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Let me start by saying I understand the Mathematics behind the twins paradox and how it is resolved. I understand that due to the acceleration of one twin, time from their subjective experience is slower and have done the Maths to verify it. Here is my problem: why is it that one twin is accelerating and the other isn't? From the perspective of the twin leaving earth, their twin on earth is the one accelerating and they are completely stationary. Hence, drawing a space-time diagram from the perspective of the twin that leaves earth (so their displacement is always 0), I can get the opposite result with the same calculations. Why is the frame of one twin more important than the other and how can one be said to be accelerating and the other not?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z May 23 at 6:47
  • $\begingroup$ I see that this question has attracted many great answers and views; and I am not complaining or any thing :) , but isn't this essentially the same as my question: physics.stackexchange.com/questions/527902/… $\endgroup$ – Deepak M S May 23 at 20:04
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    $\begingroup$ @DeepakMS Your question addresses the root of this question (more clearly ;)) but this isn't a duplicate IMO. I once asked a similar question, I'll link it to your post because it's more relevant there than here. $\endgroup$ – Dvij D.C. May 24 at 0:40
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As given in the other previous answers,
an accelerometer or simply a ball on a frictionless surface in the ship can distinguish the inertial twin from the non-inertial twin.

"Being able-to-be-at-rest"$\neq$ "Being inertial".

To more fully "work out the Mathematics and Physics" with spacetime diagrams....

Although any observer can draw a diagram (an attempted spacetime diagram) so that he is at rest, that does not mean that he is inertial. In fact, the spacetime diagram drawn by a non-inertial observer is not equivalent to a spacetime diagram drawn by an inertial observer...
indeed,

  • no Lorentz transformation of the spacetime diagram of the inertial frame
    can straighten the kink in the non-inertial worldline
  • no Lorentz transformation of the spacetime diagram of the inertial frame
    can obtain the irregularities of the non-inertial-frames's attempted spacetime diagram (as described below)

Consider these twins: inertial OPZ and non-inertial OQZ.
(Although OQ and QZ are separately inertial [geodesic],
the piecewise-inertial twin OQZ is non-inertial (non-geodesic, [somewhere-]accelerated).)

I have intentionally selected an asymmetric trip for the traveler.
However, I have chosen values so that the calculations can be done with fractions.

I've drawn it on "rotated graph paper" so that the ticks are easier to see.
The ticks are traced out by "light-clock diamonds", whose area is in an invariant, as a result of the Lorentz transformation.

RRGP-robphy-asymmetricTwins-OZ

Note in the above diagram for inertial twin OPZ can be split into two parts,
using simultaneity according to OP and according to PZ:
inertial OP and inertial PZ, then spliced together.


An attempt to draw non-inertial OQZ's "spacetime diagram"

Now...
How would the non-inertial twin OQZ attempt to construct a spacetime diagram?
inertial OQ and inertial QZ, then spliced together??

I'll draw QZ first, then OQ,
followed by the splice using simultaneity according to QZ and according to OQ:
RRGP-robphy-asymmetricTwins-QZ RRGP-robphy-asymmetricTwins-OQ
(for fullsize versions: ZQ OQ)

Now, I'll splice the two diagrams
noninertial OQZ-splice:
RRGP-robphy-asymmetricTwins-OQZ-splice

On non-inertial OQZ's attempt at a "spacetime diagram"

  • Note that event X appears twice! (In fact, every event in the green region appears twice.)
  • Note that the inertial observer OPZ has a discontinuous worldline... in fact, event P is missing!
  • noninertial-OQZ's diagram can not be obtained by a Lorentz transformation of inertial-OPZ's spacetime diagram
  • noninertial-OQZ is not equivalent to inertial-OPZ

Again, "Being able-to-be-at-rest"$\neq$ "Being inertial".


