SR only applies exactly in the absence of gravity. So in the real world it is just an approximation that works well enough for predicting things where the effect of gravity is small (such as interactions between small high velocity particles in accelerators near the Earth’s surface, or between spacecraft and small bodies like asteroids far from planets, but not for things like apples falling out of trees on Earth).
In regions where it does provide a good approximation, it works just as well for accelerated as unaccelerated frames, but for accelerated frames the formulas needed to express physical laws in terms of the observer’s coordinates are more complicated.
Tag: relativity
So the excuse used NOT to apply relativity theory in the twin paradox is a brief period of zero seconds at the turnaround point?
No one who knows what they are talking about has suggested “NOT to apply relativity theory”. On the contrary, the correct application of relativity theory leads to the conclusion that when the twins re-unite they agree on the fact that they have both seen the traveller age less. They just disagree on when during the trip the Earth-based twin aged faster. The one on Earth thinks it happened at a steady rate throughout the trip and the traveller (after actually seeing it during the return trip) thinks (after making the light travel time correction) that it happened quickly during the turn-around.
Prior to the turn around, each sees the other ageing more slowly (due to the Doppler effect) and, even after making the light travel time correction, thinks that part of that slowdown remains unexplained (and so in some sense is “really” happening).
But any claim that during the outbound journey “we know for a fact that the travelling twin is younger than the earth twin” (or vice versa) is completely false. There is nothing that is absolutely true about the relative ages of the twins until they are at rest with respect to one another.
One-way speed of light
Some authors (referenced in this Wikipedia article ) have alleged that The “one-way” speed of light, from a source to a detector, cannot be measured independently of a convention as to how to synchronize the clocks at the source and the detector – and that this somehow has important implications with regard to the interpretation and application of Special Relativity.
But I beg to differ.
Consider the scenario of two remote systems, B and C, that are stationary with respect to us who, let us say, are at position A. Now let a light signal from B be emitted towards both C and A, and let C, on receiving that signal immediately signal towards us at A. The time delay between our receipt of the two signals will be the light travel time from B to C plus the difference between the light travel times to us from B and C.
So, if we let [math]t_{XY}[/math] be the light travel time from X to Y, then the time difference between the direct and indirect signals starting at B is given by [math]\Delta t_B = (t_{BC} +t_{CA})-t_{BA}[/math].
And if we do the process in reverse for signals starting at C we get a time difference of [math]\Delta t_C = (t_{CB} +t_{BA})-t_{CA}[/math].
So, if we get the same time difference both ways, then we’ll know that [math](t_{BC} +t_{CA})-t_{BA} = (t_{CB} +t_{BA})-t_{CA}[/math],
which gives [math]t_{BC} -t_{CB} =2 (t_{BA}-t_{CA})[/math].
So in that case the difference between light travel times from B to C and from C to B is exactly twice the difference between the travel times from B to A and C to A.
Now let’s restrict to the case where B and C are separated by a distance [math]d[/math] and are both at the same distance [math]D[/math] from A, and let [math]v_{XY}[/math] be the speed of light in the direction from X to Y and [math]v_0[/math] be the speed in the perpendicular direction towards A from the midpoint of the line BC.
Then for the case of equal time differences both ways our previous result becomes [math]\frac{d}{v_{BC}} -\frac{d}{v_{CB}} =2 (\frac{D}{v_{BA}}-\frac{D}{v_{CA}})[/math].
In the case that A is midway between B and C (which is what the equidistance gives us in the case of one space dimension) we have [math]d=2D[/math] with [math]v_{BA}=v_{BC}[/math] and [math]v_{CA}=v_{CB}[/math], so the equal time differences condition is just that [math]\frac{d}{v_{BC}} -\frac{d}{v_{CB}} =2 (\frac{d/2}{v_{BC}}-\frac{d/2}{v_{CB}})[/math] – which is clearly always true even if the two one-way speeds are different.
For the case that A is not on the line BC, if we let [math]\omega=2\theta[/math] be the angle at A between AB and AC then [math]d=2D\sin{\theta}[/math] and the condition of equal time differences becomes [math]2D\sin{\theta}(\frac{1}{v_{BC}} -\frac{1}{v_{CB}}) =2D (\frac{1}{v_{BA}}-\frac{1}{v_{CA}})[/math].
