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The search based on the term "Twin Paradox" gave (today) 538 results.

In all the answers, the answerers explained the phenomenon by referring to arguments sort of falling out of the framework of special relativity. I saw answers referring to acceleration and deceleration, changing coordinate systems, etc. Even Einstein referred to General Relativity when explaining the TP...

THE QUESTION

Is it not possible to explain the phenomenon purely within Special Relativity and without having to change the frame of reference?

EDIT

... and without referring to acceleration and deceleration, without having to stop and turn around or stop and start again one of the twins?

zoli
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    The explanation is precisely that the traveling twin doesn't stay in a single inertial fame of reference. You don't need to use that to calculate the effect, but it is the explanation. – Javier Jun 01 '19 at 14:15
  • @Javier: I want to use one single frame of reference. The question is if that is possible. – zoli Jun 01 '19 at 14:18
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    "In all the answers, the answerers explained the phenomenon by referring to arguments sort of falling out of the framework of special relativity." Please give an explicit example of that. The twin paradox does not require anything outside of special relativity. Also, it's hard to believe that not a single answer on this site explains the twin paradox using just special relativity. – DanielSank Jun 01 '19 at 16:37
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    I'm with @Daniel on this. I suspect you have a mistaken idea of what indicates that you have begun using general relativity in your argument. Pointedly, special relativity is perfectly sufficient to deal with a large number of question that involve acceleration (though acceleration does generate questions that lie outside the purview of special relativity, the twin paradox simply isn't one of them). – dmckee --- ex-moderator kitten Jun 01 '19 at 22:15
  • In re your edit: there is no paradox is both parties seeing the other running slow. It is simply a fact about Minkowski space time that they don't agree on the meaning of "now" which means that they don't agree on what two events in spcae time are to be compared in order to determine whose clock is running slow/fast. That is, time dilation is an expression of the relativity of simultaneity. – dmckee --- ex-moderator kitten Jun 01 '19 at 22:19
  • If one of the twins doesn't stop and turn around, you're not talking about the twin paradox. If you can give us a detailed description of what apparent paradox you are talking about, we may be able to resolve it for you. – Harry Johnston Jun 02 '19 at 02:17

4 Answers4

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Is it not possible to explain the phenomenon purely within Special Relativity and without having to change the frame of reference?

It is possible to obtain the correct answer to the amount of time accumulated by either of the twins by using any single reference frame, without changing that frame at any point in the analysis. Whether or not such a calculation constitutes an “explanation” is a matter of opinion. I would tend to say “no” because the “paradox” is precisely about what happens when you incorrectly change reference frames.

To obtain the amount of time accumulated by any traveler we write their worldline as a parametric function of some parameter (using units where c=1), for example $r(\lambda)=(t(\lambda),x(\lambda),y(\lambda),z(\lambda))$ where $r$ is the worldline and $t$, $x$, $y$, and $z$ are the coordinates of the traveler in some reference frame whose metric is given by $d\tau$. Then, for any reference frame for any spacetime for any traveler, the amount of time is given by $\Delta\tau=\int_R d\tau$ where R is the total path of interest (i.e. all of the $r(\lambda)$ of interest). Because this is a completely general formula it applies for an inertial traveler or for a non inertial traveler, it also applies for an inertial reference frame or for a non inertial reference frame, it also applies in the presence of gravity or not.

For the specific case of an inertial frame we have $d\tau^2=dt^2-dx^2-dy^2-dz^2$ from which we can easily obtain $$\frac{d\tau}{dt}=\sqrt{1-\frac{dx^2}{dt^2}-\frac{dy^2}{dt^2}-\frac{dz^2}{dt^2}}=\sqrt{1-v^2}$$ So then $$\Delta\tau=\int_R d\tau=\int_R \frac{d\tau}{dt} dt = \int_R \sqrt{1-v^2} dt$$

Note, this last paragraph assumes an inertial frame (any inertial frame is the same). The usual mistake is to use the inertial frame expression in a non inertial frame. A similar procedure can be used in a non inertial frame, but you must use the appropriate expression for $d\tau$