Here are the diagrams next to each other

RRGP-robphy-asymmetricTwins-OZ RRGP-robphy-asymmetricTwins-OQZ-splice

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    $\begingroup$ While I am always thrilled to see a spacetime diagram in an answer, I worry that it might be possible to have too many. Speaking only for myself, this answer could be tidied up some to make it more readable. A good start would be to keep the diagrams small (and uniform in size) in the answer with, if necessary, a link to a larger version. $\endgroup$ – Alfred Centauri May 24 at 15:56
  • $\begingroup$ Thanks. I aim for readability and clarity. Unfortunately, the larger diagrams were needed to show the traveler-frames aligned, in a uniformly-sized grid. The size of the non-inertial frankensteined spacetime diagram was chosen to approximate the lab-frame diagram initially given. The diagrams at the end were reduced in order for them to appear side-by-side (since the default sizes didn't lead to a side-by-side layout). I don't think any of these diagrams can be omitted. $\endgroup$ – robphy May 24 at 16:03
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This is a good question. You're taking relativity to its "logical" conclusion and applying the idea that "all motion is relative" to also include accelerated relative motion. Now, while it's obviously correct that all motion is relative, the principle of relativity is a much stronger statement than that. It says that the laws of physics remain the same among all inertial frames. Now, if one frame is inertial then a frame accelerating with respect to it won't be an inertial frame. So, the laws of physics don't remain invariant if you go to a frame accelerating with respect to an inertial frame. Thus, only one of the twins get to claim to be in an inertial frame. So, assuming that both the twins were in inertial frame at the beginning, it can be shown that the twin who goes on a rocket trip is no longer in an inertial frame. In other words, it doesn't make sense to ask the question as to who is truly moving but it does make sense to ask the question as to who is truly accelerating (namely, the one who is accelerated with respect to an inertial frame). Don't get me wrong, you can handle accelerated frames in special relativity but the laws of physics won't look the same in an accelerated frame as they look among all inertial frames.

In general, this raises the question: how do you decide which frame is an inertial frame? Well, the answer is experimental. The twin who throws free particles and observes them to move with constant velocities is the one in possession of an inertial frame. In general relativity, there is a more satisfying answer to this question. It says that the observer who is freely falling is in an inertial frame. See, the equivalence principle.

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    $\begingroup$ Moore's relativity unit in his intro textbook Six Ideas that Shaped Physics (see also his earlier book A Traveler's Guide to Spacetime, now out of print) talks about "First Law Detectors" that beep if you're not in an inertial frame. It's a good pedagogical introduction to the idea of an inertial reference frame. $\endgroup$ – Michael Seifert May 21 at 13:51
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    $\begingroup$ @MichaelSeifert Ah, interesting. Always love the texts that don't take the first law for granted. Thanks for the recommendation :D $\endgroup$ – Dvij D.C. May 21 at 14:53
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    $\begingroup$ Nit: I think you mean "the principle of relativity is weaker". Strengthening the requirements on something weakens the thing itself, because it's less applicable. E.g. the requirement "it is Tuesday" is stronger than "it is a day," so the mandate "if it is Tuesday, it is illegal to go outside" is weaker than "if it is any day, it is illegal to go outside". Similarly, "if the frames X, Y are inertial, the laws of physics are the same in them" is much weaker than "for any frames X, Y, the laws of physics are the same in them". $\endgroup$ – HTNW May 21 at 23:51
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    $\begingroup$ The free fall remark is interesting. Can you add a paragraph about one brother being on Earth's surface at 1g and the other accelerating interstellar back and forth at 1g too? $\endgroup$ – Eugene Ryabtsev May 22 at 6:09
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    $\begingroup$ @JEB the space twin becomes non-inertial when they make their swing around Alpha C. $\endgroup$ – John Dvorak May 22 at 21:12
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From the perspective of the twin leaving earth, their twin on earth is the one accelerating and they are completely stationary.

But the accelerometers attached to each twin read the invariant proper acceleration. Thus all observers agree that the accelerometer on the twin leaving Earth shows non-zero acceleration during the journey, i.e., that the world line of that twin is not a geodesic between the two events, and that the other twin's accelerometer reads zero.

The traveling twin would need to invoke the (sudden) appearance of a (uniform) gravitational field that she is stationary in (and that the other twin is free falling in) to account for the non-zero reading on her accelerometer (and the zero reading on the other).


Update to address (and preserve) a comment:

But what is that accelerometer measuring? What is acceleration if not a change in velocity from the perspective of a stationary reference?

What you describe is coordinate acceleration which is observer dependent. Take a look at the link above regarding proper acceleration which is observer independent.

