Consider relativity of simultaneity in SR, where 2 observers S and S' are inertial. With the Lorentz transformation $$t=\gamma(t'-vx'/c^2)$$ we can see that the relativity of simultaneity comes from $vx'/c^2$ where $x'$ indicates position of event say A in S' frame. It's simple to deduce this fact, because if we drop this term, and suppose that two events are simultaneous in one frame, one can see they will be simultaneous in all frames. Let's say there are two events in spacetime such that $t'_A=t'_B$ while $x'_A\neq x'_B$. By using the equation above and dropping $vx'/c^2$, $$t_a=\gamma(t'_a)~~\mbox{and}~~t_b=\gamma(t'_b)~~\mbox{so}~~ t_a=t_b.$$ Now let's consider a special case of acceleration where observer S' moves with constant proper acceleration $g$ and observer S is inertial. This is the so called hyperbolic motion, and we have this transformation for time:

$$t=\frac{c}{g}\sinh(gt'/c)$$ As you can see, $t$ does not depend on position of events, so relativity of simultaneity can not be understood from this formula. Which is a bit off considering S' is in motion relative to S.

I think this problem comes from my misunderestanding, and this transformation does not apply for all events, rather it's merely for the events in S' origin (i.e $x'=0$) probably. However Don Koks in his book Explorations in mathematical physics the concepts behind an elegant language compares relativity of simultaneity of events for accelerated frame, where he uses the same transformation.

Also I appreciate if someone shows me a way to compare simultaneity of events for accelerated frames with an example.

So in short, what's the physical meaning of this transformation? Why is it not position dependent (from physical point of view)? Does it indicate some kind of absolute frame?

  • $\begingroup$ "it's merely for the events in S' origin" Correct. It's a useful formula, though. There are related formulae at math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html To handle general events in an accelerated frame, take a look at en.wikipedia.org/wiki/Rindler_coordinates $\endgroup$
    – PM 2Ring
    Commented Jun 23, 2019 at 17:22
  • $\begingroup$ @PM2Ring As a matter of the fact, Rindler coordinate derivation is being done by the use of transformation I've written above! (not upside down) That formula is completely general unless you define different initial velocities/position from that of rindler. See en.wikipedia.org/wiki/Hyperbolic_motion_(relativity) also I saw your first link, didn't understand how is it relevant to my question $\endgroup$
    – Paradoxy
    Commented Jun 23, 2019 at 17:31
  • $\begingroup$ I posted the Relativistic Rocket link because it collects all the basic hyperbolic equations in one place, and it explains how SR can handle constant acceleration, which may be of interest to future readers. And it's a cool page. ;) $\endgroup$
    – PM 2Ring
    Commented Jun 23, 2019 at 17:50

2 Answers 2


This question combines two aspects:

  1. what do we mean by the relativity of simultaneity, and does it always hold?
  2. what is a good way to understand the constantly accelerating reference frame (in flat spacetime)

1. Relativity of simultaneity

In special relativity, the relativity of simultaneity is the fact that if in one inertial frame two events are simultaneous, then there exist other inertial frames in which they are not simultaneous. In general relativity, the relativity of simultaneity is the fact that if two events share the same value of a temporal coordinate $t$ in some given set of coordinates used to chart a region of spacetime, then there can be other sets of coordinates in which those events do not share the same value of some other temporal coordinate $t'$. Here, by a 'temporal coordinate' I mean a coordinate such that small intervals where only this coordinate changes are timelike.

The relativity of simultaneity is an existence claim: it is the claim that there exist coordinate-charts or inertial frames which differ concerning simultaneity. Therefore no single counter-example can be called a 'violation'; the only way one to 'violate' the claim would be to show it is never true---one would have to show that there are no pairs of frames which differ about simultaneity. But this will not be possible, because it is easy to find examples which do differ about simultaneity.

The question being asked here is, therefore, really the question:

2. what is a good way to understand the constantly accelerating reference frame (in flat spacetime)

The constantly accelerating frame in flat spacetime, also called Rindler frame, is a very good platform for learning various lessons in both special and general relativity. One could write whole books about it; Wikipedia provides a useful introduction. The basic idea is to chart a large region of flat spacetime using two different coordinate systems: either ordinary Minkowski coordinates $T,X,Y,Z$, or Rindler coordinates $t,x,y,z$, related to the former by $$ T = x \sinh(\alpha t),\quad X=x\cosh(\alpha t),\quad Y=y,\quad Z=z $$ where we have set $c=1$. In terms of the quantities quoted in the question, we have $\alpha = g$ and the unprimed coordinates in the question are equal to the $T,X,Y,Z$ coordinates adopted here.

