A Geometrical View of Time Dilation and the Twin Paradox

Based on the number of questions we receive on the topic at Physics Forums, there is a lot of confusion in the general public about how time dilation works and the resolution of the twin paradox. More often than not, the confusion is based upon misunderstandings about how time dilation works and an unawareness of the relativity of simultaneity and how time is treated within special relativity. The aim of this Insight is to make an attempt at explaining these issues from a geometrical viewpoint and to see that we have an analogous effect in the geometry which we are used to, although we do not find it strange.

Table of Contents

A geometrical analogue

Consider the following two lines in normal two-dimensional space:

In order to describe the lines, we can introduce a coordinate system ##S##:

In this coordinate system, we may use the coordinate ##x## to parametrise the lines, we find that

$$y_1(x) = 0, \quad y_2(x) = kx,$$

where ##k## is some constant. The lengths of each line between ##x = 0## and ##x = x_0## are simply given by Pythagoras’ theorem ##\ell^2 = \Delta x^2 + \Delta y^2## and we find

$$\ell_1 = x_0, \quad \ell_2 = x_0\sqrt{1+k^2}.$$

Consequently we find that the ratio between these two lengths is ##\ell_2/\ell_1 = \sqrt{1+k^2}##. This is shown in the following figure:

However, there is nothing special about the coordinate system ##S##. In particular, the length of the lines between any two given points on the lines does not depend on the coordinate system chosen and we could just as well have chosen to introduce the coordinate system ##S’## according to:

In this new coordinate system, we can parametrise the lines using the ##x’## coordinate

$$y_1′(x’) = -kx’, \quad y_2′(x’) = 0.$$

For the distance between the points corresponding to ##x’ = 0## and ##x’=x_0’##, Pythagoras’ theorem now results in

$$\ell_1′ = x_0′ \sqrt{1+k^2}, \quad \ell_2′ = x_0’$$

and so the line ##\gamma_1## is longer than ##\gamma_2## for the same difference in the parameter ##x_0’##, unlike in the case when the lines were parametrised by ##x##, where we obtained the opposite result. Yet there is nothing strange going on here. If we select a point on ##\gamma_2## and draw the lines which have the same ##x## and ##x’## values as that point, it becomes clear that the points on ##\gamma_1## which share the same parameter value differs depending on whether we used ##x## or ##x’## for the parametrisation. We should therefore not be surprised to find that the length to these two different points is different. Let us call this effect the “relativity of the same ##x##-coordinate”. Naturally, this effect does not affect the actual length of any of the lines, but only what lengths are being compared. This is illustrated in the following figure:

The points ##B## and ##B’## have the same ##x## and ##x’## coordinate as ##A##, respectively, yet correspond to different points on the curve ##\gamma_1## and therefore the distance from the intersection along this curve is different.

Curves intersecting at two points

Now consider the following two curves:

Just from the figure, it should be clear that ##\gamma_2## is longer and we can verify this by introducing the coordinate system ##S## as above and computing the lengths of each of the curves, with the length of ##\gamma_2## being split into two equal contributions, one from each line segment.

We find that

$$\ell_1 = 2\ell, \quad \ell_2 = 2\ell \sqrt{1+k^2}.$$

We could also compute the length of the curves by looking at the ##S’## coordinate system:

By our previous discussion, the length of ##\gamma_2## up to the half-way point is given by

$$\ell_2’/2 = \ell’,$$

while the length of ##\gamma_1## to the same ##x’## coordinate is given by

$$\ell_1’/2 = \ell’ \sqrt{1+k^2}.$$

If we were not careful about the relativity of the same ##x##-coordinate, we might do the error of thinking that ##\ell_1’/2## would be half the length of the curve ##\gamma_1##. However, since we are aware of this, we know better than that and realise that if we multiplied the result by two, we would obtain a result that was too large.

Application to relativity

So how does this connect to relativity and time dilation? In its foundation, relativity is a theory about the geometry of space-time, which may be described using a coordinate system with four coordinates ##(t,x,y,z)##, where the first coordinate is the time coordinate and the last three are spatial coordinates. It is important to realise that, just as in the case we just described, these are just coordinated that, by themselves, do not have a direct physical interpretation. For the sake of brevity, let us consider only one spatial coordinate ##x## as the generalisation is straightforward.

In relativistic space-time, geometry works a bit differently as compared to the geometry with which we are used to working with. In particular, this manifests itself by a change in Pythagoras’ theorem, which now comes with a minus sign

$$c^2 \tau^2 = c^2 \Delta t^2 – \Delta x^2$$

as a direct consequence of the speed of light is equal in all frames. Here, ##\tau## is the proper time along a straight line in space-time with ##\Delta t## being the difference between the endpoint time-coordinates and ##\Delta x## being the difference between the endpoint space-coordinates. This also changes what kind of coordinate transformations we can do in order for the proper time to be the same regardless of the coordinates. Instead of rotating both coordinate axes in the same direction as happens for rotations, the coordinate axes are rotated by the same angle in opposite directions.

