# Twin paradox for (accelerated) dummies?

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In summary, the conversation discusses the application of the Lorentz transformation to velocity and acceleration in the context of special relativity. It is explained that the equations for constant accelerated motion can be derived from the Lorentz transformation and that this type of motion is represented by a hyperbolic worldline on a spacetime diagram. The conversation also explores a scenario involving a spaceship with constant proper acceleration and the effects on time for a twin on board compared to a stationary observer on Earth. The equations for this scenario are derived and used to calculate the time and distance traveled by the spaceship, leading to a disagreement between the traveling twin and the stationary twin about their ages and the number of birthdays missed.
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TL;DR Summary
I'm trying to express a solution to the twin paradox, without resorting too much to Minkowski diagrams, lines of simultaneous events and all that stuff, considering a constant |g| acceleration round trip. Is this solution of the twin paradox correct from the S-R point of view, considering a constant |g| round trip?
Applying the Lorentz transformation to velocity and acceleration, we can easily obtain that Aγ3 = α, where A is the acceleration measured by the stationary observer and α is the proper acceleration of the relative moving object. From this point, the equations for a constant accelerated motion can also be obtained. Is important to remark that such equations are directly derived from the Lorentz transformation itself, and so is a consequence in S-R.

In S-R, constant accelerated motion is then described in the spacetime diagram by a hyperbolic worldline.

In layman's terms this makes sense, because if the speed of light were not invariant, then the accelerated worldline would be a parabola... but then in S-R this is not possible because the c-speed limit, so a hyperbolic worldline, asymptotically converging to the c-speed, is easy to understand and visualize.

Now, imagine a spaceship that has a proper acceleration of |1g|. It comes from far, far away..., and just passes by an Earth space station with zero relative speed at that position (A). Bob-twin takes advantage of this event and promptly jumps into the spaceship, that continues with its |1g| acceleration moving away from Earth.

The trajectory of the spaceship runs far away from any massive object, and the only interest of Bob-twin is to check some S-R predictions, with its Casio© G-atomic clock in his wrist, which was a gift from his Alice-twin sister at their last mutual birthday.

At a certain intermediate position (X), the spaceship stops its rear engine, fires the front engine, and begins to slow down (relative to Earth) at |1g|. Let the proper time measured at the spaceship be τx when the switching occurs.

At some position later (B), the relative speed with Earth will be zero again. From the Earth observer, this event happens at (TB,XB), and for Bob-twin his clock reads τB when he observes such event. There's no need to be very precise with the timings, give or take a few hours.

Then, the spaceship begins moving towards Earth, always at |1g|, until a certain intermediate position is reached, at which the spaceship turns off its front engine, fires the rear engine, and begins to slow down (relative to Earth) at |1g|. When the spaceship reaches more or less the starting position (relative speed to Earth: zero), Bob-twin jumps off this spaceship, and is recovered by another, that brings him safely to Earth.

It's important to remark that the spaceship is always moving with a proper acceleration magnitude of 1g. Always!

Then, using the hyperbolic motion coordinate equations, we can derive easily the following conditions that have to be true at the event of switching the engines at proper time τx, taking into account that the spaceship will be at event (TB,XB) sometime later, as its motion continues afterwards the switching event with |1g| acceleration:

##\frac{c^2}{g} \cdot \cosh{(\frac{g \cdot \tau_x}{c})}=X_B-\frac{c^2}{g} \cdot \left[\cosh{(\frac{g \cdot (\tau_x - \tau_B)}{c})}-1\right]##

##\frac{c^2}{g} \cdot \sinh{(\frac{g \cdot \tau_x}{c})}=cT_B + \frac{c^2}{g} \cdot \sinh{(\frac{g \cdot (\tau_x - \tau_B)}{c})}##

##\tanh{(\frac{g \cdot \tau_x}{c}})=\tanh{(\frac{g \cdot (\tau_B - \tau_x)}{c}})##

Solving the system, we can get the corresponding τx, τB and XB for a given TB time in Earth:

Let's say that TB is 10 Years at Earth, so Alice-twin is about 20 years older now. Solving the equations, the total elapsed proper time τ in the spaceship is about 9.088 Years (is the reading of the Bob-twin atomic-clock elapsed time), and so Bob-twin has aged about 9 years, and the spaceship has traveled 8.246+8.246 light-years round trip.

