# Travelling 3.6 LY at $(3/4) c$, how much time passes for the traveler and how much time passes for the people in Alpha Centauri? [duplicate]

This specific thought experiment is for a story I'm writing. Assume a spacecraft can travel at 3/4 the speed of light. It leaves Alpha Centauri A and travels 3.6 LY. Being a non-physicist layperson I used the basic Time=Distance/Velocity. This made for an approximation of 4.7 years to travel 3.6 LY at $$(3/4) c$$ (please correct me if I'm wrong). However, that is not accounting for the time dilation that occurs when an object travels at near the speed of light. I know that for objects traveling at such high speeds, time will pass much slower than normal whilst the point of origin time will pass much faster. This results, I believe, in the traveling object seeming to move slowly e.g. more time passes, while the traveler will experience time more slowly e.g. less time passes.

My question(s) are: for the $$(3/4) c$$ traveler, how long would it appear to take to reach 3.6 LY from their point of origin? For the observer (Alpha Centauri), how much time would pass for them before the traveler reached their destination?

This question sadly suffers from a couple of problems, due to slightly sloppy wording. This is usually due to the fact that people are introduced to relativity using length contraction and time dilation, which make it seem as though "strange" things happen on spaceships that move close to the speed of light.

For example, when you say the following:

I know that for objects traveling at such high speeds, time will pass much slower than normal whilst the point of origin time will pass much faster.

the crucial information missing is with respect to whom? Neither the person on the planet nor the person in the spaceship will see anything change around themselves. However, the person on the planet will see time on the spaceship move slower than time seems to move around them, and likewise, the person on the spaceship will see time moving slower on the planet than time seems to move around them. This is perfectly symmetric, though admittedly a little non-intuitive, as we are used to thinking of time "moving" at the same rate for all observers.

Nevertheless, the question is well posed: if you have an observer sitting on a planet, and you have a spaceship moving past at some speed $$v$$ between the planet and a point a distance $$d$$ away (according to someone on the planet), how long will it take someone on the spaceship to get there?

Let's make the problem more precise: first, we define two frames of reference $$S$$ (the frame of the planet) and $$S'$$ (the frame of the spaceship), with $$S'$$ moving with respect to $$S$$ at a speed $$v$$. In each frame there is a fixed observer with a noting pad, who looks at different events and notes down where they occur ($$x$$ or $$x'$$) and when they occur ($$t$$ or $$t'$$). The "primed" coordinates are those measured by the person on the ship, while the "unprimed" coordinates are those measured by the person on the planet. Let us further call the time that the spaceship "passes" the planet $$t = t' = 0$$, and $$x=x'=0$$. In other words, both observers start measuring distance and time at the instant they "cross" each other.

## What does a person on the planet see?

Your "layperson" argument is perfectly correct! The person on the planet sees a spaceship moving away at a speed $$v$$, and covering a distance $$d$$. It should take a time $$t = d/v$$ to get there.

## What does a person on the spaceship see?

We would now like to relate the "events" seen by a person on the planet to those seen by a person on the spaceship. To do this, we use something called the Lorentz Transformations:

\begin{aligned} x' &= \gamma \left( x - v t\right),\\ t' &= \gamma \left( t - \frac{v}{c^2} x\right), \end{aligned}

where $$\gamma$$, called the Lorentz factor, is given by $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}.$$

So let's write out the events as they would be seen by the observer in $$S$$, and in $$S'$$:

$$\begin{array}{|c|c|c|} \hline \textbf{Event} & \textbf{Observer in S} & \textbf{Observer in S'} \\ \hline \textbf{Spaceship leaving A} & (0,0) & (0,0) \\ \hline \textbf{Spaceship reaching B} & \left(d/v,v\right) & (t',x') \\ \hline \end{array}$$

Our job is to find $$t'$$ and $$x'$$ using the equations mentioned above, and by simply substituting $$t = d/v$$ and $$x = d$$, you should be able to see that

$$t' = \gamma \left( \frac{d}{v} - \frac{v}{c^2} d\right).$$

Some elementary mathematical manipulation will allow you to see that

$$t' = \frac{d}{\gamma v},$$

which is the time an observer sitting on the spaceship will have felt elapsing between leaving the planet and reaching the destination. You can interpret this in the following way: the observer on the spaceship sees the distance between the two points as being "contracted" (less than $$d$$). Length contraction tells us that this "contracted" distance is $$d/\gamma$$. The observer on the spaceship stills sees the destination traveling towards it at a speed $$v$$, however, and so the total time is $$d/(\gamma v)$$.

I'll leave it to you to plug in the numbers.

• I apologize for the late reply. I am so grateful for your detailed and concise response. I'll have to work on figuring out how to plug it all in, but I am so much closer to the answers I need. Thank you again! Dec 30 '20 at 3:45