# Homework Help: Introductory relativity question

1. Dec 12, 2012

### CAF123

1. The problem statement, all variables and given/known data
A space probe is sent from Earth to collect samples of gas from our nearest neighbour star, Proxima Centauri, which is 4.2 ly away. The probe accelerates rapidly to a speed of 0.99c, cruises at this speed until very close to the star, slingshots around the back of the star collecting the sample and then returns back to Earth. Only when close to Earth does it rapidly decelerate.

Synchronised clocks are placed on the probe and at mission control. What times do they read the moment the probe returns to Earth?

3. The attempt at a solution

Earth's clock: Relative to Earth, distance to Proxima Centauri is 4.2 ly. This means the time for the probe to reach the star is simply (4.2)/(0.99) ≈ 4.24 yrs. It is the exact same calculation on the way back so the Earth clock reads ≈ 2(4.24) = 8.48 yrs.

Probe's clock: As the ship approaches the star, since v ~ c, in a frame stationary relative to the probe ( i.e an observer standing on the probe), the distance to the star is contracted by a factor of $\gamma$. This means $l = l_o/\gamma = 4.2/\gamma,$ where $\gamma = 1/\sqrt{1-(0.99)^2} ≈ 7.1$. So the distance to the star from the ship's perspective is just ≈ 4.2/7.1 = 0.6 ly. Is this good? Am I right in saying I can't get the time on the probe's clock by using this method? - the approach I should have taken was one involving time dilation? So on the ship, time is shorter by a factor of γ. The time to the star from Earth is ~ 4.24 yrs => time on ship ~ 4.24/7.1 = 0.6 yrs. Same on the way back so the probe clock reads 2(0.6) ~ 1.2 yrs. Is it okay?

I have some questions: Isn't time dilation and length contraction supposed to be complementary effects, that is if time dilation occurs in one frame then length contraction occurs in some other frame? Here, I have the ship experiencing both length contraction and time dilation?

Many thanks.

2. Dec 12, 2012

### TSny

This method is fine . In the ship frame, the distance is less so it takes less "ship time" to get there compared to earth time.

That's also a valid way to get the answer. From the earth frame, the ship's clocks run slow so the ship time is less than the earth time. Same result.

For two inertial frames in relative motion, each frame will measure the other frames clocks as running slow (time dilation) and each frame will measure fixed distances in the other frame as contracted. However, there's a complication that comes into the round trip scenario. The ship changes inertial frames when swinging around the distant star. This requires special care. This is an example of the famous "twin paradox". Nevertheless, the two methods that you indicated for solving the problem are valid.

3. Dec 12, 2012

### CAF123

How would I get the time elapsed on the probe's clock though using the length contraction method? What I have is 0.6 ly and not 0.6 yrs which is what I had when I considered time dilation.

Thanks!

4. Dec 12, 2012

### TSny

You can imagine a stick connecting earth and the star. From the ship frame, that stick is contracted and moving relative to the ship at 0.99c. How much time would it take a stick that is 0.6 ly long to pass you if it is traveling at 0.99c?

5. Dec 12, 2012

### CAF123

I see, so that is just $0.6/0.99$ which is approximately$0.6 yrs$ still?

6. Dec 12, 2012

### TSny

Yes. The two methods will give exactly the same result if you avoid round off error.