Age comparison to observer on Earth (time dilation)

AI Thread Summary
Traveling to Alpha Centauri at 0.780c results in significant time dilation effects. The Earth observer would measure a round trip time of 16.5 years, while the traveler would age only 10.3 years. The calculations utilize the time dilation formula, confirming that less time passes for the traveler compared to the observer on Earth. The discussion emphasizes understanding time dilation without relying on Lorentz contraction. Overall, the calculations align with the principles of relativity, illustrating the disparity in aging between the two observers.
ideaessence
Messages
2
Reaction score
0

Homework Statement



Alpha Centauri is about 4 light years from the Earth. If you were to travel to Alpha Centauri and back at the speed of the electron that you calculated in question A (0.780c), how much would you age compared to an observer on Earth?

Homework Equations



There equations were not provided in the actual problem, but so far we've used these 2 (time dilation and Lorentz contraction) equations:

T' = (T) / sqrt[1 - (v^2/c^2) ]

L' = L * sqrt[1 - (v^2/c^2) ]

The Attempt at a Solution



v = 0.780c

Both ways:
L = 8cY

L = v*T

T = L / v

T = (8cY) / (0.780c)

T = 10.3Y

T' = ?

T' = (T) / sqrt[1 - (v^2/c^2) ]

T' = (T) / sqrt[1 - (v/c)^2 ]

T' = (T) / sqrt[1 - (0.780c/c)^2 ]

T' = (T) / sqrt[1 - (0.780)^2 ]

T' = (T) / sqrt[1 - (0.608) ]

T' = (T) / sqrt(0.392)

T' = (T) / sqrt(0.392)

T' = (T) / (0.626)

T' = (10.3Y) / (0.626)

T' = 16.5Y

From what I recall, T' is the change caused by motion of the moving object, relative to the observer.
In this case, relative to the person on Earth, the travel time to Alhpa Centauri and back would be 16.5 years.

There only other time variable left is T, so I assume this applies to the person/partice traveling to Alpha Centauri.

Am I correct? (I want to avoid using Lorentz contraction to ensure I understand the time dilation equation and use it properly.)

Thanks.
 
Physics news on Phys.org
Less time passes for the guy in the spaceship. Remember the guy in Heinlein's story who comes home and marries his great-great-grand niece? (Time For The Stars)
The Earth observer says 10.3 years.
The traveler says 10.3 * .626
(ignoring general relativistic effects during the accelerations)
 
OK. Doesn't that mean I did it right so far? Don't I have half of the answer (the time for the observer on Earth)?
 
You said 16.5 years for the Earth observer; I said 10.3.
You said 10.3 for the traveler; I said 10.3*.626.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
22
Views
972
Replies
38
Views
4K
Replies
4
Views
1K
Replies
4
Views
820
Replies
2
Views
1K
Back
Top