So Alice-twin argues with Bob-twin: I missed you birthday's gift 9 times in a row, but you missed mine 20 times, so you owe me 11 gifts. Bob-twin obviously disagrees.

(As I have a lot of problems with LaTex in PF, I also attach an image of the equations, in case of...).

PeroK and Dale
Inertial frame coordinate time, ##t##, is related to the proper time, ##\tau##, of an observer starting from rest with constant proper acceleration ##a## by ##at=\sinh(a\tau)##. In your example ##a=g##, and working in years and light years, ##g\approx 1##. Thus after the 5 coordinate year acceleration phase, ##\tau\approx \sinh^{-1} 5=2.31##. The entire travel time is four times that, or about 9.25 years on the ship for 20 at home.

Given the accuracy of ##g\approx 1## that's a reasonable match to your calculation.

There's a known bug with LaTeX preview not working in new threads. Just do the best you can, post, reload the page, and LaTeX should render and you can make any corrections. It'll also work in replies.

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Sagittarius A-Star, vanhees71, member 728827 and 1 other person
Lluis Olle said:
Applying the Lorentz transformation
Between what frames? If the traveling twin is under a constant acceleration, he is not at rest in any inertial frame. And the transformation between an inertial frame and an accelerated frame is not a Lorentz transformation.

vanhees71
Lluis Olle said:
So Alice-twin argues with Bob-twin: I missed you birthday's gift 9 times in a row, but you missed mine 20 times, so you owe me 11 gifts. Bob-twin obviously disagrees.
Bob won't disagree, by the way, at least not about the maths. He can do the same computation we've done and come to the same conclusion that Alice should be older.

vanhees71, member 728827 and Dale
vanhees71, member 728827 and Ibix
PeterDonis said:
Between what frames? If the traveling twin is under a constant acceleration, he is not at rest in any inertial frame. And the transformation between an inertial frame and an accelerated frame is not a Lorentz transformation.
Is not what this guy thinks. You can read here the document that was already hyperlinked in the original post.

And, by the way, if ##\alpha=\gamma^3 \cdot A##, which is basic and well known Lorentz transformation applied to accelerations, then:

##\alpha=\gamma^3 \cdot A, \quad \gamma^3 \cdot A=\frac{d(\gamma \cdot v)}{dT} \Rightarrow \alpha=\frac{d(\gamma \cdot v)}{dT}\Rightarrow \alpha \cdot T=\gamma \cdot v##

Integrating, we get:

##v=\frac{dX}{dT}= \frac{\alpha \cdot T}{\sqrt{1+\frac{\alpha^2 \cdot T^2}{c^2}}}##

Integrating again:

##X^2 - (cT)^2 =\frac{c^4}{\alpha^2}##

From the canonical form of a hyperbola, we have:

##\frac{X^2}{a^2}-\frac{(cT)^2}{b^2}=1 \Rightarrow X=a \cdot \cosh{\beta}, \; cT=b \cdot \sinh{\beta}, \quad a=\frac{c^2}{\alpha}, b=\frac{c^2}{\alpha}##

To get what the ##\beta## is, you can consider that in S-R the proper time is the arclength of the worldline using the Minkowski metric, and you would get that ##\beta=\frac{\alpha \cdot \tau}{c}##.

vanhees71 and Dale
Lluis Olle said:
Is not what this guy thinks.
Yes, it is. He is Lorentz transforming between inertial frames. He is not Lorentz transforming between an inertial and an accelerated frame. His transformation equations between an inertial frame and his accelerated frame (Rindler coordinates) are equations 2.6 and 2.7 in the paper. Those are not Lorentz transformations.

