The explanation just says that acceleration is the way to solve the paradox. You experience time dilation even if you move at thé constant speed which is again an inertial frame of reference
People will understand the twin paradox when you say the values of the following problem (having or not similar results). Until then, you are running away from the main point. Imagine twin A on Earth. Earth and the Star (speaking about abstract points fixed in a system) are going left with v/2. Imagine twin B in a spaceship to the right with v/2. 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. I want to know the values of those stopped clocks when they meet.
In my opinion,Solving the TWIN PARADOX means solving a system of two equations with four unknowns. I consider the two Lorentz transformations: a) x_1 = gamma * (x -v * t) b) x = gamma * (x_1 + v * t_1) With gamma I obviously indicated the Lorentz factor, and I do not consider the other two Lorentz transformations because they depend on a) and b). In an empty space THE TWIN PARADOX is not resolvable, it’s impossible to determine a single solution of a system of two equations with four unknowns. BUT THE TWIN PARADOX SOMETIMES “IS NOT A PARADOX”, consider for example the astronaut twin moving with speed v to reach a star. The Earth and the star both belong to the frame of the Earth, and let the Earth-star distance be equal to d in the frame of the Earth. The TWIN PARADOX is equivalent to the CLOCK PARADOX, if the astronaut twin continues to travel and does not return to Earth. Let’s consider again the system of two equations: a) x_1 = gamma * (x -v * t) b) x = gamma * (x_1 + v * t_1) If x = v * t (x_1 = 0), then t_1 < t. (v * t = gamma * v * t_1, t_1 = t / gamma) If (x = v * t) and (x_1 = -v * t_1), we obtain “the useless solution”: x = x_1 = t = t_1 = 0. I indicate (x_1 = 0) with (c) , and I indicate (x = 0) with (d). (c) and (d) are MUTUALLY EXCLUSIVE! If it is true (c), it is not true (d). If it is true (d), it is not true (c). The spaceship actually moves with uniform rectilinear motion between any two points of the Earth’s frame (if the acceleration of the spaceship is zero). The astronaut twin leaves from the Earth to reach a star, the Earth and the star belong to the frame of the Earth! (all the other points reached by the spaceship also belong to the frame of the Earth) We cannot choose x_1 = -v * t_1, in this case THE TWIN PARADOX (THE CLOCK PARADOX) IS SOLVED! The choice x = v * t it’s obligatory: the clock of the astronaut twin slows down compared to the Earth’s clock. (even if the spaceship continues to travel and does not return to Earth)
Sir, agar ghoslo ka frame inertial nehi ha to STR nehi apply hoga... Time dilate nehi hoga,to kiu dilate hua.... Agar ghoslo ka accelerates nehi hoga to kiska age jada hoga..?
idhr half btaya hai general theory of realtivity me explained hai baaki in short goslo will be younger special theory is not complete it was later completed in general theory