If we let [math]v_{\theta}[/math] denote the speed of light in direction at angle [math]\theta[/math] from the right bisector of BC, and [math]T_{\theta}=\frac{1}{v_{\theta}}[/math], and if the equal time condition is observed in all directions, then we need to find a function [math]T_{\theta}[/math] satisfying the conditions [math](T_{\theta+\phi}-T_{-\theta+\phi})=\sin{\theta}(T_{\pi/2+\phi}-T_{-\pi/2+\phi})[/math]
Ron Davis’s answer to Who clocked the speed of light and how did they clock it? What instrument did they use and when? – Quora
Lorentz Expansion!
A Quora question asks:
Two guns on Earth D metres apart fires simultaneously at a metal plate moving by at close to light speed. Is the distance between the two bullet holes in the plate length contracted?
It depends on who thinks the guns fired simultaneously.
If, as seems most likely, the question means that those Earth-based guns were synchronized by someone on Earth with them, then the distance between the two holes, as measured by any observer stationary with respect to the Earth, is (and remains) exactly equal to the distance between the two guns. Since the plate appears contracted to these Earth-based observers, if it was marked with units of length in its own rest frame those markings would appear closer together to the Earth-based observers and so there would be more of them between the holes than the number of length units measured on Earth.
In other words the distance between the holes would appear to be
greater from the point of view of someone travelling with the plate.
It may seem puzzling that this happens despite the fact that from the traveller’s point of view the distance between the guns is “length contracted” and so appears to be less than that measured between them on Earth.
The puzzle is resolved by the fact that from the point of view of the traveller the guns did NOT fire simultaneously. The one making the front hole appears to the traveller to have gone off earlier. (And it is a worthwhile exercise for anyone seeking to learn about relativity to work through the calculation needed to show that the delay is by exactly the right amount for the forward movement of the plate to create the observed bigger distance between the holes).
Alternatively, if the traveller thinks the guns went off simultaneously, then the Earth-based observers think there is a delay. (And again, working out the details is a worthwhile exercise for any beginning student of the subject.)
P.S. The question of whether or not a length or object is “length contracted” does not really make sense without any mention of which observer is doing the measurement.
What does it mean to say that there is a distance between two events in time? (Another Quora Question)
It probably means that the speaker is taking a Galilean approach to physics.
In modern relativistic physics, the property of having a time-like separation between two events is independent of observer, but the magnitude of that separation depends on the observer. And two events which are spacelike separated, while having no time difference for some observers, will still appear to have a non-zero time difference for others.
So the concept of a time (or space) “distance” (ie a specific value of the difference) between two events in space-time does not make sense without reference to an observer.
However in the case of two time-like separated events the time difference is nonzero for all observers, and if we restrict to inertial observers it has a nonzero minimum (which corresponds to the time difference as seen by an observer who experiences both events directly without any intervening acceleration or gravitational field gradient). But although this minimum is in principle computable by any observer it does not correspond to the time difference actually “seen” by that observer.
Given that the Lorentz transformation is symmetrical with respect to interchange of space and time, how does it lead to length contraction but time dilation?
This is a question that I am surprised to not have seen before (especially since I have had to remind myself of the answer more than once – including, I suspect but can’t be sure, from way back before I entered my dotage).
It is true that in one space dimension the transformation equations
[math]x’=\gamma(x-\beta t)[/math] and [math]t’=\gamma(t-\beta x)[/math]
are completely symmetrical with respect to interchange of [math]x[/math] with [math]t[/math] and [math]x’[/math] with [math]t’[/math].
(and in the case of three space dimensions the same applies if [math]x[/math] and [math]x’[/math] are the coordinates in the same direction as the relative velocity, so it’s not got anything to do with the dimension).
So what is the difference?
Well here it is in a nutshell.
When we measure the length of a moving measuring rod, we look at both ends at the same time and so are looking at the spatial distance between two events at the same time in our frame of reference.
But when we measure the time between two ticks of a moving clock we are looking at the time difference between two events that are NOT at the same spatial position in our frame.
So the nature of the two measurements is not symmetrical with respect to interchange of space and time.
I may add some more explanation and diagrams to show how this does lead to contraction for the rod length and dilation for the tick interval, but I wanted to get this off my chest right away – and also to address a couple of natural follow-up questions.
