# Why isn't an observer on earth considered to be accelerating in the twin paradox?

This question has a duplicate, but answers to that question went miles above my head. Well, that was supposed to happen since I have just previewed special relativity that too on a high school level, but I didn't even go near general relativity. So this question has been asked from that point of view. I hope that this question won't irritate anyone.

In the twin paradox, how do we conclude that the observer in the rocket is the one responsible for breaking the symmetry by accelerating even though according to him the other observer on earth is accelerating? Is it confirmed in reference to another (third) observer which was in the same inertial frame as the other two were at the beginning?

Again if the whole incident took 10 years according to the observer on earth and 6 years for the observer in the rocket, are the 6 years out of 10 responsible for the symmetry that is due to being in the same inertial frame and the the rest 4 is due to the acceleration? Or is this calculation a bit more peculiar than it seems (I mean the calculation isn't that uniform as I have thought in the previous question)?

Any reference would be enough which would clear all my problems or even a book suggestion will be enough if this question seems to be not worth answering.

• The twin "paradox" is usually set in empty space, where the stay-at-home twin is in free-fall. Having one of them them on Earth complicates and ruins the simplicity of the scenario (except for advanced courses). Sep 21 at 16:02
• umm yeah this is why I didn't consider earth's motion to be absolute
– MSKB
Sep 21 at 16:06
• Why "him"? It could be a small fury creature from Alpha Centauri. Sep 21 at 17:38
• @MSKB I also have the same question. I can't find a valid explanation about twins being at different ages if they return to the same frame in the end. Sep 21 at 17:54
• Nah one of them should be at different age since frames are changed but my problem is that why is it the observer in the rocket, it could have happened oppositely too.
– MSKB
Sep 21 at 17:56

The key to understanding this is a concept called proper acceleration. The Wikipedia article I have linked looks formidably complicated but actually the idea is very simple. Suppose you are floating freely in space far from any source of gravity so there are no forces acting on you. Now drop an object e.g. a ball. The ball will just float beside you and will not move away.

Now suppose we put you in a rocket that is accelerating at some acceleration $$a$$. Now if you drop the ball you'll see it accelerate away from you at that acceleration $$a$$. This happens because while you are holding the ball the rocket is exerting a force on both you and the ball so you accelerate at the same rate. When you release the ball the rocket stops accelerating the ball but it is still accelerating you. The result is that you accelerate away from the ball, and this looks to you as if the ball is accelerating away from you.

We define your proper acceleration as your acceleration relative to an object (like the ball) that is moving freely, and the value of the proper acceleration is that it is an unambiguous way of detecting when a force is acting on you. Suppose in our twin paradox I am floating freely in space while you jump in the rocket and accelerate away then back again. To both of us it looks as if the other is accelerating, but my ball will remain floating freely next to me while yours will not. This means my proper acceleration is zero while yours is non-zero.

And this is what breaks the symmetry. It is the twin with the non-zero proper time who experiences less elapsed time. This is discussed in gory detail in the question What is the proper way to explain the twin paradox?, which I assume is the what you referred to in your question, but the details need not not worry us. All we need to know is that it is always possible to measure the proper acceleration for an observer, so there is never any ambiguity about which twin did the accelerating.

• Why does not the time that has elapsed during the change of inertial frames of rocket cause any change in the change in time measured by the observer in te rocket? I meant that as per my question the observer in rocket will see that 6 year has passed throughout his journey but not more than that. But why is its so? I know that this part is strictly related to general relativity of which I have no idea but the thing that is troubling me is that the change in inertial frames caused the earth observer to add 4 years to that of the rocket observer but it didn't impact the change of time of rocket
– MSKB
Sep 22 at 6:57
• @MSKB there isn't a simple way to understand this. In ordinary geometry the shortest line between two points is a straight line and any curved line must travel a longer distance. However in spacetime it turns out that the straight line is the longest distance not the shortest i.e. any curved line has a shorter length than the straight line. This seems weird but it's an essential principle in relativity. Also in relativity the length of the line is proportional to the elapsed time of an observer travelling along that line. Sep 22 at 7:06
• And it's the observer with zero proper acceleration who travels in a straight line through spacetime. Any proper acceleration causes the line to become curved, and that means the line becomes shorter than the straight line, and that means the elapsed time is less than the straight line. Sep 22 at 7:07
• 😅😅😅😅 it went miles away above my head. Anyways did get a clue. Will work on that.
– MSKB
Sep 22 at 7:38

Suppose the earth-observer and the rocket-observer each had a ball at rest on the table in front of them. During their trips between separation and reunion, only the rocket-observer’s ball would have moved since the rocket-trip is non-inertial. The rocket did the turning. The earth did nothing…it was inertial.