vanhees71
@PeterDonis i think you are nitpicking here. He didn’t say that he was performing a Lorentz transform between an inertial frame and an accelerating frame. He just said that from the Lorentz transform you can derive how acceleration transforms, even linking to the specific derivation he was using. Maybe he should have said “Deriving from the Lorentz transform” instead of “Applying the Lorentz transform”, but since he linked to the paper he was referencing it doesn’t seem like anything to bother about.

vanhees71 and Motore
Dale said:
i think you are nitpicking here
Perhaps I am, but the treatment in the paper referenced seems somewhat sloppy to me. But the hyperbolic equations being used in this thread are correct, and there are multiple ways of deriving them anyway, so for this thread I agree it's not a big issue.

vanhees71 and Dale
PeterDonis said:
Yes, it is. He is Lorentz transforming between inertial frames. He is not Lorentz transforming between an inertial and an accelerated frame. His transformation equations between an inertial frame and his accelerated frame (Rindler coordinates) are equations 2.6 and 2.7 in the paper. Those are not Lorentz transformations.
The paper referenced begins with bold letters:

"Chapter 1 Transformation of acceleration between two reference frames"

And continues:

"Before we may start with the transformation of accelerations, we of course need the transformation of velocities. Hence consider first of all the differentials

dx = γ(dX − vdT)
dt = γ(dT − dX v/c2)"

What the author does not explicitly write in the paper, I think because everybody in the targeted audience of the course "Theoretical Physics II: Electrodynamics" should be aware of, is that when in S-R an accelerated frame is considered by the inertial observer, then the Lorentz transformation is from the inertial frame to the MCRF:

"MCRF -- Momentarily Comoving Reference Frame -- An inertial frame of reference which happens to be moving in the same direction, at the same speed, as an object or an accelerated frame which we're examining. Many questions in relativity can only be addressed in an inertial frame. In many cases, when the frame under consideration isn't inertial, such questions can still be addressed by considering the MCRF of the accelerated frame."

Equations 2.6 and 2.7 in the paper are just the coordinate equations of kinematic motion from the inertial observer point of view, and that are derived from the A⋅γ3=α equation. The author doesn't say that those equations are a Lorentz transformation, but just in the previous paragraph he writes:

"This equation represents a hyperbolic path in a Minkowski diagram, i.e. a uniformly accelerated observer will follow a hyperbolic path in the stationary frame."

In another reply in this thread, I just more or less follow the same reasoning of the paper's author, to highlight how easily from a well established and known S-R equation that relates the acceleration measured by an inertial observer to the proper acceleration measured at an accelerated frame, you can come to the coordinate equations of motion of the accelerated object in the inertial frame spacetime diagram!

I also attach an excerpt of the W. Rindler book "Introduction to Special Relativity" about the subject.

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@Lluis Olle I enjoyed your scenario. Was there something you wanted? Do you need someone to check your numbers or something similar?

member 728827
Dale said:
@Lluis Olle I enjoyed your scenario. Was there something you wanted? Do you need someone to check your numbers or something similar?
Thanks @Dale, if it's correct then I'm done. Then, one more to the bag of twin paradox explanations. Hope this was somehow different.

And by the way, the spaceship-loop gives for a Sci-Fi novel... who build that spaceship, why is passing by Earth every 20 years? What the builders wanted to show us?

Lluis Olle said:
if it's correct then I'm done
The basic idea is right. I didn’t confirm the numbers.

Lluis Olle said:
when in S-R an accelerated frame is considered by the inertial observer, then the Lorentz transformation is from the inertial frame to the MCRF
Yes, this answers the question I posed in post #3.

Dale said:
The basic idea is right. I didn’t confirm the numbers.
That's the Python code I used to solve the equations.