Namely, what kind of measurements would give the symmetrical outcome? Are there situations in which these others might be relevant? And why do we instinctively prefer the ones we do?
So, for example, what kind of time measurement would be symmetrical compared to our usual rod length measurement (and so would give a “time contraction” rather than the usual time dilation)?
Since the rod length involves looking at both ends at the same time in our frame, the corresponding time measurement would involve looking at the interval between two ticks at the same place. But how can we do this if the clock is moving? Well we could if the clock was extended in space, and if we have a long train of clocks that are synchronized in their own frame, then you can easily check that observers who look at the time between the ticks right in front of them will actually see a shorter interval than that measured by the travelling system – ie a time contraction.
And going the other way, what kind of measurement would give a length dilation? Well that would have to be the symmetric version of our usual clock measurement. And corresponding to our usual measurement of the time interval between two ticks at the same place in the moving clock’s frame, interchanging space and time would have us measuring the spatial distance between events where the two ends of the rod are at the same time in the rod’s frame. For example the managers of the rod might set off flares at both ends in a way that they, travelling with the rod, perceive as simultaneous. If we measure the distance between where we see those two flares then it will indeed appear dilated relative to the length of the rod in its own frame.
So now we come to the final question. Is there anything really “wrong” about these alternative kinds of measurement? If so what is it? Or is there just something about us which makes us think of what we do as natural and the alternative as somehow, if not actually wrong, then at least rather odd?
Here’s what I think (at least for now). The thing that makes us prefer to measure lengths in terms of events at the same time in our frame but times in terms of events at the same place in the moving frame is the fact that we, as blobs of space time, are much more extended in time than in space. (This is evident in the fact that we live for many years but do not extend for many light years in our spatial extent – or equivalently that in units adapted to our own spatial and temporal extent the numerical value of c is very large.)
So here’s a follow-up question. Could we imagine an entity which was the other way around? (ie of brief duration but of great spatial extent) And from the point of view of such an entity would it make sense to define measurements differently (as suggested above to achieve the effect of time contraction and length dilation)?
OR is it more just a matter of causality?
P.S. This is a question and answer that I have been meaning to post for some time, but was prompted to do so by Domino Valdano’s excellent answer to another question (in which she covers pretty much the same ground with a slightly different way of expressing the ultimate reason for why we measure as we do – which I may yet end up deciding that I prefer to my own). Please do read that one too!
More TwinStuff from Quora
In the twin paradox where does the missing time go? If the twin turns back to Earth then turns away again their notion of now switches back to the past. What does this mean for the experience of the observer on Earth relative to the moving twin?
“In the twin paradox where does the missing time go?” I am not aware of any “missing” time. One twin experiences less time than the other but there is no gap where any time goes missing. (There is however an apparent speed-up of the Earth clock from the point of view of the traveller while turning around, and in the physically impossible case of an instantaneous turn-around that would look like a gap in the traveller’s understanding of what was happening “at the same time” on Earth; but in any possible actual scenario it would just be an apparent speed-up rather than a jump.)
“If the twin turns back to Earth then turns away again their notion of now switches back to the past. What does this mean…” Indeed! Does that sentence actually have any meaning at all?
Perhaps what that second sentence is referring to is the traveller’s idea of the time that is “now” back on Earth. It is true that when we accelerate away from something we infer a slowing down of its clock at a rate proportional to the distance. If the distance is great enough this effect can make the clock “behind” us appear to stop, but beyond that distance (called the “Rindler horizon”), rather than see it run backwards we actually just don’t see it at all. (And, yes, the Rindler horizon perceived by an accelerated observer is indeed related to the Event horizon surrounding a gravitational singularity.)
Proper Acceleration
The proper acceleration of an observer is that observer’s own sensed acceleration. Its description in terms of felt force would require an excursion into dynamics, but it can be defined kinematically if the observer is aware of some other frame which is known for some reason to be inertial. After acceleration of an observer, the previous frame of that observer appears (to that observer) to be moving in the opposite direction. So if the previous inertial frame (which matched the observer at time [math]t[/math] on the traveller’s own clock) is observed at time [math]t+\Delta t[/math] to have relative velocity [math]-\Delta v[/math], then the limiting ratio [math]a(t)=\lim\limits_{\Delta t\to 0}\frac{\Delta v}{\Delta t}[/math] is what we take as the proper acceleration. (We will ensure later that this purely kinematic definition does indeed match the dynamical definition in terms of felt force but for now just take it as “the” definition of proper acceleration.)
