# General relativity, twins & Schopenhauer

rolling stone
TL;DR Summary
A simple "thought experiment” based on a circular trajectory is investigated by elementary notions of general relativity. The result is surprising: three different observers draw three different conclusions about the same scenario.
Consider a mass ##M## that generates a curvature of the space-time and an observer ##O##, fixed and positioned at such a great distance from ##M## that the time ##t## of its clock is not affected by ##M##.
Suppose that the observer ##O##, in his polar coordinates reference system centred at ##M##, observes a spacecraft (of negligible mass with respect to ##M##) that rotates with constant linear speed ##v## in the equatorial plane at his feet. Suppose also that on this spacecraft there is spaceman ##S_1## equipped with a clock whose time is ## t_1##.

THE SCENARIO SEEN BY THE OBSERVER ##O##
According to the observer ##O##, the infinitesimal line element, in the Schwarzschild's metric and spherical polar coordinates, of the spacecraft moving in the circular path is:
(1)## ~~~~~~~~~~ds^2=-(1-r_s/R)c^2 dt^2 + R^2 dφ ^2##
where ##G## is the gravitational constant, ##r_s=2GM/c^2## is the Schwarzschild radius, ## φ## is the azimuth angle and ##R## is the radius of the circular path.
If the spacecraft rotates with constant angular speed ## dφ / dt = v /R## :
(2)## ~~~~~~~~~~ ds^2= - (1- r_s / R) c^2 dt^2 + v^2 dt ^2##
In particular if ##R=GM/v^2##, the spacecraft moves by inertia along a geodetic path, where the centrifugal acceleration is compensated by the centripetal one.
Since the same infinitesimal line element computed by spaceman ##S_1## is:
(3)## ~~~~~~~~~~ ds^2=- c^2 dτ_1^2 ##
where ##τ_1## is the proper time of ##S_1##, from eq.(2) and eq.(3) it follows that:
(4)## ~~~~~~~~~~ - c^2 dτ_1^2 =-(1-r_s/R) c^2 dt^2 + v^2 dt ^2 ##
(5)## ~~~~~~~~~~ dτ_1 = \sqrt {1 - v^2/c^2 - r_s/R } ~ dt ##
Let us now consider a second spacecraft , with the spaceman ##S_2## on board and a clock whose time is ## t_2##, moving in the same circular path as ##S_1##, with linear velocity ##-v## in the opposite direction.
NOTE: in order to avoid any clash between the two spacecraft s we can imagine, for example, that the common trajectory is a single virtual train track on which the two spacecraft s may gently slide away, face up and face down, respectively.
By the same considerations previously made for ##S_1## , the observer ##O## may deduce that for the spaceman ##S_2## too:
(6)## ~~~~~~~~~~ dτ_2 = \sqrt {1 - v^2/c^2 - r_s/R } ~ dt ##
where ##τ_2## is the proper time of ##S_2##.
Eq.(5) and eq.(6) show that, according to the observer ##O##, the time on the spacecraft s of ##S_1## and ##S_2## flows exactly at the same rate.
Therefore if we assume that ##S_1## and ##S_2## are twins who, at the initial reference time ##t_1= t_2=0##, are positioned with their spacecraft s at the same point in the circular trajectory, according to the fixed external observer ##O## they will continue to have the same age regardless of the number of laps they will run around the mass## M##.

