I have read this question and answer by John Rennie.

What is the proper way to explain the twin paradox?

Now he has an appendix

"Appendix - why the Rindler metric?"

where he states that for the accelerating twin we need to use Rindler metric and we cannot use the Minkowski metric because

"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."

"But now the fun starts. We’ll do the race again but this time you start at rest and accelerate with a constant acceleration aa."

"An observer accelerating with constant acceleration aa can outrun any ray of light starting any distance greater than c2/ac2/a behind them And that means in your coordinates there is an event horizon at a distance x=c2/ax=c2/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."

But I am not saying I disagree on it here, but I need to clarify something. He is right from the stationery observer's point of view. But I want to clarify that from the person's view who is on the spaceship, things look differently.

I need to state 3 things:

  1. John Rennie is only right in a mathematical case from the stationary observer's view where we have a theoretical spaceship that can accelerate forever.
  2. As per SR/GR we know that anything with rest mass, as it reaches near the speed of light, will not be able to physically accelerate any more, because it would need infinite energy to do so.
  3. From the view of the person on the spaceship at some FINITE time on his clock, he will then see that the spaceship is not able to accelerate anymore(because he would need infinite energy to do so), it's speed will stay constant less then c. After that, the light will catch up to him.
  4. So if this happens at some finite time on the clock of the person in the spaceship, then the stationery observer's clock will show some much bigger, but still finite time, as the spaceship, now from the stationery observer's view, will stop accelerating, and it's speed will stay constant less then c. After that the stationery observer will see light catch-up to the spaceship.

Because in theory in John Rennie's explanation of course you can outrun the light, assuming that you can accelerate forever. But in our world, since nothing can go faster then light (because after a while, near speed c the spaceship would need infinite energy to accelerate), at approx 61162111 seconds (on the person's clock on the spacecraft), he reaches almost the speed of light and stops accelerating, and the light will catch-up to him soon after. Because we know that anything with rest mass will not be able to accelerate any more as it reaches close to the speed c. So at that point from the view of the person on the spaceship, his speed will not increase anymore and the light will catch up to him.

So I would like someone to please explain to me why the accelerating twin would outrun light in our world (I proved he can't) and then why is it necessary to use the Rindler metric instead of the Minkowski metric?

My question is about what he person on the spacecraft will see at 61162111 seconds on his own clock (supposed the clocks start at 0).

And I am talking about the theoretical case when the spaceship would be able to accelerate forever.


  1. Calculating the clock difference with either the simple SR length contraction will give the same result as if we calculate using this more complicated GR acceleration Rindler metric?
  2. What will the person on the spaceship see on the ships speedometer, when his own clock shows around 61162111 seconds (supposed that his clock was set to 0 at the time of start)?
  3. According to SR we know that the person on the spaceship will see his own clock tick normally (as long as he does not compare it to the observer's clock). And he will see the accelometer showing the constant acceleration too. So at 61162111 seconds he must see his own spacecraft reaching speed c. At that point his clock must stop ticking. So time froze for him?
  4. Since he sees his own clock tick at a normal speed (as long as he does not compare it), is that time freeze sudden? At a certain point? According to SR he having rest mass cannot reach the speed c. So he must stop accelerating before that. Then the light will catch up to him?
  • $\begingroup$ The paragraph that begins with "Because in theory of course you can outrun the light." seems like absolute nonsense. In theory, you can't outrun light. This is because in reality, you can't outrun light, and theory is by construction a description of reality. $\endgroup$ – Arturo don Juan Nov 11 '16 at 21:10
  • $\begingroup$ I meant that I see John Rennie's point of view, but that assumes that the accelerating ship of the traveling twin can accelerate forever. It cannot. Because it would reach speed of light. But I will edit. $\endgroup$ – Árpád Szendrei Nov 11 '16 at 21:16
  • 1
    $\begingroup$ "It cannot. Because it would reach speed of light." - that's not true - the Rindler metric is for constant proper acceleration (the acceleration according to an accelerometer attached to the spacecraft). To an inertial observer, the coordinate acceleration of the spacecraft tends to zero as the spacecraft's speed tends to $c$. $\endgroup$ – Alfred Centauri Nov 11 '16 at 21:20
  • $\begingroup$ you are still wrong, because you say that the " To an inertial observer, the coordinate acceleration of the spacecraft tends to zero as the spacecraft's speed tends to c" I was talking about a stationary observer, just like when you look at the graphs. As an observer, you will see that the spacecraft is closing to speed c (because from your point of view it is), and the spacecraft will not be able to speed up anymore. From the observer's point of view, light will still move at speed c around the 61162111's second and then light will reach the spacecraft (from the observer's point of view) $\endgroup$ – Árpád Szendrei Nov 11 '16 at 21:33
  • 2
    $\begingroup$ @ÁrpádSzendrei wrote "you are still wrong" - I've just concluded that there is no value to be gained discussing this with you further. $\endgroup$ – Alfred Centauri Nov 11 '16 at 21:43