Here's an example of the difference: in SR, an object can have constant proper acceleration, but it can never have constant coordinate acceleration else it would eventually exceed speed c in that (inertial) coordinate system.

All observers would agree that the object has constant proper acceleration (they can all read the accelerometer attached to the object), but they would disagree on the object's coordinate acceleration.

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  • $\begingroup$ But what is that accelerometer measuring? What is acceleration if not a change in velocity from the perspective of a stationary reference? $\endgroup$ – ajax2112 May 22 at 10:29
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    $\begingroup$ @ajax2112, what you describe is coordinate acceleration which is observer dependent. Did you take a look at the linked article on proper acceleration which is observer independent? For example, in SR, an object can have constant proper acceleration, but it can never have constant coordinate acceleration (else it would eventually exceed speed $c$ in that (inertial) coordinate system. All observers would agree that the object has constant proper acceleration (they can all read the accelerometer attached to the object, but they would disagree on the object's coordinate acceleration. $\endgroup$ – Alfred Centauri May 22 at 12:19
  • $\begingroup$ @ajax2112 "But what is that accelerometer measuring?" Could be the displacement of a mass on a spring. $\endgroup$ – user76284 May 23 at 5:22
  • $\begingroup$ "the accelerometers attached to each twin read the invariant proper acceleration" How does it do that if you are accelerating under the influence of gravity? For example if one twin is falling into a black hole. $\endgroup$ – hdhondt May 23 at 5:48
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    $\begingroup$ @hdhondt - The twin paradox is a question pertaining to special relativity. The "special" in special relativity means "gravity-free". The addition of planets (where gravitation is small) orbiting small stars (where gravitational is once again small) to the twin paradox is for illustrative purposes. While special relativity isn't exactly applicable to this situation, the deviation is small. This means that using special relativity remains very close to correct -- so long as one does not posit that one of the twins is orbiting a black hole. $\endgroup$ – David Hammen May 23 at 9:46
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Since you know the mathematics of the process, I'll go to the physical perspective.

The twin that stays on Earth doesn't feel any acceleration during the process, he's just sitting on a chair, let's say.

The other twin, however, gets into a spaceship and has to accelerate to obtain a certain velocity, he feels acceleration while taking off. Therefore, the twin that leaves is not inertial during that time.

When drawing a space-time diagram from the perspective of the twin that stays, if you draw the worldline of the twin that leaves as two straight lines, that's not strictly true.

From the perspective of the twin that leaves, we simply should not describe the process from that system because it is not inertial the whole time, and special relativity restricts you to describe physics from inertial reference frames.

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  • $\begingroup$ Urb, I edited you answer (by adding whitespace) to make it more readable to me. If you're unhappy with this, please revert the edit. $\endgroup$ – Alfred Centauri May 21 at 13:43
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    $\begingroup$ Thank you. Yes, it was a little cluttered $\endgroup$ – Urb May 21 at 13:49
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The previous answers all try to focus on proving one frame inertial and the other not. That is not really what is involved here since almost all the journey can be done without acceleration at constant speed and the phases of acceleration kept so short that their influence on the clock is negligible.

While the twins are separating, the situation is completely symmetrical and their respective clocks, viewed from the middle point of the two as rest frame, show equal advance. But when one of the twins decides to return, the middle-point between the twins (which ultimately ends up meeting with both twins) occupies a different inertial frame with a different concept of simultanousness. While both twins travel towards it at equal speed and thus their clocks appear to run at the same speed as well, in this frame of reference marking their middle point after the turnaround, the clocks up to this point of time did not run at the same speed. It stuck quite closer to stationary twin and to its time and the clock of the traveler looked to be quite slower from this vantage point.

So from the vantage point of the middle-point-to-come, it was always the traveller's clock that was slow, while from the view of the middle-point-of-departure, both clocks ran at the same speed. But the inertial frame of the middle-point-of-departure will be far away by the time the twins meet again, and when the travelling twin turned around, its clock started looking a lot slower from this vantage point.

At any rate: acceleration does slow clocks down (like satellites in orbits and thus under constant acceleration show) but the effects on the twin paradox can be kept minimal by keeping the acceleration phases short.

