- #1
timpeac
- 14
- 0
Hi All,
I think that I understand (as far as you can!) the ideas behind explaining why two twins have different ages if one travels at near light speed away for a time and then returns to his twin on Earth (the return journey changing the inertial reference frame).
I have a related question where the twin doesn't actually turn around and return home.
If a spaceship leaves Earth at 0.8c towards Alpha Centauri 4 light years away then the person on Earth considers that the trip will take t = d/v = 5 years in Earth time. Someone on Alpha Centauri will also agree with this (adjusting for the 4 years time it takes for the light to travel from Earth to AC and so they physically see the ship leave at t=4). Someone on Earth, or on AC, will consider time on the ship to be running reduced by the factor ε= √(1 - v^2/c^2) = 0.6. On arrival the people on AC will therefore expect the space travellers to have aged 0.6 x 5 = 3 years during the time they themselves have aged 5 years.
So the people on AC expect the travellers to arrive at t = 5 having aged 3 years from their departure at t = 0.
For the travellers, time dilation means that the distance they are to travel is εd = 0.6d = 2.4 light years. As they are traveling at 0.8c they consider that they have aged d/v = 3 years during the trip. This agrees with what the people on AC think.
However, from the point of view of the time travellers the clocks of the people on AC are running slow by a factor of 0.6. They therefore expect the AC people to have aged 3 years x 0.6 = 1.8 years.
Can someone please explain the discrepancy between this expected age and the 5 years that the local AC people consider they have aged upon arrival? (or where my algebra has gone wrong!)
Many thanks.
I think that I understand (as far as you can!) the ideas behind explaining why two twins have different ages if one travels at near light speed away for a time and then returns to his twin on Earth (the return journey changing the inertial reference frame).
I have a related question where the twin doesn't actually turn around and return home.
If a spaceship leaves Earth at 0.8c towards Alpha Centauri 4 light years away then the person on Earth considers that the trip will take t = d/v = 5 years in Earth time. Someone on Alpha Centauri will also agree with this (adjusting for the 4 years time it takes for the light to travel from Earth to AC and so they physically see the ship leave at t=4). Someone on Earth, or on AC, will consider time on the ship to be running reduced by the factor ε= √(1 - v^2/c^2) = 0.6. On arrival the people on AC will therefore expect the space travellers to have aged 0.6 x 5 = 3 years during the time they themselves have aged 5 years.
So the people on AC expect the travellers to arrive at t = 5 having aged 3 years from their departure at t = 0.
For the travellers, time dilation means that the distance they are to travel is εd = 0.6d = 2.4 light years. As they are traveling at 0.8c they consider that they have aged d/v = 3 years during the trip. This agrees with what the people on AC think.
However, from the point of view of the time travellers the clocks of the people on AC are running slow by a factor of 0.6. They therefore expect the AC people to have aged 3 years x 0.6 = 1.8 years.
Can someone please explain the discrepancy between this expected age and the 5 years that the local AC people consider they have aged upon arrival? (or where my algebra has gone wrong!)
Many thanks.