The exercise to make everything clear

Let me take a simpler example. Suppose we're talking purely 1D classical mechanics, and you have two particles, one moving with constant speed $u$ and the other one moving with a time-varying speed parametrized by $\alpha$, $$v(t) = \frac{\alpha~u~ t}{\sqrt{u^2 + (\alpha~t)^2}}.$$ We start off the first one at position $x=0$ and the second one at position $x=L$ and set them both moving with these speeds in the $+x$ direction.

Exercise: prove that there is some $L$ such that they will never meet, that the constantly-decreasing speed difference means that the time when they actually meet goes to $\infty$.

What? Why does that clarify anything?

Now what does this have to do with relativity? Well it turns out that the above $v(t)$ is the $v(t)$ prescribed by the relativistic rocket equation, if $u = c$. Therefore from the standpoint of the person on the ground, a light ray never catches up with the relativistic rocket if it is emitted behind a certain distance.

(What does that look like from the perspective of the rocket? It looks like there is a black-hole-like event horizon behind them; these events are being emitted from behind that horizon.)

The rest of the paradox is just a standard confusion at a sort of tension between two ideas in relativity: relativity enforces some form of speed limit where you would need an infinite amount of energy to go faster than speed $c$, but nobody actually experiences this limitation as a limitation. The very principle of relativity is that nobody is ever saying, "here is my absolute motion through absolute space" and so people must experience themselves as being at rest, with light still moving away from them in all directions at speed $c$ in their rest-frame, even though it also recedes at speed $c$ in other reference frames that are moving relative to me. But this is the standard relativistic tension.

Some tension-relief: it's all about the simultaneity.

I guess my first remark that I like to make to people starting this out is, please pay careful attention to the first-order Lorentz transform. Every other Lorentz transform in relativity can be understood as the limit of "do a little Lorentz transform, then do another little Lorentz transform, then do another little Lorentz transform... repeat until you've built up your whole transform." And this is where we see a difference with Newtonian mechanics. Transform by any $\vec v = [v_x;~v_y;~v_z]$ large-or-small, and you will find that Newton prescribes that you convert time coordinates $t$ and space coordinates $\vec r = [x;~y;~z]$ according to:$$ \begin{array}{rcl} t' &=& t,\\\vec r' &=&\vec r ~-~ \vec v ~ t.\end{array} $$However Einstein's crazy idea is that this does not represent what's actually happening when we accelerate relative to each other. Instead for small velocities $\vec \epsilon = \vec v/c,~|\vec \epsilon| \ll 1,$ you should transform your time coordinate $w = c~t$ and your space coordinates $\vec r$ to the values: $$ \begin{array}{rcl} w' &=& w ~-~ \vec \epsilon \cdot \vec r,\\\vec r' &=&\vec r ~-~ \vec \epsilon ~ w.\end{array} $$ Note that the second equation is essentially unchanged but the first equation now "looks like" the second equation in that the velocity is coming in to 'desynchronize' our clocks. This is called the "relativity of simultaneity" and I am just mentioning in passing that the other two core effects of relativity, "length contraction" and "time dilation", really come from adding up the way that these little desynchronizations get transformed and retransformed by these above equations. Then it gets more useful to confine the motion to one particular direction, like $\vec v/c = [\beta;~ 0;~ 0],$ in which case you can write that the general result is: $$ \begin{array}{rcl} w' &=& (w ~-~ \beta~x)~/~\sqrt{1 - \beta^2},\\ x' &=& (x ~-~ \beta~w)~/~\sqrt{1 - \beta^2},\\ y' &=& y,\\ z' &=& z. \end{array} $$ So, that's what happens when you propagate a ton of these little desynchronizations from little transforms through each other into a bigger transform. These terms $1/sqrt{1 - \beta^2}$ appear that start to affect "here is how long I think your spaceship is, it's much shorter than you think!" You can derive them a totally different way, by seeing that the cross term $2~\beta~w~x / (1 - \beta^2)$ is the same for the square of both, so that $(w')^2 - (x')^2 = w^2 - x^2.$ This is very important because this property that $\big(w^2 - x^2 - y^2 - z^2 \big)' = w^2 - x^2 - y^2 - z^2$ is really a statement about everyone agreeing on the speed of light receding from you: imagine that an event happens suddenly at your location, the light notifying the rest of the cosmos about that event now recedes from you as a thin spherical shell at the speed of light, which means obeying $\sqrt{x^2 + y^2 + z^2} = c~t.$ Well we defined earlier that $w = c~t$ and so this number $w^2 - x^2 - y^2 - z^2$ is zero for this expanding-light-bubble: and what we've just proven is that zero-for-me means zero-for-everyone; nobody sees this "light cone" that notifies the universe about this event as anything other than a perfect sphere expanding with velocity $c$. So this is what these little desynchronizations are really doing in the background, they're making everyone agree on the structure of these light bubbles, which makes everyone agree on things like causality and stuff.

