How Does Relativistic Deceleration Affect Velocity Calculations in Space Travel?

[Al68]

I think it's fair to say that, early on, Einstein did not make use of the geometric interpretations suggested by Minkowski. Formulated geometrically, the "paradox" is easily resolvable... in particular, by direct calculation (i.e. spacetime arc-length) without issues of "transformations" and "reference frames".

One should also note that modern relativists (who emphasize the geometric structure and not much on "issues of reference frames") interpret "SR" and "GR" differently from the early physicists and relativists and from many textbooks and pop-sci books that haven't caught up yet.

The key word missing in your post is "inertial" (which is not the same as "at rest"). The traveling twin can "regard himself at rest"... but he cannot regard himself as "inertial".

I'm not sure what you mean by direct calculation. I've seen it explained many ways. But in every case I have seen, the explanation, the drawings, and the calculations only describe the assumptions and conclusions made. For example, one could easily just assume that the ship was at rest, and draw a spacetime diagram just like the normal ones, except labeled the other way around. The same could be done with the calculations. I'm not saying that the accepted explanations are wrong. Just that the ones I've seen explain what happens, but not why it happens that way instead of a different way. And of course you're right, the ship's twin is not inertial during the turnaround. But, if one object accelerates away from another, the mathematics and kinematics look the same on paper regardless of which object is said to accelerate relative to the other.

I'm not saying that it's not relevant that only one twin accelerates. Maybe I just haven't seen the right explanation. That's why I ask so many questions. Is there an explanation available on the internet that shows why it's important that only one twin accelerates? Instead of just assuming that it's important.

Thanks,
Alan

 

[robphy]

I'm not sure what you mean by direct calculation.

By direct calculation, I mean the computation of the path-dependent integral \int_A^B ds, where A and B are events and ds is a timelike arc-length element in spacetime [analogous to the Euclidean arc-length used when measuring the length of a curve]. This always works. In certain simplified situations, like a piecewise-inertial path, one often computes proper-time intervals by Lorentz-boosting displacements between end-events (endpoints)...then summing. In the latter case, one often makes use of "time dilation" formulae or "length contraction" formulae, which are okay... although incomplete in my opinion if the key idea of spacetime-arc-length is not addressed.

I've seen it explained many ways. But in every case I have seen, the explanation, the drawings, and the calculations only describe the assumptions and conclusions made. For example, one could easily just assume that the ship was at rest, and draw a spacetime diagram just like the normal ones, except labeled the other way around. The same could be done with the calculations.

I'm not saying that the accepted explanations are wrong. Just that the ones I've seen explain what happens, but not why it happens that way instead of a different way. And of course you're right, the ship's twin is not inertial during the turnaround. But, if one object accelerates away from another, the mathematics and kinematics look the same on paper regardless of which object is said to accelerate relative to the other.

If all you drew were the worldlines of the two twins, you could be misled into thinking they were equivalent. The key point is that "the inertial observer between the two events logs the most proper time". [Mathematically, inertial-worldlines are geodesics.] To see this in your diagrams, you should draw in (say) a family of inertial-worldlines traveling with the same velocity. In the inertial-twin's diagram those worldlines are all straight (and are past and future extendible as such). In the rocket-twin's diagram, those inertial-worldlines will have a kink, break, or other irregularity in them... which no Lorentz transformation can transform away. [Draw the worldline of a ball sitting on the frictionless floor of the rocket. At the turn around event, according to the law of inertia, does the ball's worldline follow the rocket's worldline?] Another way to express this is to say that the coordinate system of the rocket-twin (which needs to be fully specified) does not have the same properties of the coordinate system of the inertial-twin.

I'm not saying that it's not relevant that only one twin accelerates. Maybe I just haven't seen the right explanation. That's why I ask so many questions. Is there an explanation available on the internet that shows why it's important that only one twin accelerates? Instead of just assuming that it's important.

