The Twin Paradox Explained

How special relativity predicts time dilation, length contraction and simultaneity loss. One way Einstein’s special relativity messes with time.

Virtutes Vocis by Kevin MacLeod

[paulomanuelsendimairespere]

People will understand the twin paradox when you say the values of the following problem (having or not similar results).

Imagine twin A on Earth. Earth and the Star (speaking about abstract points fixed in a system) are going left with u/2. Imagine twin B in a spaceship to the right with u/2 só that they have a relative velocity v. When twin A calculates twin B to reach the Star, he reverses (and the Earth and the Star go together with him.) When twin B sees reaching the Star, he reverses too. Twin B stops a clock just after leaving the Earth, another just before reaching the Star, another just after leaving the Star, and another just before reaching the Earth. Twin A also stops one clock just after leaving Twin B, another one just before reversing, another one just after reversing, and another one when they meet.

The values of those stopped clocks should be compared with the respective ones of the classical twin paradox, in which only twin B reverses.

[massimilianodellaguzzo]

Hi MathAdam, nice video!
I think that in the twin paradox, if there is a stationary star (or planet) with respect to the Earth, the paradox vanishes.
If we consider the presence of the star, the time for the astronaut is a waiting time and not a travel time.
According to the astronaut twin's perspective, it is also not said that for the terrestrial twin the time must also be waiting (in this case the twins are the same age), the astronaut twin must also consider that the time of the twin left on Earth can be (rightly!) a travel time and therefore greater than the waiting time. The travel time with respect to the Earth's frame is t = d / v (while the spaceship travels a distance d at speed v) and the waiting time for the arrival of the star is t_1 = (1 / gamma) * t, because the Earth - star distance (in motion) is contracted. (t_1 < t)
Congratulations again for having considered the distance between one of the twins and Alpha Centauri contracted while moving relative to the other twin, in other videos this very important detail is not mentioned!

[ABC-jqve]

Adam, I was wondering if you could do a video or an article about what you observed from tutoring math. Are there any commonalities to the students that do well? Is there anything a beginning math learner should know about? Any stories about a student who used to be bad at math, but got good at it (and what they did to get there)?

[jmile]

Good for someone who is learning about SR for the first time. But dude; I hate the music, a lot.

[richardlinsley-hood]

Now create a reference frame which is symmetrically placed between Bob and Alice. Which clock is then slower/the same and why?

[chasemarangu]

Great video, better than most twin paradox explanation, but I still don't get 1:26 when alice fires her thrusters in the reverse direction makes her move closer to bob, how is that not the same as bob slowing down to meet alice

[ABC-jqve]

Wow this video blew up. Congratulations!

[vesuvandoppelganger]

Ship1 at rest on top:



Ship2 at rest at bottom:



Here we have 2 ships passing each other and each diagram shows one of the ships at rest and the other one moving and length contracted. There is 1 moment when T of ship1 is lined up with N of ship2. This moment must be the same moment for both ship1 and ship2. If it isn't the same moment for both T of ship1 and N of ship2 then that means that T of ship1 is located somewhere else, according to ship1, when, according to ship2, T of ship1 is lined up with N of ship2, which means that T of ship1 has already passed or will pass N of ship2 so that is 2 events of T of ship1 passing N of ship2 which is clearly impossible. It also means that T of ship1 is simultaneously in 2 different places which is impossible. So this proves that the moment that T of ship1 and N of ship2 is lined up is the same moment in both the reference frame of ship1 and the reference frame of ship2. At this moment N of ship1 is located on opposite sides of T of ship2 simultaneously which is clearly impossible.

[rouxenophobe]

I'd like to see the explanation but without Alpha Centauri. In the first instance as Bob and Alice travel away from each other they could both be said to be stationary w.r.t. each other. The difference must be that Bob must turn around and accelerate after Alice in order to synchronise frames of reference. To me it seems the explanation must involve acceleration (although having watched a few explanation now this seems to be a contentious point).

[roberthansen]

Unfortunately, what was a beautiful thought experiment, has been overwhelmed by an alleged paradox that never existed. The suspicious student will ask "But if each observer sees the other observer's clock as being slow, can't we look at this in the spaceship's frame of reference, and then the Earth twin will have aged less?" And the teacher replies "No, because it is the spaceship that changes reference frames (or accelerates) which makes the experiment asymmetric". And we have been conditioned to believe that we avoid the paradox by creating explanations of how the experiment in asymmetric.

The experiment is not asymmetric. If you consider that the spaceship is stationary, and the earth and star (together) move to the left, till the star is at the spaceship, and then to the right, till the earth is at the spaceship, you get the same result. The twin on the spaceship ages less. So whether you consider that the spaceship does all the moving or that the earth and star do all the moving, you get the same result. The twin on the spaceship ages less. The experiment is symmetric and the principle of relativity holds.

So, why do we not get the expected paradoxical result that the Earth twin ages less when we move the earth and star instead of the spaceship? Each observer should see the other observer's clock as being slow, right? No, not right.

Imagine two observers, A and B, traveling past each other.

If A pulls out a ruler and times how long B takes to traverse it, end to end, and B also times how long it took to traverse A’s ruler, end to end, then A will record a longer time than B, and B a shorter time than A.

But if B pulls out a ruler and times how long A takes to traverse it, end to end, and A also times how long it took to traverse B’s ruler, end to end, then B will record a longer time than A, and A a shorter time than B.

Thus, while it is true that each observer can see the other observer's clock as being slow, they can't both do this at the same time. In order to compare clocks you have to choose a ruler. If we choose A's ruler, then A and B will agree that A's clock is running faster. But if we choose B's ruler, then A and B will agree that B's clock is running faster.

So which ruler are we using in the twin paradox? The distance from Earth to the star is in Earth's frame of reference, so we are using Earth's ruler. Since we are using Earth's ruler, Earth's clock will be the fast clock and the Twin's clock will be the slow clock. Which is why it doesn't matter if the Earth and star (together) move to the left and then the right, or they stay still and the spaceship moves to the right and then to the left. Since we are always using Earth's ruler (the distance from Earth to the star) to measure time, the twin on the spaceship will be the one who ages less.

The next time the suspicious student asks "Why can't we...", the teacher should reply "Yes, we can, and we get the same result, and here is why."

Unfortunately, we have been conditioned to believe that a paradox awaits us if we go down that path, and we reply "No we can't, because the experiment is asymmetric...." And those explanations, fall an inch short of convincing the student (or even us), and so we try to make better explanations of the asymmetry, and as hard as we try, they to come up an inch short, because the experiment is not asymmetric.

Note: Working the problem with the spaceship moving or stationary is pretty simple. In either case, whether the spaceship is moving to the star or the star is moving to the spaceship, the distance is contracted the same, and the trip is shorter in years. Likewise, to Earth's perspective, whether the Earth and star are stationary or moving, the distance between them is the same, the original distance, thus the trip takes more years. It is easy to see why the math in both scenarios (moving spaceship or stationary spaceship) gives the same result. Because the distance being used is always in Earth's frame of reference.

[tommytheoctopus]

This is an old way of thinking. You would need to calculate the absolute frame of reference to determine which twin was older at the end and this can be done by calculating all accelerated frames of reference. Once an absolute frame of reference is determined each twins maneuvers can be calculated.