The real difference comes from a change of reference frames for the traveling twin, and those different reference frames have different notions of simultanity meaning that the required change of view (to be able to join the meeting of both twins) comes with a change of clock. Like a timezone change on your pocket watch, it does not really make a difference on just when you adjust your clock while in a plane. You just have to do it before arrival.

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Talking about twins and acceleration just makes the issue harder to understand. In my opinion it's more illuminating to consider three triplets, each having its own inertial frame:

• The world-line of triplet 1 intersects event A and event B.

• The world-line of triplet 2 intersects event A and event C.

• The world-line of triplet 3 intersects event C and event B.

Note that no accelerations are involved. Each time two triplets' world lines intersect, they exchange their clock times with each other. This makes it possible to calculate the proper times between:

• Events A and B for triplet 1. Call this $\Delta \tau_{AB}$.

• Events A and C for triplet 2. Call this $\Delta \tau_{AC}$.

• Events C and B for triplet 3. Call this $\Delta \tau_{CB}$.

Now consider the triangle inequality, which is a well-known theorem in euclidean geometry, which says that the sum of two sides of a triangle is larger than the third side. This inequality is reversed in Minkowski space since the minkowskian metric is not positive-definite, so we always have $||\mathbf{AC} + \mathbf{CB}|| = ||\mathbf{AB}|| \ge ||\mathbf{AC}|| + ||\mathbf{CB}||$. For our triplets this means that $\Delta\tau_{AB} \ge \Delta\tau_{AC} + \Delta\tau_{CB}$, and equality holds only when $\mathbf{AC}$ and $\mathbf{CB}$ are parallel 4-vectors.

Now if we transfer this to the case of two twins, where the first twin moves just like triplet 1, and the second twin first accelerates to match triplet 2, then accelerates again to match triplet 3, and then finally accelerates to match triplet 1 again, it is easy to see that the accelerated parts of the second twin's world line are just irrelevant details that only make the proper time calculations more tricky, without adding any insight.

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  • $\begingroup$ FWIW, this is, I believe, a true statement (in the context of SR): The greatest elapsed time between two events, as recorded by any clock with world line through those events, is recorded by that one clock that is inertial, i.e., unaccelerated according to an accelerometer attached to the clock. $\endgroup$ – Alfred Centauri May 21 at 20:23
  • $\begingroup$ @Alfred Centauri Yes, that can very likely be proven fairly easily for all smooth world lines (in SR), and I am quite sure that non-smooth world lines are unphysical, so I agree it's a true statement. $\endgroup$ – Cuspy Code May 22 at 16:55
  • $\begingroup$ This isn't fair, since there is no paradox in four dimensional Minkowski space (M4). The problem is that M4 itself is paradoxical to those with E3 + 1 intuition, and in this version of the paradox you still have the 3+1 problem that when triplet 2 transfers his time to triplet 3, they disagree on triplet 1's age by $\frac {2L} v (1-1/\gamma^2)$. How can it be 2 totally different times back on Earth at the same distant event? $\endgroup$ – JEB May 23 at 1:17
  • $\begingroup$ @JEB, I was thinking about this yesterday during a walk. Going back to the twins (or better, two identical clocks), the sudden acceleration required for the traveling clock to jump off the outgoing IFR (triplet 2 IFR) to the ingoing IFR (triplet 3 IFR) doesn't change the time on that clock or the rate at which it ticks (according to the inertial stay at home clock). However, as (I think) you point out, the sudden change in the lines of simultaneity means that the traveling clock observes (not sees) a sudden jump forward of the time reading on the inertial clock. $\endgroup$ – Alfred Centauri May 23 at 12:20
  • $\begingroup$ @JEB The $\gamma vx/c^2$ term (in the Lorentz transformation of $t$) has opposite signs for triplet 2 and triplet 3, so they will never agree on the coordinate time for a distant event (i.e. for an event with $x \neq 0$). $\endgroup$ – Cuspy Code May 23 at 13:11
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Let us try to answer this in a simple way using Physical arguments here. First up what is meant by Symmetry, and the answer lies in the fact that, a physical system is invariant under some transformation which leads us to say that the system itself is symmetric under that particular transformation. Th second important thing we would have to consider here is, how would we recognize or the system itself recognize that the symmetry is broken.

