Homework Help: Special relativity of two spaceships

1. Mar 5, 2007

Brewer

1. The problem statement, all variables and given/known data
Two spaceships, each of length 100m in their rest frames, pass each other travelling in opposite directions. Instruments on ship A find that the front end of B requires a time $$5x10^{-6}$$s to transverse the full length of A.
What is the relative velocity of the spaceships?
A clock in the front end of B records 1am as it passes the front end of A. What will it read when it passes the rear end?

2. Relevant equations
$$\Delta x\prime = \gamma (\Delta x - v\Delta t)$$

3. The attempt at a solution
I assumed that spaceship A was the stationary frame.

As a result I said that $$\Delta x$$ = 100m, and $$\Delta t$$ = $$5x10^{-6}$$s.

I then said that $$\Delta x\prime = \gamma \Delta x$$. With this information I put it back into the equation given above, and ended up with v = 0. Which I would assume is quite clearly wrong for the question, or it wouldn't have been given to me.

So some questions I have about this, where I'm confused. I assumed that $$\Delta x$$ and $$\Delta x\prime$$ have different values, as I assumed that $$\Delta x\prime$$ would be affected by length contraction of some form or the other. However because it says in that the two lengths are equal in their own rest frame, does that make both $$\Delta x$$ and $$\Delta x\prime$$ equal for the question? I think as I went about it, I had too many variables and not enough enquations.

2. Mar 5, 2007

Brewer

I think I may have been over complicating this problem.

The way I see it now, is that spaceship A sees the front of B travel a distance of 100m in the time specified above. Therefore using simple equations of motion (v = s/t) I can plug the values I have (s = 100, t = 5*10^-6) into this equation and obtain a relative velocity of c/15.

Correct?

From here I can then use Lorentz transformations to solve the rest of the problem.

3. Mar 5, 2007

Staff: Mentor

Exactly right.

4. Mar 5, 2007

Brewer

For the second part of the question I have been using
$$\Delta t\prime = \gamma (t - \frac{vx}{c^2})$$ in order to find the time elapsed according to B.

But when I plug the numbers that I have into this I get an answer of 4.99*10^-6 seconds. I'm sure they wouldn't ask what time was showing on the clock if it was so miniscule would they? I also get the feeling that there is more to it than this. I'm sure they've given at least one more piece of information that I'm yet to use (the length of B in its rest frame). I always get worried when I have seemingly redundant information - things like that never happen in questions!

5. Mar 5, 2007

Staff: Mentor

Looks right to me.
Apparently they would. (Just for fun, compare the time that it would take if there were no special relativistic effects.)
True. The length of B is irrelevant.
Often that's true, but not always.

6. Mar 5, 2007

Brewer

Thank you for that.

Do you think that you could help interpret the following question for me?

A member of a colony on a moon of Jupiter is required to salute the UN flag at the same time as it is being done on Earth, which is at noon in New York. If observers in all inertial frames are to agree that he has performed his duty, for how long must be salute?
The distance from Earth to Jupiter is 8*10^9 km. The relative motion of Earth and jupiter's moon may be ignored.

I can't even gleam what they're asking me to do here. Surely for the light to reach every inertial frame possible then the man would have to salute indefinately in order to overcome time dilation effects of the photons travelling at c?

7. Mar 5, 2007

Staff: Mentor

Think in terms of the relativity of simultaneity. If there were no relativistic effects, then Jupiter and Earth could synchronize clocks and be done with it. The man on Jupiter could wait for the right moment, snap out a salute at the appointed time, and everyone would agree he did his duty.

With relativity, moving frames will see the clocks on Jupiter and Earth to be out of synch. Some will see the Jupiter clocks ahead of Earth's, while others will see the opposite. Hint: How far out of synch can those clocks be?

8. Mar 5, 2007

denverdoc

Brewer, not to butt in, but can you tell me where this problem came from, prof, textbook, etc?

9. Mar 6, 2007

Brewer

I don't know. We were given it on a sheet to do in a problems class with the professors help. Except in the class our normal guy was ill, and the stand in was only in for 10 mins before disappearing, which wasn't too much help.

10. Mar 6, 2007

Brewer

From a bit of research on the web (our notes are very vague on relativity of simultaneity), is this to do with "local time"?

I.e.

This seems to be the only equation that I can find where I get an answer that doesn't equate to be infinite (as it doesn't use the gamma factor)

11. Mar 6, 2007

Brewer

Using this I get that the worker must salute the flag for 7 hours and 25 minutes (based upon all observers seeing him salute for 5 seconds).

Now if I were in charge (and assuming that I've gotten the answer correct), I would scrap this practice in order to let my workers do some work rather than just saluting all day.

12. Mar 6, 2007

Staff: Mentor

All we care about here is the time according to the man on Jupiter's own clock, not what other observers zooming by would say the time was.

Careful. Since the flag must be saluted at 12:00 Earth time, what time on Jupiter must he begin and end saluting to ensure that any observer would agree that he was saluting at the same time as the folks on Earth? (Hint: Are the Jupiter clocks ahead or behind the Earth clocks?)