# Understanding the spacetime diagram for an accelerated observer moving farther away

This video on the Twin Paradox shows the following spacetime diagram to explain the scenario. While accelerating back to Earth, the travelling twin perceives time on Earth as speeding up, which is represented by the simultaneity lines being more spaced out (in the inertial observer's frame of reference).

How would the same spacetime diagram look if the travelling twin accelerated even farther away from Earth? I believe the simultaneity lines, instead of being more spaced out (like the original case), would get even more close together, more or less like the drawing below:

The problem is, during acceleration, you perceive your clock as slower (much like the travelling twin, during acceleration, perceives the time on Earth as speeding up).

If my drawing above was correct, the travelling observer would still perceive the other observer's clock as slower during acceleration (that's what the simultaneity lines suggest). That would be wrong, because he must still notice the other observer's clock as speeding up (just like the original case). I mean: it shouldn't depend on the direction of acceleration, correct?

What is wrong with my understanding? Is there an intuitive way of representing this situation, much like what is represented on the mentioned video on YouTube? I have no grasp of complex math, so I'd appreciate easy, intuitive answers, if they even are possible for this case.

• I don't think you will get a good picture without maths, so you should either choose to spend time with maths, or choose not to understand. Simple statement I would make is that time measured by the clock of either twin is essentially the length of its worldline. It so happens that an accelerated observer, in Hyperbolic motion there and back, travels between the events of twins splitting and joining along a shorter route (in spacetime).
– Cryo
Commented Jan 5, 2021 at 22:36
• I don't think you should discuss what one twin will think happens to the other without them communicating. If they are communicating you need to specify how. The way I look at it is this. Two observers have two clocks, one each. One observer stays put, second observer accelerates (along a straight line) for a while, then decelerates, then stops, then accelerates in the opposite direction, then decelerates and stops. The two observers are then in the same place and can compare the clocks each one has carried. The clocks will show different time.
– Cryo
Commented Jan 5, 2021 at 23:00
• The observer that has gone through accelerations/decelerations will have travelled shorter route as measured by integrating $ds=cd\tau=\sqrt{c^2dt^2-dr^2}$, where $\tau$ is the proper time (of either observer)
– Cryo
Commented Jan 5, 2021 at 23:02
• You are asking the same thing again :-). How do you tell if accelerated observers clock is running slower or faster? How do you compare these two clocks? Imagine two curves coming out of the same point and meeting in another point, you want to know which curve is longer, but you want to know this before the two curves meet again. How do you decide when to 'cut'? That's why you need to specify how observers communicate or wait for them to meet again
– Cryo
Commented Jan 5, 2021 at 23:12
• If no-one else sweeps in with a link to proper explanation, I will try to give an explanation... later
– Cryo
Commented Jan 5, 2021 at 23:16

# Intro

I think your question is a bit misguided because you are trying to draw a straight axis along which the time would be measured. Instead time measured by any observer, the only time that matters for that observer in most cases, is it's proper time. To measure proper-time you need to measure the arc-length of a curve, that is the world-line of the observer. Once you understand that you will no longer need to draw scores of dashed lines. Space-time diagrams will still remain useful of course.

So the way I will answer your question is to very quickly introduce proper time and how it relates to arc-length. I will then resolve the twin paradox by showing that if the two twins ($$A$$ and $$B$$) meet at one event, then split, then meet again, the twin that did not accelerate will measure greatest time lapse (will be older).

# World-line

Position of any point-like object in 4d spacetime can be given by $$x^\mu=\left(ct, \mathbf{r}\right)^\mu=\left(ct,x,y,z\right)$$. Here I will always use Cartesian coordinates. This is the position assigned to an object by some inertial (lab-frame) observer.

We tend to want to discuss objects that do not suddenly disappear, so instead of a single position, all objects will have a world-line, a continuous line that gives the position of the object at all times. You drew some world-lines in your sketches.

Because world-line is indeed a line, it can be parametrized by a single (real) number. One possible way to parametrize world-line is by lab-time, i.e.

$$x^\mu\left(t\right)=\left(ct,\,\mathbf{r}\left(t\right)\right)^\mu$$

This is easy, and often necessary to tie mathematics to observable phenomena, but such description is theoretically inconvenient since your time might be different to mine if we are moving relative to each other.