For this, let us consider a ferromagnet and us being creatures living inside of that ferromagnet. Initially we would conclude that out world is rotationally symmetric as there is no preferred direction by the little dipoles inside the ferromagnet. So, the system is initially rotationally invariant. But then some creature from outside applies an external magnetic field in a specific direction, all the dipoles in each domain start to align themselves in the direction of the field in such a way that it would be energetically minimal. Us, as the creatures inside the ferromagnet, by doing the experiments now would conclude that the system is rotationally invariant as there is clearly a preferred direction. So, we as creatures (if we are clever enough) would conclude that there might be an external magnetic field influencing the preferred direction which broke the symmetry (keeping the subtleties of Spontaneous Symmetry Breaking conclusion aside).

Now, let us come back to the question under consideration. Let us say the spaceship of the twin that left the planet is analogous to the little ferromagnet that we have considered above and let us take the perspective of the twin inside this spaceship. And the symmetry we are considering here is the Physical laws themselves, that no matter what inertial frame we switch to, the laws of Physics remain the same. So, we as the creatures in this spaceship now, devise a way to test for this symmetry. And the test is as follows. We take a ball, keep it on the floor beside us and initially we see that there is no motion in the ball. So, we conclude that the laws of Physics are in full working order, no motion, so the ball is also stationary with respect to us. Now, we do the experiment again, then the ball suddenly moves backwards or forwards(depending or acceleration or deceleration and let us say this experiment is done when the spaceship turns) and then we say, oh this is interesting now because the ball has moved and that is against the laws and so we conclude the symmetry of Physical laws under test is broken. Just like in the previous case where we inferred that it maybe because of some external magnetic field, we here infer that, it maybe that we are the ones under motion and even accelerating. We feel it in the form of a broken symmetry. And acceleration is responsible for that. But now if we do the same experiment by being on the planet, the ball would remain stationary all along and we would perfectly conclude that the laws of Physics are what we need them to be. There is no broken symmetry of any sort in this frame of reference.

So, now clearly there has been this asymmetry established where in in one frame of reference, you report one sent of conclusions which are clearly different from the other frame's conclusions. It is in other words, the frames of reference are not equivalent, which means the two systems are no longer identical w.r.t the measured physical quantities as the laws themselves have manifested differently in each reference frame.

This is the physical argument, I think would be helpful to understand this situation better and I feel, a deep intuition about acceleration, not just numbers, is needed to have a more profound understanding of what is going on here.

Punchline : Two systems which would generically differ in the description of the laws of Physics in their frame of reference would therefore obviously differ in the measured physical quantities attached to those frames of reference.

Mistake made : Your interpretation of acceleration is purely mathematical, but there is more to it than just shift of origin in this problem as you are wanting to do. There is the difference in Physical laws itself which if I may use can be labelled as Inertial and Non-Inertial reference frames. So, once you are part of a Non Inertial reference frame, this symmetry which you are assuming to hold by simple shift of origin is no longer there, the symmetry is genuinely broken. To be more precise, it is explicitly broken by the acceleration of the system which makes it a system under true motion.

I hope this answer helps!

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It is not the acceleration that creates the time difference. The acceleration is necessary to create a path difference, but the acceleration is not the cause of the time difference. The time difference is do to the path difference that they traversed in spacetime. Their accelerations may be symmetrical, but their paths travelled through spacetime are not symmetrical. If you drew their paths on a spacetime graph it would be clear that one twin traveled a longer path through spacetime and that’s why his clock ran slower than his twin.