In relativity, the proper time defined above is the only time that will be measured by any observer. A curve in space-time is called a world line and will generally describe the position of the observer as a function of time. The proper time of a world line, i.e., the “length” of the world line with the modified Pythagorean theorem, is the time a clock following that world line will measure, the time coordinate ##t## has no particular meaning other than that it happens to numerically be the same as the time measured by an observer for which the spatial coordinates are not changing. This is completely analogous with our previous example, where the value of the ##x##-coordinate was equal to the length of a line for which the ##y##-coordinate was not changing.

We can now do the exact same exercise as we did use regular geometry. Consider the following two world lines described in the coordinate system ##S##:

The first world line ##\gamma_1## has a constant ##x##-component while the second world line describes an observer travelling with velocity ##v##. The world lines may be parametrised as

$$x_1(t) = 0, \quad x_2(t) = vt.$$

We can now compute the proper time from the time coordinate taking values ##t=0## to ##t=t_0## for both world lines using the modified Pythagorean theorem and find that

$$\tau_1 = t_0, \quad \tau_2 = t_0\sqrt{1-\frac{v^2}{c^2}}$$

or, equivalently,

$$\frac{\tau_1}{\tau_2} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \equiv \gamma,$$

where ##\gamma## is the famous gamma-factor of special relativity. In other words, the proper time elapsed for an observer at rest between two time coordinates will differ from the proper time elapsed for a moving observer between the same time coordinates by a factor ##\gamma##. This is the essence of time-dilation but, just as for the example we started with, it crucially depends on the coordinate system.

Taking a different coordinate system ##S’## where ##\gamma_2## describes a world line with constant ##x’##

the exact same argumentation will now lead us to conclude

$$\frac{\tau_1′}{\tau_2′} = \sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}.$$

This is the exact opposite relation as that we found in the coordinate system ##S##, just as was the case in our first example! The resolution of this apparent mismatch is also completely analogous. We should not be surprised to find a different relation between the times elapsed between different ##t## coordinates and that elapsed between different ##t’## coordinates simply because the point on the curve ##\gamma_1## with the same ##t## coordinate as the point ##A## on ##\gamma_2## will not be the same as the point on ##\gamma_1## with the same ##t’## coordinate as ##A##:

This effect is called the “relativity of simultaneity” and simply states that two events which are simultaneous in one set of coordinates are generally not simultaneous in a different set of coordinates (simultaneous simply refers to having the same time-coordinate). As such, there is really nothing strange about the proper times to ##B## and ##C## along with ##\gamma_1## being different since they are simply not the same point.

The twin paradox

The twin paradox is the analogue of our second example with two curves starting and ending at the same points. In the case of two twins, where one leaves to travel to a faraway land and later returns, the world lines in the coordinate system ##S## would look like:

Just applying the modified Pythagorean theorem it becomes clear that the proper times along the world lines are given by

$$\tau_1 = 2\tau, \quad \tau_2 = 2\tau \sqrt{1-\frac{v^2}{c^2}} \quad \Longrightarrow \quad \frac{\tau_1}{\tau_2} = \gamma$$

and therefore more proper time will pass for the twin staying behind, i.e., the red curve. Going to the coordinate system ##S’##, we can compute the relation between the proper time ##\tau’## taken for the travelling twin to reach the turnaround point and the proper time elapsed for the staying twin during the same ##t’## interval as

$$\frac{\tau_1′}2 = \tau’ \sqrt{1-\frac{v^2}{c^2}}, \quad \frac{\tau_2′}2 = \tau’.$$

However, just as in our example above, the relativity of simultaneity tells us that ##\tau_1’## is not equal to half the proper time elapsed for the staying twin until the reunion:

Instead, the change of geometry indicates that we are not counting part of the world line if we make this assumption and therefore underestimate the elapsed proper time. If taken properly into account, we would find that the time elapsed for the staying twin is longer also as computed by the travelling observer. This resolves the twin paradox and shows that it only appears due to not taking the relativity of simultaneity properly into account.

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90 replies
1. whatif says:

Your experience is much more closely related to your past light cone than your frame.Rest frame was, perhaps, a bad choice of wording. What I meant was all frames in which I was stationary at the time, including non inertial frames of acceleration, which I guess correlates to the light cone. Yes/no?
To be more precise about the twin paradox. So long as each twin occupies a different inertial frame continuing to separate apart from the other's frame then each continues to age faster than the other. That, I think, exemplifies what requires insight; or just plane acceptance because it seems to be implicitly verified by experiment.

2. Dale says:
whatif

Only for the spacetime interval.The choice of reference frame or coordinates is not only arbitrary for the spacetime interval. It is also arbitrary for the expression of any physical law and for the prediction of the outcome of any physical measurement for any experiment. That is the meaning of the first postulate.

whatif

Except that my rest frame is my experience.Not really. Your experience is much more closely related to your past light cone than your frame. And your past light cone is the same in all reference frames.

3. whatif says:

Of course the choice of axes is arbitrary.A choice of inertial system is exactly analogous to an arbitrary choice of (mutually orthogonal) coordinate axes.It is completely arbitrary.Only for the spacetime interval. As I see it, time is not independent of the location in space, but a restriction on the 3D location that can be occupied. In the 2D analogue there is no such interdependence/restriction. The article introduces the rotation of axes and ‘modified Pythagorean theorem’ without explanation, which, to me, reduces the idea to an arduous mathematical how to, rather than an insight.

