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Length Contraction of the Universe Surpassing Length of Moving Object? 
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#1
Jun1711, 08:43 PM

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A spaceship is of length 1000 meters (to the observer standing in the spaceship).
It is travelling so close to the speed of light that to the observer in the spaceship, the 'length' of the universe has contracted to 900 meters. Q1: If I am the observer in the spaceship, what do I think the front 100 meters of my ship is 'in'? This is probably only sensible (if any of this is sensible) if we assume the universe is finite. I just thought of a photon which as I understand it has the universe at 0 length, and this question popped into my head. If the answer is "we don't know, what ever is outside the universe" does this mean there is definitely an outside of the universe? Q2: Is a photon (in its frame) simultaneously in all universes as they are all length 0 and the photon is moving? I find understanding of relativity extremely hard to keep a hold of, so forgive me if the basic premise of this question is wrong altogether. EDIT: Oh I see this is my first post. I think I joined a long time ago but I obviously just read. Hi! 


#2
Jun1711, 09:27 PM

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#3
Jun1711, 09:50 PM

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Hi, Ricky116,
Welcome to PF! This is a version of the ladder paradox: http://en.wikipedia.org/wiki/Ladder_paradox . I think the resolution is the same. BTW, we don't know whether the universe is finite or infinite: http://physicsforums.com/showthread.php?t=506986 Ben 


#4
Jun1711, 10:29 PM

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Length Contraction of the Universe Surpassing Length of Moving Object?
This was a fun problem to think about. My conclusion: In an inertial coordinate system that's comoving with the ship, two observers on the ship, one near the back and one near the front, will be next to the same galaxy at the same time. But in the inertial coordinate system that's comoving with that galaxy, those two events will be very far apart in time. The observer near the back would see a galaxy full of stars, but the observer near the front would see the same galaxy at an age when all the fissionable* material has been used up and all that remains are black holes, neutron stars and cold dwarf stars**.
To see this, I suggest thinking of "space" as 1dimensional, specifically as a circle. If we pretend that there's no expansion of space, the universe would appear as a cylinder in a spacetime diagram. Uh, that sounds like a weird thing to say. How about this: A spacetime diagram depicting the motion of the front and back of the rocket would have to be drawn on a cylinder. To see how the diagram answers the question, we must see that a simultaneity line from the world line of the back of the ship to world line of the front can spiral around the cylinder before it gets there, even though the world lines are very close together on the cylinder. To be honest, I can't visualize this well enough in my head to see this, but I believe it's true. I used some nonrigorous arguments to convince myself, so it's kind of hard to explain. I haven't tried to figure out what would change (other than the shape of the "cylinder") if we try to include the cosmological expansion in the answer. I suspect that the answer would be essentially the same, except that the big crunch might end the experiment early. *) Not sure if that's actually a word. **) I didn't do any calculations. I'm just guessing it would be a very long time. 


#5
Jun1811, 10:35 AM

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I have read about the ladder paradox before, but I think I need to give it a good think  it seems to be quite key to my understanding of length contraction here! Thanks! 


#6
Jun1811, 12:10 PM

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#7
Jun1811, 01:55 PM

P: 3,967

There is much more to this than the ladder paradox and I think it is a very good question.
One aspect that has not been touched on is the visible horizon of the universe. Even if the universe is finite, but much larger than the visible horizon, it seems very unlikely that an observer moving at very close to the speed of light relative to the CMB will see beyond the visible horizon. Any thoughts on this? 


#8
Jun1811, 02:50 PM

P: 351




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