Thanks,
Alan

The bottom line is that the "proper-time logged between two events" depends on the spacetime path under consideration... In Galilean/Newtonian relativity, for two given events, those time-intervals are equal regardless of the choice of path. In relativity, the clock effect could be seen even when each twin undergoes some acceleration... certainly there may situations when they are equal... but in general, they will differ. It's just easier to analyze if one case is inertial... as well as potentially confusing because now one has an opportunity to misuse and misinterpret the SR results up to that point... e.g. the "twin paradox".

 

[Al68]

The key point is that "the inertial observer between the two events logs the most proper time".

Well, it looks like if we assume that the inertial observer will always log the most proper time, then of course we will find that the observer who logs the most proper time will turn out to be the inertial observer. And if we assume that the observer who logs the most proper time is the inertial observer, we will find that the inertial observer will turn out to be the one who logs the most proper time.

And if we draw inertial worldlines as straight, and put kinks in non-inertial worldlines, we will end up with inertial worldlines that are straight and non-inertial worldlines that are kinked.

My question is, if the relative velocity between two objects changes, how do we know that the inertial worldline should be drawn as straight, and the non-inertial worldline should be drawn as kinked?

I'm not saying you are wrong, I'm just saying that the mathematics and drawings don't really seem to answer the question. They just seem to explain the answer afterward.

(edit) There are also examples where the twins could both be inertial and experience differential aging. The twins could be in intersecting orbits around earth, one nearly circular and the other very elliptical. Each twin would be following a geodisic the entire time, yet everytime they met, they would see that they experienced different lapses of proper time since the last time they met, even though they are both in freefall.

Thanks,
Alan

 

[jtbell]

Well, it looks like if we assume that the inertial observer will always log the most proper time,

No, we do not assume this. It is a consequence of our definition of an inertial observer as one who remains at rest (or moving at constant velocity) in an inertial reference frame. You can test it by calculation on some simple examples.

For example, consider a simple twin paradox scenario in which the traveling twin goes out in a straight line for 5 years at a speed of 0.8c, traveling a distance of 4 light-years, turns around "instantaneously" and returns at the same speed. In the Earth's reference frame (assumed to be inertial), let us calculate the total spacetime path length for the stay-at-home twin (A) and the traveling twin (B).

Event #1 is the traveling twin's departure, at x = 0 and t = 0.

Event #2 is the traveling twin's turnaround, at x = 4 ly and t = 5 yr.

Event #3 is the traveling twin's return, at x = 0 and t = 10 yr.

The spacetime path of twin A has just one straight-line segment, with length

\Delta s_A = \sqrt{(t_3 - t_1)^2 - (x_3 - x_1)^2} = \sqrt{10 - 0)^2 - (0 - 0)^2} = 10[/itex]<br /> <br /> The spacetime path of twin B has two straight-line segments, with total length<br /> <br /> \Delta s_B = \sqrt{(t_2 - t_1)^2 - (x_2 - x_1)^2} + \sqrt{(t_3 - t_2)^2 - (x_3 - x_2)^2} = \sqrt{(5 - 0)^2 - (4 - 0)^2} + \sqrt{(10 - 5)^2 - (0 - 4)^2} = 6[/itex]&lt;br /&gt; &lt;br /&gt; This calculation is easiest in the inertial reference frame in which twin A is at rest throughout, but we can use any other inertial reference frame, and get the same values for \Delta s_A and \Delta s_B.&lt;br /&gt; &lt;br /&gt; If we let twin B follow any other path we like, except of course one which simply duplicates twin A&amp;#039;s path, we will likewise calculate that \Delta s_B &amp;amp;lt; \Delta s_A. I&amp;#039;m sure this can be proven mathematically, but I&amp;#039;m not up to mathematical proofs at 3:00 am. &lt;img src=&quot;https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61b.png&quot; class=&quot;smilie smilie--emoji&quot; loading=&quot;lazy&quot; width=&quot;64&quot; height=&quot;64&quot; alt=&quot;:-p&quot; title=&quot;Stick Out Tongue :-p&quot; data-smilie=&quot;7&quot;data-shortname=&quot;:-p&quot; /&gt; &lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; My question is, if the relative velocity between two objects changes, how do we know that the inertial worldline should be drawn as straight, and the non-inertial worldline should be drawn as kinked? &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;&lt;br /&gt; It seems to me that you&amp;#039;re asking basically, &amp;quot;why are inertial reference frames special?&amp;quot; or &amp;quot;why is inertial motion special?&amp;quot;&lt;br /&gt; &lt;br /&gt; These two assumptions go all the way back to Newton and his laws of motion. They&amp;#039;re not a new invention in special relativity. In classical mechanics, we define an inertial reference frame as one in which Newton&amp;#039;s laws of motion hold, e.g. the First Law which states that an object that has zero net external force acting on it moves in a straight line at constant speed. In special relativity we do pretty much the same thing, except that we have to tweak the Second Law (F = ma) a bit in order to get it into relativistic form.&lt;br /&gt; &lt;br /&gt; In a non-inertial reference frame, Newton&amp;#039;s Laws don&amp;#039;t work. We have to add terms corresponding to &amp;quot;fictitious&amp;quot; or &amp;quot;inertial&amp;quot; forces that arise from our non-inertialness. (For example, the force that apparently pushes you into your car seat when you press your car&amp;#039;s accelerator pedal.)

