# Does changing direction change the frame?

In looking at explanations of the twin paradox, two examples are given to show that acceleration is not a factor: First, where one rocket flies out past the star and a second rocket flies back to earth.
Second, where the rocket flies out and instantly turns back, with no acceleration.

In the explanations I've seen, it says that the reason why there is no paradox is because we are dealing with three frames in total, the person on earth, the outbound person and the inbound person.

But why would the person who flies out and instantly turns back be considered as two frames? From his point of view, he is completely stationary and it is the earth that flies out and back towards him. So why is this considered two frames?

And if so, would this mean that a person orbiting a planet would be constantly changing frames?

• "Frame" is implied to be an inertial frame, in this context. Inertial frames of reference are at constant relative velocities to other inertial frames of reference. So, from the point of view of the Earth-bound twin's "inertial" frame of reference (close enough), the rocket-bound twin's motion is via two distinct inertial frames since they will have two distinct velocities relative to the Earth-twin. From the perspective of the rocket-twin, it is the Earth-twin that has 2 distinct inertial reference frames. Oct 8 '20 at 13:45
• A body in orbit around a planet would indeed have a constantly changing inertial frame of reference from the perspective of the inertial frame of the planet. Special relativity can handle such accelerated frames of reference, naively, quite fine by taking each instantaneous frame to be inertial and then integrating over the motion. Oct 8 '20 at 14:28

The twin paradox is just a spacetime version of the triangle inequality.

In special (and general) relativity, elapsed proper time (i.e., the amount you age) is the length of your worldline. The stay-at-home twin has a straight worldline in spacetime. The out-and-back twin has a bent worldline. The worldlines have different lengths because they have different shapes. The twins age different amounts because they traveled different "distances" (actually times) through spacetime.

Whether the bent worldline changes direction suddenly at a point (like the triangle) or bends more gradually only matters inasmuch as it slightly affects the total length.

The spacetime version of the Pythagorean theorem has a minus sign in it, and so it turns out that a straight worldline in spacetime is the longest time between two points, instead of the shortest. Other than that, the twin effect is no different from the fact that if one person drives straight from point A to point C, and another takes a detour via point B, their tripmeters will show different values when they arrive.

Reference frames are just coordinate systems. You don't need reference frames to understand the twin effect, and if you do want to use reference frames, you only need to use one. I don't know why there are so many people who believe that you need to use a different reference frame for each different velocity. You certainly don't, just like in Euclidean geometry you don't need a Cartesian coordinate system for each different direction. If you try to use multiple coordinate systems, you're only making a simple problem far more difficult.

There is no "turns back instantly" in relativity. He needs to decelerate, stop, turn, and accelerate again. Acceleration (for special relativity) is a change in frame of reference (so yeah, a person orbiting a planet is constantly changing frames. This is explored in GR).

There is a way to go past this problem and study the real core of the twin paradox (I think this is what you were referring to): in the "modified twin paradox" one person stays on earth (1), one leaves (2), and one comes from far away (3). When (2) and (3) meet they exchange data on distance and time, and then compare them with the ones for (1). You still need three frames of reference but there is no acceleration involved: the "paradox" is still there, and the reason why it works is (again) the change of frame of reference. This Fermilab video is useful to visualize the problem.

• Why intuitively would we think a priori that one twin leaving, turning around and returning would give the same result as in the modified paradox where the leaving twin exchanges data with person (3)? Oct 5 '20 at 18:24
• @Mauro Giliberti it is exactly because of that Fermilab video, in combination with another video, that my question arose. In the Fermilab video he uses a rocket going out and a different rocket coming back to illustrate two frames. But in the other video they use the same rocket going out and coming back. So why would this one rocket have two frames just because it turns around? Oct 5 '20 at 19:01
• @Not_Einstein Well, it wouldn't give the exact same result. If the trip is long enough, what happens at the furthest point is neglegible in comparison to the main part (going back and forth): since this latter part is in common, we expect about the same result. The interesting thing isn't how these situations differ, but what they have in common: the change in the frame of reference, that ultimately leads to the "paradox". Oct 5 '20 at 19:08
• @foolishmuse because turning around means changing (the direction of the) velocity, and a change in velocity means change in the ineriale reference frame. Oct 5 '20 at 19:48
• @MauroGiliberti thanks. so it is a change of frame. Just what I wanted to know. So traveling in a circle would be a constant change of frame as well. Oct 5 '20 at 20:06