THE SCENARIO SEEN BY THE SPACEMAN ##S_1##
Moving in a geodetic path, each of the two spacemen may believe that his spacecraft is stationary while the twin's one is in motion. No physical experiment, on either of the two spacecraft s, will ever demonstrate the opposite, since their motion is inertial with constant speed. Consequently, according to spaceman ##S_1## , ##S_2## is moving away with relative speed ##–2v##.
In order to express, according to ##S_1## , the infinitesimal line element traveled by ##S_1## , the eq.(1) must be rewritten in terms of the clock time ##t_1##, remembering that ##(1-r_s/R) dt^2= dt_1^2## (because the clock of ##S_1## is located at distance ##R## from the centre of ##M##) and that ## dφ / dt_1 = - 2 v /R ## .Thus:
(7)## ~~~~~~~~~~ds^2=-c^2 dt_1^2 + R^2 ( dφ/dt_1)^2~dt_1 ^2= - c^2 dt_1^2+4v^2 dt_1 ^2##
But, according to spaceman ##S_2##:
(8)## ~~~~~~~~~~ ds^2= - c^2 dτ_2^2 ##
Now combining eq.(7) with eq.(8) :
(9)## ~~~~~~~~~~- c^2 dτ_2^2 = - c^2 dt_1^2+4v^2 dt_1 ^2##
one obtains:
(10)## ~~~~~~~~~~ dτ_2 = \sqrt {1 - 4~v^2/c^2 } ~ dt_1 ##
So, according to spaceman ##S_1## , the time of the clock on board of the spacecraft of ##S_2## flows more slowly than his own; therefore after one lap his twin ##S_2## will be younger than him.

THE SCENARIO SEEN BY THE SPACEMAN ##S_2##
By repeating, on behalf of the spaceman ##S_2##, the same previous reasoning one obtains:
(11)## ~~~~~~~~~~ dτ_1 = \sqrt {1 - 4~v^2/c^2 } ~ dt_2 ##
So, according to spaceman ##S_2##, the time of the clock on board of the spacecraft of ##S_1## flows more slowly than his own; therefore after one lap his twin ##S_1## will be younger than him.

DISCUSSION
In conclusion: after the two spacecraft s have run one lap around the mass ##M##,
- according to the external observer ##O##: “the twins ##S_1## and ##S_2## will continue to have the same age" [cfr. eq.(5) and eq.(6) ]
- according to spaceman ##S_1##: “##S_2## will be younger than ##S_1##” [cfr. eq.(10)]
- according to spaceman ##S_2##: “##S_1## will be younger than ##S_2##” [cfr. eq.(11)].

Questions:
- What is wrong in the previous reasoning?
- If nothing is wrong, may we assume that in the theory of relativity the description of the same scenario depends on the reference frame system from which the scenario is observed?
- If so, isn't Schopenhauer right when he says that "the world is a representation"?

2022 Award
By symmetry, the spacemen are the same age when they meet up. I haven't followed through your maths in detail, but I suspect that what you've done is the same basic problem as trying to synchronise clocks in a rotating frame in flat spacetime. By extending one of your spacemen's local inertial frame's planes of simultaneity into more global structures as you have done, you've constructed a helical plane, not a closed plane. Then you've forgotten to account for the discontinuity in the planes when the orbits close and you can compare clocks directly.

Some insight might be obtained by considering direct observation of the other spaceman's clock as he orbits.

• vanhees71
Mentor
Summary:: A simple "thought experiment” based on a circular trajectory is investigated by elementary notions of general relativity. The result is surprising: three different observers draw three different conclusions about the same scenario.

What is wrong in the previous reasoning?
##ds^2## is invariant. Your calculations fail to reflect that.

The reason is that you did not actually do a valid coordinate transform from the Schwarzschild coordinates to the astronauts frame. You need to write down an actual transform and calculate the metric in those coordinates.

• Ibix
...
Moving in a geodetic path, each of the two spacemen may believe that his spacecraft is stationary while the twin's one is in motion. No physical experiment, on either of the two spacecraft s, will ever demonstrate the opposite, since their motion is inertial with constant speed. Consequently, according to spaceman ##S_1## , ##S_2## is moving away with relative speed ##–2v##.
...
Why is that? In flat spacetime sure, but they are orbiting and passing each other every half orbit. How are they moving away from each other with constant velocity ##-2v##!