 

[robphy]

Well, it looks like if we assume that the inertial observer will always log the most proper time, then of course we will find that the observer who logs the most proper time will turn out to be the inertial observer. And if we assume that the observer who logs the most proper time is the inertial observer, we will find that the inertial observer will turn out to be the one who logs the most proper time.

It's not an assumption... it's a key result obtained from Einstein's postulates for Special Relativity, which can be mathematically formalized as a Minkowskian geometry (the geometry of an R⁴ with a Minkowskian metric... such a geometry is preserved under Lorentz Transformations, whose eigenvectors are the lightlike vectors).

And if we draw inertial worldlines as straight, and put kinks in non-inertial worldlines, we will end up with inertial worldlines that are straight and non-inertial worldlines that are kinked.

We do this in Euclidean geometry and in Galilean/Newtonian relativity.
In Euclidean geometry, a straight line is a geodesic.
In relativity, inertial observers (which are "free particles", not being subjected to forces... and thus have no acceleration and thus no worldline curvature) travel on a http://en.wikipedia.org/wiki/Geodesic" in spacetime.

While geodesics are often characterized by minimizing (or in the Minkowskian case, maximimzing) the arc-length, there is another way to characterize a geodesic. Along a geodesic, the tangent vector is parallelly transported along itself. Crudely, "it doesn't turn" according to its geometry.

My question is, if the relative velocity between two objects changes, how do we know that the inertial worldline should be drawn as straight, and the non-inertial worldline should be drawn as kinked?

As a said earlier, if all you draw is the worldlines, you can draw them how you like. However, if you wish to attach a coordinate system to span spacetime (so that observer can assign a unique set of coordinates for each event in spacetime), then you'll find that the inertial observer can do it, however, a non-inertial one can't generally do it... in some cases, the non-inertial one will not be able to assign coordinates to an event (e.g. the uniformly accelerated observer has a horizon), or else will have multiple assignments to an event (e.g. associated with the instantaneous turn-around event in the twin paradox, events beyond the turn-around event may have multiple coordinate assignments [draw the lines of simultaneity to see]).

I'm not saying you are wrong, I'm just saying that the mathematics and drawings don't really seem to answer the question. They just seem to explain the answer afterward.

Of course, the result of mathematical formulation is just that:
the result of a mathematical formulation. The one used in special relativity is used because that seems to correspond well with the experimental result (e.g. the muon lifetime experiments) as well as lead to further predictions that can be tested in the real world. Had that muon experiment and others like it not yielded their non-Newtonian results, we probably would not be discussing special relativity now.

