twins has to accelerate to reverse his velocity, otherwise they would never meet. Acceleration is not relative, both twins will agree on who accelerated. Thus, symmetry is broken and the apparent paradox resolved. $\endgroup$ – Jannik Pitt Nov 25 '17 at 13:12
  • $\begingroup$ A great answer! I just want to say that in the beginning of this last semester I have tried to study SR from Naber's book, but I couldn't understand these things that you have explained. Thanks a lot sir! $\endgroup$ – onurcanbektas Feb 6 '18 at 4:14
12
$\begingroup$

The paradox is about twins, one of which goes on a journey to outer space where he is observed by the Earth twin to be moving with great speed. Finally, the traveler twin returns to meet his twin on Earth. According to special relativity, when the twins meet, the traveler twin has aged much less than the Earth twin.

The paradox is supposedly in that both of the two twins see the other one undergo the same kind of fast travel. If both see the other one move similarly, how is it possible that the twin that got off the Earth and got back aged much more?

Answer: equations of special relativity refer to quantities and coordinates measured in inertial frames only. Consequently the calculation of proper time elapsed on the moving watch corresponding to measured coordinate time can be done only if this coordinate time was measured in an inertial frame.

Only the Earth twin can be in an inertial frame all the time, while the traveler twin can't, as he needs to accelerate and decelerate to get back to the Earth.

So there is actually no paradox - although the two observations of speed are the same, other things are not and this destroys the symmetry of the twins. The traveler twin has no way to use the time dilation formula because he has no coordinate time measurements that would qualify.

Given the trip took $T$ seconds of the Earth time, time dilation formula implies that total proper time that the traveler experienced on his trip is ($v$ is speed of the traveler):

$$ \int_0^T \sqrt{1-\frac{v^2}{c^2}}\,\mathrm dt $$ which is lower or equal to $T$.

Morale of the story: if you don't want to age quickly, get yourself moving.

$\endgroup$
  • $\begingroup$ When could it be equal to T? $\endgroup$ – Árpád Szendrei Nov 10 '16 at 20:46
  • $\begingroup$ When velocity is zero for all times. $\endgroup$ – Ján Lalinský Nov 12 '16 at 18:27
  • $\begingroup$ thank you, does this formula work when v is changing throughout the travel (so not just at the returnpoint, but let's say constantly accelerating)? $\endgroup$ – Árpád Szendrei Nov 12 '16 at 22:41
  • $\begingroup$ Yes, in theory the integral gives total time elapsed on the moving clock for any possible motion. $\endgroup$ – Ján Lalinský Nov 14 '16 at 0:48
8
$\begingroup$

Appendix - why the Rindler metric?

After reading my answer you could be forgiven for feel a bit cheated because it all depends on my claim that the accelerating twin observes a spacetime described by the Rindler metric not the Minkowski metric and I did kind of pull this out of the air.

It isn’t hard to prove that the accelerating twin’s metric has the Rindler form, but unless you are a fan of algebra the proof isn’t very exciting (if you are a fan of algebra see this article or Gravitation chapter 6). What I’m going to do instead is demonstrate a reason why the accelerating twin’s metric cannot be Minkowski, and in the process hopefully illustrate just how fascinating special relativity can be.

Suppose you try to outrun a light beam by travelling at a constant velocity $v$, where obviously your velocity has to be less than the speed of light. We’ll give you a head start by allowing you to start at $x = d$ while the light has to start at the origin. The spacetime diagram for the race looks like this:

Minkowski race

Figure 1

Note that in this diagram the $y$ axis shows $ct$ not just time. I’ve done this because for a beam of light $x = ct$ and therefore on my diagram the trajectory of a beam of light is a line at 45°. Anything travelling slower than light follows a line at an angle $\theta$ greater than 45°.

Hopefully it’s obvious that you cannot outrun the light and it will always catch you eventually. The faster you go the nearer the angle $\theta$ of your line gets to 45° degrees but since you can never reach $c$ the angle $\theta$ must always be more than 45° and therefore the world line of the light ray must eventually cross yours. And this makes sense. In your inertial frame light travels at $c$ so it doesn’t matter what distance $d$ the light ray starts, it will always reach you at a time $t = d/c$ so the light will always catch you.