• Ibix
2022 Award
Why is that? In flat spacetime sure, but they are orbiting and passing each other every half orbit. How are they moving away from each other with constant velocity ##-2v##!
Actually the ##2v## is not correct to start with, since relativistic velocity addition applies. The speed measured by either astronaut is not the sum of the speeds measured by a Schwarzschild hovering observer.

I've looked at the maths properly now and I think I'm correct about the OP implicitly adopting a notion of space that is discontinuous somewhere around the orbit. I drew spacetime diagrams illustrating this point in this post and subsequent ones (note that the diagram in the linked post has a slight bug, corrected in a subsequent post). This was a discussion about trains on circular tracks but applies here too. The cyan lines are the worldlines of train carriages and the green helix around the cylinder is a naive "line of now" as I believe has been implicitly adopted in this thread. The failure of the helix to close accounts for the "missing" time in each spaceman's calculation of the other's elapsed time.

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Mentor
- What is wrong in the previous reasoning?
So, if I were working this problem I would do the following. First, clearly state the transform that we are using to go into the astronaut's frame:
$$\begin{equation} \label{transform} t'=\sqrt{1-\frac{r}{R}-\frac{v^2}{c^2}} t \end{equation}$$ $$\phi' = \phi - \frac{v}{R} t$$

Notice that the expression for ##\phi'## involves ##t##. So the two terms will no longer be independent. Missing this is the key mistake above that happened from the "handwaving" approach rather than a rigorous approach. So next I would differentiate ##(\ref{transform})## and solve for the unprimed variables to obtain
$$\begin{equation} \label{dtransform} dt=\frac{1}{\sqrt{1-\frac{r}{R}-\frac{v^2}{c^2}}} dt' \end{equation}$$ $$d\phi = d\phi' + \frac{v}{R\sqrt{1-\frac{r}{R}-\frac{v^2}{c^2}}} \ dt'$$

Notice again, the additional term in the expression for ##d\phi##. Substituting this ##(\ref{dtransform})## into your (1) gives
$$\begin{equation} ds^2 = -c^2 dt'^2 + R^2 d\phi'^2 + \frac{2 R v}{\sqrt{1-\frac{r}{R}-\frac{v^2}{c^2}}} dt' d\phi' \end{equation}$$ which has some cross terms and is too painful for me to want to evaluate to confirm that ##ds^2## is indeed invariant.

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Mentor
What is wrong in the previous reasoning?

First, you have made obviously invalid assumptions, such as (as has already been pointed out) the assumption that the spacemen are always moving away from each other at relative speed ##2 v##.

Second, more generally, you have not defined a valid "rest frame" for either of the spacemen. If you try to do so, you will find that it is impossible to define a "rest frame" for either spaceman that has the intuitive properties you are assuming and also includes the other spaceman for an entire half-orbit (i.e., the entire region of spacetime between two meetings of the spacemen). So the basic method you are trying to use will not work.

You can define a coordinate chart in which one of the spacemen is at rest for all time (this looks like what @Dale is doing in post #6), but in this chart the metric will not be diagonal and intuitive assumptions that you are implicitly relying on will not be valid (for example, surfaces of constant coordinate time in the chart will not be orthogonal to the worldline of the spaceman that is at rest in the chart). Of course you can use such a chart to calculate the same answer that you get in the standard Schwarzschild chart, but such a chart does not constitute the kind of "rest frame" for a spaceman that you are implicitly thinking of, since it lacks properties that your intuition says such a frame ought to have.

• Dale and martinbn
Mentor
clearly state the transform that we are using to go into the astronaut's frame

This looks like an analogue of the transformation from Langevin to Born coordinates in Minkowski spacetime. It would be instructive to write down the line element in this chart.

• Dale
Mentor
This looks like an analogue of the transformation from Langevin to Born coordinates in Minkowski spacetime. It would be instructive to write down the line element in this chart.
Yes, even what I wrote down is not the full line element because I assumed ##dr=0## and ##\theta=\pi/2##. And it is already a big hairy mess with cross terms.

It is important to understand that coordinate transformations should always be done to make the math simpler. In this case, it does not.