Solving the system:
from mpmath import *
import numpy as np
import math
# ............
c =mpf('299792458')  # speed of light
c2=power(c,2)        # square of speed of light
g =mpf('9.8')        # desired (comfortable) proper acceleration

YTS=mpf(365*24*3600)

mp.dps = 32
mp.pretty=True
# ............
def EqTwins(T_b):

Tb=T_b*YTS

eq1 = lambda ttx,ttb,xxb: (c2/g)*sinh(g*ttx/c)-c*Tb-(c2/g)*sinh(g*(ttx-ttb)/c)
eq2 = lambda ttx,ttb,xxb: (c2/g)*cosh(g*ttx/c)-xxb+(c2/g)*(cosh(g*(ttx-ttb)/c)-1)
eq3 = lambda ttx,ttb,xxb: tanh(g*ttx/c)+tanh(g*(ttx-ttb)/c)

Jc = lambda ttx,ttb,xxb: ( (c*(cosh(g*ttx/c)-cosh(g*(ttx-ttb)/c)),c*cosh(g*(ttx-ttb)/c),0),
(c*(sinh(g*ttx/c)+sinh(g*(ttx-ttb)/c)),-c*sinh(g*(ttx-ttb)/c),-1),
((g/c)*(power(sech(g*(ttx-ttb)/c),2)+power(sech(g*ttx/c),2)),-(g/c)*power(sech(g*(ttx-ttb)/c),2),0) )

tx,tb,xb=findroot([eq1,eq2,eq3],(Tb/4,Tb/2,(Tb/2)*c),solver='mdNewton',J=Jc)

print('tx=%.3e tb=%.3e xb=%.3e'%(tx/YTS,tb/YTS,(xb-(c2/g))/(c*YTS)))

return tx,tb,xb
# ............
EqTwins(10)

haushofer and Dale
I'm worried about Bob-twin getting hurt in the sudden acceleration change when switching the front and rear engines all at once. To mitigate this issue, now the engines will follow a progressive switch on/off and off/on programmed sequence.

(Not true, I'm just bored...)

Lets define then a time variable acceleration that we call g(T). We know from S-R theory that:

##\gamma^3 \cdot A=\frac{d}{dT}(\gamma \cdot \upsilon)=g(T)##

From WolframAlpha© (God bless it) we obtain the following solution for the differential equation:

##\upsilon(T)=\frac{c \cdot (\int_{T_1}^{T}g(t)dt+k_1)}{\sqrt{c^2+\int_{T_1}^{T}g(t)dt)^2+k_1^2+2k_1\int_{T_1}^{T}g(t)dt}}##

We can get the integration constant k1 when v(T)=v1 at T=T1 and so ##\int_{T_1}^{T_2}g(t)dt=0##:

##k_1=\frac{c \cdot \upsilon_1}{\sqrt{c^2-\upsilon_1^2}}=\gamma_1 \cdot \upsilon_1##

I prefer that there is no change in the speeds between the start and the end of the switching sequence. So, after inverting the acceleration we just end having the same speed as we started with. And so, our programmed acceleration sequence must be such as:

##\int_{T_1}^{T_2}g(t)dt=0##

For example, the following programmed sequence will do it, and be quite progressive also:

##g(T)=g_0 \cdot cos{(\frac{\pi}{ΔT} \cdot (T-T_1))} \quad T=T_1..T_1+ΔT##

We can consider a switching proper time duration of 10" (as measured in the spaceship), and so ΔT is of course greater because of time dilation (about 5x at halfway in our original example).

The distance traveled during the switch sequence, can be obtained integrating the obtained speed function:

##\int_{T_1}^{T_2}\upsilon(T)dT##

But for now, I'm only interested in the approximate value on how much more advances the spaceship at its original g0 acceleration, compared with the programed variable sequence. I know that:

##\upsilon=\frac{1}{\gamma}\int_{T_1}^{T_2}g(T)dT##

where cheating in a solitaire game, I take the γ out of the integral, and approximately:

##ΔX \simeq \frac{1}{\gamma} \cdot (\frac{1}{2}g_0 \cdot ΔT^2 - \frac{2g_0}{\pi^2} \cdot ΔT^2)##

Considering a γ about 5, then ΔT is 50", and the difference is not greater than 1,500 m.

So, in our original system of equations, we don't need to worry about X related function, because the difference in position is so small, and we only change - easily - the condition about speeds, considering now that the ζx proper time is the ending time of the switching event.

##\tanh{(\frac{g \cdot (\tau_x - Δ\tau)}{c}})=\tanh{(\frac{g \cdot (\tau_B - \tau_x)}{c}})##

where Δζ is the proper duration of the switching sequence (10" in this example).

