# Homework Help: Ship and Dock - Special Relativity

1. Oct 4, 2008

### kehler

1. The problem statement, all variables and given/known data
A relativistic ship is undergoing testing in a station. This involves it flying through a station at speed v/c = (3/4)-1/2, corresponding to the lorentz factor, gamma = 2. Once inside the dock, laser doors simultaneously close at each end. After 1.54 x 10-6s, the testing is finished and the laser doors simultaneously open, and the ship flies out.
The proper length of the ship is 800m and the proper length of the dock is 1000m. According to the station, the ship length is contracted to 400m and hence comfortably fits in the dock. The doors are closed when the ship is 100m away from the back door and hence 500m from the front door. While the doors are closed for 1.54 x 10-6s, the ship travels 400m and hence it is 100m from the front door when it opens.
The captain of the ship is nervous because for him the dock is length contracted to 500m and hence he can't see how his 800m ship can fit in when the doors close. Should the captain panic, or is all well?

To answer this, find the times and positions of the following four events in the ship's frame (S'). Let the station be the S frame in standard configuration with S', and let the common coordinate origin be event A, the closing of the back door.
Event A - Closing of the back door. xA = 0, tA = 0
Event B - Closing of the front door. xB = 1000m, tB = 0
Event C - Opening of the back door. xC = 0, tC = 1.54 x 10-6s
Event D - Opening of the front door. xD = 1000m, tD = 1.54 x 10-6s

a)Use the Lorentz transformations to find the positions and times of these events in the ship's frame. Use c = 3 x 108
b)What are the positions of the front and back of Red Dwarf in its rest frame?
c) Should the captain panic? Explain using (a).

2. Relevant equations
Lorentz transformations:
x' = gamma(x-vt)
t' = gamma(t - (vx/c2))

3. The attempt at a solution
(a) I just used the equations above and plucked the numbers in
Event A - xA' = 0, tA' = 0
Event B - xB' = 2000m, tB' = -5.77 x 10-6s
Event C - xC' = -800m, tC' = 3.08 x 10-6s
Event D - xD' = 1200m, tD' = -1.34x 10-6s

(b)This is where I'm having trouble.
At t=0, the back of the ship is 100m from x=0.
In the S' frame, it will be
x' = gamma (x - vt) = 2(100 - 0) = 200m from x'=0.

But if I use t= 1.54 x 10-6s where the back of the ship is 500m from x=0
In the S' frame, it will be
x' = 2(500 - (1.54 x 10-6)(.75c)) = 307m from x'=0

I don't get it why :S. The ship shouldn't move in its rest frame.

(c)I'm guessing the ship gets through fine but I don't know how to explain it :(

I know the question is really long, but I would really really appreciate any help with this :)

Last edited: Oct 4, 2008
2. Oct 4, 2008

### granpa

its usually called the pole barn paradox

3. Oct 4, 2008

### kehler

I looked that up but it didn't quite help how I can calculate (b).
My problem is finding the position of the ship with respect to its rest frame.

4. Oct 4, 2008

### granpa

0,100
0,500
1.54 x 10-6, 500
1.54 x 10-6,900

just transform all 8 numbers

5. Oct 4, 2008

### granpa

the origin of the 2 coordinate frames will coincide at t=0 in both frames

6. Oct 4, 2008

### granpa

v is not .75c
its the square root of that

7. Oct 4, 2008

### kehler

Ohh ok :). I get the same values for both times now. Thanks granpa!
One thing I don't understand is why the ship is 200m from the back door at t'=0. If it's 100m from the back door in the S frame at t=0, shouldn't the ship see the length contract to 50 instead?

8. Oct 4, 2008

### granpa

I'm not allowed to do your homework for you. transform all 8 numbers and you will immediately see why.

9. Oct 4, 2008

### kehler

It's not really my homework ;). Just trying out the question.
I've done the transformations. I get the ships coordinates to be at 200 and 1000.
Maybe I'm slow but I just don't understand why the length from the back door to the ship didn't contract to 50...

10. Oct 4, 2008

### granpa

i forgot the lasers

0,0
1.54 x 10-6,1000

11. Oct 4, 2008

### gabbagabbahey

What is t' at x=100 and t=0? How far has the back of the barn moved in frame S' in that time?

12. Oct 4, 2008

### kehler

exactly?

13. Oct 4, 2008

### granpa

i was responding to 'What is t' at x=100 and t=0?' which i thought you wrote.

14. Oct 4, 2008

### gabbagabbahey

Basically, the position of the back of the barn in frame S' is not x'=0 when t=o and x=100m. It should be at x'=150m so that the length of the gap between the back of the ship (Red Dwarf=awesome BTW :0) ) and the back of the barn is 200m-150m=50m in S'.

15. Oct 4, 2008

### granpa

the back of the barn would be the lasers.

16. Oct 4, 2008

### kehler

t'= -5.77x10-7 at x=100 and t=0... And I think it's moved 150m. Oh ok, 200m-150m= 50m :)

I'm still a bit confused. Aren't t=0 and t'=0 coincident when x=0? So the times are the same only when x=0 but different everywhere else?
Ugh, special relativity confuses the heck outta me :(

17. Oct 4, 2008

### granpa

18. Oct 4, 2008

### granpa

another trick is to think in terms of long lines of perfectly synchronized clocks all evenly spaced at one light second apart and each sending out radio pulses simultaneously at one second intervals. one line of clocks for each frame. you con immediately see why simultaneity is lost.

19. Oct 4, 2008

### kehler

Thanks heaps. I'll read up on that. You guys are awesome :). Did it take long for you to get special relativity?
I normally grasp physics concepts fast but I'm finding this rather confusing :(

20. Oct 4, 2008

### granpa

well time dilation and length contraction are simple. its the idea that a moving observer sees the stationary object as contracted thats confusing. thats where simultaneity comes in. it started to make sense to me when i read a translation of a russian childrens book. it compared length contraction to the apparent shrinking of objects due to change in perspective. it all seemed less confusing then. reality doesnt change when you speed up. you change. your perspective changes.

Last edited: Oct 4, 2008