A different approach is to tie the parametrisation to arc-length of the curve. Imagine there are two events:

\begin{align} x^\mu_1&=\left(ct,\mathbf{r}\right)^\mu \\ x^\mu_2&=\left(ct+c\delta t,\mathbf{r}+\delta\mathbf{r}\right)^\mu \end{align}

that are close to each other. I would be glad to be corrected here, but as far as I understand, definition of 'close' here would rely on $$\mathbb{R}^4$$ topology of spacetime.

Next, we can compute the following quantity:

$$ds = \sqrt{c^2 \left(\delta t\right)^2 - \left(\delta\mathbf{r}.\delta\mathbf{r}\right)}$$

which would be the invariant 'distance' between $$x^\mu_1$$ and $$x^\mu_2$$. Invariant because all inertial observers would agree on this distance. Here, for simplicity we shall constrain ourselves to world-lines of objects moving at sub-liminal speeds. Which would mean that for all closely-spaced events on the world-line of the object $$ds>0$$.

We can then define some point on the world-line, $$x_0^\mu$$ to be at $$s=0$$, and find the invariant distance from that point to any other point, $$x^\mu_1$$, on the world-line, by integrating:

$$s_1 = \int_{x_0}^{x_1} ds = \int_{x_0}^{x_1} \sqrt{c^2 \left(d t\right)^2 - \left(\delta\mathbf{r}.\delta\mathbf{r}\right)}$$

Don't worry about actually doing the integral, it is enough to understand it is possible to do it. The beuaty of this approach is that any two observers will agree that 'distance' from $$x_0^\mu$$ to $$x_1^\mu$$ along the curve, as outlined above, which I will call arc-length is $$s_1$$.

So now we can simply parametrize all points on the curve by this arc-length.

$$x^\mu\left(s\right)=\left(ct\left(s\right),\,\mathbf{r}\left(s\right)\right)^\mu$$

# Proper time

Now imagine an observer sitting in some rocket or something like that and holding on to clock. Irrespective of whether the observer (and the clock) is inertial or accelerating, one can always imagine an inertial frame in which observer is instantaneously at rest (and next second observer may no longer be at rest in this frame). Lets call it the instantaneous rest frame (I recall it being called the tangential rest frame as well). In that instantaneous rest-frame, $$\bar{S}$$ the world-line of the clock will be

$$\bar{x}^\mu\left(s\right)=\left(c\bar{t}\left(s\right),\mathbf{0}\right)^\mu$$

Of course this is only valid for the time at which the clock remains nearly at rest in $$\bar{S}$$. Consider the arc-length between two closely spaced events $$\bar{x}^\mu\left(s\right)$$ and $$\bar{x}^\mu\left(s+\delta s\right)=\bar{x}^\mu\left(s\right)+\left(c\delta\bar{t},\mathbf{0}\right)^\mu$$. Here $$\delta \bar{t}$$ is the time-difference that would be measured by observer's clock (same as time in $$\bar{S}$$ since clock is instantaneously at rest in $$\bar{S}$$). By definition of the arc-length, it is:

$$\delta s=\sqrt{c^2\left(\delta\bar{t}\right)^2}=c\delta\bar{t}$$

Since speed of light, $$c$$, is constant one can see that observer's clock is measuring the arc-length in 'light-years' (i.e. multiply the observer's time by speed of light to get the arc-length). Based on this insight we define the proper time $$\tau=s/c$$. This is the time measured by observer that is moving along some world-line, but it is also the arc-length of that world-line measured in units of time. We can therefore parametrize any world-line by it's proper time:

$$x^\mu\left(\tau\right)=\left(ct\left(\tau\right),\,\mathbf{r}\left(\tau\right)\right)^\mu$$

# Curved world-lines & twin paradox

Lets imagine two events $$x_1^\mu$$ and $$x_2^\mu$$ and assume that it is possible to get from $$x_1^\mu$$ to $$x_2^\mu$$ at sub-luminal speed. Lets imagine two paths between these two points. One path is that of an object $$A$$ that is not accelerating, another of object $$B$$ that is accelerating, and then decelerating (or it would not meet with $$A$$). The first world line will be straight, the second one will be curved (I will not explain line curvature here any further since it can be avoided).