Here are some spacetime diagrams showing why the non-inertial observer is not equivalent to the inertial one.

https://physics.stackexchange.com/a/434193/148184 (symmetric rocket trip .. down in the update)

https://physics.stackexchange.com/a/553751/148184 (asymmetric rocket trip)

• Well, remove the table and the ball will fall down :-) Thus the observer on Earth's surface is accelerated because he is not in free fall. I never did a calculation, and presumably in the most (if not all) settings that 1$g$ of acceleration can be ignored. Sep 21 at 18:03
• @emacsdrivesmenuts For simplicity, we can "restrict the motion to one dimension". So, leave the table in place to balance the gravitational force...and we can focus on motion in an idealized plane. (To introduce general relativity, overcomplicates the situation.) What is proposed is to follow what we do In PHY 101: We use carts on low-friction tracks to study motion in 1-dimension. We complicate the study that we are interested in if we remove the tracks. Sep 21 at 19:08

The "paradox" part of the twin paradox is that there seems to be a symmetry: each twin sees the other moving away and then coming back, so shouldn't everything else, including the readings on their watches, be the same at the end?

You can resolve that paradox by pointing out that one twin accelerates, and that can be locally detected (one of them is pushed against the wall of their ship, the other isn't). So they don't have entirely the same experience.

The key point is: that resolution is not a way of calculating the elapsed time for each twin. The only point of it is to counter the incorrect symmetry argument, and show it's possible for the times to be different.

This is very similar to the following situation: Alice and Bob drive from P to Q, leaving simultaneously and arriving simultaneously, but Alice drives in a straight line while Bob doesn't. Each sees the other getting farther away, then closer, but their odometers have different readings at the end. How is that possible? This situation is also not symmetrical because, for one thing, Bob accelerates, and can detect that (he's pushed sideways in his seat).

It would be completely wrong to conclude that the extra distance that Bob traveled happened "during the acceleration". It's also wrong to conclude that about the twin paradox, for the same reason.

The correct way to calculate the travel distance in the driving problem is to measure the length of each straight segment of driving with the formula $$\sqrt{Δx^2+Δy^2}$$, and add them (if they're driving on a plane). The correct way to calculate the travel time in the twin paradox is to do the same thing, but with the formula $$\sqrt{Δt^2-Δx^2/c^2-Δy^2/c^2-Δz^2/c^2}$$.

[...] which was in the same inertial frame as the other two were at the beginning?

This is another important point. Everybody is "in" every inertial frame.

A bit of nonsense that's unfortunately very common in introductions to special relativity is the idea that everyone has to use their own inertial rest frame to do calculations, as though they all inhabit different private universes.

The actual meaning of the principle of relativity is the exact opposite of that: all inertial frames are equivalent, so you can use any one you want, regardless of your state of motion. The elapsed-time formula that I mentioned above works in every inertial frame. You will get the same answer for the elapsed time for both twins no matter which frame you pick. The principle of relativity guarantees that. You don't need to use three different reference frames. You don't need the Lorentz transformation. If an introduction to special relativity claims otherwise, stop reading it and find a different one.

• In the end, whatever happened, Bob is at the same position with Alice. The amount of net distance traveled in the x direction is 0. As I already commented above, to return to the same frame, the 'accelerated' twin should also 'decelerate' after some time. Or the stationary twin must accelerate too (and undergo the same effects) so that they can end up in the same frame/position. By symmetry I would say, "if acceleration results in a difference, the deceleration should annul/rewind it". If both accl & declr contribute to the age difference in the 'same' direction, what's the math behind? Sep 21 at 18:49
• If Bob and Alice drive with the same speed (the acceleration is a change in direction) then Alice arrives earlier, so they are not in the same place in the end. If they keep driving, Bob will always be behind Alice. In space-time you don't stop. Sep 22 at 6:44

One of the assumptions of special relativity (and this assumption is confirmed by many, many experiments) is that there exist a set of privileged reference frames, inertial frames, which are not accelerating. You can tell a frame is inertial experimentally because an object at rest will remain at rest in an inertial frame.

Now, as you said, the observer on a rocket ship will observe the second time derivative of Earth observer's trajectory to be non-zero, when using coordinates at rest in the rocket ship's frame. However, the rocket ship observer can still be confident that the rocket ship is in fact in a non-inertial frame, because of the presence of fictitious forces. We usually use the word "accelerating reference frame" to refer to the fact that the rocket ship's frame is non-inertial, even though the second time derivative of the rocket ship's motion in the frame of the rocket ship is zero. While this might seem a bit weird at first, this language is useful because it is independent of the choice of observer and reference frame -- everyone will agree on which observers are inertial, and which are accelerating.