But now the fun starts. We’ll do the race again but this time you start at rest and accelerate with a constant acceleration $a$. As before we’ll give you a head start and this time we’ll start you at $x = c^2/a$. You’ll see why I chose this starting point in a moment.

Both of the articles I linked above give the equation for your trajectory in my rest frame. If we start you at $x(0) = c^2/a$ the equation for your world line is:

$$ x = \frac{c^2}{a}\sqrt{1 + \left(\frac{at}{c}\right)^2} $$

This time the spacetime diagram looks like this (this is a real calculation for a constant acceleration of 9.81 m/s$^2$):

Rindler race

Figure 2

As before the light travels along a straight line at 45° but this time your line is a curve because of course you are accelerating not travelling at constant velocity. What’s more, your world line is a curve that tends asymptotically to the line $x = ct$ so your world line and the world line of the light ray never meet.

Hang on, let’s take a step back for a moment, your world line and the world line of the light ray never meet so you can outrun a ray of light.

An observer accelerating with constant acceleration $a$ can outrun any ray of light starting any distance greater than $c^2/a$ behind them

And that means in your coordinates there is an event horizon at a distance $x = c^2/a$ behind you. Your spacetime geometry contains an event horizon just like a black hole does, and this fact alone shows that your spacetime cannot be described by the Minkowski metric.

If we go back to the Rindler metric we can show how the event horizon arises. The Rindler metric is:

$$ ds^2 = -\left( 1 + \frac{ax}{c^2}\right)^2c^2dt^2 + dx^2 $$

A light ray follows a null geodesic that has $ds^2=0$, and if we set $ds^2=0$ and rearrange the equation above we can get an expression for the velocity of light:

$$ \frac{dx}{dt} = c\left(1 + \frac{ax}{c^2}\right) $$

So in your rest frame not only is the velocity of light not constant but it goes to zero at $x = -c^2/a$. That’s why there is an event horizon there.

Proof that you can outrun light

just for completeness let's prove you can outrun the ray of light. The equation for your trajectory is:

$$ x = \frac{c^2}{a}\sqrt{1 + \left(\frac{at}{c}\right)^2} $$

We take a factor of $at/c$ out of the square root to get:

$$\begin{align} x &= \frac{c^2}{a}\frac{at}{c}\sqrt{1 + \left(\frac{c}{at}\right)^2} \\ &= ct\sqrt{1 + \left(\frac{c}{at}\right)^2} \end{align}$$

For large times $t \gg c/a$ we have $c/at \ll 1$ so we can approximate this using the binomial theorem:

$$ x \approx ct\left(1 + \tfrac{1}{2}\left(\frac{c}{at}\right)^2\right) $$

The trajectory of the light is given by $x_\text{light}=ct$ so the distance between you and the light beam is:

$$\begin{align} x - x_\text{light} &\approx ct\left(1 + \tfrac{1}{2}\left(\frac{c}{at}\right)^2\right) - ct \\ &\approx \tfrac{1}{2}\left(\frac{c}{at}\right)^2 \end{align}$$

And this shows $x - x_\text{light} \gt 0$ is always greater than zero i.e. the light can never catch you.

$\endgroup$
6
$\begingroup$

I would like to add something on the simpler version known as the Symmetric Twin Paradox without acceleration, which has been mentioned in a comment and always pops up sooner or later anyways.

Although it is constantly pointed out that the "inertial" setup is not physically consistent, for reasons of acceleration/deceleration, non-Minkovsky metric etc, beginners are often left with a nagging sense that it should be working as a toy model, hence something is missing etc, and the discussion continues ad infinitum. I think there is a useful lesson to be gained in making explicit the source of inconsistency, and some additional insight on the main question, so I'd rather add this here.