As a general rule, adding complications to a scenario seldom aids understanding,

vanhees71
As a general rule, general rules doesn't usually work very well.

Dale, vanhees71 and PeroK
Lluis Olle said:
As a general rule, general rules doesn't usually work very well.
This one, however…..

Let's analyze Bob's point of view in his non-inertial accelerated frame aboard the spaceship's laboratory. For him, the coordinate equations that describe Alice's motion are as follows:

##t(T)=\frac{c}{g} \cdot \ln{(\sqrt{1+(\frac{g \cdot T}{c})^2}+\frac{g \cdot T}{c})}##

##x(T)=\frac{c^2}{g} \cdot (\sqrt{1+(\frac{g \cdot T}{c})^2}-1)##

Then, for the same conditions described in the original scenario of this thread, the following set of simultaneous equations must apply to fulfill such stated conditions:

##\frac{c}{g} \cdot (\ln{\sqrt{1+(\frac{g \cdot T_x}{c})^2}}+\frac{g \cdot T_x}{c})-t_b-\frac{c}{g} \cdot (\ln{\sqrt{1+(\frac{g \cdot (T_x-T_b)}{c})^2}}+\frac{g \cdot (T_x-T_b)}{c})=0##

##\frac{c^2}{g} \cdot \sqrt{1+(\frac{g \cdot T_x}{c})^2}-x_b+\frac{c^2}{g} \cdot (\sqrt{1+(\frac{g \cdot (T_x-T_b}{c})^2}-1)=0##

and for the speed:

##g \cdot T_x+g \cdot (T_x-T_b)=0##

where Tx is the Alice's time in which the first switching (acceleration) event will happen, Tb is the Alice's time in which the relative speed gets to zero, tb is the time for the same event in Bob's time, and xb is the distance traveled so far by Bob at his time tb. Solving the system, we get:

Tx=5.000 Years tb=4.544 Years xb=8.246 Light-years

So Bob will give the same answer according to his non-inertial frame, although for him the first switching event happens exactly at Year 5 in Alice's time.

Solving the system:
from mpmath import *
import numpy as np
import math

# ............
c =mpf('299792458')  # speed of light
c2=power(c,2)        # square of speed of light
g =mpf('9.8')        # proper acceleration

YTS=mpf(365*24*3600)

mp.dps = 48

def EqTwins_Inertial(T_b):

Tb=T_b*YTS

eq1 = lambda ttx,ttb,xxb: (c2/g)*sinh(g*ttx/c)-c*Tb-(c2/g)*sinh(g*(ttx-ttb)/c)
eq2 = lambda ttx,ttb,xxb: (c2/g)*cosh(g*ttx/c)-xxb+(c2/g)*(cosh(g*(ttx-ttb)/c)-1)
eq3 = lambda ttx,ttb,xxb: tanh(g*ttx/c)+tanh(g*(ttx-ttb)/c)
# Jacobian
Jc = lambda ttx,ttb,xxb: ( (c*(cosh(g*ttx/c)-cosh(g*(ttx-ttb)/c)),c*cosh(g*(ttx-ttb)/c),0),
(c*(sinh(g*ttx/c)+sinh(g*(ttx-ttb)/c)),-c*sinh(g*(ttx-ttb)/c),-1),
((g/c)*(power(sech(g*(ttx-ttb)/c),2)+power(sech(g*ttx/c),2)),-(g/c)*power(sech(g*(ttx-ttb)/c),2),0) )

tx,tb,xb=findroot([eq1,eq2,eq3],(Tb/4,Tb/2,(Tb/2)*c),solver='mdNewton',J=Jc)

print('INE tx=%.3e tb=%.3e xb=%.3e'%(tx/YTS,tb/YTS,(xb-(c2/g))/(c*YTS)))

return tx,tb,xb

def EqTwins_Non_Inertial(T_b):