Now be brutally honest about spaceships.

My second remark is that we have to think very brutally and physically about what we are describing, if we want to understand what's really going on. What happens to the spaceship that keeps accelerating faster and faster but never hits the speed $c$? Well, let me model its constant acceleration in a simple way: it drops some sort of "beacon" and then accelerates until it detects that beacon is travelling away from it with speed $u,$ then it throws down another beacon and accelerates until that beacon travels away with speed $u,$ and so on, and so on. To accelerate with constant acceleration, the clock where these beacons are thrown down must be throwing them down at constant intervals.

Combine those two last paragraphs and you realize that trying to outrace a beam of light is a real-life Zeno paradox! You accelerate to $c/4$ relative to this beacon, measure the speed that the beam of light is moving away from you, and it's still moving at speed $c$ away. You throw down another beacon, accelerate to $c/4$ relative to that, measure the speed that the beam of light is moving away from you, and it's still moving at speed $c$ away. You cannot win! Therefore nobody can travel faster than the speed of light. These bubbles of light then enforce a very powerful form of causality that's not technically present in Newtonian mechanics; it says: no form of matter can get to a destination faster than the information about that matter's history, which moves at speed $c$.

Then, learn to love rapidities.

Now when the spaceship tries to measure all of the beacons as a whole, it does not see a straightforward linear progression of speeds: those speeds would go off to infinity, therefore they would exceed $c$! Rather, it sees a progression which goes like a function called the hyperbolic tangent, $\tanh x = (e^x - e^{-x})/(e^x + e^{-x}).$ In fact if we stick with motion in one dimension we find out that each velocity can be characterized by a rapidity, $\beta = \tanh \phi,$ and that the velocity addition formula actually then becomes linear in rapidities. So that's why you see this hyperbolic tangent pattern in the speeds of the beacons! It's because your single-beacon Lorentz boost shifts its rapidity by some $\phi_1$, then every other rapidity must be $\phi_n = n ~ \phi_1,$ therefore every velocity must be $v = c \tanh (n ~\phi_1),$ for cases of constant acceleration.

Finally, one integral gets the original formula.

Actually rapidities are even more useful than that, because our expression $\gamma = 1/\sqrt{1 - \beta^2} = (e^\phi + e^{-\phi})/2 = \cosh \phi$ by the variant of the Pythagorean theorem that holds for hyperbolic trigonometry, $\cosh^2\phi - \sinh^2\phi = 1.$ This lets us perform the following integral. We know that the spaceship emits beacons at some interval of proper time $\delta \tau$ and can easily reconstruct the time on the clock $\tau = n~\delta \tau$ from that plus the number of emitted beacons. Since each time the spaceship gains a constant boost to its rapidity $\delta \phi,$ making everything exact to first-order gives the straightforward result that the speed of the spaceship is $v = c~\tanh(a~\tau/c).$ But our time coordinates $t$, watching all of this, are going to be different than these $\tau$ coordinates in the spaceship! Well that's where we just integrate: the basic time-dilation formula says that $dt = \gamma~d\tau = \cosh(a~\tau / c)~d\tau.$ Integrating that and discarding the constant gives $t = (c/a) ~ \sinh(a~\tau/c),$ therefore its velocity using our time coordinates for our reference frame is simply:$$v(t) = c\tanh\big(\sinh^{-1}(a t/c)\big).$$But, one of these basic results from $\cosh^2 \phi - \sinh^2 \phi = 1$ is that $\tanh \phi = \sinh \phi / \cosh \phi = \sinh \phi / \sqrt{1 + \sinh^2 \phi},$ so therefore $\tanh\big(\sinh^{-1} x\big) = x / \sqrt{1 + x^2},$ and the above simplifies to,$$ v(t) = \frac{a t}{\sqrt{1 + (at/c)^2}}. $$ Multiplying both top and bottom by $c$ gives precisely the form in the above exercise I provided with $u = c.$ Then we have a perfectly valid reference frame in our ground reference frame, and we know that this spaceship moving with speed $v(t)$ will not be hit by some incident beams of light with speed $c$, if they start out a distance $L$ behind it. All of the rest of the reasoning is identical between classical mechanics and special relativity; special relativity only affects how you change between reference frames.