For convenience of notation, I restate the problem as follows:

Let twins A and B move at the same speed $v/c = \beta$ in opposite directions relative to inertial observer O: A in the negative direction of O's $x$-axis, B in the positive direction. Both synchronize their clocks to O when they pass by O's origin at $x = 0$, such that for this event $ct = ct_A = ct_B = 0$. Subsequently A and B continue on their respective ways towards planets $P_A$ and $P_B$, both at rest wrt to O, at locations $x(P_A) = - x_0$, $x(P_B) = x_0$. As soon as they reach their planets, both twins jump ship into new inertial frames A' and B' moving at the same speed $\beta$ wrt O, but in opposite directions. That is, A' moves now in the positive direction of O's $x$-axis, while B' moves in the negative direction. At the moment of the transfer, A' synchronizes its clock to A, such that $ct_{A'} = ct_A$, while B' synchronizes with B, such that $ct_{B'} = ct_B$. When A' and B' pass again by the origin of O, everybody compares clocks. The question is, as before, if A' and B' report the same time when they meet again at O, how is this compatible with the assertion that A/A' and B/B' must see each other undergoing time dilation and conversely?

The answer lies in a serious discontinuity brought in by the transition from frames A and B to frames A' and B'.

One way to see this is to note that although coordinate transforms between O, A, and B are immediate, given their standard Einstein synchronization, other frame pairs no longer synchronize in this usual way and the corresponding transforms are slightly different. For instance, we know that A' synchronizes with A when both pass by $P_A$, but neither clocks time at $t=0$ for this event. So how do we write the coordinate transformation between A and A'? Simple: we just account for the coordinate shifts between the respective origins by means of Poincaré transformations:

Take frame O of coordinates $(x, ct)$ and frame O' of coordinates $(x', ct')$ moving at relative velocity $\beta$. If an event observed by O' at coordinates $(x'_0, ct'_0)$ is observed by O at coordinates $(x_0, ct_0)$, then the (Poincaré) transformations between O and O' read simply $$ x' - x'_0 = \gamma\left[\left(x-x_0\right) - \beta\left(ct - ct_0\right) \right]\\ ct' - ct'_0 = \gamma\left[\left(ct-ct_0\right) - \beta\left(x - x_0\right) \right] $$ and $$ x - x_0 = \gamma\left[\left(x'-x'_0\right) + \beta\left(ct' - ct'_0\right) \right]\\ ct - ct_0 = \gamma\left[\left(ct'-ct'_0\right) + \beta\left(x' - x'_0\right) \right] $$ for $\gamma = 1/\sqrt{1-\beta^2}$ as usual. All we have to do now is identify the correct relative reference coordinates and velocities of the various frames.

It is easy to show that at the final rendezvous O-s clock shows $ct = 2ct_0 = 2x_0/\beta$, while the clocks of A' and B' both show $ct_{A'} = ct_{B'} = 2ct_0/\gamma$, as expected from symmetry and time dilation in O's view. But let us consider for example the way A and A' observe B and B':

  • Between the "departure" at O and the "turning" at $P_A$, A observes B moving in the positive x direction at velocity (see relativistic addition of velocities) $\bar\beta = \frac{2\beta}{1+\beta^2}$, and the corresponding Lorentz transforms are $$ x_A = {\bar\gamma}(x_B + {\bar\beta} ct_B) \;\;\; ct_A = {\bar\gamma}(ct_B + {\bar\beta} x_B)\\ x_B = {\bar\gamma}(x_A - {\bar \beta} ct_A) \;\;\; ct_B = {\bar\gamma}(ct_A - {\bar \beta} x_A) $$
    with $\bar\gamma = \frac{1}{\sqrt{1-\bar\beta^2}} = \frac{1+\beta^2}{1-\beta^2}$. When A reaches $P_A$ at $ct_A = ct_0/\gamma$, it observes B at location $x_A(B) = \bar\beta\; ct_A = \bar\beta\; ct_0/\gamma$. However, according to the 2nd of the above transforms, at this point B-s clock only shows time $ct_B = \frac{1}{\bar\gamma}\frac{ct_0}{\gamma} < \frac{ct_0}{\gamma}$. This is the expected time dilation as observed by A, which is good. But this also shows that when arriving at $P_A$, A observes that B has not yet reached $P_B$.