Tb=T_b*YTS

eq1 = lambda TTx,ttb,xxb: (c/g)*log(sqrt(1+power(g*TTx/c,2))+g*TTx/c)-ttb-(c/g)*log(sqrt(1+power(g*(TTx-Tb)/c,2))+g*(TTx-Tb)/c)
eq2 = lambda TTx,ttb,xxb: (c2/g)*sqrt(1+power(g*TTx/c,2))-xxb+(c2/g)*(sqrt(1+power(g*(TTx-Tb)/c,2))-1)
eq3 = lambda TTx,ttb,xxb: g*TTx+g*(TTx-Tb)

Tx,tb,xb=findroot([eq1,eq2,eq3],(Tb/4,Tb/2,(Tb/2)*c))

print('NIE Tx=%.3e tb=%.3e xb=%.3e'%(Tx/YTS,tb/YTS,(xb-(c2/g))/(c*YTS)))

return Tx,tb,xb

EqTwins_Inertial(10)
EqTwins_Non_Inertial(10)

I attach both Alice and Bob's worldlines. For the Alice worldline, remark that the coordinate speed of Alice is faster than the coordinate speed of c, but I think that the "real" speed of light line should be drawn as ##c \cdot (1+\frac{g \cdot x}{c^2})##, as for Bob in the origin of a non-inertial frame, the "speed of light" will be dependent on the position in which "the photon" is located relative to Bob. Someone could confirm this point?

Also, Bob and Alice has agreed beforehand to celebrate some birthday parties during the travel, and the attached figure shows the proposed schedule.

PeroK and Dale
Lluis Olle said:
For him, the coordinate equations that describe Alice's motion are as follows
Since Bob is non-inertial there is no standard coordinate system that is regarded as "Bob's frame". So it is usually a good idea to be specific. You can either describe them by name if you are using a well-known coordinate system, like Rindler or radar coordinates. Or if you are using something unique then you can simply explicitly state the coordinate transform. It looks like you are using radar coordinates, is that correct?

The code and the plots look good. Having some programmed routines for solving these sorts of problems is quite useful.

member 728827 and PeroK
@Dale , I forgot to mention that the twin names are Alice and Bob Rindler. So, go figure.

Well, I think Bob is a Rindler observer in constant accelerated hyperbolic motion.

Dale
Lluis Olle said:
@Dale , I forgot to mention that the twin names are Alice and Bob Rindler. So, go figure.

Well, I think Bob is a Rindler observer in constant accelerated hyperbolic motion.

It'd be helpful, then, to say that Bob is using Rindler coordinates, and to name them. As there are a couple of different variations of Rindler coordinates, the best practice would be to give Bob's line element or metric tensor.

As is discussed by Misner in "Precis of General relativity", for the purposes of GR, giving a line element is both necessary and sufficient to completely describe a coordinate system.

However, as I think about the problem, it seems to me that the fact that Bob turns around means that Bob actually has a coordinate system that is divided into two different patches or charts. This is consistent with the rules for manifolds - while in many cases we can cover a manifold with one chart, we have the option to cover it with multiple charts as long as certain overlap conditions are met. In this case, I believe Bob's description of the manifold has two charts.

Basically, the meta level point is that one can replace the idea of a "frame of reference" with the idea of a coordinate chart, which as Misner suggests can be specified by its line element.

member 728827
pervect said:
It'd be helpful, then, to say that Bob is using Rindler coordinates, and to name them. As there are a couple of different variations of Rindler coordinates, the best practice would be to give Bob's line element or metric tensor.
Bob in his onboard laboratory is the Rindler observer of this story, and is using Rindler coordinates, described by the line element

##ds^2=-(1+\frac{g \cdot x'}{c^2}) \cdot d(ct')^2+(dx')^2##
pervect said:
the fact that Bob turns around means that Bob actually has a coordinate system that is divided into two different patches or charts
Both Bob and Alice use "two patches", because first they travel at a constant acceleration towards the destination turning point, but in a halfway position they must reverse the acceleration, just to arrive to the destination turning point with zero speed relative to Earth.

The only two values given in the scenario are the acceleration and the time in Alice's (Earth) frame in which the Bob spaceship reaches the turning point, which is 10 years (in Alice's time).