Dale
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    "It is possible to obtain the correct answer to the amount of time accumulated by either of the twins by using any single reference frame..." Would you, please, show me how? – zoli Jun 01 '19 at 13:58
  • I will add to the answer to include an example. – Dale Jun 01 '19 at 14:53
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    I thought the "paradox" was more about the contradiction implied by the fact that each observes the other's clock to slow down, and by the fact that without a preferred frame, there is no grounds for justifying why the clocks should not be synchronise later. – Steve Jun 01 '19 at 14:55
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    @Steve The apparent paradox comes from trying to describe things from the traveling twin's point of view. The traveller accelerates, so if you try to treat their perspective as an unaccelerated frame under SR, you're guaranteed to get nonsense. You can describe things from the traveller's perspective in GR, because it can handle accelerated frames. – Gordon Davisson Jun 01 '19 at 23:41
  • @GordonDavisson, but points of view are not the nub of the paradox - the nub is that the clocks, as a matter of experimental fact, don't show the same elapsed time after the journey is complete, and attempting to reconcile this with the principle that no preferred frame exists. The absolute forces of acceleration do not explain the discrepancy, because the discrepancy is proportional to the length of the journey, and it can be shown that two clocks that experience the same forces of acceleration, but travel different distances overall, show a discrepancy that is proportional to the distance. – Steve Jun 02 '19 at 00:02
  • @Steve you said “attempting to reconcile this with the principle that no preferred frame exists”. There is no such principle. The first postulate is that all inertial frames are equivalent, meaning there is no preferred inertial frame. Also, an observer’s “point of view” is commonly understood to refer to a reference frame wherein the observer is at rest. I think Gordon is on track here. – Dale Jun 02 '19 at 00:46
  • @Dale, but my point is that both clocks eventually return to the same inertial frame - to be compared side-by-side. In that frame, they display different readings. The difference is proportional to the length of the journey undertaken. The travelling clock is (having accelerated and whilst it is moving at a steady speed) in an inertial state where it operates more slowly than the home clock. In an inertial state, the travelling clock should be able to employ SR in the same way as the home clock - but that is in contradiction to the assertion that the inertial frames would be "equivalent". – Steve Jun 02 '19 at 01:19
  • Ah @Steve you are killing me here. You said “both clocks eventually return to the same inertial frame - to be compared side-by-side. In that frame, they display different readings”. This is exactly as predicted in said frame. Indeed, it is exactly as predicted in any frame since it is an invariant. – Dale Jun 02 '19 at 02:19
  • @Dale, but the fact of the elapsed time being different for each clock, and the fact of the difference being proportional to the travel time, establishes that some of the discrepancy is accrued during the period in which the travelling clock was inertial. If they are both inertial, but one is losing time against the other (and equivalently, the other is gaining time against the first), then they cannot make identical predictions. There has to be some grounds for distinguishing their circumstances even when inertial, and they must both agree which of them will finally be ahead and which behind. – Steve Jun 02 '19 at 02:42
  • @Steve you incorrectly claim “If they are both inertial, but one is losing time against the other (and equivalently, the other is gaining time against the first), then they cannot make identical predictions.” And yet they do make identical predictions, using the method shown in my answer above. Just because the traveler spends some time traveling inertially does not mean that a reference frame where he is always at rest is inertial. – Dale Jun 02 '19 at 02:47
  • Let us continue this discussion in chat. – Dale Jun 02 '19 at 02:56
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In Minkowski space, the worldline of one twin is a geodesic. The worldline of the other is not, although it can be approximated or idealized as piecewise geodesic. This asymmetry explains why there is no paradox.