Assume that clocks of $$A$$ and $$B$$ are synchronized at $$x_1^\mu$$ ($$\tau_A=\tau_B=0$$ at $$x_1^\mu$$).

One can show that proper time measured by $$A$$ at point $$x_2^\mu$$, i.e. $$\tau_{A@2}$$ will be the longest possible for all world-lines that go through both $$x_1^\mu$$ and $$x_2^\mu$$ and are time-like at all points (i.e. only sub-luminal speeds of travel)

How? Consider the world-line of observer $$A$$ in its own rest-frame ($$S'$$):

$${x'}_A^\mu \left(t'\right) ={x'_1}^\mu + \left(ct',\mathbf{0}\right)^\mu$$

Where $$t'=\tau_A$$ ($$A$$'s rest frame). Therefore $${x'_2}^\mu=\left(c\tau_{A@2},\mathbf{0}\right)^\mu$$. What is the world-line of $$B$$ in $$S'$$? Lets say that velocity of $$B$$, as seen by $$A$$ is $$\mathbf{v}_B=\mathbf{v}\left(t'\right)=d\mathbf{r}_B/dt'$$, and its position is $$\mathbf{r}_B$$, therefore $$B$$-s world-line is:

$${x'}_B^\mu \left(t'\right) ={x'_1}^\mu + \left(ct',\mathbf{r}_B\left(t'\right)\right)^\mu,\quad \mathbf{r}_B\left(0\right)=\mathbf{r}_B\left(\tau_{A@2}\right)=\mathbf{0}$$

What is the arc-length of $$B$$?

$$c\tau_{B@2}=\int_{0}^{\tau_{A@2}} \sqrt{c^2 \left(dt'\right)^2-\left(d\mathbf{r}_B\right)^2}=\int_{0}^{\tau_{A@2}} dt'\,\sqrt{c^2 -\left(\mathbf{v}_B\right)^2}$$

Now you should see that any non-zero speed $$\mathbf{v}_B$$ along the path of $$B$$ will lead to reduction in $$\tau_{B@2}$$, therefore unless $$B$$ is at rest relative to $$A$$, the time $$\tau_{B@2}$$ will be less than that of $$\tau_{A@2}$$ when they meet again.

That's the twin paradox resolved. $$x_1$$ is when the two twins meet for the first time, $$x_2$$ is when they meet again and compare clocks.

There seems to be a lot of questions on 'accelerated reference frames'. I will try to address it, but happy to be corrected. A proper reference on this is chapter 6 'Accelerated Observers' from Misner, Thorne, Wheeler Gravitation (first part of the book is on Special Relativity).

So what do we mean by reference frame? It would seem that Wiki's definition is appropriate:

A frame at a point x ∈ X is an ordered basis for the vector space Ex

the key here is this basis of vectors. Lets say you have an object that is moving at some, possibly non-constant, speed relative to lab frame. The world line is $$x^\mu=x^\mu\left(\tau\right)$$, where $$\tau$$ is the proper time. Consider some point $$x^\mu_1=x^\mu\left(\tau_1\right)$$. At that point you can always identify the tangent to the world-line, $$u_1^\mu=\left(dx^\mu/d\tau\right)_{@\tau_1}$$. This is known as four-velocity, but the name is not important. Once you have the four-velocity, you can find three more vectors that are perpendicular to it, and this tetrad of basis vectors, call them $$\hat{\tau}^\mu=u^\mu/c,\,\hat{x}^\mu,\,\hat{y}^\mu,\,\hat{z}^\mu$$ will allow you to express any vector at that point $$x^\mu_1$$. What you could try to do next is draw four sets of straight lines, parallel to your basis vectors, to get something like the white grid on the picture below, but in 4d.