In other words -- the word "accelerating" really means "accelerating relative to an inertial frame," not "position having a non-zero second time derivative in a non-inertial frame."

In differential geometry, you introduce a new type the derivative (the covariant derivative) that is more suited to handling different frames. You can then define acceleration as the second covariant derivative of the trajectory. The covariant derivative has the property that if the usual derivative is zero in an inertial frame, then it is zero in any frame, while if the usual derivative is not zero in an inertial frame, it is not zero in any frame. So the acceleration defined in terms of the covariant derivative is non-zero in the non-inertial frame of the rocket ship. Because of this, the covariant derivative formalism represents the physics of the situation better than the usual partial derivative. This is a more sophisticated way to answer your question.

• It would have made sense if the calculations weren't much of a headache I mean we could have considered earth to be in absolute motion just like we consider k=1 in Newton's law of gravitation. But its quite harder for me to assume other way round that the calculations provide a concrete proof that the rocket is the one which is accelerating but not earth though both of them ar not absolute.
– MSKB
Sep 21 at 17:52
• @MSKB No no no -- the calculations don't prove anything in physics. The physics is that there exist inertial reference frames. The rocket ship is not in an inertial frame during the time when it is reversing its velocity. This can be verified experimentally -- if you hang fuzzy dice from the roof of the rocket ship, the fuzzy dice will move in the direction of the acceleration. Every observer will agree the fuzzy dice move relative to the rocket ship. If the observer on earth hung up fuzzy dice, they would not move when the rocket turned around. Sep 21 at 17:57
• Note the above comment does not refer to any calculations, but only statements that can be confirmed experimentally. Sep 21 at 17:58
• ahh I get it. I never considered the resolution of "which one is accelerating?" in that way
– MSKB
Sep 21 at 18:00
• why isn't the time elapsed for the observer in the rocket more than 6 years and less than 10 years? I know that general relativity comes into play this time which is out of my bounds at least for now. Still it seems quite awkward that the acceleration of the rocket didn't have any impact on its time
– MSKB
Sep 21 at 18:25

The asymmetry arises because the travelling twin changes direction at the end of the outbound voyage, which means that she changes her reference frame, while the twin on Earth stays in the same frame throughout.

The difference in the overall passage of time occurs as a consequence of the outbound twin's change of reference frame at the turning point. Until that point, each twin's clock runs slowly relative to the frame of the other, so they both age the same amount. When the outbound twin reverses direction, they change their reference frame, as a consequence of which their plane of simultaneity undergoes a significant tilt, which means that 'now' at their location in space becomes a much earlier time at the Earth's location. After the switch of reference frame, the arrangement becomes symmetrical again, with each twin's clock running slower when viewed from the frame of the other.

In response to the OP's comments, here is an example of the twin paradox with no acceleration, illustrating that it is the change of reference frame that causes the effect.

Imagine Person A is stationary on a platform.

At time t=0 on Person A's clock, they are passed by Person B heading east on a train. As they pass, Person B sets their clock to t=0 too.

At some later time t1 on Person B's clock, Person B passes Person C, who is on a different train heading westbound back towards the platform. Person C sets their clock to t1.

When Person C's train reaches the platform where Person A has remained, Person C and Person A compare their clock readings as they pass. The time on Person C's clock will be less than the time on Person A's clock.

So, the total time that has passed for Person A, who has remained stationary on the platform, is greater than the sum of the time spent by Person B on the eastbound leg and by Person C on the westbound return leg.

The time dilation effect was symmetrical for each observer on the two legs. What causes the overall time to be less for the observers on the trains is the shift in reference frame from that of Person B to that of Person C- that causes an abrupt change in the planes of simultaneity from that of Person B to that of Person C, and it is that switch which accounts for the time difference.

• how would I know which observer in the same inertial frame because apparently each of them are in rest with respect to themselves?
– MSKB
Sep 21 at 17:37
• does that mean answer to my second question is "yes", according to the second para of your answer?
– MSKB
Sep 21 at 17:39
• The outbound twin reverses the direction of their motion at the limit of their outbound journey- that's how you can tell they have switched from one reference frame to another. The stay-at-home twin doesn't change the direction of their motion, nor do they speed up or slow down relative to their original state of motion, so they remain in the same reference frame. Sep 21 at 18:26
• The actual calculation of the twin paradox can be quite complicated, as you should take into account the time it takes for the outbound twin to slow to a halt at the end of their outbound journey and to speed up again to head back. However, there is a simplified version of the scenarios which I will post as an addendum to my answer. Sep 21 at 18:34
• doesn't the time taken while changing the frames have any impact on the clock in the rocket?
– MSKB
Sep 21 at 18:37