The twin paradox has a resolution when the twins reunite. However, so long as the twins separate in different inertial frames each is older than the other, forever in each's own frame. That, I imagine, is time dilation.

It is completely arbitrary. There is no requirement that I do my calculations using the frame in which I am rest.Except that my rest frame is my experience.

4. whatif says:
Mister T

Are you talking about the lengths of the line segments that are drawn on the spacetime diagrams?I was talking about the lengths of the projections of the line (object) onto the axes in the analogue, which was just about 2 D space.

5. Nugatory says:
whatif

it still seems to me that a significant flaw in the analogue is that the different coordinate systems come down to an arbitrary choice of axes but the transformations between the different moving frames is not about an arbitrary choice of axes; I think.It is completely arbitrary. There is no requirement that I do my calculations using the frame in which I am rest, and no matter which frame I use I will get the same results for anything and everything that happens – for example, the time on my wristwatch when light from some event reaches my eyes.

Often I choose to use the frame in which I am at rest, but that's not because that frame is special, it's because the problem at hand is most easily solved in that frame.

6. Orodruin says:
whatif

but it still seems to me that a significant flaw in the analogue is that the different coordinate systems come down to an arbitrary choice of axes but the transformations between the different moving frames is not about an arbitrary choice of axes; I think.Then this is where your intuition fails you. It is exactly what it comes down to. A choice of inertial system is exactly analogous to an arbitrary choice of (mutually orthogonal) coordinate axes. The geometry of Minkowski space just implies that the corresponding transformation is a hyperbolic rotation rather than a normal rotation.

7. Ibix says:
whatif

I think that I understand the resolution of the twin paradox using a Minkowski diagram, but this geometrical analogue and explanation does not help me better understand time dilation. I would say that maybe it was not intended to if that was not stated as an aim.The ticks of a clock are regular marks on the timelike coordinate axis. If you are in motion with respect to me, your timelike coordinate axis is at an angle to mine. The spacing between your regular ticks, as measured by me, is the projection of the spacing you measure onto my coordinate axis. That's time dilation. The Euclidean analogy is two rulers at an angle to each other. Assuming the rulers cross at their zero mark, both rulers agree that the 1m mark on the other ruler is level with their 95cm mark (or whatever – depends on the angle). Minkowski geometry is just more mathematically interesting.

whatif

Also, no analogue is perfect but it still seems to me that a significant flaw in the analogue is that the different coordinate systems come down to an arbitrary choice of axes but the transformations between the different moving frames is not about an arbitrary choice of axes; I think.Of course the choice of axes is arbitrary. How is an experiment affected if I observe it sutting still, or walking past, or flying by in a spaceship at 0.95c? Each of those options would naturally choose a different frame. The transform between any pair of frames is well defined, but no frame is right or wrong – the choice is completely arbitrary.

8. Mister T says:
whatif

I think the lengths in the different coordinate systems of the geometric analogue is just a matter of linear scaling with different base vectors.Are you talking about the lengths of the line segments that are drawn on the spacetime diagrams?

9. whatif says:

You need to distinguish carefully between the two cases Orodruin was talking about in the article, and I don't think you are doing so.I think that I understand the resolution of the twin paradox using a Minkowski diagram, but this geometrical analogue and explanation does not help me better understand time dilation. I would say that maybe it was not intended to if that was not stated as an aim.

Also, no analogue is perfect but it still seems to me that a significant flaw in the analogue is that the different coordinate systems come down to an arbitrary choice of axes but the transformations between the different moving frames is not about an arbitrary choice of axes; I think.

10. Ibix says:
whatif

That was my point, except that that it applies in the same inertial reference frame. While the transformations are mathematically linear I understood that length varied according to relative motion of the frame in which it is measured.You need to distinguish carefully between the two cases Orodruin was talking about in the article, and I don't think you are doing so. One case is Euclidean geometry, in which length is the same in all coordinate systems and time is no part of geometry. The other case is the Minkowski geometry that underlies relativity, which includes time as a dimension. In this case, a concept called "interval" is analogous to length in Euclidean geometry while the spatial length of an object is analogous to the difference in x coordinates between the ends of a rod in Euclidean geometry.

The interval is invariant between frames in relativity, but length is not. In Euclidean geometry, length is invariant and interval isn't a concept. The coordinate transforms between frames are linear in both cases.

11. whatif says:
the Lorentz transformations are linear transformations.Sorry, yes they are.

Length has nothing to do with base vectors. It is invariant and does not depend on the coordinate system.That was my point, except that that it applies in the same inertial reference frame. While the transformations are mathematically linear I understood that length varied according to relative motion of the frame in which it is measured.

12. Orodruin says:
whatif

I think the lengths in the different coordinate systems of the geometric analogue is just a matter of linear scaling with different base vectors.Length has nothing to do with base vectors. It is invariant and does not depend on the coordinate system.

whatif

As I understand it, the lengths and time intervals for special relativity are different, not as a result of arbitrary scaling but as a result of relative motion between of the inertial frames and the transformations are nonlinear.No, the Lorentz transformations are linear transformations. The entire point is that spacetime intervals do not depend on the coordinate system, just as little as the distance between you and your neighbour does not depend on the coordinate system.