Quick recap

To recap, because the spaceship is emitting these beacons at constant time intervals, with no concern about its own state; and because the universe has the particular desynchronizing structure it has, we discover that we at rest don't think that the beacons are emitted at constant time intervals, nor is the spaceship linearly accelerating. Instead we measure the beacons as actually emitted slower and slower, and we measure the velocity of the spaceship as a $v(t)$ curve which as we learn from the exercise above, cannot be hit by light rays from a certain distance $L$ before where it started. Therefore we realize that when you accelerate, there is some strange event horizon created at a very long distance behind you, and it gets closer and closer as you accelerate harder and harder. Events from behind this "wall of death" contain information that cannot possibly reach you; objects that you pass appear to slow down as they approach the wall of death, getting slower and slower as they fade closer and closer into the wall, never quite falling all the way through. You can only get rid of it by ceasing to accelerate, which will push that wall of death further and further away.

Just to complete the reasoning, John Rennie is claiming that one of the twins must see this wall of death and cannot use an inertial reference frame with special relativity to describe his/her situation; instead he/she needs to use general relativity with the Rindler metric, which basically forces him/her to acknowledge a form of gravitational time dilation is applying specially to him/her. Thereby, both twins agree again.


But in our world, since nothing can go faster then light, at approx 61162111 seconds, you reach almost the speed of light and you stop accelerating, and the light will catch-up to you soon.

I'm afraid you misunderstand the setup. One must distinguish between proper acceleration and coordinate acceleration in SR.

Imagine a spacecraft that is accelerating with constant acceleration $\alpha$ according to an accelerometer in the spacecraft. This is the proper acceleration and it is constant.

In an inertial reference frame, the spacecraft has a coordinate acceleration - the second derivative of the spacecraft's position coordinate(s) with respect to the time coordinate $a = \frac{d^2 x}{dt^2}$ - that equals $\alpha$ only for the instant that the spacecraft is instantaneously at rest (in this frame).

At all other times, the spacecraft has coordinate acceleration $a$ that is less than $\alpha$ and further, $a \rightarrow 0$ as the spacecraft's speed approaches $c$.

That is to say, there is no inertial reference frame in which the spacecraft's speed reaches $c$.

So I would like someone to please explain to me why the accelerating twin would outrun light in our world

In an inertial frame, the spacecraft's worldline is hyperbolic and thus asymptotically approaches a null (light-like) world-line in the spacetime diagram. This is the world-line of a photon that approaches but never catches the spacecraft:

enter image description here

Image credit

  • $\begingroup$ correct. But the twin paradox is viewed from a third observer's point of view, like when you look at the graphs of the two twins' clocks. What you will see is both twins' clocks as they meet again. So since you are a third observer, from your point of view, the spacecraft will stop accelerating after a point, and at that point from your point of view, the light still travels at speed c, and after that light has to catch up to the spacecraft from the observer's point of view. $\endgroup$ – Árpád Szendrei Nov 11 '16 at 21:38
  • $\begingroup$ this is also only when you set it up so that the acceleration from the observer's point of view is not a (with time decreasing) function of the speed of light. because in that case you are trying to create an acceleration for the spaceship, that from a statinary observer's point of view, looks like if the spaceship would accelerate at decreasing level with time. In that case as the spaceship accelerates, with time it will always get closer to speed c. Since the universe is infinite, and time is too, even then the spaceship's speed must converge to speed c. $\endgroup$ – Árpád Szendrei Nov 11 '16 at 21:41
  • $\begingroup$ So there must be a point in time from the observer's (let's say he lives forever) point of view where the spaceship stops accelerating (reaches it's converge speed which must be calculatable). If that speed would come out to be c at a certain time (even if that time if infinity) , then this theory does not hold. Because the two things either #1 speed of the spaceship would reach speed of light (impossible even at t=infinity). #2 Spaceships speed will reach a converge speed<c (even if that will happen at t=infinity). $\endgroup$ – Árpád Szendrei Nov 11 '16 at 21:46
  • $\begingroup$ Then light will catch up to the spaceship at a t=infinity+some constant time. Which will still happen. Light will have to catch up to that spaceship. $\endgroup$ – Árpád Szendrei Nov 11 '16 at 21:46
  • 1
    $\begingroup$ Another point is that in whichever reference frame (and whichever acceleration) accelerating an object (or keeping it such) requires an amount of energy that diverges as $v\to c$ (standard example is a rotating ball attached to a rope with large $L$, such that its rotational velocity could in principle be greater than $c$: the rotational energy will then diverge). As such, one should be careful to even state constant $a$ as that cannot really be kept on forever (it can be kept only in those metric whose corresponding energy doesn't diverge). $\endgroup$ – gented Nov 11 '16 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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