If you are moving in one direction, then turn around and move in another direction, at some point you accelerated. Consider a particle going in the positive direction (towards the sun, say) at $$t_1$$, and later at a different time going in the negative direction (away from the sun).

\begin{align} v(t_1) &= + 5\ \text{m}/\text{s} \\ v(t_2) &= - 5\ \text{m}/\text{s} \end{align}

The net acceleration between $$t_1= 0$$ and $$t_2 = 1$$ is: \begin{align} a_\text{net} = \frac{ \Delta v }{ \Delta t} = \frac{ v(t_2) - v( t_1) }{ t_2 - t_1} = \frac{ \left( - 5 \text{m} /\text{s} \right) - \left( + 5 \text{m}/\text{s} \right) }{ 1 \text{s} } = \frac{ - 10 \text{ m }/\text{s} }{ 1 \text{s} } = - 10 \text{ m }/ \text{s}^2 \end{align}

For the instantaneous acceleration, you need more information. However! You know $$a(t)$$ could not equal zero for the entire duration from $$t_1$$ to $$t_2$$ because if that were the case, $$a_\text{net}$$ would be zero, and we just calculated that it is not zero.

So, changing directions (meaning moving in one direction, and then later moving in a different direction) does necessarily mean acceleration occurred.

ETA: the fact that one twin accelerated means their POV is different. One way to see this is that they would have felt a fictitious force in their own rest frame, whereas the twin on Earth in a not accelerating frame of reference would have felt no forces. This breaks the symmetry between them and resolves the paradox.

• In the videos I've watched, they go to great lengths to say that acceleration is NOT the solution for the twin paradox. Rather the solution is because there are different frames for the rocket leaving and the rocket returning, even if that is the same rocket. So my question is, why does a single rocket have two frames just because it turns around? Oct 5 '20 at 19:03

The statement "There is no paradox" is false. A paradox is not a contradiction, it is an apparent contradiction, and there is always an apparent contradiction in the twin paradox when we use our non-relativistic intuition (because it is wrong). Of course, there is no contradiction because special relativity is logically consistent.

At the core of the twin paradox is the Andromeda paradox, or as David Mermin puts it, "That no inherent meaning can be assigned to the simultaneity of distant events is the single most important lesson to be learned from relativity".

For concreteness, consider 100 year legs at $$\gamma=10$$. If space-twin leaves on Jan 1, 1900 AD, he arrives at the turn around at Jan 1, 2000 AD. What time is it on Earth? Obviously, Earth thinks its 2000 AD, but space-twin has only experienced 10 years. Meanwhile, has sees Earth's clocks dilated by 10x for 10 years. For him, it is only 1901 AD on Earth.

Then he turns around. Boom, it is now 2099 AD on Earth.

From here, he can go home, aging 10 more years, seeing Earth age 1 year to 2100 AD, for a total of 200 years passing on Earth, while he ages 20 years.

But let's go back to the turn around. What is the date on Earth at that moment? It can be anything between 1900 AD and 2100 AD, depending on radial velocity. Moreover, you can freely switch between them, forward or backward, just by changing your radial velocity. You simply cannot assign a definite date on Earth, from the turn around point, which is nearly 100 light years distant.

Here, instantaneous acceleration highlights the paradox: it takes $$\Delta t = 0$$ in all reference frames, yet for space-twin, the date on Earth jumps forward 198 years. One cannot call this time dilation for 2 reasons: (1) it takes 0 seconds, there is nothing to dilate. (2) It is reversible, if he turns around again, the date on Earth goes backwards.

In the end, Earth sees space-twin age 20 years, while space twin sees Earth age 1 year on each leg, with a 198 year discontinuity that he cannot observe (it all happens outside his light cone, after all).

So there is no contradiction; whether that is still a paradox depends on how comfortable you are with Minkowski space.