Also, in the "overlap" Event position, we need that the speeds of both trajectories to be equal. This also makes the piecewise function that describes the whole trajectory as smooth as it can be, so in fact there's no such "multiple charts", but a continuous function that is described with different mathematical equations in different time intervals, but reflects a continuous reality of events.

This describes half the voyage, but the return story is believed to be symmetric.

Also, I discussed in this thread the possibility of switching the accelerations gradually (to avoid hurting Bob during the transient phase), using a sinusoidal function that reverses the acceleration in 10" Bob's time (for example). That would make for a lot of "patches" indeed!

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Dale
That is a much better description.
Lluis Olle said:
Also, in the "overlap" Event position, we need that the speeds of both trajectories to be equal. This also makes the piecewise function that describes the whole trajectory continuous, so in fact there's no such "multiple charts", but a continuous function that is described with different mathematical equations in different time intervals, but reflects a continuous reality of events.
Regarding multiple charts. It is certainly possible to describe it using multiple charts or using a single chart. We call a “function that is described with different mathematical equations” a piecewise function. Such a function is typically written as $$f(t,x,y,z)=\begin{cases} f_1(t,x,y,z) & x\leq x_0 \\ f_2 (t,x,y,z) & x_0 < x \end{cases}$$
If you are using a single piecewise chart then your metric will be a piecewise function. More challenging however is that a coordinate chart must be invertible and smooth. Ensuring smoothness at the boundary between pieces is challenging and requires care.

It is also possible to use multiple overlapping charts. Then the metric need not be piecewise, but you will have to specify it separately for each chart. There will also need to be a coordinate transform between the two charts, but there is no need to ensure smoothness or even continuity between the charts.

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member 728827
Dale said:
If you are using a single piecewise chart then your metric will be a piecewise function.
The metric in this case doesn't change between the two trajectories and it's the same in both, because when g is towards the turning point, the x'-coordinate that Bob considers is positive, but when the spaceship inverts acceleration (let's say "-g"), then the x'-coordinate is negative, because it's then referenced to a turning point in the future, so it's at the negative x'-side of the new trajectory.

Consider that in order to be as comfortable as possible in his laboratory, Bob choose to accelerate at 1g, and the laboratory is oriented so the floor is orthogonal to the acceleration. When the acceleration is inverted, the whole laboratory turns around 180º, and that's why his x' coordinate is inverted also.

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Lluis Olle said:
Both Bob and Alice use "two patches", because first they travel at a constant acceleration towards the destination turning point, but in a halfway position they must reverse the acceleration, just to arrive to the destination turning point with zero speed relative to Earth.
This actually requires three coordinate patches, correct? The first patch for accelerating away from the starting point, the second after reversing the acceleration halfway to the turning point--and the third after reversing acceleration again halfway back to the starting point on the return leg, so as to arrive back at the starting point with zero speed relative to Earth.

Also, the description I just gave is only for Bob, correct? Alice remains in free fall at the starting point location the whole time, correct? That means Alice does not need to use multiple coordinate patches; she can just use the single inertial frame in which she is at rest the whole time.

PeterDonis said:
This actually requires three coordinate patches, correct?
Not quite, because the third one and the first one is the same thing. This is a closed cycle. It starts and ends in the same place, at different times of course, and repeats forever. You only switch the engines two times per cycle.
PeterDonis said:
Alice remains in free fall at the starting point location the whole time, correct?
Not quite. Alice follows a ballistic trajectory, much like a cannon that shoots vertically (against the g acceleration) a projectile with some initial speed.

At every of the two switching events (acceleration flip), we can consider that Alice is "shoot" with an initial speed against the g acceleration, so a time later will get to the maximum height in which the speed relative to Bob will be zero, and afterwards will "fall" to the Bob's laboratory floor and "crash" with the same but inverted speed.

Alice describes Bob motion using an Inertial Frame. Bob describes Alice's motion using a non-inertial frame. Every twin is moving with respect the other. Each of the twins worldlines would have two switching events. That two events happens at B1(2.272y, 4.123ly), B2(6.816y,4.123ly) and A1(5.0y,4.123ly), A2(15.0y,4.123ly), so they happen at the same spot but the twins don't agree in the time of the events.