If both worldlines were geodesics, then the twins would never meet again.

user52817
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  • "If both worldlines were geodesics, then the twins would never meet again." There is a whole little industry in making arguments about these questions in space times with a periodic space dimension (where they can meet again even if they both follow a geodesic). – dmckee --- ex-moderator kitten Jun 01 '19 at 22:22
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I believe that in search for a better explanation, it is best to turn to the Lorentz ether theory. It is often said, that this theory is empirically equivalent to the SR. Since the same mathematical formalism occurs in both, it is not possible to distinguish between LET and SR by experiment. The introduction of length contraction and time dilation for all phenomena in a "preferred" frame of reference, which plays the role of Lorentz's immobile ether, leads to the complete Lorentz transformation

It should be noted, that there is no "paradox" related to the Lorentz theory. In the Lorentz theory, the paradox that has arisen in the depths of special relativity is resolved by means of elementary algebraic methods, staying within the same frame of reference and not taking into account the acceleration or deceleration.

Let's consider resolution of the "paradox" in the framework of the Lorentz theory.

1) Let’s consider what would happen if one of two twins who are at rest in the ether at one point, flies at speed $v$ to a distant point and then after a while returns to twin $A$ remaining at rest.

If for the twin flying in the ether his “local time” characterizing the rate of physical processes in his body and the pace of the movement of his clock on both segments of his flight (there and back) slows down due to interaction with the ether, then the lapse of his “local time” will be $1/\sqrt {1-v^2/c^2}$ times less than for the twin at rest in the ether, and the “travelling” twin will get less “old”. The turn of the travelling twin, provided it is virtually instantaneous, has no practical effect on the ratio of times of both twins.

2) Now let's calculate, what will happen if the two twins are flying side by side in the ether at speed $v$ – with their “local time” passing slower – then one of them stops, staying at rest in the ether for some time, then catching up with the travelling twin.

The twin who continued his flight in the ether with no information about the fact of his motion in the ether perceives this maneuver of his brother as a round trip to a distant point.

An obvious answer is that, since according to the ether theory after the twin’s stop in the ether his time will pass faster than the “local time” of his twin brother who continues his flight, and then when the twin stopping in the ether after some time catches up with the missing brother, he will age more than the latter. The “local” time of the twin catching up with his flying brother will actually flow slower than for the flying brother. This is due to the faster speed of the twin catching up with his brother. As a result, the brother making a stop in the ether will age not more, but less than his twin brother who has not interrupted his flight.

Let us demonstrate that if the proper times of the motion there and back of the non-inertial twin relative to the inertial twin are equal, then for the non-inertial twin it will take $1/\sqrt {1-v^2/c^2}$ times less time than for the moving inertial twin, and the non-inertial twin will age less.

Let at the time of stop of one of the twins in the ether the clocks of the parting twins show zeros. Suppose that after making a stop for some time, the twin who has lagged behind, at the moment $t_1$ of the ether time when his clock (because of the stop) was showing this time, left at speed $u$, such that $v<u<c$, following his brother flying away from him. The distance between the twins at the start of the twin who has left behind is equal to $vt_1$. Setting out, the twin left behind will catch up with the twin flying at a constant speed $v$ at the point in time $t_2$, having spent the time equal to $vt_1/(u-v)$. During this period, by the clock of the twin following the flying away brother at speed u, there will be a lapse of proper time, which is $1/\sqrt {1-v^2/c^2}$ times less than the ether time and equals $vt_1\sqrt{1-(u/c)^2}/(u-v)$. Let us assume the velocity $u$ such that the proper time $t’_2-t’_1$ of the catching up twin is numerically equal to the time $t_1$ of his stay at rest relative to the ether, i.e. $t’_2-t’_1=t_1$ or

$$t_1=vt_1\sqrt {1-(u/c)^2/(u-v)} (1)$$

This equation meets the condition under which the twin spends the same proper time on a trip to a distant point and back. By elementary transformations of the equation (1) we can obtain the value of velocity $u$, which is equal to $\frac {2v}{1+(v/c)^2}$ . Substituting this value in the expression for the time $vt_1/(u-v)$ required for the return of the twin, and summing the time $vt_1/(u-v)$ and the time $t_1$, we obtain the ether time spent by the lagging behind twin on the stop and return to the flying twin. This time is equal to $2t_1/(1-v^2/c^2)$. Since the clock of the inertial twin flying at a speed $v$ go $1/\sqrt {1-v^2/c^2}$ times slower than the clock at rest in the ether, the flying twin will determine the time spent by the lagging behind twin on the stop and return to the flying twin as a quantity meeting the equality:

