Geometry and Speed of Light

[blike]

Say we have a spacestation and a rocket.

 

The spacestation has a docking port, but at rest, the rocket is too long to fit into the docking port. However, at 90% the speed of light, the rocket will fit inside.

 

When the rocket is traveiling at 90% speed of light the space station appears contracted (from the rocket's point of view), thus making even less room in the docking port for the rocket to fit.

 

So, will the rocket fit?

[fafalone]

Good luck docking while travelling 90% of c.

[blike]

Hypothetical situation.

[Radical Edward]

this sounds like the pole and cloakroom paradox:

 

John Smith, the famous English polevaulter has developed, alongside his Scientific Chums from Imperial College London, a new running technique which can get him up to extremely high speeds, approaching the speed of light. In the olympic games, he tests out his new technique. Unfortunately however it requires a rather long run up which takes him through the changing rooms, which are 20 metres long. His pole is 20m long, but he is running so fast, that according to an outside observer, his pole appears to have contracted to a mere 10m, so he should fit easily.. However, the evil Australian Sports Magnate, Teddy Murdoch, concocts a cunning scheme, whereby the two doors of the changing room will shut instantaneously when Mr Smith is inside the changing rooms.

Unable to disarm the device, the Engineers from Imperial College despair, but fortunately they manage to construct a device which will immediately open the doors once again, so Jim Smith Breathes a sigh of relief and begins the run up.....

 

As he approaches the speed of light, Jim Smith looks towards the doors of the changing room, and sees the room which was 20m long contract to a mere 10m.... he realises suddently that his pole will be chopped to a fraction of it's length.... The problem is, that the Audience will see him pass through the changing room unscathed....

 

what will happen, and what is the solution to this infamous paradox....

[blike]

^ exactly.

 

So yea, anyone have the answer??

[dragoon]

he'll be fine... it only looks as if the room is 10 but it's still really 20 so it shouldnt matter what it looks like but what it really is

[blike]

but "what it really is" is relative.

[aman]

It depends on how fast the doors are first. If it is sub-light speed then relative to Jim the closest door will operate at a different time then the farthest door so if he times it right he will get through both

Just my observations.

Also on the rocket question the rocket could enter the dock and fit, meaning the nose would hit the back wall at the same relative time that the fins cleared the entrance as observed by the station.

Just aman

[Radical Edward]

aman wins a cookie . did you look that up?

[blike]

It depends on how fast the doors are first. If it is sub-light speed then relative to Jim the closest door will operate at a different time then the farthest door so if he times it right he will get through both

 

Very nice.

 

btw ignore my rocket post, radical edward's is the same concept, but better explained.

 

Also, what happens if the doors ARE faster than light.

[Sayonara]

Originally posted by blike

Also, what happens if the doors ARE faster than light.

The pole is chopped into three parts.

 

Except that it isn't.

[aman]

If the doors close as soon as he's in and open again faster the the speed of light he should theoretically still not be touched.

Just aman

[Guest arigordita]

the paradox and its solution are reviewed here (the site is also a great site for beginners trying to understand relativity...

 

http://www.phys.virginia.edu/classes/109N/lectures/sreltwins.html

 

The key to understanding what is happening here is that we said the bandits closed the two doors at the ends of the tunnel at the same time. How could they arrange to do that, since the doors are far apart? They could use walkie-talkies, which transmit radio waves, or just flash a light down the tunnel, since it's long and straight. Remember, though, that the train is itself going at a speed close to that of light, so they have to be quite precise about this timing! The simplest way to imagine them synchronizing the closings of the two doors is to assume they know the train's timetable, and at a prearranged appropriate time, a light is flashed halfway down the tunnel, and the end doors are closed when the flash of light reaches the ends of the tunnel. Assuming the light was positioned correctly in the middle of the tunnel, that should ensure that the two doors close simultaneously.

 

Now consider this door-closing operation from the point of view of someone on the train. Assume he's in an observation car and has incredible eyesight, and there's a little mist, so he actually sees the light flash, and the two flashes traveling down the tunnels towards the two end doors. Of course, the train is a perfectly good inertial frame, so he sees these two flashes to be traveling in opposite directions, but both at c, relative to the train. Meanwhile, he sees the tunnel itself to be moving rapidly relative to the train. Let us say the train enters the mountain through the "front" door. The observer will see the door at the other end of the tunnel, the "back" door, to be rushing towards him, and rushing to meet the flash of light. Meanwhile, once he's in the tunnel, the front door is receding rapidly behind him, so the flash of light making its way to that door has to travel further to catch it. So the two flashes of light going down the tunnel in opposite directions do not reach the two doors simultaneously as seen from the train.

 

The concept of simultaneity, events happening at the same time, is not invariant as we move from one inertial frame to another. The man on the train sees the back door close first, and, if it is not quickly reopened, the front of the train will pile into it before the front door is closed behind the train.