Any observer, [math]\mathcal{O}[/math], lives through a sequence of inertial frames [math]\mathcal{F}_{\mathcal{O}}(t)[/math], labelled by the “proper time” [math]t[/math] showing on [math]\mathcal{O}[/math]’s clock, and each such frame associates any other frame with a relative velocity and a space-time displacement.
Let us use [math]v_{t}[/math], [math]t_{t}[/math], and [math]x_{t}[/math] to represent possible velocities and coordinates relative to the inertial frame [math]\mathcal{F}_{\mathcal{O}}(t)[/math], with [math]v_{t}(\mathcal{F})[/math], [math]t_{t}(\mathcal{F})[/math], and [math]x_{t}(\mathcal{F})[/math] being the values of those variables corresponding to some other frame [math]\mathcal{F}[/math] as seen from [math]\mathcal{F}_{\mathcal{O}}(t)[/math].
If [math]\mathcal{O}[/math] has proper acceleration [math]a(t)[/math], then the frame at time [math]t+\Delta t[/math] is thus seen from that at time [math]t[/math] as having velocity and coordinates given by [math]v_{t}=v_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=a(t)\Delta t +O((\Delta t)^{2})\approx a(t)\Delta t[/math], [math]x_{t}=x_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=a(t)\frac{(\Delta t)^{2}}{2}+O((\Delta t)^{3})\approx a(t)\frac{(\Delta t)^{2}}{2}[/math], and [math]t_{t}=t_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=\Delta t +O((\Delta t)^{2})\approx \Delta t[/math] [where the reason this last is not exactly [math]\Delta t[/math] is because, as soon as the velocity changes, the clock measuring that [math]\Delta t[/math] for the accelerated observer is moving (and so slowed down) relative to that of the observer at time [math]t[/math]].
[The inertial frame of the observer at time [math]t+\Delta t[/math] thus has time axis [math]x_{t+\Delta t}=0[/math] given in terms of the time [math]t[/math] coordinates approximately by [math]x_{t}-a(t)\frac{(\Delta t)^{2}}{2}=(a(t)\Delta t)(t_{t}-\Delta t)[/math], and the spacelike axis of simultaneity at [math]t_{t+\Delta t}=0[/math] is given by [math]t_{t}-\Delta t=(a(t)\Delta t)(x_{t}-a(t)\frac{(\Delta t)^{2}}{2})[/math].So the simultaneity space [math]t_{t+\Delta t}=0[/math] intersects [math]t_{t}=0[/math] at the event where [math]-\Delta t=(a(t)\Delta t)(x_{t}-a(t)\frac{(\Delta t)^{2}}{2})[/math], i.e. [math]x_{t}=-\frac{1}{a(t)}+O((\Delta t)^2)[/math].
For the case of constant acceleration [math]a[/math] this makes all the simultaneity spaces pass through the same “fulcrum event” and this can be used to get a geometric proof that the worldline is hyperbolic and that events beyond the “Rindler horizon” (which is always at [math]x_{t}=-\frac{1}{a}[/math]) remain forever inaccessible to the accelerated observer.
This approach is discussed in more detail in the Wikipedia page on Rindler Coordinates (see also this version on the anonymously authored ‘mathpages’ website, and this discussion by Greg Egan). But in the discussion below I shall take what I think is a more intuitively direct path, and try to say something about the case of non-constant proper acceleration as well.]
As judged by the “birth” frame (corresponding to [math]t=0[/math]), a difference of [math]\Delta t[/math] in the value of [math]t_{t}(\mathcal{F})[/math] corresponds to a difference of [math]\gamma(t)\Delta t[/math] in [math]t_{0}(\mathcal{F})[/math](where [math]\gamma(t)=\frac{1}{\sqrt{1-{v_{b}(t)}^{2}}}[/math]).