(edit) There are also examples where the twins could both be inertial and experience differential aging. The twins could be in intersecting orbits around earth, one nearly circular and the other very elliptical. Each twin would be following a geodisic the entire time, yet everytime they met, they would see that they experienced different lapses of proper time since the last time they met, even though they are both in freefall.

Thanks,
Alan

Due to the presence of spacetime curvature, this is not special relativity any more. In Minkowski spacetime, distinct timelike geodesics can't intersect more than once. In your case, you seem to be describing a situation with http://planetmath.org/encyclopedia/ConjugatePoints.html".)

 

[jtbell]

From a purely mathematical, descriptive point of view, non-inertial reference frames are just as "valid" as inertial reference frames, except perhaps that in non-inertial frames you can have two or more sets of coordinates for the same event, as robphy points out. But it might be possible to invent rules for choosing consistently which set of coordinates to use in such cases.

But from a physical point of view, we have the experimental observation that inertial frames are simpler to work with than non-inertial ones. Not only do inertial reference frames not have the problem of multiple coordinate sets (in relativistic situations), we also don't have to invent "fictitious" forces that don't have any apparent physical source, in order to explain certain changes in velocity of objects. In an inertial reference frame, any change in an object's velocity, so far as we know, can be explained as the result of a physical interaction with some other object. And there are a relatively small number of possible physical interactions, each with relatively well-defined rules.

Going back to the twin-paradox scenario, in the stay-at-home twin's inertial reference frame, the traveling twin's worldline develops a "kink" because his rocket engines fire, thereby creating a reaction force on the rocket which changes its velocity. We can easily calculate that reaction force and the resulting change in velocity, in terms of the nozzle speed of the rocket engine's exhaust, and the amount of fuel burned. The traveling twin knows this is happening, even if the engines were fired by remote control without his advance knowledge, because he can feel it happening.

In the traveling twin's non-inertial reference frame, when he fires his rocket engines, the stay-at-home twin suddenly changes his motion. What is the physical cause of that change? What force acts on the Earth to change its velocity in that non-inertial reference frame? What laws allow us to calculate that force? And why doesn't the stay-at-home twin feel any effect from this?

 

[JesseM]

I don't think we could say that the velocity addition formula actually causes v = at/\gamma to be correct.

I also mentioned time dilation though, which means that in the stepped-velocity-increase scenario I was discussing, if each constant-velocity interval lasts for the same amount of time in the co-moving inertial rest frame of the ship during that interval, each step will appear longer and longer in the Earth's inertial frame. With the velocity addition formula and the time dilation formula I think it would be possible to figure out just how my stepped-velocity-increase scenario would look in the Earth's frame, and I think in the limit of smaller and smaller time-intervals (and velocity increases) this scenario reduces to the continuous acceleration scenario. Do you disagree?

One might even say that Lorentz contraction causes the velocity addition formula to be correct.

No you couldn't. The velocity-addition formula depends on Lorentz contraction and on time dilation and the relativity of simultaneity. If you imagined an alternate universe with no time dilation and no relativity of simultaneity, but where there was still Lorentz contraction relative to a preferred frame, the relativistic velocity-addition formula would not be correct in such a universe. And I don't think the constant proper acceleration scenario would lead to the same functions as on the relativistic rocket page in this universe either, which is why I don't think it makes sense to explain things like v(t) = at/ \gamma (t) and a_c (t) = a / \gamma (t) solely in terms of Lorentz contraction.

But I do have another question. In Einstein's own writings about the Twins Paradox it is clear that he did not agree with the resolutions commonly accepted today. He even said that the ship's twin should be able to consider himself at rest the entire time, and that the Twins Paradox could not be resolved in SR. This was supposedly one of his reasons for pursuing GR. He tried to resolve it in GR by considering the ship's twin to be at rest, but that resolution is considered faulty today. If one could argue that Einstein understood SR as well as anyone, why would he still view the Twins Paradox as a problem for SR?

