# Homework Help: Special Relativity: Time Dilation and Length Contraction

1. Aug 10, 2012

### Peter G.

Hi,

A super fast spacecraft is moving at a speed of 0.80c with respect to the observers on Earth. The spacecraft leaves Earth in May 2004 on its way to a distant solar system.

a) i) According to the observers in the spacecraft, 6.0 years have elapsed since leaving Earth. Calculate the time elapsed according to an observer on Earth

From what I understand, the observers on the spacecraft measure proper time. Hence, I performed 6 times the Lorenz factor, using 0.80c as my v. I got 10 years as an answer.

ii) Explain whether either measured time interval can be considered to be correct:

I know that, according to the theory of relativity, both measurements should be correct. This requires two marks though and I do not know how to complement my answer.

iii) Calculate the distance of the spacecraft from Earth according to an observer on Earth

Here lies my central doubt. Should I simply perform velocity multiplied by time (the time measured by observers on Earth) or should I find the distance travelled according to the observers on the spacecraft and apply the length contraction formula?

2. Aug 10, 2012

### Simon Bridge

Are they asking for the time on the Earth clock when the spaceship clock reads +6years to the Earth people or for the time on the Earth clock as read by the spacecraft crew when the ship clock says it's 2010?

It says "observer on Earth".
What they've literally asked is what time is on the Earth clock to an earth observer when the spacecraft clock reads 2010 to a crewmember... not sure that makes sense...

You have to do this in the context of your course ... one of the times can be considered "proper" for eg. So it is down to what you mean by "considered correct". I'd guess you just need to write more than one sentence showing you understand.

You want to say why both should be considered "correct" for the extra mark. "According to the theory of relativity..." is not it ... what is it about the theory of relativity that tells you this?

You could do it both ways and see ...

But how does the observer on the Earth work out the distance?
Well, by knowing the speed and the time of travel of course :)
All that basic stuff still has to work.

If you do it right, should you get the same answer either way?

3. Aug 10, 2012

### Tobbin92

According to Einstein no point in space can be considered central or correct, the same thingst goes for time. Both times are equally correct.

4. Aug 10, 2012

### Simon Bridge

... appeal to authority? Really?
The test is to demonstrate an understanding of special relativity - so what is it about SR that would lead Einstein, or anyone else for that matter, to make that assertion?

You'd get one mark for saying that both clocks are equally correct.
You'd get another mark for explaining how come.

5. Aug 10, 2012

### Tobbin92

Time along with space is not absolute and depends on the observers speed. This comes from the fact that the speed of light is a constant and involves both time and space (meters per second). Therefore time and space has to adapt in order to please the law of a constant speed of light.

The speed of light is therefore the only thing that can be considered correct involving the dimensions time and space.

6. Aug 10, 2012

### Simon Bridge

Better - I'd want to put "...speed in relation to the clock" in there to show you know what the "relative" part of SR means. May want to talk about proper time a bit just to be on the safe side. But we are not supposed to be doing OPs homework for him. ;)

We also are not doing him any favors if the marker googles the answer given (it happens), finds this page, and marks him down for "cheating": every correct answer we give him is one he cannot use.

There is no "one correct answer" to this question. Also see:
... OP needs to come up with one that makes sense in terms of the course so far.

7. Aug 11, 2012

### Peter G.

Hi,

Thanks for the help guys. Don't worry about the fact Tobbin 92 put that answer up. I did my research and had come up with an answer which I will check with my teacher.

Now, regarding the third question. I think both methods do not give the same answer because length is not being contracted in this case. I think we can agree (provided I interpreted the first question correctly) that the observers at Earth measure a longer time. Since light has a constant speed in a vacuum for every intertial observers, they must measure a longer time and not a contracted time (what the formula I have is for), correct?

8. Aug 11, 2012

### Tobbin92

Why should length not be contracted in this case? Think that part through again and make sure that such a statement can be justified.

9. Aug 11, 2012

### Peter G.

This is how I thought of it:

The clock on the spacecraft measures proper time and it claims that the astronauts have been travelling for 6 years.

The observers on Earth experience a time dilation. On their clocks, the astronauts have been travelling for 10 years.

In both cases the spacecraft is travelling at 0.8c.

That means that for the people on the Earth, the spacecraft was travelling at 0.80 c for 10 years, which yields a longer (thus not contracted) length than that measured by the astronauts, who have been travelling at 0.80 c for only 6 years.

Is this wrong?

10. Aug 11, 2012

### cepheid

Staff Emeritus
Note that the distance travelled by the spacecraft, according to observers on the spacecraft, is 0.

I'm not sure what you're asking here, but here is my take on the situation. According to observers on Earth, the spacecraft is receding from Earth at 0.8c, and after travelling for 10 years, is 8 light years away. According to observers on the spacecraft, the Earth is receding from the spacecraft at 0.8c, and after travelling for 6 years, is 4.8 light years away from them. For the spacecraft observers, distances are contracted in the direction of motion by a factor of 1/γ = √(1 - (v/c)2) = √(1 - 0.64) = √(0.36) = 0.6

Notice that this is the factor by which the distance between Earth and the spacecraft is contracted (comparing the measurement in the spacecraft frame to the measurement in the Earth frame).

11. Aug 11, 2012

### Peter G.

Hi Cepheid,

The question asked me to: iii) Calculate the distance of the spacecraft from Earth according to an observer on Earth

My first attempt was to apply the length contraction formula, which didn't work. I believe the simple application of speed * time is what yields the correct answer.

What you quoted was me trying to convince myself as to why the length contraction formula does not work.

I think you confirmed that for me. The length is NOT contracted for the observers on the Earth (that is what the question asks) it is contracted for those in the spacecraft. Correct?

12. Aug 11, 2012

### BruceW

That's right. If you think about it, the speed of 0.8c and the time of 10 years are both measurements according to the earth's frame. So you can use these in the normal way to find the distance according to the earth's frame. It is only when you are using measurements from several different reference frames that you need to be careful about length contraction, time dilation e.t.c.

13. Aug 12, 2012

### Simon Bridge

Yep:
0.8c for 10years puts the spaceship 8ly away.
The Earth observer knows SR and figures that the spacecraft crew will measure the contracted distance of 8/gamma = 4.8ly

Back on the ship - they see the Earth receding at 0.8c for 6years, which puts the Earth 0.8x6=4.8ly away.

So they agree - but disagree about why.
The confusion comes in when you wonder what happens when the ship crew attempt the same reasoning ... that the Earthsiders would measure the contracted distance they just worked out, 4.8/gamma = 2.9ly ... erm... or would they figure that the 4.8ly is a distance in the earth reference frame, which is moving wrt them, and therefore they are the one's seeing the contracted length?

Thing is - the whole thing was from the Earth observers POV.
When the earth observer see +6 on the ship clock, the earth clock reads +10 and the distance is 8ly measured by the Earth observer and 4.8ly measured by the ship observer.

When the ship observer sees +6 on the ship clock, they see +3.6 on the Earth clock and the distance is 4.8ly by the ship observer and 3.6x0.8=2.9ly by the Earth.

... this is why I wanted to see care about who is observing what, where.

Last edited: Aug 12, 2012