[Note: It is also true that, as judged by the frame at time [math]t[/math], a difference of [math]\Delta t[/math] in the value of [math]t_{0}(\mathcal{F})[/math] corresponds to a difference of [math]\gamma(t)\Delta t[/math] in [math]t_{t}(\mathcal{F})[/math] (since the question of which frame appears to be time dilated depends on which frame is making the comparison). And if this still seems confusing then you may need to re-visit the earlier discussion of time dilation.]So, if we use [math]t_{b}(t)[/math] as shorthand for [math]t_{0}(\mathcal{F}_{\mathcal{O}}(t))[/math], and similarly for [math]x_{b}(t)[/math] and [math]v_{b}(t)[/math], then the relations above can be written in terms of the birthframe coordinates as follows:
[math] t_{b}(t+\Delta t)=t_{b}(t)+\gamma(t) \Delta t+O((\Delta t)^{2})[/math], [math] x_{b}(t+\Delta t)=x_{b}(t)+v_{b}(t)(\gamma(t) \Delta t)+O((\Delta t)^{2})[/math], and [math] \begin{align}v_{b}(t+\Delta t)&=v_{b}(t)[+]a(t)\Delta t =\frac{v_{b}(t)+a(t)\Delta t}{1+(v_{b}(t))(a(t)\Delta t)}\\&=(v_{b}(t)+a(t)\Delta t)(1 – (v_{b}(t))(a(t)\Delta t))+O((\Delta t)^{2})\\&= v_{b}(t)+a(t)\Delta t – (v_{b}(t))^{2}(a(t)\Delta t)+O((\Delta t)^{2})\end{align}[/math](where [+] is just shorthand for the relativistic velocity sum).
So we have the differential equations
[math]t_{b}'(t)=\gamma(t)=\frac{1}{\sqrt{1-{v_{b}(t)}^{2}}}[/math], [math]x_{b}'(t)=\gamma(t)v_{b}(t)=\frac{v_{b}(t)}{\sqrt{1-{v_{b}(t)}^{2}}}[/math], and [math]v_{b}'(t)=a(t)(1-{v_{b}(t)}^{2})[/math].The last of these is separable and so is easily solved (by cross-multiplying and integrating) to get [math]\int \frac{v_{b}'(t)dt}{(1-{v_{b}(t)}^{2})}=\int a(t)dt[/math].
This then gives [math]\ln{\sqrt{\frac{1+v}{1-v}}}=\int a(t)dt[/math], which gives [math]\frac{1+v}{1-v}=e^{2\int a(t)dt}[/math] and so [math]v_{b}(t)=\frac{e^{2\int a(t)dt}-1}{e^{2\int a(t)dt}+1}=\tanh(\int a(t)dt)[/math].
Note: Here the integrated proper acceleration [math]\int a(t)dt[/math] is the subjective velocity change, which from now on we’ll call [math]v_{s}(0,t)[/math], and [math]v_{b}(t)=\tanh(v_{s}(0,t))[/math] is the apparent velocity of the traveller as inferred by the birthframe observer at the birthframe time when the traveller’s clock reads time [math]t[/math].
For the coordinates, with respect to the birthframe, of [math]\mathcal{O}[/math]’s worldline event at proper time [math]t[/math] we now get
[math]\begin{align}t_{b}(t)&=\int\frac{dt}{\sqrt{1-{v_{b}(t)}^{2}}}=\int\frac{dt}{\sqrt{1-{\tanh(v_{s}(0,t)}^{2}}}\\&=\int\cosh(v_{s}(0,t))dt\end{align}[/math]and
[math]\begin{align}x_{b}(t)&=\int\frac{v_{b}(t)dt}{\sqrt{1-{v_{b}(t)}^{2}}}=\int\cosh(v_{s}(0,t))\tanh(v_{s}(0,t))dt\\&=\int\sinh(v_{s}(0,t))dt\end{align}[/math].Note: By FTC&ChainRule, [math]x_{b}'(t)=\sinh(v_{s}(0,t)dt)\frac{d}{dt}v_{s}(0,t)[/math] and [math]t_{b}'(t)=\cosh(v_{s}(0,t)dt)\frac{d}{dt}v_{s}(0,t)[/math], so the apparent velocity as seen by the birthframe observer is given by [math]v_{b}=\frac{dx_{b}}{dt_{b}}=\frac{x_{b}'(t)}{t_{b}'(t)}=\tanh(v_{s}(0,t))[/math].