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  • $\begingroup$ The point isn't that acceleration plays a significant role in how the clock of the traveling twin behaves, the point is that acceleration is what tells us that we can't consistently use the proper frame of the traveling brother as an inertial frame. The comment applies to a bunch of other answers who are trying to point out that acceleration doesn't cause much of a time dilation. $\endgroup$ – Dvij D.C. May 23 at 5:09
  • $\begingroup$ Lambda, this seems (to me) to downplay the role of acceleration too much. The greatest elapsed time between two events, as read by a clock with world line through those events, is given by the inertial (unaccelerated) clock. All other clocks, with a different world line through those events, will read less time. What do all those other clocks have in common? It is this: the accelerometer attached to the clock shows acceleration over some or all of the world line segment between those events. In this sense, it is acceleration that creates the time difference. $\endgroup$ – Alfred Centauri May 24 at 13:43
  • $\begingroup$ For example, the stay-at-home twin observes that her traveling twin's clock is behind her clock during the entire journey. During the outbound leg, the traveling twin observes that her twin's clock is behind hers but, and this is crucial, just after the turnaround, her twin's clock is ahead of hers - so much so that it is ahead of hers for the entire journey back despite her twin's clock running slower. Behind before the turnaround, ahead after the turnaround (acceleration). Can it really be said that "it is not the acceleration that creates the time difference"? $\endgroup$ – Alfred Centauri May 24 at 15:41
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You are asking how we can tell that the traveling twin's frame is the one that is accelerating, and why is that not symmetrical from the view of that twin?

The answer is that according to the numerous experiments that we made and built our theories around them, SR/GR tells us that speed is relative, but (proper) acceleration is absolute.

Within the context of Newtonian mechanics, there's a simple answer: velocities are not absolute, but differences in velocities are. So you can state that acceleration occurs unambiguously. In special relativity, this is a bit more complicated because of relativistic velocity addition, but all observers can unambiguously compute a "proper" acceleration for every object, which is the acceleration in that object's momentary rest frame. In fact, the same logic still works in general relativity; acceleration is unambiguous even in a universe without matter.

If absolute velocity does not exist, how can we say a rocket accelerates in empty space?

Now how can we tell, that the traveling twin's frame is the one that is accelerating, and the answer is the equivalence principle. The equivalence principle tells us, that according to the experiments, acceleration can have the same effect on the passage of time (the temporal component of the four vector) as being in a gravitational field.

In the theory of general relativity, the equivalence principle is the equivalence of gravitational and inertial mass, and Albert Einstein's observation that the gravitational "force" as experienced locally while standing on a massive body (such as the Earth) is the same as the pseudo-force experienced by an observer in a non-inertial (accelerated) frame of reference.

https://en.wikipedia.org/wiki/Equivalence_principle

Now in your case, the traveling twin, as it turns around at half way, will have to undergo acceleration/deceleration. This phenomenon has the same effect as being in a gravitational field, that is, causes time dilation. A third observer will specifically be able to tell, that during the turn around, the traveling twin's clock's ticks slow down (relative to the observer's clock), while the other twin on Earth will not show the same effect. This tells the observer that the traveling twin is (either in a changing gravitational field or is) accelerating.

Now you are asking, how can it be that from the traveling twin's frame, the other twin on Earth will not seem to be symmetrically accelerating? Let's assume that there are no other objects in the universe to compare to. How will the traveling twin still be able to tell that it is not the other twin on Earth accelerating? Now this is because the traveling twin is able to release beacons, at similar temporal coherence. What the traveling twin will see during the turn around, is that the released beacons seem to become different distances apart, and this distance seems to be changing during the turn around. The beacons will seem to recede from the traveling twin with different/changing speed.

Kinematically, yes. In terms of describing the positions of objects, it is equivalent to say "A is accelerating away from B" and "B is accelerating away from A". However, it is an observed fact that the universe treats these two situations differently. A and B can check whether they feel artificial gravity in their reference frame. If so, it's accelerating.

How are accelerated reference frames non-symmetrical?

It is very important to understand the difference between proper and coordinate acceleration. You need proper acceleration, and that is only in the traveling twin's frame.

Imagine a spacecraft that is accelerating with constant acceleration α according to an accelerometer in the spacecraft. This is the proper acceleration and it is constant.

Explanation on how can't we really use Rindler metric on twin parad. and that you really can't outrun light

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  • $\begingroup$ Shouldn't it be "SR/GR tells us that speed is relative, but (proper) acceleration is absolute". I don't think I'm being picky here - much of the confusion surrounding the twin paradox is due to not distinguishing coordinate acceleration from proper acceleration. $\endgroup$ – Alfred Centauri May 24 at 15:50
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    $\begingroup$ @AlfredCentauri thank you I edited. $\endgroup$ – Árpád Szendrei May 24 at 15:52

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