Twin Paradox: Acceleration-Free Solution

In this video, I use the Minkowski spacetime diagram to solve what's called the "twin paradox" of Einstein's special relativity, without the need to use acceleration.
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[@aniksamiurrahman6365]

I got something to say. The acceleration is what causes the 'travelling' twin to switch from the 'B outbound' frame to 'B inbound' frame. So, by all definition of causality, acceleration Is the 'cause'. So, why does everyone keeps denying it? I know of other thought experiments where both twin accelerates and yet has been aged differntly upon reunion. I konw that the ultimate factor deciding who ages the more is whose worldline is the longest. But even there, the intensity and duration of acceleration is what 'causes' the worldline to differ. Without acceleration, there's no way two world line will differ. So, why does everyone keeps ignoring the role of acceleration here? Acceleration is always associated with energy. There's no way you accelerate something without spending free energy. So, every observer from every frame of reference will agree whether something is accelerating or not, even though they may disagree on the magnitude and duration. AKA acceleration isn't relative. So, why do y'll so eager to push such an important factor out of equation?

[@ankidokolo]

Acceleration makes only a small difference actually, and not a difference by years as the paradox says. So, in this video, I show how the paradox can be solved without even involving acceleration. Changing the frame of reference alone, causes a time jump in A's frame for B. Hence, at the end of the journey, A will always be older than B, no matter which frame of reference we're measuring from. ♥

[@TheAnkFreeArt]

Acceleration makes a small difference only.

[@aniksamiurrahman6365]

@@ankidokolo And what causes the change of frame in the single clock/observer the first place? What is the causative agent behind that?

[@ankidokolo]

@@aniksamiurrahman6365 There is no acceleration and no forces at all. Think of 2 spaceships moving in opposite directions at equal and constant speed. The traveller B just jumps between the two when they pass close to each other. This solution shows that even when the change happens instantaneously, observer A who didn't travel will be older than B, no matter how we look at it. Hence, there is no paradox in the first place, that needs acceleration to be solved.

[@aniksamiurrahman6365]

@@ankidokolo Again, you're shunning my question. Any physical process tranfering twin B from outbound frame to inbound one is impossible without acceleration (say jumping from one spaceship to other moving at 0.96c, enourmous acceleration and deceleration).

[@massimilianodellaguzzo8571]

" A does NOT change his frame of reference at any time ": ??? ..... I do not agree! If you consider twin B stationary, there are also these three frames: B, A outbound, A inbound. I believe that the twin on Earth is younger, if we consider the motion of the Earth in the frame of the spaceship. (round trip) And this is not a defeat, the motion of the spaceship (in the frame of the Earth) is different from the motion of the Earth. (in the frame of the spaceship) I understand that we can disagree on this, but time dilation is symmetrical.

[@ankidokolo]

Actually no. You have 3 frames of reference (You don't have only 2 frames). Once B turns back his, he lands in an entirely new frame of reference (inbound), and for the entire trip, A always lives longer. (3 frames of reference) we don't compare the same 2.

[@massimilianodellaguzzo8571]

@@ankidokolo If you consider twin B stationary, there are: B, A outbound, A inbound...

[@ankidokolo]

@@massimilianodellaguzzo8571 No. There isn't. I considered B stationary when I changed once to B outbound, and once to B inbound. I took 2 different frames for B. It requires some thinking, I admit, but then you'll see that the paradox is no more.

[@massimilianodellaguzzo8571]

@@ankidokolo I would like to tell you about this scenario. A spaceship moves in the frame of the Earth, and travels a distance L. (to reach a star S, suppose the Earth-star distance is L in the frame of the Earth) The nose of the spaceship (the astronaut twin) and the Earth twin occupy the same position at the initial times (t = t’ = 0) ... and the twins are in relative motion to each other at speed v. If we consider the uniform linear motion of the spaceship in the frame of the Earth: 1) the astronaut twin reaches the star S at time t = L/v, in the frame of the Earth 2) the star S reaches the astronaut twin at time t’ = L / (gamma*v) in the frame of the spaceship In the spaceship frame the Earth-star distance is CONTRACTED ! But in my opinion if we consider the uniform linear motion of the Earth in the frame of the spaceship, then we need to consider SOMETHING ELSE: in this case it is necessary to consider the uniform linear motion of the star S in the frame of the spaceship. And if we consider the uniform linear motion of the star S in the frame of the spaceship: 1) the star S reaches the astronaut twin at time t ’ = L/v in the frame of the spaceship 2) the astronaut twin reaches the star S at time t = L / (gamma*v) in the frame of the Earth In my opinion the uniform linear motion of the star is hidden but it exists and happens. If we wait for the arrival of the star, the astronaut twin is younger, but if we wait for the arrival of the spaceship's tail, the earth twin is younger. (And the return journey is similar to the outward journey)

[@ankidokolo]

@@massimilianodellaguzzo8571 You can tell your opinion of course, but what I made in the video is not an opinion. Just separate the 2 frames in- and outbound, from each other. Look at time 04:48 , you'll see that in the outbound frame itself, A's frame and B inbound frame move in the (same) direction. In the inbound frame 05:18 , A's frame and B outbound now move in the (same) direction. You see how A's worldline maintains the same angle (speed) between it and the outbound worldline and between it and the outbound worldline, in both B's frames. This makes the entire worldline of A a single, long straight line. Therefore, I said A doesn't change his frame at any time.