This is in agreement with our previously derived formula for [math]v_{b}(t)[/math], but to get it in terms of the [math]t_{b}[/math] coordinate we need to compose with the function giving the proper time [math]t[/math] of [math]\mathcal{O}[/math] when it is at birthframe time [math]t_{b}[/math] – which is the composition inverse of the function [math]t_{b}(t)=\int\cosh(\textstyle\int a(t)dt)dt=\int\cosh(v_{s}(0,t))dt[/math] .
In general, the above integral is not easy to compute explicitly, but in the special case that [math]a(t)=a[/math] is constant, then we have [math]v_{s}(0,t)=at+C_{v}[/math], so [math]v_{b}(t)=\tanh(at+C_{v})[/math] (with [math]v_{b}(0)=0[/math] giving [math]C_{v}=0[/math]), and in that case we can complete the integrals for the coordinates to get
[math]t_{b}(t)=\int\cosh(at+C_{v})dt=\frac{1}{a}\sinh(at+C_{v})+C_{t}[/math] with [math]t_{b}(0)=0[/math] giving [math]C_{t}=0[/math], so [math]t_{b}(t)=\frac{1}{a}\sinh(at)[/math], and [math]x_{b}(t)=\int\sinh(at+C_{v})dt=\frac{1}{a}\cosh(at+C_{v})+C_{x}[/math] with [math]x_{b}(0)=0[/math] giving [math]C_{x}=-\frac{1}{a}[/math], so [math]x_{b}(t)=\frac{1}{a}\cosh(at)-\frac{1}{a}[/math].And in this case we can invert to get [math]t(t_{b})=\frac{1}{a}[\sinh^{-1}(at_{b})][/math] which gives
[math]\begin{align}v_{b}(t_{b})&=\tanh(a[\frac{1}{a}[\sinh^{-1}(at_{b})])\\&=\tanh(\sinh^{-1}(at_{b}))=\frac{at_{b}}{\sqrt{1+(at_b)^2}}\end{align}[/math]Which, as expected, remains less than [math]c[/math] (which is 1 in our units) and so the acceleration observed by the birthframe observer is decidedly NOT constant but rather decreasing to zero as the speed gets closer and closer to [math]c[/math].
[insert brief discussion of Rindler coords and horizon for the case of const accel for all time]For the case of piecewise constant proper acceleration, say [math]a(t)=a_{i}[/math] between proper times [math]t_{i}[/math] and [math]t_{i+1}[/math], we can use the foregoing analysis to follow how the frame changes in each interval. But in this case the period of constant acceleration is starting at [math]t=t_{i}[/math] rather than [math]t=0[/math], so the conditions for the constants of integration (which depend on [math]i[/math]) are now as follows:
[math]\tanh(a_{i}t_{i}+C_{v,i})=v_{b}(t_{i})[/math] gives [math]C_{v,i}=\tanh^{-1}(v_{b}(t_{i}))-a_{i}t_{i}=v_{s}(0,t_{i})-a_{i}t_{i}[/math], but we’ll often just use [math](a_{i}t_{i}+C_{v,i})=\tanh^{-1}(v_{b}(t_{i}))=v_{s}(0,t_{i})[/math] directly; [math]\frac{1}{a_{i}}\sinh(a_{i}t_{i}+C_{v,i})+C_{t,i}=t_{b}(t_{i})[/math] gives [math]\begin{align}C_{t,i}&=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(a_{i}t_{i}+C_{v,i})=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(v_{s}(0,t_{i}))\\ \text{[also}&=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(\tanh^{-1}(v_{b}(t_{i})))\\&=t_{b}(t_{i})-\frac{1}{a_{i}}v_{b}(t_{i})\sqrt{1+v_{b}(t_{i})^2}\;\; \text{];}\\ \text{and}\end{align}[/math] [math]\frac{1}{a_{i}}\cosh(a_{i}t_{i}+C_{v,i})+C_{x,i}=x_{b}(t_{i})[/math] gives [math]\begin{align}C_{x,i}&=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(a_{i}t_{i}+C_{v,i})=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(v_{s}(0,t_{i}))\\ \text{[also}&=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(\tanh^{-1}(v_{b}(t_{i})))\\&=x_{b}(t_{i})-\frac{1}{a_{i}}\frac{v_{b}(t_{i})}{\sqrt{1+v_{b}(t_{i})^2}}\;\; \text{].}\end{align}[/math]So, for [math]t_{i}<t<t_{i+1}[/math] we have
[math]\begin{align}v_{b}(t)&=\tanh[\tanh^{-1}(v_{b}(t_{i}))+a_{i}(t-t_{i})]\\&=\frac{v_{b}(t_{i})+\tanh(a_{i}(t-t_{i}))}{1+(v_{b}(t_{i}))(\tanh(a_{i}(t-t_{i})))}=v_{b}(t_{i})[+]\tanh(a_{i}(t-t_{i}))\end{align}[/math]with
[math]\begin{align}t_{b}(t)&=\int\cosh[a_{i}t+C_{v,i}]dt=\frac{1}{a_{i}}\sinh[a_{i}t+C_{v,i}]+C_{t,i}\\&=\frac{1}{a_{i}}\sinh[a_{i}t+v_{s}(0,t_{i})-a_{i}t_{i}]+\{t_{b}(t_{i})-\frac{1}{a_{i}}\sinh[v_{s}(0,t_{i})]\}\\&=t_{b}(t_{i})+\frac{1}{a_{i}}\{\sinh[a_{i}(t-t_{i})+v_{s}(0,t_{i})]-\sinh[v_{s}(0,t_{i})]\}\text{,}\end{align}[/math](giving
[math]\begin{align}t_{b}(t_{i+1})-t_{b}(t_{i})&=\frac{1}{a_{i}}\{\sinh[a_{i}(t_{i+1}-t_{i})+v_{s}(0,t_{i})]-\sinh[v_{s}(0,t_{i})]\}\\&=\frac{1}{a_{i}}\{\sinh[v_{s}(0,t_{i+1})]-\sinh[v_{s}(0,t_{i})]\}\\&=\int_{t_{i}}^{t_{i+1}}\cosh(v_{s}(0,t))dt\end{align}[/math] since [math]\frac{d}{dt}\sinh[v_{s}(0,t)]=\cosh[v_{s}(0,t)]\frac{d}{dt}v_{s}(0,t)[/math] and [math]\frac{d}{dt}v_{s}(0,t)=a_{i}[/math] for [math]t_{i}<t<t_{i+1}[/math] ),and
[math]\begin{align}x_{b}(t)&=\int\sinh[a_{i}t+C_{v,i}]dt=\frac{1}{a_{i}}\cosh[a_{i}t+C_{v,i}]+C_{x,i}\\&=\frac{1}{a_{i}}\cosh[a_{i}t+v_{s}(0,t_{i})-a_{i}t_{i}]+\{x_{b}(t_{i})-\frac{1}{a_{i}}\cosh[v_{s}(0,t_{i})]\}\\&=x_{b}(t_{i})+\frac{1}{a_{i}}\{\cosh[a_{i}(t-t_{i})+v_{s}(0,t_{i})]-\cosh[v_{s}(0,t_{i})]\}\text{.}\end{align}[/math](giving
[math]\begin{align}x_{b}(t_{i+1})-x_{b}(t_{i})&=\frac{1}{a_{i}}\{\cosh[a_{i}(t_{i+1}-t_{i})+v_{s}(0,t_{i})]-\cosh[v_{s}(0,t_{i})]\}\\&=\frac{1}{a_{i}}\{\cosh[v_{s}(0,t_{i+1})]-\cosh[v_{s}(0,t_{i})]\}\\&=\int_{t_{i}}^{t_{i+1}}\sinh(v_{s}(0,t))dt\end{align}[/math] since [math]\frac{d}{dt}\cosh[v_{s}(0,t)]=\sinh[v_{s}(0,t)]\frac{d}{dt}v_{s}(0,t)[/math] and [math]\frac{d}{dt}v_{s}(0,t)=a_{i}[/math] for [math]t_{i}<t<t_{i+1}[/math]).Spacetime diagrams from POV of both twins in the finite acceleration versions of the twin “paradox”.
If properly worded this would have been a good question. From the comments attached to the question we see that the questioner is really asking for two diagrams, one showing the point of view of each of the twins rather than a single diagram showing the coordinate systems of both. And by the ambiguous condition of “constant acceleration” he means constant acceleration as perceived by the stationary observer rather than constant proper acceleration as felt by the traveler.