13. whatif says:

I am a novice about relativity, but I found this convoluted and very difficult to follow. I think the lengths in the different coordinate systems of the geometric analogue is just a matter of linear scaling with different base vectors. The lengths are identical in each frame when accounting for the scaling. As I understand it, the lengths and time intervals for special relativity are different, not as a result of arbitrary scaling but as a result of relative motion between of the inertial frames and the transformations are nonlinear.

14. greswd says:

[QUOTE=”Orodruin, post: 5447359, member: 510075″]But exactly here lies the problem. I am not arguing that the coordinates are abad idea, I just question the pedagogical value to people who are just learning relativity. I do not see the point of introducing non-inertial coordinate systems on top of the struggles they already have.[/QUOTE]
That’s true. But the moment they understand time dilation, they will demand to know things from the travelling twin’s perspective.

15. Orodruin says:

[QUOTE=”the_emi_guy, post: 5448229, member: 415127″]Again, I am sharing that there is a way to introduce the clock skew phenomena in a way that does not require proper time, Minkowsky diagrams, etc.[/QUOTE]
This to me is counter productive for the real understanding of SR, which should be the actual goal. You will still require the use of the Lorentz transformations, the application of the relativity of simultaneity, etc. This completely defeats the pedagogical purpose as it is essentially just giving the formulas and letting the student figure out their meaning. It is like giving a student the formula for the coordinate change due to a rotation and expect them to figure out the rest.

Proper time and Minkowski diagrams are fundamental tools in understanding SR. You should teach them, not try to sweep them under the carpet.

16. Dale says:

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]On the other hand, the “traveling” twin is undergoing acceleration bringing him into a different inertial frame. This is outside the scope of SR unless we assume a non-physical observer[/QUOTE]I disagree with both of these sentences.

In the first, no object goes into or out of any frame. All objects or observers are in every frame at all points of the journey. An object may be at rest in one frame and moving in another, but it is in both frames.

In the second, SR applies as long as gravity is negligible (spacetime is flat). An accelerating observer is certainly within the scope of SR.

17. PeterDonis says:

[QUOTE=”the_emi_guy, post: 5448229, member: 415127″]but the explanation seems to be a stumbling block for many (thus the existence of this very thread)[/QUOTE]

This thread is about Orodruin’s post describing the geometric method, which, as he says, he uses to teach the subject.

[QUOTE=”the_emi_guy, post: 5448229, member: 415127″]You have never heard of this approach? I thought it was fairly well known.[/QUOTE]

I don’t mean solving the problem yourself; of course that can be done. I mean using it to teach someone else. That’s what you’re claiming: that your approach will work better for teaching. Have you tried that? Has it worked?

18. the_emi_guy says:

[QUOTE=”PeterDonis, post: 5448209, member: 197831″]But the single traveling twin doesn’t do this. He just goes out, comes back, and never adjusts his clock at all during the whole trip–yet when he comes back, his clock has less elapsed time than his stay-at-home twin. So it seems to me that your “explanation” is introducing extraneous factors that aren’t there in the original scenario. That seems more likely to confuse than to enlighten a person who is struggling to understand the scenario.[/QUOTE]

Sure, the two observer problem statement is simpler (2 < 3), but the explanation seems to be a stumbling block for many (thus the existence of this very thread). And even if they learn how to draw the diagrams, they may not have an intuitive feel for what is going on. Again, I am sharing that there is a way to introduce the clock skew phenomena in a way that does not require proper time, Minkowsky diagrams, etc. As soon as a student gets time dilation he/she can discover and understand clock skew behavior with three observers.[QUOTE="PeterDonis, post: 5448209, member: 197831"]Have you ever actually tried this? Has it worked?[/QUOTE]You have never heard of this approach? I thought it was fairly well known. Section 4 of the attached journal is one reference and begins:"In this section we will demonstrate a method that can be used in the “clock paradox” problem without the need to consider the effects of various accelerations and decelerations, at least in principle. Such a method was first proposed by Lord Halsbury in 1957 [17], as a “triplet” or a “three clocks” problem, and can be briefly stated as follows..."

19. PeterDonis says:

[QUOTE=”the_emi_guy, post: 5448196, member: 415127″]I think that there is unrecognized value in seeing that an outbound (non-returning) observer sees Earth clock running slow, later synchronizes his “ahead” clock with an inbound clock, then watches the inbound clock go from being ahead to being behind Earth do to the greater inbound vs. outbound relative velocity.[/QUOTE]

But the single traveling twin doesn’t do this. He just goes out, comes back, and never adjusts his clock at all during the whole trip–yet when he comes back, his clock has less elapsed time than his stay-at-home twin. So it seems to me that your “explanation” is introducing extraneous factors that aren’t there in the original scenario. That seems more likely to confuse than to enlighten a person who is struggling to understand the scenario.