This is nice, but you must remember that you generated this grid from a single tangent at a specific point. Will you get the same grid if you use a tangent to the world-line at some other point? For inertial observers you will - their world-lines are straight. For accelerated observers you will not - their world-lines are curved. Coming back to the definition of frame, for accelerated observers different points on the world-line have different frames, i.e. different basis vectors.

You can pretend that the curved line is actually straight, but only if you limit your interest to the region where the curvature of the world-line is too small to notice. In this sense you can talk about the rest-frame for an accelerating observer, but it is local.

Another approach is to have a global reference frame, but your accelerating observer will only be at rest in that frame for a short period of time. This would be the instantaneous rest-frame. I prefer the second approach.

• This doesn't actually "resolve the twin paradox" because it goes straight to proper-time in Minkowski space, and there is no paradox in that formulation. The apparent contradiction that is the paradox never appears.
– JEB
Commented Jan 6, 2021 at 15:29
• @JEB :-) I claim this as an advantage, it gives a correct description of the end result without generating a debate on inertial frames and without forcing one compare the clocks of the two spatially separated observers that have no defined way to communicate between each other.
– Cryo
Commented Jan 6, 2021 at 17:16

How would the same spacetime diagram look if the travelling twin accelerated even farther away from Earth? I believe the simultaneity lines, instead of being more spaced out (like the original case), would get even more close together, more or less like the drawing below:

Well the answer is simple: the more the worldline tilts to the right, the more the line of simultaneity tilts upward. I can see that happening in your picture, so it's correct.

As your drawing is correct, the traveling observer perceives the other observer's clock as slower during acceleration (that's what the simultaneity lines suggest). It depends on the direction of acceleration.

Low clock is slow. Which clock is the fastest inside an accelerating body? I mean, in an accelerating frame of reference a low clock is slow.

• Thanks. I previously thought that, during acceleration, in whatever direction, the accelerated observer would always perceive it's own clock as slower in respect to other observers (duo to time dilation). Apparently, the answer is not that simple, although I am not totally convinced my drawing is an adequate representation.
– user137288
Commented Jan 6, 2021 at 0:32
• @stuffu. Accelerating frame of reference is a subtle thing, only makes locally, instantaneous rest-frame might be a better concept to invoke
– Cryo
Commented Jan 6, 2021 at 1:29
• @Cryo A Dude making jumps between inertial frames perceives some clocks making forward-jumps in time, and some clocks making backward-jumps in time, depending on both the frame-jump direction, and the direction of the clock. Like that? :) Hmm that sounds actually ok. But there were no mention of jumps in the question, so maybe not quite the most relevant answer . Commented Jan 6, 2021 at 2:50
• @Cryo do you mind explaining what you meant by this comment? I'm curious and I don't think I fully understand
– user137288
Commented Jan 6, 2021 at 2:53
• @RobertoValente, Misner-Thorne-Wheeler have a long section on it in the Special Relativity part of their book ("Gravity"). When you are talking about the frame of reference you are probably thinking about four basis vectors that define the direction of your time-axis, and the three spacial axis. Time-axis will be tangent to the world-line of the observer. The problem with accelerated observer is that its world-line is not straight, so the tangent to this line has to change direction as a function of space. So you cannot define a single frame anymore.
– Cryo
Commented Jan 6, 2021 at 3:11

Don't confuse the plane of simultaneity changing with time dilation: they are different.

The Lorentz transform from the space ship back to the Earth looks like:

$$t = \gamma(t' + \frac{vx'}{c^2})$$

Time dilation is:

$$\frac{dt}{dt'} = \gamma$$

and depends only on the relative velocity.

The plane of simultaneity is independent from $$\gamma$$, and is defined by:

$$c^2t' + vx' = {\rm const.}$$

If the turn around occurs at a time:

$$t = \frac L v$$

on Earth, then before/after turn around the rocket twin's "now" back on Earth (in the Earth frame) goes from:

$$\frac L v - \frac{v(v\frac L v)}{c^2}= \frac L v(1-\beta^2)= \frac L v \frac 1{\gamma^2}$$

to

$$\frac L v - \frac{-v(v\frac L v)}{c^2}= \frac L v(1+\beta^2)= \frac L v(2-\frac 1{\gamma^2})$$

which exactly accounts for the "missing time", according to the space traveling twin.