Of course the case of constant proper acceleration would be more realistic in the sense that it just requires the traveler to experience a constant g-force, whereas constant observed acceleration requires an increasing applied force (which would actually become unbounded as the speed got closer and closer to c). But for a limited time it is possible to keep adjusting the applied force so as to create a constant acceleration relative to the Earth’s frame and in that case the relevant part of the world line (in any inertial frame) is a simple parabolic segment (rather than the hyperbolic segment that would correspond to constant proper acceleration).
With the assumption of constant accelerations in the stay-at-home inertial frame, the spacetime diagram in terms of stay-at-home coordinates is just this:
Here we have a parabolic segment taking the traveler from the start event to where he reaches a cruising speed of , followed by a straight line segment or the bulk of the trip, then a parabolic segment for deceleration, a vertical segment for time spent at the destination, another parabolic segment for acceleration back towards home, straight line for the cruise, and the final parabolic deceleration phase.
In this diagram the coordinates are [math]t_H[/math] for the time on the stay-at-home clock and [math]x_H[/math] for the position in the stay-at-home coordinate system, and we will use the name [math]x^{I}_{HT}[/math] for the function which gives the traveler’s position in stay-at-home coordinates in terms of the time [math]t_H[/math] that the stay-at-home observer perceives as concurrent with the traveler’s arrival at that position. (The superscript I on the function name is to indicate that this is what he infers rather than what he actually sees). So the graph of [math]x_H=x^{I}_{HT}(t_H)[/math] shows what the stay-at-home thinks is the position of the traveler when his (stay-at-home) clock shows time [math]t_H[/math]. This is one interpretation of the homie’s “point of view” but it is not what he actually sees.
What the homie actually sees is delayed by the light travel time from the traveler (just as what we see of a distant star many light years away is not what it is actually happening there now but what happened that many light years ago).
So to get the graph of what the homie actually sees we must look at the point on the previous graph that is the source of a light signal reaching home at time [math]t_H[/math].
We can get a graph of what the homie actually sees by tracing down each light-line from the [math]t_H[/math] axis to where it meets the [math]x_H=x^{I}_{HT}(t_H)[/math] graph and plotting the [math]x_H[/math] value of that event as [math]x^{O}_{HT}(t_{H})[/math] (with the superscript [math]O[/math] identifying the position actually observed at time [math]t_{H}[/math] rather than that which was inferred to be simultaneous).
Now let’s look at things from the point of view of the traveler.
The vertical axis now corresponds to the traveler’s clock time [math]t_T[/math] and the horizontal lines either to distances that we want to associate with that time. If we want to plot what is actually seen by the traveler then for each [math]t_T[/math] we plot the position coordinate corresponding to the distance from which the signal is coming (as determined, eg, by parralax). and if we want to plot where the traveller infers that the homie actually is at the time [math]t_T[/math] we attribute the distance of the source seen at time [math]t_T[/math] to the earlier time [math]t_T-\frac{|x_T|}{c}[/math]
What the traveler actually sees at any event on his worldline is exactly the same as what is seen by an inertial traveler whose world line passes through that event with zero relative velocity (ie for which the worldline is tangent to that of the traveler at that event). Such a tangential traveler sees the values of [math]x^O_{TTE}[/math] and [math]t^O_{TTE}[/math] corresponding to a time [math]x^O_{TTE}/c[/math] earlier in his own frame – so that [math]x^O_{TTE}=[/math] and [math]t^O_{TTE}[/math]
events that are seen by him at the time his clock shows time [math]t_T[/math] with position along that line corresponding to the distance he measures (eg by parallax) to that event; or, in the case of the inferred view those that are inferred to be happening simultaneously with that [math]t_T[/math] click of the clock or to those distance [math]x_T[/math] which he measures (eg by parallax) to whatever event we are talking about.
When the traveler’s clock reads time [math]t_T[/math] he is at the event for which [math]t_H[/math] is such that [math]\int_{0}^{t_H}\frac{1}{1-v(t)^2}dt=t_T[/math] where [math]v(t)=at,v_f,v_f-a(t-t_f),0,-a(t-t_r),-v_f,-v_f+a(t_h-t)…[/math]
To find [math]x^{I}_{TH}(t_T)[/math] we have to use the [math]t_T=[/math]constant simultaneity space for the traveler and find its intersection with the [math]x_H=0[/math] worldline of the homie
The light signal that the traveler is receiving from homie at this event can be seen from the above diagram to come from [math]t_H=[/math] and we have [math]x_H=[/math]
And