• Nugatory
rolling stone
The reason is that you did not actually do a valid coordinate transform from the Schwarzschild coordinates to the astronauts frame. You need to write down an actual transform and calculate the metric in those coordinates.
In the scenario seen by spaceman ##S_1## he may legitimately consider himself - on board of his spacecraft that moves by inertia along a geodetic path (if ##R=GM/v^2##) - at rest with respect to the mass ##M## (if ##M## is a homogeneous sphere). So ##S_1##’s metric is the Schwarzschild metric generated by ##M## , exactly as the metric that observer ##O## in his own scenario is allowed to use.
The first difference - with respect to the scenario seen by ##O## - is that the clock of ##S_1## is located at distance ##R## from ##M## (and that's why ##(1-r_s/R) dt^2= dt_1^2## ) , while the second difference is that the angular velocity of ##S_2## is twice (that's why ## dφ / dt_1 = - 2 v /R ##).

rolling stone
Why is that? In flat spacetime sure, but they are orbiting and passing each other every half orbit. How are they moving away from each other with constant velocity −2v!
This problem is really tough, but I think it may be overcome, first by remembering that spacecraft s are moving along a geodetic path in a “curved" space, and then by resorting to the definition of speed as the fraction between the space that has been traveled (not the space that will be traveled) and the time spent to travel it. Obviously, to proceed with the measurement of speed, it is first necessary to establish an origin of space and time, then from this common origin one may count the number of meters traveled and the number of seconds used to travel them.
This "retroactive" conception of speed becomes clearer when one thinks about average speed instead of instantaneous speed.
Alternatively, one may imagine that the spacecraft s are not moving (in opposite direction) along a single circle, but along a solenoid with negligible height compared to its radius.

Homework Helper
Gold Member
2022 Award
In the scenario seen by spaceman ##S_1## he may legitimately consider himself - on board of his spacecraft that moves by inertia along a geodetic path (if ##R=GM/v^2##) - at rest with respect to the mass ##M## (if ##M## is a homogeneous sphere). So ##S_1##’s metric is the Schwarzschild metric generated by ##M## , exactly as the metric that observer ##O## in his own scenario is allowed to use.
The first difference - with respect to the scenario seen by ##O## - is that the clock of ##S_1## is located at distance ##R## from ##M## (and that's why ##(1-r_s/R) dt^2= dt_1^2## ) , while the second difference is that the angular velocity of ##S_2## is twice (that's why ## dφ / dt_1 = - 2 v /R ##).
If you look at it from a purely geometrical point of view:

1) The Schwarzschild metric defines the differential geometry of the spacetime region in question.

2) The spacetime distance along a given worldline is invariant (under any coordinate transformation). This can be proved using the axioms of geometry.

3) You cannot get different results by using different reference frames or different coordinate systems.

4) Therefore, you must have made a mistake in the mathematics.

In effect, you have tried to show that axiomatic differential geometry (in this case of Schwarzschild geometry) is inconsistent. That can't be right, as that is simply the pure mathematical geometry underpinning the physical theory.

To confirm GR the results of an experiment must match the predictions of the (mathematical) theory. There is no question of the underlying mathematics being inconsistent, as that is just pure maths.

For example, because we live in a non-Euclidean spacetime does not mean that Euclidean geometry itself is wrong and can be shown to be mathematically inconsistent. Both Euclidean and Schwarzschild geometries and perfectly valid mathematically. You cannot disprove either. You can only do an experiment to determine which geometry represents our spacetime.

• Dale
2022 Award
In the scenario seen by spaceman ##S_1## he may legitimately consider himself - on board of his spacecraft that moves by inertia along a geodetic path (if ##R=GM/v^2##) - at rest with respect to the mass ##M## (if ##M## is a homogeneous sphere). So ##S_1##’s metric is the Schwarzschild metric generated by ##M## , exactly as the metric that observer ##O## in his own scenario is allowed to use.
Hm. An inertial body that is initially at rest in Schwarzschild coordinates will immediately start to fall straight down. You contend that an orbiting observer may use the same coordinate system on the same spacetime and regard themselves as at rest in it, and hence you imply that an orbiting observer immediately begins to fall straight down. Does this not seem problematic to you?