  • Between the "turning" at $P_A$ and the final rendezvous at O, A' observes B' moving in the negative x direction at relative velocity $\bar\beta=-\frac{2\beta}{1+\beta^2}$. Since we know that the 2nd rendezvous takes place at coordinates $(x_{A'} = 0, ct_{A'} = 2\frac{ct_0}{\gamma})$ for A' and, likewise, at $(x_{B'} = 0, ct_{B'} = 2\frac{ct_0}{\gamma})$ for B', the coordinate transformations between A' and B' read $$ x_{B'} ={\bar\gamma}\left[x_{A'} + {\bar \beta} \left(ct_{A'} - 2 \frac{ct_0}{\gamma}\right)\right], \;\;\;ct_{B'}-2 \frac{ct_0}{\gamma} = {\bar\gamma}\left[\left(ct_{A'}-2 \frac{ct_0}{\gamma}\right) + {\bar \beta} x_{A'}\right]\\ x_{A'} = {\bar\gamma}\left[x_{B'} - {\bar \beta} \left(ct_{B'}-2 \frac{ct_0}{\gamma}\right)\right],\;\;\; ct_{A'} - 2 \frac{ct_0}{\gamma} = {\bar\gamma}\left[\left(ct_{B'}- 2\frac{ct_0}{\gamma}\right) - {\bar \beta} x_{B'}\right] $$ But now we can check that when A' "takes over" at $P_A$ for $ct_{A'} = \frac{ct_0}{\gamma}$, it observes B', $x_{B'} = 0$, at location $x_{A'}(B') = {\bar\beta}\;\frac{ct_0}{\gamma}$, and sees B'-s clock showing time $ct_{B'} = \left(2 - \frac{1}{\bar\gamma}\right)\frac{ct_0}{\gamma}$. Since for B' the time difference to the final rendezvous is $2\frac{ct_0}{\gamma} - \left(2 - \frac{1}{\bar\gamma}\right)\frac{ct_0}{\gamma} = \frac{1}{\bar\gamma}\frac{ct_0}{\gamma} $, this is consistent with time dilation as observed by A'. But at the same time we find that $ct_{B'} = \left(2 - \frac{1}{\bar\gamma}\right)\frac{ct_0}{\gamma} \ge \frac{ct_0}{\gamma}$, which means that according to A' as it passes $P_A$, B' has already passed by $P_B$!! Conversely, due to the time jump introduced this way, the time A' observes on B'-s clock at the final rendezvous is $\left(2 - \frac{1}{\bar\gamma}\right)\frac{ct_0}{\gamma} + \frac{1}{\bar\gamma}\frac{ct_0}{\gamma} = 2\frac{ct_0}{\gamma}$, consistent with both its own clock and the point of view of O. Needless to say, the same result obtains for A and A' as seen from B an B', and even for O as seen from either A/A' or B/B'. So we are left with the following conclusion:

A/A' does observe B/B' undergoing time dilation, and conversely, as expected. But, the fortuitous switch of inertial frames to simulate a $180^o$ reversal of velocity introduces an artificial discontinuity in worldviews and time measures for the frames involved. For example, at the turning point $P_A$, A claims that B has not yet turned at $P_B$, while A' claims that B' passed past $P_B$ long ago. Neither A nor A' ever witness the actual turning of B/B' at $P_B$ (it can also be verified that A observes B' before reaching $P_B$, while A' observes B past $P_B$). Technically, it is this discontinuity of worldview that allows twins A/A', B/B' to display the same time at the final rendezvous at O, despite mutual time dilation.

As an exercise, check that a similar discontinuity is responsible for the discrepancy between the clocks of O and A'/B' at the final rendezvous, although O does appear to undergo time dilation as observed from both A/A' and B/B'.

$\endgroup$
1
$\begingroup$

Wikipedia: Eventually, Lord Halsbury and others removed any acceleration by introducing the "three-brother" approach. The traveling twin transfers his clock reading to a third one, traveling in the opposite direction.

