Thanks to Andrew Steane and Pulsar in this topic I understood that in a frame with constant proper acceleration, each hyperbola in $T-X$ diagram demonstrates a constant position while each line passing through origin demonstrates constant time.

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That is, from viewpoint of an accelerated frame, two simultaneous events in different locations are actually two points in the same line (with constant t) but two different hyperbolas.

What bothers me however is that to my understanding, each of these hyperbolas demonstrates an accelerated frame with different proper acceleration. As Pulsar put it:"Note that each hyperbola represent worldlines of travelers with different constant accelerations".

So what are we doing here? Assume a frame with constant proper acceleration $g_0$. Let say worldline of this frame is $x_0 = \frac{c^2}{g_0} = 0.4$. Now if from his viewpoint two events happen at different locations $x_1 = 0.2$ and $x_2 = 0.6$, he has to use two different hyperbolas $x_1$ and $x_2$.

  1. But it's as if he has assumed two different constant proper accelerations for those points. Even worse, different from his acceleration as well. Let say our accelerated frame resides in a spaceship. Center of mass spaceship is in $x_0$. Two simultaneous events $x_1$ and $x_2$ happen at the different end points of spaceship. If from viewpoint of our observer, these points have different proper accelerations, it would mean spaceship should be torn apart! Because every point of spaceship would have different acceleration. It's as if there is some kind of tidal force here. But why is that? I mean physically speaking. For example in Classical mechanics, an accelerated frame will feel a fictitious force due to the inertia. What actually happens in special relativity that we have a tidal force (if any)?

  2. Two observers with constant velocity can not use each other coordinates, unless they use Lorentz transformations first. Our accelerated frame however, use other worldlines (which corresponds to other observers with different accelerations) without using any kind of transformations. How he can do that? I mean if each of these hyperbolas demonstrates a point with different acceleration, how does it make sense to put all of them in the same diagram and make this grid to begin with? Instead of comparing two different hyperbolas, you can compare two different lines as well. Two different lines corresponds to two different inertial observers with different velocities, and we use these lines without doing any kind of transformations.

Edit: Regarding my first question, I think I am reading this diagram incorrectly. Maybe the observer in spaceship don't see a tidal force, rather it's actually the inertial observer outside of spaceship who sees every point on spaceship has different acceleration? At least it makes more sense from what I know from Lorentz transformation.

Edit 2: I changed title to make it more interesting for people.


2 Answers 2


Accelerated coordinates are just a spacetime analog of polar coordinates. The curves of constant $x$ in accelerating coordinates have different curvatures (different accelerations) for the same reason the circles of constant $r$ in polar coordinates do.

Consider a bend in a road of constant width, as seen in this road sign:

Note that the sides of the road at the bend are (at least approximately) circular arcs with a common center and different curvatures. This is the most natural way to bend a road while preserving its width. For the same reason, the most natural way to accelerate an extended object in spacetime involves accelerating its ends at different rates. It doesn't tear the object apart; in fact it may be the least stressful way to accelerate it.

  • $\begingroup$ I think more than an analogy I need a direct answer for my questions. In polar coordinates, acceleration and velocity are orthogonal, while in my question they are parallel. And while I understand (mathematically speaking) that's is the least stressful way to accelerate an object, I do not understand (physically) why is that? $\endgroup$
    – Paradoxy
    Sep 16, 2020 at 10:19
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    $\begingroup$ @Paradoxy Spacetime acceleration and velocity are orthogonal when expressed as 4-vectors. To be honest I don't know how to demonstrate that "concentric" acceleration is the least stressful, and I don't really know if it's true, so I added some weasel words to the answer. $\endgroup$
    – benrg
    Sep 17, 2020 at 2:50

The straight line from the origin crossing all hyperbolas is the line of simultaneity of an inertial frame with a given velocity. If tangents are made at any hyperbola at the crossing point with this line, they are parallel. That is why it is not necessary a Lorentz transformation. All that points of spacetime (along the straight line) are in the same (momentarily) inertial frame. It would not happen in the case of parallel hyperbolas (same acceleration).

So, the opposite will happen regarding stress. Objects inside the ship are under compressive stress due to the artifical gravity. When it is turned off, that stress vanishes. (well, without damping, everything would oscillate around a new greater equilibrium distance).

It is more clear for 2 ships keeping a distance $d$ between then. They have different accelerations. When they reach the same velocity of a $3^{rd}$ ship that is moving in the same direction, but with constant velocity, both turn off their motors. The $3^{rd}$ ship records that it happens at the same time. In this case, they keep the same distance $d$, all 3 ships now at rest.

Edit at $16^{th}$ about the comment "v increases at the same rate for 2 points if their acceleration is different.":

It can be seen in your diagram. For $\tau = 0$ the distance between $x=0.4$ and $x=0.6$ is $\Delta x = 0.2$. The velocity is zero for the inertial frame of the diagram.
For the next $\tau$ of the diagram (the line with the smallest angle with the $x$-axis), the distance between the hyperbolas is also $0.2$. The velocities at each point are the same (tangents to the hyperbolas are parallel). So the rate of increase of velocities with respect to $\tau$ is the same. Of course it is not true for the rate of increase with respect to $t$. And that is why the local acelerations are different for the points.

  • $\begingroup$ 1. I think I was looking for a more intuitive answer, I mean yes I got it it's a requirement to satisfy SR mathematically. But why, physically speaking, there is a tidal force (if any) in accelerated frame? In Newtonian mechanic an accelerated frame feel fictitious force due to the inertia which is obvious, but what actually happens in SR? $\endgroup$
    – Paradoxy
    Sep 16, 2020 at 10:13
  • $\begingroup$ 2. I was comparing two different hyperbolas in my second question. Yes, "a" line belongs to "an" inertial frame. But accelerated frame coordinate is made up infinite lines, which corresponds to "different" inertial frames with different velocities! How is it possible? And we use two lines (or two different hyperbolas) without using any kind of transformations $\endgroup$
    – Paradoxy
    Sep 16, 2020 at 10:16
  • $\begingroup$ There is no tidal force because the separating distance and relative speed of two points of a ship are always the same during the acceleration. Their acceleration and the coordinate time are however different. I think our intuition is based in maths. We are used to think of $v = at$. Then it seems strange that $v$ increases at the same rate for 2 points if their acceleration is different. $\endgroup$ Sep 16, 2020 at 15:02
  • $\begingroup$ "v increases at the same rate for 2 points if their acceleration is different." Can you please elaborate on this a bit more? I don't think their velocity increases at the same rate, because after all, those two events (points) due to simultaneity have same time rate (they are in the same time axis) and different accelerations. $\endgroup$
    – Paradoxy
    Sep 16, 2020 at 18:37
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    $\begingroup$ The hiperbolas come from differential equations relating the components of the 4-velocity and its derivatives with respect to $\tau$, under the assumption that the local acceleration is constant and parallel to the velocity. It is well analyzed in the book Gravitation of Misner, Thorne and Wheeler in the chapter 6. The clocks doesn't show the same time for different points. I deleted the first paragraph where I wrote that. $\endgroup$ Sep 17, 2020 at 14:32

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