$$t’_2=2t_1/\sqrt {1-(v/c)^2}$$

Since the time elapsed for the non-inertial twin by the moment of his return is numerically equal to $2t_1$, and the time of the inertial twin is numerically equal to $2t_1/\sqrt{1-(v/c)^2}$, then the lapse of time for the non-inertial twin is $1/\sqrt {1-v^2/c^2}$ times shorter, and he has aged less than the inertial twin has.

This way we can see, that non–inertial twin (or clock) will show $1/\sqrt {1-v^2/c^2}$ less time than inertial one, despite of direction of its motion in ether.

We get absolutely the same result as in the Special Relativity, but resolution of the paradox is very simple.

  • I upvote your answer, and would give more up votes if that was possible. However, I do not accept your answer since it just refers to acceleration. The "stopping" twin has to decelerate and then accelerate. Some (not me) would say: "See that is when the time difference gets smuggled in." – zoli Jun 01 '19 at 14:10
  • @Albert you said “what would happen if one of two twins who are at rest in the ether at one point”, but you cannot know that. You also said “then one of them stops, staying at rest in the ether for some time”, but you also cannot know that. If you really want to push a discarded interpretation then at least do it right. The ether frame is unknown and unknowable, so you need to have the speed wrt the ether as an unknown variable, which must later drop out. – Dale Jun 01 '19 at 14:49
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    @zoli, you can completely get rid of acceleration in any theory this way: the non-inertial twin does not actually stop. Let's say that the second twin (let's call him non - inertial ) does not stop, but meets the third twin. At this moment the "second" passes readings of his clock to the "third" and the "third" goes back to the "first". The fact is, that clock of the "third" twin will show $\gamma$ times less time than the "first".. Actually even in SR you get explanation exactly without changing frames: https://physics.stackexchange.com/questions/383337/twins-clock-slowing1-down/383412#383412 –  Jun 01 '19 at 14:56
  • @zoli, the acceleration cannot be when the difference sneaks in, because the difference is related to the overall time spend moving, not the time spent accelerating or the forces sustained. That is, a twin that travels further before returning will sustain a greater difference, notwithstanding that the acceleration profile would be exactly the same on a shorter trip. – Steve Jun 01 '19 at 14:59
  • I have never said that I know which frame is "preferred" or that it is "detectable". I simply assumed existence of this frame, and modeled twin paradox in that frame. I believe that I can assume existence of that frame. However, the problem disappears very quickly, if only we assume the existence of a preferred frame of reference or even if we choose one frame and stay within that frame. I believe that even if we cannot find which frame is preferred it does not give us grounds to jump from frame to frame like monkeys on trees. Preferred frame brings order, absence of it brings nonsense. –  Jun 01 '19 at 15:03
  • @Albert: Your comment "you can completely get rid of acceleration in any theory this way..." It seems that this was the answer I was looking for. Thank you. – zoli Jun 01 '19 at 15:11
  • I agree with you Albert. The key point in LET is that any ether is not itself detectable - it's conceptual function is only in what it logically entails (i.e. it entails the behaviours we observe which are consistent with SR). Or to put it the other way around, the correctness of SR implies the existence of the ether, because it is not correct without the implication. All other methods of making sense of SR are, as you say, a nonsense. – Steve Jun 01 '19 at 15:41
  • @Albert you said “I simply assumed existence of this frame” which is false. You also assumed that is had a specific velocity wrt the entities in the problem, as clearly shown by direct quotes above. There is nothing wrong with using LET, but you should use it right. The speed wrt the ether is an unknown. – Dale Jun 01 '19 at 15:55
  • @Dale, it seems to me plausible to argue that one's speed relative to the universal centre of mass, is equivalent to one's "speed wrt the ether". The question of one's speed relative to the ether would then be the same question as one's speed relative to the universal centre of mass - and the problem is shifted from detecting the ether itself (apparently impossible), to detecting the universal centre of mass (a tractable problem). – Steve Jun 02 '19 at 00:35
  • @Steve, really? 1) That is not a claim made by LET. 2) There is no universal center of mass in modern cosmology. 3) There is no and can be no experimental evidence to support the argument. 4) That wasn’t mentioned by Albert, so it isn’t a particularly relevant suggestion. 5) while there is no universal center of mass, there is a CMB center of momentum frame at each point but the frames at different points are moving wrt each other which violates LET. Overall, the argument seems less plausible to me than it apparently does to you. – Dale Jun 02 '19 at 01:27
  • @Dale, (1) agreed that is not a claim made by LET. (2) whether modern cosmology employs the concept is immaterial. The question is whether it can be thus employed. (3) the experimental evidence would be whether any clock travelling at an appreciably higher speed relative to the majority of mass in its vicinity (or the whole universe), experienced a different rate of time than clocks stationary relative to the majority of mass (4) I'm making a separate point to Albert, addressing yours about the speed wrt ether. (5) "at each point" of what? – Steve Jun 02 '19 at 01:51
  • @Steve, this is absurd. You said “whether modern cosmology employs the concept is immaterial. The question is whether it can be thus employed”. You want to employ something that doesn’t exist (the center of mass of the universe) to make something impossible (detecting the ether) into “a tractable problem”. No, it cannot be this employed. May I ask what sources you have been studying? I would recommend finding some better sources, maybe Taylor and Wheeler’s excellent book. – Dale Jun 02 '19 at 02:29
  • @Dale, there must be a centre of mass. Even if the universe is treated as infinite in all directions, that would just mean any place can be treated as the centre, and we would need only refer to the mass in the vicinity to establish (at least to an approximation) whether you were stationary or not. And I would worry about your own sources before mine. – Steve Jun 02 '19 at 02:53
  • Let us continue this discussion in chat. – Dale Jun 02 '19 at 02:56
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It's possible that the twin paradox is the most misunderstood thing on the internet. First, there are those who claim it isn't a paradox when they mean it's not a contradiction. Rest assured it is a paradox, it is several paradoxes, and at the end of the day, it is the Andromeda paradox.