Sad to say, but the best explanation I've ever seen, gives preferred frame, or I would better say - a frame of mutual co-operation.

$\endgroup$
  • $\begingroup$ You would have a real impossible situation.... Look, Can you describe the other frame of reference? you depart from a static twin2 and after some time, twin3 starts with high velocity your direction you with clock set at x=0. There is your dilatation, his dilatation, velocity summing... $\endgroup$ – jaromrax Feb 20 '17 at 15:11
  • 1
    $\begingroup$ The trouble with transferring readings between clocks like this is that it makes the situation non-symmetric in an obvious way in the statement of the scenario, so it is no longer paradoxical that the outcome is also non-symmetric. $\endgroup$ – Andrew Steane Nov 18 '18 at 15:28
1
$\begingroup$

It seems to me that the existing answers contain far more information than the average novice is looking for when s/he asks about the twin paradox.

So with that novice in mind:

Alice stays home on earth. Bob flies from earth to Betelgeuse, and then from Betelgeuse to earth. The rookie mistake is to think that there are two relevant frames here --- Alice's and Bob's. Instead there are three --- Alice's, Outbound Bob's and Inbound Bob's.

In Alice's frame, Bob leaves earth at noon, arrives at Betelgeuse at (say) 3:00, having aged two hours, and arrives back home at 6:00, having aged another two hours. Bob's clock, obviously, runs slow by a factor of 2/3.

In Outbound Bob's frame, Bob leaves earth at noon, and arrives at Betelgeuse at 2:00, just as Alice's clock is striking 1:20. Alice's clock, obviously, runs slow by a factor of 2/3.

In Inbound Bob's frame, Bob leaves Betelgeuse at 2:00, just as Alice's clock is striking 4:40, and arrives home at 4:00, just as Alice's clock strikes 6:00. Alice's clock, obviously, runs slow by a factor of 2/3.

The thing many beginners overlook is that Outbound Bob and Inbound Bob disagree about what Alice's clock says at 2:00.

$\endgroup$
  • 1
    $\begingroup$ I think it's worth mentioning that at 3g it takes about 3 years to reach 0.99 c and at 5g it takes about 2 years to reach 0.99 c. Source: madsci.org/posts/archives/2000-05/959189602.Ph.r.html -- Also you probably want to slow prior to 'turning on a dime' and heading the opposite direction, to avoid being crushed. Traveling at 1g in a large circular route (like a rollercoaster loop) is convenient since your craft will have normal gravity (the vector of forward and the centrifugal force). You also need to calculate your route carefully or look where you're going. $\endgroup$ – Rob Dec 22 '17 at 16:05
0
$\begingroup$

There is a mental picture I've always used for this. There are two people, the traveler and the couch-potato. As the traveler is moving away at constant velocity, he sees the couch-potato's clock advancing more slowly. Then suppose there is an event where the traveler's velocity reverses, like bouncing off a big trampoline in space.

In that instant, the couch-potato's clock is seen by the traveler to suddenly jump forward. That's what acceleration does - it makes the clocks of unaccelerated objects appear to run faster, because the lines of simultaneity are changing direction. Then, during the return trip, the couch-potato's clock appears to run more slowly than the traveler's clock, but not enough to make up for the turnaround in the center. So at the end of the trip, the couch-potato's clock is ahead (the couch-potato is older).

$\endgroup$
  • $\begingroup$ On the topic of easy thoughts that grow intuitions, it's worth asking, "why exactly is it a paradox if I think my twin's clock is running slow and he thinks mine is, too?" I don't think the "which one is older?" objection really gets to the heart of it: instead I want to say "well, let me just call him up on my phone and one of us will be speaking in slow motion! Who is it?" And the great thought here is, "how is your phone transmitting information...?" Presumably by light: aha! Instantaneous information transfer could answer that question, but the speed limit of relativity prohibits that. $\endgroup$ – CR Drost Nov 17 '16 at 21:04
  • $\begingroup$ Final paragraph, couch-potato's clock: better to say not "is seen by" but "is observed by", and then add a sentence saying "Here 'observed' means what is deduced from the evidence and from the nature of distance and time measurements." $\endgroup$ – Andrew Steane Nov 18 '18 at 15:33
0
$\begingroup$