The spacetime interval between two events separated by $dT,dX,dY,dZ$ is $$ ds^2 = - dT^2 + dX^2 + dY^2 + dZ^2 $$ (the Minkowski metric). The spacetime interval between two events separated by $dt,dx,dy,dz$ is $$ ds^2 = -(\alpha x)^2 dt^2 + dx^2 + dy^2 + dz^2 $$ (the Rindler metric).

The events along any given straight line through the origin in the $T,X$ plane (with slope less than 45$^\circ$) are simultaneous in the Rindler coordinates: they all have the same $t$. But they are not simultaneous in the Minkowski coordinates, so far from avoiding the relativity of simultaneity, this case illustrates that aspect of relativity perfectly well.

The following diagram shows the lines of constant $t$ (straight lines through origin) and the lines of constant $x$ (hyperbolae) in the $T,X$ plane.

By Dr Greg, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=22635537

The equation which is quoted in the question, namely $$ ``\,t = \frac{c}{g} \sinh (g t'/c)\," $$ is, in my notation, $$ T = \frac{1}{g} \sinh (g t) . $$ This is the equation for one of the hyperbolae: it is the one with $x = 1/g$. So no wonder it makes no mention of $x$! But perhaps the question has arisen from another aspect of this case. Each hyperbola crosses the $T$ axis at some given $X$ (in fact at $X = x$), and the proper acceleration of a particle whose worldline is that particular hyperbola is itself proportional to $1/x$. So the equation $$ T = x \sinh(\alpha t) $$ can also be written $$ T = \frac{c}{a_0} \sinh(\alpha t) $$ where $a_0 = c/x$ is the proper acceleration for the given worldline. This hides the fact that $T$ depends on $x$, and perhaps this is the reason for the confusion that gave rise to the question.

  • 1
    $\begingroup$ A simple way of getting at the OP's question is to zoom in one small part of the diagram showing the transformation from Minkowski coordinates to Rindler coordinates. In such a small region, we see that the pink graph-paper grid looks like parallelograms. This is identical to what we see in a Lorentz transformation, so it has all the same features, including relativity of simultaneity (because the horizontal lines tilt). $\endgroup$
    – user4552
    Commented Aug 5, 2019 at 17:34
  • $\begingroup$ I already know $x_0\sinh(gt)$ is worldline of accelrated frame in inertial frame's point of view, and I already have seen this diagram too. But Assume two events on one of these hyperbolaes, which are simultaneous w.r.t accelerated frame (or Rindler coordinate). So $t_a=t_b$, Because $T=x_0\sinh(gt)$ we have $T_a=T_b$, my question is are they necessarily simultaneous w.r.t any arbitrary inertial frame (minkowski coordinate)? But it shouldn't be the case right? $\endgroup$
    – Paradoxy
    Commented Aug 5, 2019 at 18:02
  • $\begingroup$ @BenCrowell, that's a very nice way too look at it, however, can we actually compute time difference between two simultaneous events in accelerated frame from inertial frame's point of view? (Please read my comment above) $\endgroup$
    – Paradoxy
    Commented Aug 5, 2019 at 18:20
  • 2
    $\begingroup$ @Paradoxy Two events on one of the hyperbolas are not simultaneous in the accelerated frame. Rather, a pair of events on two different hyperbolas can be simultaneous in the accelerated frame. And they have $T_a \ne T_b$. $\endgroup$ Commented Aug 5, 2019 at 19:48
  • $\begingroup$ Thank you professor, this makes sense now. In other words, diagonal lines are indicating line of simultaneity of accelerated frame, and thus if we assume two simultaneous events in accelerated frame at different locations we should use two different hyperbolas to find their time in inertial frame. It's too easy, i wonder why it took too long for someone to answer me $\endgroup$
    – Paradoxy
    Commented Aug 5, 2019 at 20:40