[QUOTE=”the_emi_guy, post: 5448196, member: 415127″]Doing this in all three observer rest frames and getting the same clock skew I think is compelling to a beginner, and the concept of proper time was not even needed or required.[/QUOTE]

Have you ever actually tried this? Has it worked?

20. the_emi_guy says:

Peter, thanks for the thoughtful response, I appreciate it.

I think that there is unrecognized value in seeing that an outbound (non-returning) observer sees Earth clock running slow, later synchronizes his “ahead” clock with an inbound clock, then watches the inbound clock go from being ahead to being behind Earth do to the greater inbound vs. outbound relative velocity. Note that this clock skew is relative to outbound’s local clock, Earth’s clock ended up being behind, and inbound’s clock even more behind.

Doing this in all three observer rest frames and getting the same clock skew I think is compelling to a beginner, and the concept of proper time was not even needed or required.

21. PeterDonis says:

[QUOTE=”the_emi_guy, post: 5448087, member: 415127″]These folks are better served by showing them how simple application of time dilation to all three inertial frames (see post #18) shows this in a trivial and intuitive manner (and I do mean “inertial frames” here, with each observer at rest in his frame).[/QUOTE]

Whether this is “trivial and intuitive” is debatable (see below), but it seems to me that these folks would be served better still by giving them the most general method of all: spacetime geometry and lengths of curves. What we have here is a triangle in spacetime: one side is the stay-at-home twin’s worldline, and the other two sides are the traveling twin’s outbound and inbound worldlines. Then all we need is the Minkowski spacetime analogue to the Pythagorean theorem–of which the time dilation equation you are implicitly using in your method is just a special case.

[QUOTE=”the_emi_guy, post: 5448087, member: 415127″]As a [I]follow on[/I] we could consider “what if there were only two observers?”[/QUOTE]

That’s not a follow-on; it’s how the original scenario was posed. When you formulate it in terms of three observers, you are, on the face of it, talking about a [I]different[/I] scenario. You then have to explain how the result you derive with three observers gives the right answer for the case of only two, which was the actual scenario posed. This is certainly doable, but I question whether it is “trivial and intuitive” to a person who is struggling with time dilation and differential aging.

[QUOTE=”the_emi_guy, post: 5448087, member: 415127″]what did the second observer do? Did he turn around instantaneously experiencing infinite acceleration (a non-physical physics problem)? What if he accelerates and decelerates gradually over time? What if he slingshots himself around another planet experiencing weightlessness the whole way? It may very well turn out that all of these produce the same result because spacetime does not get curved and SR still applies (or maybe not).[/QUOTE]

The first case is the one that is equivalent to your analysis–although I would prefer to phrase it in terms of taking the limit as the turnaround time (the time elapsed on the traveling twin’s clock during the period when he is accelerating to turn around) goes to zero. The key point is that, if the answer we are interested in is elapsed proper time for the two twins, we don’t really care about the details of how the traveling twin turns around, as long as the time elapsed on his clock while doing so is negligible compared to the time elapsed on his clock during the inbound and outbound legs.

The second case, accelerating and decelerating gradually over time, is [I]not[/I] equivalent to your analysis. And the simplest way to state the reason why is by using spacetime geometry and worldlines: the worldline of the traveling twin is [I]different[/I] in this case than it is in the idealized case you are analyzing (which is equivalent to the idealized case of instantaneous turnaround). So you would not expect the traveling twin’s elapsed time to be the same. It will still be shorter than the stay-at-home twin’s elapsed time when they meet up again, but by a different amount (which will depend on the details of how the twin accelerates and decelerates gradually). The spacetime geometry method adapts easily to this case; it just amounts to evaluating arc length along a curve instead of a pair of straight lines.

The two cases above can both be formulated in flat spacetime. The third, by contrast, requires curved spacetime, because in flat spacetime there is no way for the traveling twin to turn around without feeling acceleration, whereas in this third “slingshot” case, the traveling twin is weightless, feeling no acceleration, the whole way. The spacetime geometry method carries over to this case with no problem at all, since, as I noted, it is completely general.

22. the_emi_guy says:

[QUOTE=”PAllen, post: 5447959, member: 275028″]
What if there are only two observers? How is it useful, pedogogically, to say we must introduce a third to explain what happens? Especially, if the spacetime geometry viewpoint is very simple and requires only one frame (not necessarily one in which any observer is at rest).[/QUOTE]

Thanks for asking this, it gets right to my point.

Many folks who visit PF have learned about time dilation between observers who are in relative motion but want to get a handle on how perfectly good clocks can wind up sitting side-by-side but out of sync due to some motion that occurred. These folks are better served by showing them how simple application of time dilation to all three inertial frames (see post #18) shows this in a trivial and intuitive manner (and I do mean “inertial frames” here, with each observer at rest in his frame).

As a [I]follow on[/I] we could consider “what if there were only two observers?”.
Well, what did the second observer do? Did he turn around instantaneously experiencing infinite acceleration (a non-physical physics problem)? What if he accelerates and decelerates gradually over time? What if he slingshots himself around another planet experiencing weightlessness the whole way? It may very well turn out that all of these produce the same result because spacetime does not get curved and SR still applies (or maybe not). Volumes have been written about all these various scenarios.