Lluis Olle said:
the third one and the first one is the same thing
No, they aren't. For the third and first patches to be "the same thing", you would have to be able to continue the first patch into the region of spacetime occupied by the third, so that Bob's spatial coordinates would remain the same and Bob's time coordinate would be just right to account for the "gap" between the patches (where the second patch was used). You can't do that.

In other words, the fact that the metric and the direction of the ##x## coordinate are the same in the first and third patches is not enough to make them "the same thing". They are still distinct and disconnected coordinate patches.

Lluis Olle said:
Alice follows a ballistic trajectory
In other words, she is in free fall the whole time. That's what I was asking.

Lluis Olle said:
The metric in this case doesn't change between the two trajectories and it's the same in both, because when g is towards the turning point, the x'-coordinate that Bob considers is positive, but when the spaceship inverts acceleration (let's say "-g"), then the x'-coordinate is negative, because it's then referenced to a turning point in the future, so it's at the negative x'-side of the new trajectory.
This doesn't work. The Rindler frame is for a constant acceleration, it is not valid for a varying acceleration. Flipping the coordinates is a clever idea, but then it is not a single chart but multiple charts with a coordinate transform (e.g. ##x=-x'##) between the two charts.

Either approach is viable, but as written you are incomplete either way. Either you need to revise the metric for a single chart or you need to identify the transform between the multiple charts you are using.

Note, this is not a negative reflection on your work here. For new posters this is far better than anyone else I have seen in a very long time. It is just letting you know some of the difficulties that make doing this sort of thing rigorously more difficult than many people appreciate.

PeterDonis said:
In other words, she is in free fall the whole time. That's what I was asking.
Ok, so for Bob perspective Alice is in free falling the whole time. Then at the start position in Earth at T=t'=0, when Bob jumps into the spaceship, in Bob's non-inertial reference Alice is in free falling and is moving AWAY from Bob, but after some time after the acceleration switch event, Alice begins "free falling" TOWARDS Bob. Seems a Sesame street clip.

So is always "free falling", but changed from "going away" to "coming towards", what is not something you can explain changing the sign of the x'-coordinate.

If you mean that "free falling" means that the only force that acts over Alice is "gravity", or using GR equivalence principle an acceleration, then neither is true, because the "net" acceleration changed its direction, and during that transient phase, even when Bob really feels the effect of the change, will describe what is happening to Alice as a new force acting on her, that was not there before. You can think that the original "cause" of the "+g" acceleration is still there, but now there's a new "cause" that produces "-2g".

And is not free falling by any definition you try to apply. It's quasi-quasi-free falling, but not quite, that was what I answered.

Lluis Olle said:
And is not free falling by any definition you try to apply. It's quasi-quasi-free falling, but not quite, that was what I answered.
No, it's always free falling by the only measure that actually matters: what do accelerometers attached to Alice show? They always show zero. Thus she is in free fall - that is the definition. Bob trying to interpret her motion in terms of time varying gravitational fields does not change that her weighing scale always reads zero.

Dale said:
This doesn't work. The Rindler frame is for a constant acceleration, it is not valid for a varying acceleration.
The Rindler frame is derived for the infinite series of attached CMRF that at every instant has the same velocity as the non-inertial frame, and the "clock-hypothesis". If that hypothesis is true, the patch applied to transit from one worldline with +g, to the other with -g if there's continuity in the speed of the MCRF is correct.

Lluis Olle said:
The Rindler frame is derived for the infinite series of attached CMRF that at every instant has the same velocity as the non-inertial frame, and the "clock-hypothesis". If that hypothesis is true, the patch applied to transit from one worldline with +g, to the other with -g if there's continuity in the speed of the MCRF is correct.
Yes, what you've done works, I think (not sure off the top of my head, but you might need to worry about discontinuity in your X coordinates - however that's an annoyance rather than a problem). I think @Dale's point is that it's not one Rindler coordinate system with the sign of ##g## flipping, it's two coordinate systems with opposite ##g##s stitched together.

Dale

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