$\let\D=\Delta$ I only recently became aware of this post. An important contribution to an ever-ready question. Incidentally, it always makes me wonder to see how hard is this matter, how many questions and answers it still raises after over a century. At least in part, this shows IMHO that there should be much to be thought about the teaching of relativity, apparently worldwide. But let me leave general reflections and come to the real motivation of my post.

There are some points, not of minor importance, which could be improved in John' post. I'll try to give some contribution.


1) I read a phrase

"the geometry of spacetime looks different for the two twins.

I would like to mark a distinction between geometry and metric, or better between metric and the form it takes in different coordinates. I'm afraid it may be taken as a puristic criticism, but I believe that the lack of a clear distinction may harm understanding and also be a cause (one of several) to make the subject difficult.

Geometry and metric - an element of the whole structure we call geometry - are intrinsic properties of spacetime (let alone more general mathematical structures). They can and should be defined independently of reference frames and coordinates. As an example: euclidean geometry implies a special metric, and both can be defined without introducing cartesian coordinates.

For our subject this becomes relevant as an easy misunderstanding should be avoided: that geometry (and metric) of spacetime be different if it is viewed from an inertial frame or from an accelerated one. To be more precise: spacetime of SR is flat (Minkowskian) and such remains even though we reason and measure in an accelerated frame. To be sure, John never says anything different, but not even says that explicitly. There is a concrete risk that a not well-educated reader may draw a wrong conclusion.

To be honest, I'm not sure I myself never incurred in the same flaw. It's rather common among physicists, who are generally accustomed not to pay much attention to issues of language accuracy. It's usually understandable as a shortening of discourse. But in critical points it can have unwanted consequences.


2) Another general point. Once you have shown that what we are doing is just measuring lengths of two distinct paths, there is nothing more to be proven. Spacetime length is invariant (John is very clear on that) and we are free to choose the most easy way to measure it. In the present case an inertial frame, with its "natural" coordinates $(x,t)$, is the right one.

Nothing forbids, of course, to choose another way, i.e. a different (accelerated) frame. As an exercise it's very useful. Unfortunately John can only mention how it may happen that the relevant integral, in spite of its intimidating look, will give the same result. He can't go further, as he didn't give - with good reason! - $x(t)$ and $v(t)$ of the standing observer as seen from the accelerated one... Unfortunately this weakens the argument. The reader has to take John's word. Sure, he's not writing a book. But this leaves a question open about relativity teaching, not to be dealt with here.


3) Now for a technical (but highly physical) issue. John in his Appendix explains Rindler's "metric". All OK but...

We'll do the race again but this time you start at rest and accelerate with a constant acceleration a.

Unless I have missed the point John never says what a constant acceleration is (we know that it's the proper acceleration). To be sure, he writes

It isn't hard to prove that the accelerating twin's metric has the Rindler form, but unless you are a fan of algebra the proof isn't very exciting

This is not a matter of more or less algebra. There is an important piece of physics involved and I would have appreciated a deeper approach. To understand proper acceleration is far from trivial and requires thinking about basic aspects of SR.

Another point. John shows figures where Rindler's coordinates are named $(x,t)$. He doesn't write the transformation equations from or to $(x,t)$ of an inertial frame and this is understandable, as those equations involve exponential (hyperbolic) functions and logarithms. But to use equal names is objectionable. Not only because they are in any case different beasts, but above all because their physical meaning is different. More precisely: whereas Rindler's $x$ and inertial $x$ both measure proper lengths of standing objects (standing in respective frames) this true for the $t$'s.

Not a word about that. Rindler's $t$ isn't the time a clock in the accelerated frame marks. Equal $\D t$ at different $x$ don't correspond to equal proper times of standing clocks (this is written in the metric). It has a far-reaching consequence: a sort of "gravitational redshift" in the accelerated frame. Clocks in different positions can't be synchronized with each other.