That was for instant acceleration, which leads to a sudden change in the plane of simultaneity. If you drag it out of a period of $$\delta t$$ with acceleration $$a$$, you can make it look just like pseudo gravitational time dilation; do not let that fool you. It is not time dilation, it is a change in the bias of the clock because of a changing hyperplane of simultaneity.

The root cause is that when the rocket twin is far from Earth, there is no unique way to define "now" back on Earth; at fixed distance, $$L$$, it depends on the local speed of the observer, and can vary by $$t_0\pm L/c$$ where $$t_0$$ is the time in Earth's rest frame.

• When the travelling twin accelerates to return, he perceives time on Earth as "speeding up" (duo to a sudden change in the plane of simultaneity - not time dilation). However, this particular change depends on the direction of acceleration, correct? If he were to accelerate even farther away from Earth, he might as well perceive time on Earth as ticking even slower than before (as my drawing suggests), if they were able to compare each other's clocks at that precise moment, correct?
– user137288
Commented Jan 6, 2021 at 15:16
• You can make Earth's clock change to anything between $t_0\pm L/c$, which includes going fast, going slow, and going backwards. For example, "now" at the Andromeda galaxy oscillates $\pm190$ years on a yearly basis (190 years = 2.5M light years $\times \cos{42^{\circ}}\times$ 0.01% $c$), and imho, you can't call that time dilation; rather, it's a clock bias thanks to the Earth's orbital motion.
– JEB
Commented Jan 6, 2021 at 15:40
• Do you mind telling me what "cos 42°" stands for in this formula you mentioned? I mean, what is the meaning of this angle? Also, why are you multiplying by 0,01% c, and not dividing by c, like the original formula you mentioned (L/c)? -- Sorry, I tried to find this information elsewhere on the internet, and also some of your other answers, but wasn't able to.
– user137288
Commented Jan 8, 2021 at 13:49
• the R.A. of Andromeda. It's not in the plane of the ecliptic. Come to think of it, it's the angle relative to the ecliptic, not the celestial equator, so it needs a tweak. Of course, it's near Pisces, the sign during the equinox, so it's a small tweak. (Astrology can be useful.)
– JEB
Commented Jan 8, 2021 at 16:20
• Thanks! That stands for right ascension, correct? Regarding my second comment, why did you multiply by 0,01% c instead of dividing by c?
– user137288
Commented Jan 8, 2021 at 17:25

All you really need to know to understand the twin paradox is that clocks (including biological clocks) measure the lengths of their worldlines, and different worldlines that begin and end at the same events (spacetime points) can have different lengths, if they're different shapes.

Drawing planes of simultaneity is no help in solving twin-paradox problems. On a Euclidean plane, if you drive from point A to B in two cars by two different routes then the tripmeters may show different total distances at the end. You could draw planes perpendicular to the path of each car and look at where they intersect the path of the other car. When one car makes a turn, the perpendicular turns by the same angle, and may sweep over a large part of the other car's path, and you could say that the other car's tripmeter changes by that length "during the turn". If you're careful you'll even get the correct tripmeter readings at the end, because Euclidean geometry is internally consistent. But you won't learn anything from it. It's simply not true that there is a correct matching point on the other car's track that you can find by this or any other geometric construction. The two cars just took different paths, end of story.

If you really want to draw these perpendiculars on spacetime diagrams, you can use the fact that if your worldline has slope $$m$$ then the perpendicular plane of simultaneity has slope $$1/m$$. This is similar to Euclidean geometry (where the perpendicular slope is $$-1/m$$) but with a sign flip, as is often the case with these things.

• "you can use the fact that if your worldline has slope m then the perpendicular plane of simultaneity has slope 1/m" -- Does that mean my drawing is generally correct? I mean: the more the worldline tilts to the right, the more the line of simultaneity tilts upward (as stated by the other answer to this question)?
– user137288
Commented Jan 6, 2021 at 14:28
• Excellent answer by the way, as usual. Thanks.
– user137288
Commented Jan 6, 2021 at 14:29