23. PeterDonis says:

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]I can have an observer (real human observer) in an single inertial frame[/QUOTE]

You are misusing the term “inertial frame”. The correct way of stating what you are trying to state here is: I can have an observer in the same inertial [I]state of motion[/I]… The observer’s state of motion is the physical thing. The “inertial frame” is a mathematical construct that we use to help us describe what is going on. And, as others have remarked, the same observer’s worldline, with the same state of motion, can be described using as many different inertial frames–or non-inertial frames, for that matter–as we like.

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]the “traveling” twin is undergoing acceleration bringing him into a different inertial frame.[/QUOTE]

No. The traveling twin undergoes acceleration which [I]changes his state of motion[/I]. (Even this terminology has limitations, but it will do for this discussion.) But there is no need to switch inertial frames to describe this; you can describe it all perfectly well using a single inertial frame.

[QUOTE=”the_emi_guy, post: 5447914, member: 415127″]So acceleration (and presumably gravity by equivalence property)[/QUOTE]

You are misstating this as well. “Gravity” is an ambiguous term. The correct statement, which has already been given in this thread, is that SR deals with any scenario in which spacetime is flat. GR deals with scenarios in which spacetime can be curved.

24. PAllen says:

[QUOTE=”the_emi_guy, post: 5447914, member: 415127″]So acceleration (and presumably gravity by equivalence property) is withing the scope of SR, I’ll have to take your word for this. Still, I’ll repeat from earlier:
[/QUOTE]
No, per modern understanding, gravity involves curved spacetime, while SR involves flat spacetime. The equivalence principle is local,and is analogous to the statement that a small region of sphere is well approximated by a plane. Tidal gravity effects distinguish gravity from acceleration, but they become very small in a small region.
[QUOTE=”the_emi_guy, post: 5447914, member: 415127″]
Why bother with all of this complexity when formulating the problem with three observers (first proposed by Lord Halsbury) gets the expected result with simple plug ‘n chug application of time dilation applied twice to of the three observers.[/QUOTE]

What if there are only two observers? How is it useful, pedogogically, to say we must introduce a third to explain what happens? Especially, if the spacetime geometry viewpoint is very simple and requires only one frame (not necessarily one in which any observer is at rest).

25. the_emi_guy says:

So acceleration (and presumably gravity by equivalence property) is withing the scope of SR, I’ll have to take your word for this. Still, I’ll repeat from earlier:

Why bother with all of this complexity when formulating the problem with three observers (first proposed by Lord Halsbury) gets the expected result with simple plug ‘n chug application of time dilation applied twice to of the three observers.

26. Orodruin says:

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]If we are not talking about actual masses traveling in these inertial frames then we are not talking about physics.[/QUOTE]
Wrong again. Please read my previous post.

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]Nope, I can have an observer (real human observer) in an single inertial frame and change coordinates all day. While traveling through New York my position can be rendered in Cartesian coordinates, then when I cross the border into New Jersey we decide to render my position in some polar coordinates. Nothing about this change in coordinates implies a change in inertial frame. On the other hand, the “traveling” twin is undergoing acceleration bringing him into a different inertial frame. This is outside the scope of SR unless we assume a non-physical observer.[/QUOTE]
Again, a very common misconception. What you fail to realise is that the changes from Cartesian to polar coordinates is just the same as applied to space as a Lorentz transformation is to space-time (not completely true, a Lorentz transformation is the equivalent of a change between different [I]Cartesian[/I] coordinate systems in space). You can easily use a curvilinear coordinate system in SR as well and there are many examples. Unless you accept that SR is not a theory about observers but a theory about the geometry of space-time, you will be very hard pressed to try to learn GR.

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]My question is why bother? With three observers, each in his own unchanging inertial frame, we get the expected result with simple plug ‘n chug application of time dilation applied twice to of the three observers, all comfortably within SR.[/QUOTE]
Everything in this thread is SR. That SR cannot deal with acceleration is a myth.

27. PAllen says:

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]If we are not talking about actual masses traveling in these inertial frames then we are not talking about physics.

[/QUOTE]
This is a nonsensical claim, and the rest of your claims follow from it. So far as I know, there is no reputable modern theoretical physicist who would agree with the claim that acceleration and accelerated observers are outside the scope of SR.

28. the_emi_guy says:

[QUOTE=”Orodruin, post: 5447808, member: 510075″]No, this is not true. You can use any actual observer to define a reference frame, but it is not necessary for the theory to make sense. [/QUOTE]
If we are not talking about actual masses traveling in these inertial frames then we are not talking about physics.

[QUOTE=”Orodruin, post: 5447808, member: 510075″]Changes of inertial frames are just changes of how you assign coordinates. What you are claiming is essentially the equivalent of saying that you cannot use spherical coordinates unless you have an actual sphere, which is obviously absurd. .[/QUOTE]
Nope, I can have an observer (real human observer) in an single inertial frame and change coordinates all day. While traveling through New York my position can be rendered in Cartesian coordinates, then when I cross the border into New Jersey we decide to render my position in some polar coordinates. Nothing about this change in coordinates implies a change in inertial frame. On the other hand, the “traveling” twin is undergoing acceleration bringing him into a different inertial frame. This is outside the scope of SR unless we assume a non-physical observer.