You are still waving your hands instead of writing down a valid coordinate transform. Your problems come from the disconnect between your assumptions about the transformed coordinates and the reality of them. They cannot be the same as Schwarzschild coordinates. Also you may choose between them being orthogonal and having a discontinuity somewhere (my interpretation of what you were doing), or not being orthogonal (@Dale's interpretation). You cannot have both.

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• Dale
Mentor
In the scenario seen by spaceman S1 he may legitimately consider himself - on board of his spacecraft that moves by inertia along a geodetic path (if R=GM/v2) - at rest with respect to the mass M (if M is a homogeneous sphere). So S1’s metric is the Schwarzschild metric generated by M , exactly as the metric that observer O in his own scenario is allowed to use
No, this is not correct. He may indeed consider himself to be at rest wrt M, meaning that he can use a coordinate system where both he and M are at rest. However, it does not follow that the metric is the standard Schwarzschild form of the metric in those coordinates.

This should be obvious by the fact that the astronaut is moving inertially and in the standard Schwarzschild coordinates a stationary observer is not inertial, but experiences an outward proper acceleration. Furthermore, the central mass is not spherically symmetric in these coordinates as the mass has angular momentum wrt these coordinates.

You can consider the astronaut at rest, but to do so requires an actual legitimate coordinate transform. The resulting coordinates are not simply a time-rescaling of Schwarzschild. You must transform the spatial coordinates also to get the astronaut at rest or you get a metric that does not match the astronaut’s own experience.

• PeterDonis, Ibix and PeroK
rolling stone
No, this is not correct. He may indeed consider himself to be at rest wrt M, meaning that he can use a coordinate system where both he and M are at rest. However, it does not follow that the metric is the standard Schwarzschild form of the metric in those coordinates.

This should be obvious by the fact that the astronaut is moving inertially and in the standard Schwarzschild coordinates a stationary observer is not inertial, but experiences an outward proper acceleration. Furthermore, the central mass is not spherically symmetric in these coordinates as the mass has angular momentum wrt these coordinates.
OK Dale
Unfortunately this is not obvious to me, therefore I understand that I have still a lot to learn about the Schwartschild metric.

rolling stone
So, if I were working this problem I would do the following. First, clearly state the transform that we are using to go into the astronaut's frame:
(1)t′=1−rR−v2c2t ϕ′=ϕ−vRt
And thanks also for this suggestion of yours. I will try to work it out

Mentor
OK Dale
Unfortunately this is not obvious to me, therefore I understand that I have still a lot to learn about the Schwartschild metric.
Sorry about that. You are right, nothing in GR is “obvious”. I should have said “unambiguous”. What I meant is that the astronaut can unambiguously tell that they are not at rest in the Schwarzschild coordinates by looking at their accelerometer. Their accelerometer reads 0 and if they were at rest in Schwarzschild coordinates then it would not read 0.

• vanhees71
2022 Award
I'd recommend studying the flat spacetime case of circular motion. The only complication that it doesn't have that this scenario does is gravitational time dilation between the orbiting observers and the observer at infinity. The maths is a good bit simpler as a result, since ##t'=t/\gamma##.

Mentor
nothing in GR is “obvious”

Reminds me of the old joke about the professor who starts class by writing an equation on the board; then he writes a second equation and says, "Now it is obvious that the second equation follows from the first." Then he pauses, stares at the board, and mutters, "Wait a bit, I may be wrong..."

He then spends the whole class scribbling equations all over the board, and at the end of the class, right before the bell rings, he steps back and smiles and says, "I was right! It is obvious that the second equation follows from the first!"

• • vanhees71, martinbn and Dale