I think you may have misunderstood what Einstein was arguing. I am sure he would have agreed that 1) the standard non-tensor laws of special relativity only work in inertial frames, and 2) as long as you stick to an inertial frame, you will always get the same prediction about the ages of the twins when they reunite. He probably did not see this as satisfactory because the fundamental distinction between inertial and non-inertial frames in 1) went against his "Machian" view of physics (are you familiar with principle[/url] and the influence it had on Einstein's thinking?), so although he wouldn't disagree that 1) and 2) would "resolve" the twin paradox in SR, I think he saw SR itself as problematic for this reason, and wanted a new theory that would apply the same laws to non-inertial observers as inertial ones. He was ultimately successful in this, since GR has the same laws in all coordinate systems, inertial and non-inertial alike. And with GR you can understand the twin paradox in terms of a coordinate system where the traveling twin is at rest throughout the journey, it just means that the traveling twin will see a gravitational field during the period where inertial observers see him accelerating--see this section of the Twin Paradox page.

 

[jtbell]

And with GR you can understand the twin paradox in terms of a coordinate system where the traveling twin is at rest throughout the journey, it just means that the traveling twin will see a gravitational field during the period where inertial observers see him accelerating--see this section of the Twin Paradox page.

I was about to jump on you with the question "but where does the gravitational field come from?", but then I saw that the page that you linked to does address this point, very well. In the spirit of that page, both "GR" and "gravitational field" would be in scare quotes in your statement.

 

[George Jones]

What logical fallacy are you talking about? I never said anything about proof or authority. And if you assumed that I intended to challenge SR, you are not correct.

Are you saying that the logical fallacy of proof by authority can occur only when someone uses the words "proof" and "authority"?

I've gone back and reread the passage of yours that I quoted, and, even if you did not intend it that way, I don't think it's too big a leap to intrepret what you wrote as

[(Einstein understood SR as well as anyone) /\ (Einstein viewed the twin paradox as a problem for SR] => (the twin paradox is a problem for SR)

He then says the cause of the asymmetry between the twins is a pseudo-gravitational field created during the acceleration of the ship. He then uses the gravitational time dilation of GR to try to resolve the paradox.

It might be the case that, in spite of what he writes, Einstein's analysis never leaves the cozy confines of SR.

there is also a link at the bottom of the Wikipedia article that discusses Einstein's resolution of the paradox.

Unfortunately, I can't get this link to work.

It also seems strange that Einstein's thought's on the Twins Paradox are rarely mentioned in discussions of the Twins Paradox.

Which thoughts? Einstein also analysed the twin paradox in terms of the relativity of simultaneity, and this concept is often used in discussions of the twin paradox.

It seems that you want an answer to the question, "Why are some frames of reference inertial?"

I think this is similar in kind to the question "Why is the magnitude of force in Newton's law of gravity given by F = GmM/r^2?"

My answer to both questions is "I don't know."

Why questions tend to be very deep and difficult, but this doen't mean that we shouldn't ask them.

 

[pervect]

By direct calculation, I mean the computation of the path-dependent integral \int_A^B ds, where A and B are events and ds is a timelike arc-length element in spacetime [analogous to the Euclidean arc-length used when measuring the length of a curve].

This is a very important point. I don't know if Al68 appreciates how the Lorentz interval is calculated (as above), the significance of this calculation, or the fact that this quantity (unlike time, or distance) is the true invariant that is shared between different observers. He (Al68) seems to be very focused on coordinates and not particularly interested in the Lorentz interval.

The Lorentz interval defines "proper time" and "proper length", i.e. how much time a clock reads, and how long a ruler measures. So when one wants to calculate the amount of time that a clock takes to traverse its path, you just calculate the Lorentz interval of that path. This automatically gives you the time that a clock traveling that path would read, by definition.

I think that calculating the Lorentz interval has mainly been presented in terms of SR. It's possible to generalize this to GR by adding in the concpet of a metric (another rather important idea that hasn't been talked about much). The metric is the key to calculating the Lorentz interval in arbitrary/accelerated/noninertial coordinate systems.