I am not just a “novice” making this up. I have seen 30 page dissertations on the subject of how to properly handle the acceleration problem inherent in the two observer twin paradox. Just go to Wikipedia under “twin paradox” and search for the word “acceleration”.

My question is why bother? With three observers, each in his own unchanging inertial frame, we get the expected result with simple plug ‘n chug application of time dilation applied twice to of the three observers, all comfortably within SR.

29. Orodruin says:

[QUOTE=”the_emi_guy, post: 5447761, member: 415127″]We have to allow the observers to have physical bodies, and the clocks to have actual mass, otherwise this is not a physics problem but a mathematical exercise.[/QUOTE]
No, this is not true. You can use any actual observer to define a reference frame, but it is not necessary for the theory to make sense. Changes of inertial frames are just changes of how you assign coordinates. What you are claiming is essentially the equivalent of saying that you cannot use spherical coordinates unless you have an actual sphere, which is obviously absurd. It may not be the most convenient coordinate system unless you have some sort of spherical symmetry, but you certainly can use it.

A physical exercise is any exercise which makes experimentally verifiable predictions. You can use any coordinate system which might be convenient to make those predictions. Like many novices, you are putting way too much relevance into coordinates instead of focusing on the actual issue in the twin paradox problem – the geometry of space-time. You can use any coordinate system you wish, the geometry will not change.

[QUOTE=”the_emi_guy, post: 5447761, member: 415127″]More importantly, why would we take a problem that when formulated with three real human observers is trivial to analyze, and formulate it in a way that is very complicated (look at how many questions PF gets on this topic), and requires hypothetical massless observers?[/QUOTE]
There are no massless hypothetical observers in this example. Just different coordinate systems on the same geometrical object.

30. the_emi_guy says:

We have to allow the observers to have physical bodies, and the clocks to have actual mass, otherwise this is not a physics problem but a mathematical exercise.

More importantly, why would we take a problem that when formulated with three real human observers is trivial to analyze, and formulate it in a way that is very complicated (look at how many questions PF gets on this topic), and requires hypothetical massless observers?

31. Orodruin says:

[QUOTE=”the_emi_guy, post: 5447701, member: 415127″]Seems to me that SR cannot be applied to a problem that involves three inertial frames and only two
observers. Isn’t an observer moving from one inertial frame to another is out of the scope of SR?[/QUOTE]

No. It is perfectly within the scope of SR. You do not need an actual observer to define an inertial frame in SR.

32. the_emi_guy says:

Seems to me that SR cannot be applied to a problem that involves three inertial frames and only two
observers. Isn’t an observer moving from one inertial frame to another is out of the scope of SR?
Seems that any solution would require a footnote indicating that
the presumably infinite acceleration experienced by the “travelling” twin is assumed to have no implication.

On the other hand, allowing three observers, one in each frame, is a valid SR problem, and its solution is a trivial
application of time dilation and removes all of the complexity of the two observer problem.

Observers:
Observer1 (Earthbound observer) is on earth.
Observer2 (outbound observer) passes earth traveling toward Alpha Centauri (and never comes back)
Observer3 (inbound observer) is traveling from Alpha Centauri and first passes outbound observer, then later passes earth.

Events:
Earth passes his time to outbound observer when they meet.
Outbound observer passes his time to inbound observer when they meet.

Analysis:
Outbound observer’s point of view (exclusively): Earthbound clock is running slow (simple time dilation)
thus inbound observer’s clock is set ahead of Earth’s clock when outbound and inbound meet.

Now continuing along with outbound observer (*not* switching inertial frames). Earth’s clock is still running slow, but the
inbound observer’s clock is running even slower since this relative velocity is larger.
Thus, by the time that the outbound observer sees the inbound observer reach earth, he will see that the inbound observer’s clock will have gone from being ahead of Earth’s to being behind Earth’s by simple application of time dilation applied to this larger relative velocity.

This can be worked using simple SR time dilation from any of the three inertial frames to obtain identical results.

33. Dale says:

[QUOTE=”Orodruin, post: 5447359, member: 510075″]But exactly here lies the problem. I am not arguing that the coordinates are abad idea, I just question the pedagogical value to people who are just learning relativity. I do not see the point of introducing non-inertial coordinate systems on top of the struggles they already have.[/QUOTE]I agree. Pedagogically, for a novice I would encourage them to stick with inertial frames. Then if they persist use the paper in the spirit of “your question does have an answer, here it is, come back to it when you are ready”.

34. Orodruin says:

[QUOTE=”Dale, post: 5447343, member: 43978″]I like it. It is one of my favorites.

That said, I wouldn’t worry too much about details like the part of the trip where the distance is constant. It is a non-inertial frame so distance and speed don’t have their ordinary meanings anyway. The point is mostly that the travelling twin’s “perspective” is not symmetrical to the home twin.[/QUOTE]
But exactly here lies the problem. I am not arguing that the coordinates are abad idea, I just question the pedagogical value to people who are just learning relativity. I do not see the point of introducing non-inertial coordinate systems on top of the struggles they already have.

