-4

A young-looking astronaut has just returned home after a long trip in outer space. He found an old lady in his house and the old lady introduced herself as his daughter. How could this be possible?

Qmechanic
  • 201,751
Safiya Han
  • 7
  • 3
    This is the standard twin paradox found in every relativity textbook. What is your actual question? Why the "paradox" is true? – Anders Sandberg Jun 11 '21 at 10:15
  • 2
    You can look here https://en.wikipedia.org/wiki/Twin_paradox or watch the movie Interstellar for explanation of the exact same situation (regarding daughter and father). :-) – Koschi Jun 11 '21 at 10:21
  • Also, general relativity is not needed here (in the title). – User123 Jun 11 '21 at 11:08
  • 1
    Please, do some basic research! This looks like a homework question . . . – m4r35n357 Jun 11 '21 at 12:02

1 Answers1

0

In 1905, Einstein stated two postulates (assumptions):

  1. speed of light is always the same
  2. laws of physics are everywhere the same

The theory is called special relativity. The consequences are extraordinary. One consequence of this is time dilation (change of measured and even perceived time).

Some human moves with a velocity $v$ and measures time $t$, but some stationary human measures time $t'$. These values are related by the equation:

$$t'=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$$

Let's say that an astronaut moved with velocity $v=0.998c=299192873.1\frac{m}{s}$ and he traveled for 5 years. He started his journey when he was 30 and that's when his wife gave birth to his daughter who was 0 years old. By the equation:

$$t'=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{5\rm\, years}{\sqrt{1-0.998^2}}=79\rm\,years$$ Thus, his daughter measured around 80 years, while he measured only 5 years. He is now 5 + 30 = 35 years old, but she is 80 + 0 = 80 years old. We get so called "Twin Paradox".

This can also be solved using spacetime interval invariance. It basically says that in spacetime world, the distance between the two events stays the same, independently of the frame of reference used. Basic equation:

$$I=\sqrt{t_1^2-d_1^2}=\sqrt{t_2^2-d_2^2}$$ where both $t$ and $d$ are measured in same units (for example both in years; yes, we can measure distance in years (in spacetime, space and time are basically the same)).

Let's say that 1 = daughter and 2 = astronaut. Then $d_2=0\rm\,a$ as astronaut didn't move in his frame of reference. Also, $t_2=5\rm\,a$ as he measured 5 years on his clock. Let's say that his daughter measured that he moved for $d_1=79.844\rm\,a$. Then she measured time: $$t_1=\sqrt{t_2^2-d_2^2+d_1^2}=\sqrt{(5\rm\,a)^2-(0\rm\,a)^2+(79.844\rm\,a)^2}=80\rm\,a$$ Same as with first method: she is now around 80 years old.

User123
  • 1,173