Step one in understanding it is make it as simple as possible, where twin A stays at home, and twin B travels. (If you stumble on a explanation where B stays at home: keep clicking).

Ignore gravity and acceleration problems: twin B flies straight out, and straight back. Keep it simple. Pick a nice $\gamma$, let's do $\gamma=10$. For 100 year legs.

Paradox 1: Is that twin B is younger than twin A at the end. Fine, that's time dilation. A paradox in 1905, not in 2019.

Paradox 2: Is that all motion is relative. So A see's B's clock run slow (see paradox (1), but B also sees A's clock run slow. That's weird:

A ages 100 years outbound, 100 inbound. He sees B ages 10 + 10. B experiences 10 + 10, but sees A age only 1 year on each leg (1 + 1). A paradox, indeed. Many will claim acceleration or GR at this point.

Paradox 3: Of course the resolution of (2) is that A instantly ages 198 years when B changes directions. This isn't acceleration or GR, because in both A and B's frames, the turn around takes ZERO time, so NO amount of time dilation can fix it. $0 \times \gamma' = 0$, after all.

A sees B change direction instantly, and the only change for him is that he computes B's definition of "now on Earth" going from 49 years in his past to 49 years in his future.

Meanwhile, B turns around in zero time. The only difference is that his definition of "now on Earth" (which is well outside his light cone) jumps forward 198 years...and that accounts for ALL the aging difference. That pretty much is the Andromeda paradox.

JEB
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