35. Dale says:

[QUOTE=”greswd, post: 5446549, member: 445614″]What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

As described in this paper: [URL]http://arxiv.org/abs/gr-qc/0104077v2[/URL][/QUOTE]I like it. It is one of my favorites.

That said, I wouldn’t worry too much about details like the part of the trip where the distance is constant. It is a non-inertial frame so distance and speed don’t have their ordinary meanings anyway. The point is mostly that the travelling twin’s “perspective” is not symmetrical to the home twin.

36. PAllen says:

[QUOTE=”Orodruin, post: 5446561, member: 510075″]This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.[/QUOTE]
Note that radar coordinates are exactly Minkowski coordinates if the defining world line is eternally inertial. They are asymptotically the same as Fermi-Normal coordinates near the defining world line. They have the specific feature that as long as the defining world line is inertial for all proper time before some event, and also inertial for all proper time after some event, no matter how complex the motion in between, and however ‘long’ such complex motion occurs, they give a complete coordinate chart of Minkowski space (unlike Fermi-Normal coordinates). Thus, they can be seen as a generalization of Minkowski coordinates to describe the ‘experience’ of a non-inertial observer, having several nice properties (including also they are built from more achievable measurements than positing very long rulers).

All the same, one must never attach ‘more reality’ to a description in radar coordinates than in some other coordinates.

37. Nugatory says:

[QUOTE=”greswd, post: 5446574, member: 445614″]This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time[/quote]
[quote]By “same frame”, I mean same inertial reference frame. Both being at rest w.r.t. each other.[/QUOTE]
Whether the twins are at rest relative to one another has nothing to do with whether they are in the same frame. Both twins are always in all frames all the time.

38. greswd says:

[QUOTE=”Orodruin, post: 5446561, member: 510075″]This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.[/QUOTE]
By “same frame”, I mean same inertial reference frame. Both being at rest w.r.t. each other.

I’m not sure what you mean by “special status”, I don’t think I mentioned that any frame had a special status.

The “time compression” may be viewed as a form of gravitational blueshift. This and the “same frame” effect are described as being due to “shock like scale discontinuity”.

Speaking of additional pedagogical value, I was just curious about the travelling twin’s perspective. I think many people have never seen such a diagram before too.

39. Orodruin says:

[QUOTE=”greswd, post: 5446557, member: 445614″]I think many people are interested in the travelling twin’s perspective.

This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin’s frame.

And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).[/QUOTE]
This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.

40. greswd says:

[quote=”greswd, post: 5446549″]What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

[img]https://www.physicsforums.com/attachments/triplets10-png.55802/[/img]

As described in this paper: http://arxiv.org/pdf/gr-qc/0104077v2.pdf%5B/quote%5DI think many people are interested in the travelling twin’s perspective.

This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin’s frame.

And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).

41. greswd says:

What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

[img]https://www.physicsforums.com/attachments/triplets10-png.55802/[/img]

As described in this paper: [URL]http://arxiv.org/abs/gr-qc/0104077v2[/URL]

42. Smattering says:

[QUOTE=”Orodruin, post: 5287000, member: 510075″]Orodruin submitted a new PF Insights post[/QUOTE]

This is indeed a very good read. Thanks a lot for the effort.

43. Orodruin says:

[QUOTE=”SlowThinker, post: 5287442, member: 572662″]Perhaps you’re going a bit fast near the end.

It seems the twin staying behind should have more proper time, since ##frac{tau_1}{tau_2}=gammageq1##.[/QUOTE]
Youre right of course, fixed.

44. SlowThinker says:

Perhaps you’re going a bit fast near the end.
[QUOTE]Just applying the modified Pythagorean theorem it becomes clear that the proper times along the world lines are given by
$$tau_1=2tau, tau_2=2tausqrt{1-frac{v^2}{c^2}} implies frac{tau_1}{tau_2}=gamma$$
and therefore [B]less[/B] proper time will pass for the twin staying behind, i.e., the red curve.[/QUOTE]
It seems the twin staying behind should have more proper time, since ##frac{tau_1}{tau_2}=gammageq1##.

45. Orodruin says:

[QUOTE=”DaleSpam, post: 5287092, member: 43978″]The geometrical view does help clear up a lot of the confusing aspects of SR, but it seems to be a big leap for new students.[/QUOTE]
I find the geometrical comparison to Euclidean space helps students and it is how I start the presentation of the subject in my SR class, by introducing Lorentz transforms as the hyperbolic rotations leaving the Minkowski line element invariant. Admittedly, these are students who are in their fourth year of university and they may be more ready for it. I had the idea of writing this Insight three days ago when I was giving the corresponding lecture, although I have tried to make it slightly more accessible than the level I present it on in class.

46. Dale says:

Thanks Orodruin. This is definitely worth referencing.

The geometrical view does help clear up a lot of the confusing aspects of SR, but it seems to be a big leap for new students. I wonder which is more difficult for most students: understanding the geometrical view or understanding the relativity of simultaneity.

47. greswd says:

I think many people are interested in the travelling twin's perspective.This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin's frame.And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).

48. RJLiberator says:

Excellent write up. Well-written! Thanks.

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