Question

If astronauts could travel at v = 0.950c, we on Earth would say it takes (4.20/0.950) = 4.42 years to reach Alpha Centauri, 4.20 light-years away. The astronauts disagree. (a) How much time passes on the astronauts’ clocks? (b) What is the distance to Alpha Centauri as measured by the astronauts

Answer 1

Imagine a light clock with two facing mirrors. Make a photon particle "bounce" (reflect) between the two mirrors.

From the perspective of someone on the ship, the photon particle only has to bound straight up and down a distance h, if we make the distance between the two mirrors h.

So, h = ct => t = h/c

From the perspective of someone watching the ship go by, the photon particle must bounce diagonally to hit the mirrors. Specifically, if the spaceship is going v, then the total distance the photon particle must travel is (ct)^2 = (vt)^2 + h^2.

Solve for t. This gets you t = h / sqrt(c^2 - v^2).

Now, let's make a ratio out of these two time periods.

[h / sqrt(c^2 - v^2)] / [h/c] = c / sqrt(c^2 - v^2)

So, this gets you the ratio. Let v = 0.95c and substitute that into the formula. Then, let the astronaut's time be equal to [c / sqrt(c^2 - v^2)] * 4.42.

Then, once you know how long it takes, from their perspective, you can solve for distance (0.95c * t = d), where t is now known.

Answer 2

I made a mistake. The astronaut's time should be equal to sqrt(c^2 - v^2) / c * 4.42, not [c / sqrt(c^2 - v^2)] * 4.42.

Answer 3

I still blur.. so who is be the observer right now..
is this use the time dilation which t=y(t')