Chat GPT, just now. Question posed verbatim:
How much time would pass on Earth while a passenger in a spaceship made a fifty light year round trip journey at constant 1g acceration, experiencing only one subjective year of time?
Answer 1:
Approximately
59 Earth years would pass while a spaceship undergoing 1g constant acceleration and deceleration over a 50-light-year round trip allows the passenger to experience only 1 year of proper time.
Exit chatbot, so it forgets. Restart.
Answer 2.
About
2.09 Earth years will pass.
(Oh look ChatGPT just invented faster than light travel!!)
Exit chatbot, so it forgets. Restart.
Answer 3.
To make a 50 light-year round trip at 1g while experiencing only 1 year, the Earth time would be astronomically huge —
millions of years.
davewhite04: Post that question, verbatim in your chatbot, see what answers you get.
Close the tab, re-open ChatGPT and repeat.
You must of read my mind, I was going to ask you to try it.
So... this is what I get back first time. I posted the whole thing as not sciency at
If a passenger experiences only
1 year of subjective time on a spaceship accelerating constantly at
1g during a
50 light-year round trip, then:
Only ~1.045 years would pass on Earth.
Wait — But This Doesn't Seem Right?
You're correct to feel skeptical. A
50 light-year round trip is a vast distance. If only
1 year passes for the astronaut, then
far more than 1 year should pass on Earth due to relativistic effects.
Let’s correct this:
We should invert the problem:
You gave:
- Proper time τ=1\tau = 1τ=1 year (astronaut’s clock)
- Acceleration a=1ga = 1ga=1g
- Solve for distance traveled and Earth time for a round trip
Let me recalculate correctly using the equation:
D=2a(cosh(aτ2)−1)D = \frac{2}{a} \left( \cosh\left(\frac{a \tau}{2}\right) - 1 \right)D=a2(cosh(2aτ)−1)
Then use that to compute Earth time.
Thanks for your patience. Here's the corrected interpretation:
If an astronaut travels under
constant 1g acceleration, experiencing
1 year of subjective time, then:
- They would cover only about 0.26 light-years total round trip.
- Earth time elapsed would be approximately 1.045 years.
Key Insight:
To make a
50 light-year round trip under constant 1g while experiencing only
1 year of time onboard is physically
impossible — because:
- Covering such a large distance in such a short proper time requires acceleration beyond 1g, or
- Reaching speeds extremely close to light, which would drastically increase Earth’s elapsed time.
Answer 2 after restart.
To answer this question, we need to use the equations of
relativistic rocket travel — specifically, the
relativistic equations of constant acceleration. You’re asking: if a spaceship accelerates constantly at
1g (Earth gravity, 9.81 m/s29.81 \, \text{m/s}^29.81m/s2), how much
Earth time passes while the traveler experiences just
1 year of subjective time, during a
50 light-year round trip (i.e., 25 light-years each way)?
Let’s go step-by-step:
Key Concepts and Equations
When a ship accelerates at constant aaa, the relativistic equations for proper time (τ\tauτ) and coordinate time (ttt) for a round trip are:
- Total proper time (τ\tauτ) (time experienced by traveler):
τ=2ca arcosh(aD2c2+1)\tau = \frac{2c}{a} \, \text{arcosh}\left( \frac{aD}{2c^2} + 1 \right)τ=a2carcosh(2c2aD+1)
- Total Earth time (ttt) (as measured in Earth’s frame):
t=2ca sinh(aτ2c)t = \frac{2c}{a} \, \sinh\left( \frac{a\tau}{2c} \right)t=a2csinh(2caτ)
Where:
- DDD = total distance (50 light-years)
- aaa = acceleration (1g)
- ccc = speed of light
- τ\tauτ = proper time (1 year)
- ttt = Earth time (what we want to find)
Let’s compute this numerically.
If a spaceship makes a 50 light-year round trip while accelerating constantly at
1g, and the
passenger experiences just 1 year of subjective time, then approximately:
1.045 years would pass on Earth.
This surprisingly small value shows how dramatic
time dilation can be under constant relativistic acceleration. The ship gets very close to the speed of light quickly, which compresses the subjective time drastically.
.
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