Answer Relative Time Travel Question: What Speed to Remain 21 for 10 Years?

[Procrastinate]

A physicist decides she would like to remain 21 for 10 years. What is the minimum constant speed relative to Earth at which she would have to take a rocket trip into outerspace and back in order to achieve this?

Let t=10 and $$t_{o}=1$$

$$t=\frac{t_{o}}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$10=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

$${\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{10}$$

$$1-\frac{v^2}{c^2}=\frac{1}{100}$$

$$\frac{v^2}{c^2}=\frac{99}{100}$$

$$v^2=8.91\times10^{16}$$

$$v=298496231.1=0.9949874371c$$

This was the correct answer when I looked up the solutions.

What I do not understand is why the normal time is considered to be 1 year since ten years of proper time has actually past. If I do this problem the other way, a negative answer is obtained. Could someone please explain why this occurs, or have I just not grasped the concepts of what is the proper time and what is the relative time?

 

[Doc Al]

What I do not understand is why the normal time is considered to be 1 year since ten years of proper time has actually past.

No, her proper time is only one year, not ten. She only wants to age 1 year (proper time) while 10 years pass on earth.

 

[Procrastinate]

Since t represents the interval between two events as measured in a reference frame moving with speed v with respect to the first reference frame, wouldn't this be 1? This is because the physicist is traveling at high velocity and thus time would be shorter relative to earth.

This doesn't seem to coincide with the values I put as t and $$t_o$$.

 

[Doc Al]

Since t represents the interval between two events as measured in a reference frame moving with speed v with respect to the first reference frame, wouldn't this be 1?

The physicist represents the "clock" (her body) that we are interested in. t is the time between events (her birthdays, perhaps) as seen on earth. t0 is the proper time, which is measured from a frame in which the clock is at rest. t0 is the time that the physicist measures in her rocket ship; t0 is her actual age.

This is because the physicist is traveling at high velocity and thus time would be shorter relative to earth.

The time as measured by her will be shorter. Only 1 year compared to 10 years on earth. Seen from a moving observer (the earth), her biological clock appears to run slow: she only ages 1 year when 10 years have passed on Earth clocks.

 

[rdl]

The proper time, denoted as t_{o}, is the time experienced by an observer who is at rest relative to the event being measured. In this case, the physicist is the observer and she is at rest on Earth. The relative time, denoted as t, is the time experienced by an observer who is in motion relative to the event being measured. In this case, the physicist is in motion on the rocket trip into outer space and back.

When we solve for t in the equation t=\frac{t_{o}}{\sqrt{1-\frac{v^2}{c^2}}}, we are finding the time experienced by the physicist during the 10 years of proper time (t_{o}) on Earth. Therefore, it makes sense to set t_{o}=1, since we are looking for the minimum speed for the physicist to experience 10 years of proper time while only 1 year passes on Earth.

If we were to solve for t_{o} instead, we would be finding the proper time experienced by the physicist while she is in motion on the rocket trip. This would give us a negative answer because the proper time would be less than the 10 years of relative time experienced on Earth.

In summary, the proper time is the time experienced by an observer at rest, while the relative time is the time experienced by an observer in motion. In this problem, we are looking for the minimum speed for the physicist to experience 10 years of proper time while only 1 year passes on Earth, so it makes sense to set t_{o}=1.