I will use a slightly different approach as Andrew Steane's answer. I will call $(t,x)$ the coordinate system of a stationary inertial observer (suppressing the $y$ and $z$ coordinates), and $(\bar{t},\bar{x})$ the coordinate system of a traveler with constant proper acceleration. The four-velocity of the traveler w.r.t. the stationary frame is then $$ \frac{\text{d}\boldsymbol{x}}{\text{d}\bar{t}} = \left(c\frac{\text{d}t}{\text{d}\bar{t}},\ \frac{\text{d}x}{\text{d}\bar{t}}\right) = (\gamma c,\, \gamma v). $$ Let's introduce a parameter $\eta\,$, defined as $v = c\tanh\eta$. It follows that $$ \left(c\frac{\text{d}t}{\text{d}\bar{t}},\ \frac{\text{d}x}{\text{d}\bar{t}}\right) = (c \cosh\eta,\, c\sinh\eta).\tag{1} $$ The four-acceleration is then $$ \frac{\text{d}^2\boldsymbol{x}}{\text{d}\bar{t}\vphantom{t}^2} = \left(c \sinh\eta\,\frac{\text{d}\eta}{\text{d}\bar{t}},\, c\cosh\eta\,\frac{\text{d}\eta}{\text{d}\bar{t}}\right). $$ The proper acceleration $g$ of the traveler is the Lorentz scalar associated with this four-vector: $$ g = \sqrt{c^2\cosh^2\eta\,\left(\frac{\text{d}\eta}{\text{d}\bar{t}}\right)^{\!2} - c^2\sinh^2\eta\,\left(\frac{\text{d}\eta}{\text{d}\bar{t}}\right)^{\!2}} = c\frac{\text{d}\eta}{\text{d}\bar{t}}. $$ Since we consider motion with constant $g$, we can therefore write $$ \bar{t} = \frac{c}{g}\eta,\tag{2} $$ and from $(1)$ $$ \frac{\text{d}t}{\text{d}\eta} = \frac{c}{g}\,\frac{\text{d}t}{\text{d}\bar{t}} = \frac{c}{g}\cosh\eta,\\ \frac{\text{d}x}{\text{d}\eta} = \frac{c}{g}\,\frac{\text{d}x}{\text{d}\bar{t}} = \frac{c^2}{g}\sinh\eta, $$ so that we obtain the equations $$ ct = \frac{c^2}{g}\sinh\eta,\\ x = \frac{c^2}{g}\cosh\eta,\tag{3} $$ which is the parameter equation of a hyperbola (blue line in the left figure).

enter image description here

This only defines a single worldline, namely the path of the traveler on which $\bar{x} = \text{const}$. How can we extend this to a $(\bar{t},\bar{x})$ coordinate grid? First, note that the traveler is momentarily at rest at $\bar{t}=t=0$. It is therefore a natural choice to define $\bar{x}\equiv x$ at $\bar{t}=0$. As a consequence, the $\bar{x}$-coordinate of the traveler is $c^2/g$.

We can use a similar procedure for arbitrary $\bar{t}$: at any moment, we can let the $\bar{x}$-coordinate coincide with the $x'$ coordinate of a momentarily comoving inertial frame (i.e. an inertial frame moving with the same velocity as the traveler at time $\bar{t}$).

For example, take an arbitrary spacetime point $A$ on the traveler's path, with coordinates $(t_A, x_A)$ in the stationary frame and $(\bar{t}_{\!A}, \bar{x}_{\!A})$ in the accelerating frame. We already know that $\bar{x}_{\!A}=c^2/g$. Now we introduce the $(t',x')$ coordinates of a momentarily comoving inertial frame at time $\bar{t}_{\!A}$.

The time axis of this frame, defined as $x' \equiv \bar{x}_{\!A}$ (purple line) is given by the tangent line of the traveler's path at $A$. From $(3)$, we obtain a tangent vector $(\cosh\eta_A, \sinh\eta_A)$, so that the time axis can be expressed as $$ ct = ct_A + c(t' - \bar{t}_{\!A})\cosh\eta_A,\\ x = x_A + c(t' - \bar{t}_{\!A})\sinh\eta_A. \tag{4} $$ The spatial axis ($t' \equiv \bar{t}_{\!A}$) of the comoving frame (red line) is perpendicular to the time axis, and therefore defined by $(t_A, x_A)$ and the vector $(\sinh\eta_A, \cosh\eta_A)$. Indeed, the Minkowski dot product of $(\cosh\eta_A, \sinh\eta_A)$ and $(\sinh\eta_A, \cosh\eta_A)$ is zero. We thus find $$ ct = ct_A + (x' - \bar{x}_{\!A})\sinh\eta_A = x'\sinh\eta_A,\\ x = x_A + (x' - \bar{x}_{\!A})\cosh\eta_A = x'\cosh\eta_A. \tag{5} $$ We now define $\bar{x}\equiv x'$ on this spatial axis at $\bar{t}_{\!A}$. But we can do this for any point on the traveler's worldline, so we can immediately generalize $(5)$ to $$ ct = \bar{x}\sinh(g\bar{t}\!/c),\\ x = \bar{x}\cosh(g\bar{t}\!/c). \tag{6} $$ These are the transformations we sought. Curves of constant $\bar{t}$ are straight lines passing through the origin, curves of constant $\bar{x}$ are hyperbolae. Note that the latter represent worldlines of travelers with different constant accelerations; this is closely related to the concepts of Born rigid motion and Bell's spaceship paradox. Finally, note that none of these travelers can communicate with the origin: light rays passing through the origin do not intersect with the hyperbolae, and thus define a horizon.


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