But I don't want to go into this too much (the metric) unless I'm reassured that Al68 understands the Lorentz interval and why we want to calculate it in the first place.

 

[Al68]

I don't think it makes sense to explain things like v(t) = at/ \gamma (t) and a_c (t) = a / \gamma (t) solely in terms of Lorentz contraction

Jesse, I guess I have to agree, since length contraction and time dilation are so intertwined, that we really can't separate them.

I think you may have misunderstood what Einstein was arguing. I am sure he would have agreed that 1) the standard non-tensor laws of special relativity only work in inertial frames, and 2) as long as you stick to an inertial frame, you will always get the same prediction about the ages of the twins when they reunite. He probably did not see this as satisfactory because the fundamental distinction between inertial and non-inertial frames in 1) went against his "Machian" view of physics (are you familiar with principle[/url] and the influence it had on Einstein's thinking?), so although he wouldn't disagree that 1) and 2) would "resolve" the twin paradox in SR, I think he saw SR itself as problematic for this reason, and wanted a new theory that would apply the same laws to non-inertial observers as inertial ones. He was ultimately successful in this, since GR has the same laws in all coordinate systems, inertial and non-inertial alike. And with GR you can understand the twin paradox in terms of a coordinate system where the traveling twin is at rest throughout the journey, it just means that the traveling twin will see a gravitational field during the period where inertial observers see him accelerating--see this section of the Twin Paradox page.

Yes, I agree that Einstein saw SR as problematic because inertial frames were considered different from non-inertial frames. But according to several sources on the net, including this one, http://en.wikipedia.org/wiki/Twin_paradox Einstein's GR resolution of the Twins Paradox is considered to be faulty today. So I would disagree that he was ultimately successful. At least in this respect. It's my understanding that he never considered any of the SR resolutions of the Twins Paradox to be satisfactory.

Are you saying that the logical fallacy of proof by authority can occur only when someone uses the words "proof" and "authority"?

I've gone back and reread the passage of yours that I quoted, and, even if you did not intend it that way, I don't think it's too big a leap to intrepret what you wrote as

[(Einstein understood SR as well as anyone) /\ (Einstein viewed the twin paradox as a problem for SR] => (the twin paradox is a problem for SR)

That is not what I intended, although I can understand why it could be interpreted that way, given all the internet discussions on this issue. I am trying to understand the Twins Paradox better. I consider SR to be correct, so even if I were to ultimately choose to disagree with the accepted Twins Paradox resolution, it will not cause me to view the Twins Paradox as a problem for SR. It would cause me to view SR as a problem for the accepted Twins Paradox resolutions.

And I certainly don't believe in proof by authority, if I did, I wouldn't have so many questions.

Why questions tend to be very deep and difficult, but this doen't mean that we shouldn't ask them.

Thanks, I agree.

No, we do not assume this. It is a consequence of our definition of an inertial observer as one who remains at rest (or moving at constant velocity) in an inertial reference frame. You can test it by calculation on some simple examples.

For example, consider a simple twin paradox scenario in which the traveling twin goes out in a straight line for 5 years at a speed of 0.8c, traveling a distance of 4 light-years, turns around "instantaneously" and returns at the same speed. In the Earth's reference frame (assumed to be inertial), let us calculate the total spacetime path length for the stay-at-home twin (A) and the traveling twin (B).

Event #1 is the traveling twin's departure, at x = 0 and t = 0.

Event #2 is the traveling twin's turnaround, at x = 4 ly and t = 5 yr.

Event #3 is the traveling twin's return, at x = 0 and t = 10 yr.

