At a slow speed, a space craft would get through a given enough light years (X light years) in many real to earth and many real to the craft's inhabitants years. At a fast enough speed, the real years are there, fewer, but the inhabitants would experience a short time. At just the right speed, the number of years experienced by the inhabitants would exactly equal the light years distance. So for 8 light years traveled, the inhabitants (crew/passengers/whatever) of the craft would experience a voyage of 8 years, for 10, 10, etc. What is the speed, and is it the same for each distance or is there a formula to find such? (expect same speed, just checking)
-
There are two frames of reference: Earth's and the craft's. I will use primes for the craft. We know many things just by reading the problem...
x = 8 ly
t' = 8 y
v = ?
We then remember that light travels a ly in a y, so the speed of light c is...
c = 1 ly/y = x/t'
and we remember that gamma is defined by...
γ = 1 / √(1 - v² / c²)
Using the definition of velocity as distance over time we get...
t = x/v
where the time is Earth's time when the craft arrives at it destination.
We are now ready to do the problem. We use the time dilation equation...
t = γ t'
where t' is called the proper time of the craft. Plugging things in gives...
x/v = t' / √(1 - v² / c²)
so...
√(1 - v² / c²) = v t' / x = v / c
so...
1 - v² / c² = v² / c²
so...
v = 1/√2 c ≈ 0.7071 c
We can get the same result using the Lorentz transformations...
x' = γ (x - v t) = γ (x - v x / v) = 0
t' = γ (t - v x / c²) = (1 - v² / c²) (x/v) / √(1 - v² / c²) = √(1 - v² / c²) x/v
The first equation tells us that the craft is always at x' = 0 in its own frame, which is why t' is called the proper time. The second equation can be manipulated into...
√(1 - v² / c²) = v t' / x = v / c
which is what we got for time dilation.
This solution never required the use of 8 or 10 for X. It only used the fact that x/t' = 1 ly/y = c. So, you are right that the speed is the same for all distances as long as the ratio of x/t' is c.
x = 8 ly
t' = 8 y
v = ?
We then remember that light travels a ly in a y, so the speed of light c is...
c = 1 ly/y = x/t'
and we remember that gamma is defined by...
γ = 1 / √(1 - v² / c²)
Using the definition of velocity as distance over time we get...
t = x/v
where the time is Earth's time when the craft arrives at it destination.
We are now ready to do the problem. We use the time dilation equation...
t = γ t'
where t' is called the proper time of the craft. Plugging things in gives...
x/v = t' / √(1 - v² / c²)
so...
√(1 - v² / c²) = v t' / x = v / c
so...
1 - v² / c² = v² / c²
so...
v = 1/√2 c ≈ 0.7071 c
We can get the same result using the Lorentz transformations...
x' = γ (x - v t) = γ (x - v x / v) = 0
t' = γ (t - v x / c²) = (1 - v² / c²) (x/v) / √(1 - v² / c²) = √(1 - v² / c²) x/v
The first equation tells us that the craft is always at x' = 0 in its own frame, which is why t' is called the proper time. The second equation can be manipulated into...
√(1 - v² / c²) = v t' / x = v / c
which is what we got for time dilation.
This solution never required the use of 8 or 10 for X. It only used the fact that x/t' = 1 ly/y = c. So, you are right that the speed is the same for all distances as long as the ratio of x/t' is c.
-
dt' = dt sqrt(1 - (v^2/c^2))
dt' = (x/v) sqrt(1 - (v^2/c^2))
If you set the subjective time and the distance to the same value and solve for "v", you end up with
v = c / sqrt(2).
Let's check that for your example of 8 lightyears.
dt' = (8ly/ (c/sqrt(2))) sqrt(1 - (c/sqrt(2))^2 / c^2))
dt' = (11.31 year) sqrt(1 - (c^2/2 / c^2))
dt' = (11.31 year) sqrt(0.5)
dt' = 8 years
dt' = (x/v) sqrt(1 - (v^2/c^2))
If you set the subjective time and the distance to the same value and solve for "v", you end up with
v = c / sqrt(2).
Let's check that for your example of 8 lightyears.
dt' = (8ly/ (c/sqrt(2))) sqrt(1 - (c/sqrt(2))^2 / c^2))
dt' = (11.31 year) sqrt(1 - (c^2/2 / c^2))
dt' = (11.31 year) sqrt(0.5)
dt' = 8 years