# Clarification regarding Special Relativity

Empty the universe of all matter. Place the Earth in that universe sitting in an inertial frame. Take a pair of twins. One twin gets into a space ship and takes off from Earth (t=a). He accelerates rapidly and obtains .9c within a day of Earth's time and then turns off his engines (t=b). He then glides for 4 Earth years. After 3 Earth years 364 earth days (t=c), he starts to decelerate the ship, turns it around and returns to Earth the exact same way he left (1 earth day to decelerate 1 earth day to accelerate). After 4 years, he decelerates for one Earth day as he approaches Earth and then lands the ship 8 years to the day after he took off.

My understanding is that the stay-at-home twin has now aged 8 years. The spaceship twin has aged less than that (about 7 years). If we assume linear acceleration the 4 Earth days of acceleration account for 2 days equivalent of time discrepancy. The remaining 7 years 361 days account for the majority of the discrepancy.

Question 1: Is that analysis correct?

Now, I repeat the experiment, spaceship twin takes off in the spaceship as before, but now as soon as he hits Earth orbit he parks his ship. stay-at-home twin attaches a really big rocket engine to the Earth itself and blasts the planet away from his brother. The distances versus time between the ship and Earth is exactly as before.

I assume now the spaceship twin will have aged 8 years and the Earth twin will have aged about 7 years.

Question 2: Is that correct?

I repeat the experiment one last time. As before the spaceship twin blasts off of Earth. This time however he accelerates to .45c; his brother again attaches the rockets to the Earth and accelerates it .45c. Again they maintain the exact same distance to time relationship.

My guess now is that when the brothers meet they will be the exact same age.

Question 3: Is this correct?

Question 4: In each of these scenarios from t=b to c the systems appear to be exactly equal to one another. The ship and the Earth are receding from one another at .9c. Where is the difference in their situations stored? How would an outside observer differentiate the 3 scenarios?

Question 5: It appears that time dilation, length contraction is really caused by acceleration not velocity. Why does acceleration not show up in the special relativity formulas? Is perhaps velocity in these equations not really velocity, but rather the integral of acceleration?

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1. Yes. I haven't verified your numbers precisely, but the general conclusion is right: the majority of the time discrepancy is due to the time the traveling twin spends moving at near-lightspeed relative to the stay-at-home one. Since the acceleration takes only a small fraction of the overall travel time, the accelerating periods are also responsible for only a small fraction of the time dilation. This is the standard "twin paradox".

2. Yes. The situation is symmetrical to that in case 1.

3. Yes; since the two brothers have undergone identical acceleration profiles (as seen by an observer who remains inertial throughout), when they meet again they have experienced the same proper time and are therefore the same age. (Note however that although the brothers' relative velocity is 0.9c in the initial inertial reference frame, it is not 0.9c in their reference frames, since the composition of velocities is nonlinear in relativity.)

4. It's true that from $t = b$ to $t = c$ the relative velocity between the two brothers is 0.9c in all three cases, as seen by an observer who remains inertial throughout. The difference here is not any specific leg of the journey but in who remains inertial and who doesn't. Much like a line is the shortest distance between two points in space, an inertial reference frame is the longest proper time between two events in spacetime. (It's longest instead of shortest because of the sign flip in the Minkowski interval, $d\tau^2 = dt^2 - (dx^2 + dy^2 + dz^2)/c$.)

In all three cases the brothers depart from some spacetime event and meet again at some later event; in cases 1 and 2, one of the brothers remains inertial while the other doesn't, so the inertial brother experiences more time. In case 3, neither brother is inertial, but as judged by an observer who remains in the inertial frame from which the journeys started, the brothers undergo identical acceleration profiles and thus experience the same proper time.

5. No, time dilation and length contraction are definitely caused by velocity, as shown in every introduction to relativity, where one considers two observers with a constant relative velocity - no acceleration in sight. But time dilation is a function of your reference frame - someone else in a different reference frame would not generally agree with you about the amount of time dilation experienced by any particular object or observer. The proper time along a trajectory between two identifiable events is observer-independent, though. Given two trajectories that meet at two events - i.e. the brothers' journeys which meet at the beginning and end - all observers will agree on the proper time experienced by each brother, and therefore they will agree on the difference between the brothers' ages.

So I think it would be fair to say that discrepancy in proper time between two such trajectories is caused by acceleration - more specifically, by differing acceleration profiles over the lengths of the trajectories.

As for "Is perhaps velocity in these equations not really velocity, but rather the integral of acceleration?", well, velocity is always the time integral of acceleration, since acceleration is defined as the time derivative of velocity. So it makes no sense to say it's "not really velocity". :)

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Thanks for your thorough answer. I wanted to get more clarification on point 4. Let's say a new observer arrives at t=b. I show him the situation and tell him that one of the three scenarios occurred and ask him to tell me which. Will he be able to determine which of the three have occurred and if so how? –  aepryus Mar 29 '13 at 3:05
As for point 5, the issue is that in all three situations the velocity of both brothers is the same. So, it becomes difficult to differentiate their situations. However, if I take the Integral of acceleration instead the inertial brother will be zero while the non-inertial brother will be equal to v. –  aepryus Mar 29 '13 at 3:08
@aepryus If a new observer arrives after the initial acceleration phase, he will not be able to tell who previously accelerated and who, if anyone, remained inertial. But if he continues watching he will of course see the later acceleration phases and be able to tell the difference that way. –  Nathan Reed Mar 29 '13 at 3:14
Yes, but from t=b to t=c will he be able to determine the situation and if so how? –  aepryus Mar 29 '13 at 3:15
As I said, he will not be able to tell who previously accelerated and who, if anyone, remained inertial. –  Nathan Reed Mar 29 '13 at 3:16
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