The spacetime path of twin A has just one straight-line segment, with length

\Delta s_A = \sqrt{(t_3 - t_1)^2 - (x_3 - x_1)^2} = \sqrt{10 - 0)^2 - (0 - 0)^2} = 10[/itex]<br /> <br /> The spacetime path of twin B has two straight-line segments, with total length<br /> <br /> \Delta s_B = \sqrt{(t_2 - t_1)^2 - (x_2 - x_1)^2} + \sqrt{(t_3 - t_2)^2 - (x_3 - x_2)^2} = \sqrt{(5 - 0)^2 - (4 - 0)^2} + \sqrt{(10 - 5)^2 - (0 - 4)^2} = 6[/itex]&lt;br /&gt; &lt;br /&gt; This calculation is easiest in the inertial reference frame in which twin A is at rest throughout, but we can use any other inertial reference frame, and get the same values for \Delta s_A and \Delta s_B.

&lt;br /&gt; &lt;br /&gt; I think this last statement is the key. You will only get these values if you use the numbers from an inertial frame for the calculation. If we were to assume that the ship could be viewed the same way, we would get different values.&lt;br /&gt; &lt;br /&gt; Like this: We consider the ship&amp;#039;s frame (B) to be preferred instead of the inertial frame (A). And we say that the distance is given as 4 ly in the ship&amp;#039;s frame. And:&lt;br /&gt; &lt;br /&gt; Event #1 is the traveling twin&amp;#039;s departure, at x&amp;#039; = 0 and t&amp;#039; = 0.&lt;br /&gt; &lt;br /&gt; Event #2 is the traveling twin&amp;#039;s turnaround, at x&amp;#039; = 4 ly and t&amp;#039; = 5 yr.&lt;br /&gt; &lt;br /&gt; Event #3 is the traveling twin&amp;#039;s return, at x&amp;#039; = 0 and t&amp;#039; = 10 yr.&lt;br /&gt; &lt;br /&gt; Now the spacetime path of twin B has just one straight-line segment, with length&lt;br /&gt; &lt;br /&gt; \Delta s_B = \sqrt{(t&amp;amp;#039;_3 - t&amp;amp;#039;_1)^2 - (x&amp;amp;#039;_3 - x&amp;amp;#039;_1)^2} = \sqrt{10 - 0)^2 - (0 - 0)^2} = 10&lt;br /&gt; &lt;br /&gt; And the spacetime path of twin A has two straight-line segments, with total length&lt;br /&gt; &lt;br /&gt; \Delta s_A = \sqrt{(t&amp;amp;#039;_2 - t&amp;amp;#039;_1)^2 - (x&amp;amp;#039;_2 - x&amp;amp;#039;_1)^2} + \sqrt{(t&amp;amp;#039;_3 - t&amp;amp;#039;_2)^2 - (x&amp;amp;#039;_3 - x&amp;amp;#039;_2)^2} = \sqrt{(5 - 0)^2 - (4 - 0)^2} + \sqrt{(10 - 5)^2 - (0 - 4)^2} = 6.&lt;br /&gt; &lt;br /&gt; And of course this would add up, since the distance traveled would now be 4/\gamma or 2.4 light years each way in Earth&amp;#039;s frame. &lt;br /&gt; &lt;br /&gt; The only difference here is, I considered the non-inertial frame as the &amp;quot;preferred frame&amp;quot; and said that the Earth turned around relative to the ship&amp;#039;s frame at a distance specified in the ship&amp;#039;s frame.&lt;br /&gt; &lt;br /&gt; &lt;b&gt;I&amp;#039;m not saying this is correct, just that the math can be done either way.&lt;/b&gt; Of course, I know that I did not use an inertial frame for my numbers. And by definition, the inertial frame is the one with the longest spacetime interval. And this will always result in the most time passing for the inertial frame (between any two events). I have to repeat that I do not consider my above calculations to be correct. I just wanted to illustrate that mathematics doesn&amp;#039;t provide the answer, it only describes the answer. And it is very easy to use mathematics to describe a wrong answer, as I think I have shown above.&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;jtbell&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-title&quot;&gt; jtbell said: &lt;/div&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; It seems to me that you&amp;#039;re asking basically, &amp;quot;why are inertial reference frames special?&amp;quot; or &amp;quot;why is inertial motion special?&amp;quot; &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;&lt;br /&gt; Yes, that is my question.