4) About time dilation. I have an objection to John's post [1]. In a first part he defines time dilation as the difference in length between two paths sharing their extremes (figs 1,2). But later (fig. 4) he changes his mind and time dilation becomes the difference between AB and AC. IMHO the latter is the right meaning of the term "time dilation". It refers to the difference of time interval between two given events when one measures it in the frame where both events happen in the same place (proper time $\D t_0$) and when the measurement is done in any other frame (generic time interval $\D t$). It's widely known that in these conditions we find $$\D t = \gamma\,\D t_0.$$ The former effect - different lengths (proper times) of paths with common extremes - is related to the twin paradox and also more general thereof. Actually time dilation may also be redrawn as a case of path difference (please excuse me if I don't show how). But the link is rather artificial and I find it better to keep the distinction.

[1] https://physics.stackexchange.com/posts/241773

$\endgroup$
-6
$\begingroup$

What is the proper way to explain the twin paradox?

By explaining that time is merely a measure of local motion, that time dilation is a reduced rate of local motion, and that being separated by relative motion can be likened to being separated by distance.

The paradox in the twin paradox is that the situation appears symmetrical so each twin should think the other has aged less, which is of course impossible.

When we're separated by distance, we are subject to perspective, such that you look smaller than me and I look smaller than you. But we don't claim that this is impossible We do not cry paradox. In similar vein when we pass each other at some relativistic speed, I say that the light in your light clock appears to be moving zigzag fashion like this /\/\/\/\/\ at a slower rate than my own, which appears to be moving straight up and down like this ||. However you say that the light in my light clock appears to be moving like zigzag fashion like this /\/\/\/\/\ at a slower rate than yours, which appears to be moving straight up and down like this ||. There is no paradox. It's just a form of perspective, caused by the fact that we are separated by relative motion rather than distance. See the simple inference of time dilation on Wikipedia where you can find these images by user Mdd4696:

enter image description here

There are a thousand explanations out there for why this doesn't happen, but they all end up saying something vague like it's because one twin is accelerating or you need general relativity to understand it.

No, there are correct explanations for this, which succinctly and clearly describe why there is no paradox at all. The Wikipedia explanation isn't wrong, but it's so long-winded it rather bores the reader to death.

Please will someone give a simple and definitive explanation for why both twins agree on which twin is younger when they meet for the second time?

No problem. But that isn't the paradox. The paradox is that two twins in relative motion each claim that the other's clock is running slower than his own. As I said, this is merely the "perspective" associated with relative motion.

Anyway, when one of the twins slows down and turn round and comes back to Earth, both twins concur that he was the one who was moving, that he was the one who accelerated and decelerated, and that his light "really" moved like this /\/\/\/\/\ in his light clock, which ran at a lower rate than that of the stay-at-home twin. Due to the wave nature of matter the same effect occurred in the body and brain of the travelling twin, who didn't notice this time dilation locally, and who has fewer grey hairs as a result. He suffered time dilation because the rate of local motion is of necessity reduced by the macroscopic motion through space, because the maximum rate of motion is c. Not because "the geometry of spacetime looks different for the two twins". To calculate the elapsed time we do not need a function called the metric, all we need is Pythagoras's theorem, wherein the hypotenuse is the light path, and the base is the speed as a fraction of c, and the height is the Lorentz factor:

$$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$

It really is rather simple. Just focus on the motion and the light-paths, and you can't go wrong.

$\endgroup$
  • 1
    $\begingroup$ "...Not because "the geometry of spacetime looks different for the two twins..." well, except that it does in fact. Accelerated reference frames cannot be reduced to Minkowski at all points (i. e. globally), hence the geometry is indeed different. As a consequence you cannot use the above formula, rather you have to integrate the appropriate equation of motion (of either observer) over the space-time path $\gamma$ (of either observer). $\endgroup$ – gented Nov 11 '16 at 22:27

protected by Qmechanic Mar 8 '16 at 12:51

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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